Vector Calculus Lecture Notes - David M. McClendon

Vector CalculusLecture Notes

David M. McClendon

Department of MathematicsFerris State University

Fall 2021 editionc©2021 David M. McClendon

1

Contents

Contents 2

1 Preliminary material 51.1 What is this class about? . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.4 Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.5 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.6 Homework exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2 Vectors and matrices 322.1 Some basic set theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.2 Introducing vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.3 Vector geometry and dot product . . . . . . . . . . . . . . . . . . . . . 452.4 Matrices and matrix operations . . . . . . . . . . . . . . . . . . . . . . 552.5 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 632.6 Cross products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 672.7 Equations of lines, planes and hyperplanes . . . . . . . . . . . . . . . 712.8 Polar, cylindrical and spherical coordinates . . . . . . . . . . . . . . . 842.9 Some basic topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . 952.10 Mathematica and calculator commands for vector and matrix opera-

tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1012.11 Summary of Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1032.12 Homework exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

3 Functions and limits 1303.1 Functions from Rn → Rm . . . . . . . . . . . . . . . . . . . . . . . . . . 1303.2 Graphs of functions Rn → Rm . . . . . . . . . . . . . . . . . . . . . . . 1353.3 Conic sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

2

Contents

3.4 Quadric surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1653.5 Limits of functions Rn → R . . . . . . . . . . . . . . . . . . . . . . . . 1733.6 Limits and continuity of functions Rn → Rm . . . . . . . . . . . . . . . 1823.7 Homework exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1843.8 Review material for Exam 1 . . . . . . . . . . . . . . . . . . . . . . . . 200

4 Differentiation 2094.1 The total derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2094.2 Partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2154.3 Linear approximation, tangent lines and tangent planes . . . . . . . . 2284.4 Higher-dimensional Chain Rule . . . . . . . . . . . . . . . . . . . . . . 2354.5 Directional derivatives and gradients . . . . . . . . . . . . . . . . . . . 2414.6 Mathematica commands for derivatives . . . . . . . . . . . . . . . . . . 2514.7 Summary of Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 2524.8 Homework exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254

5 Analysis of motion 2765.1 Old ideas in higher dimensions . . . . . . . . . . . . . . . . . . . . . . 2795.2 New ideas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2895.3 Arc length parametrization . . . . . . . . . . . . . . . . . . . . . . . . 2995.4 Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3095.5 Summary of Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . 3155.6 Homework exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317

6 Optimization 3276.1 Finding and classifying local extrema . . . . . . . . . . . . . . . . . . . 3276.2 Absolute extrema . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3426.3 Lagrange’s method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3466.4 Homework exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3556.5 Review material for Exam 2 . . . . . . . . . . . . . . . . . . . . . . . . 360

7 Multiple integration 3657.1 Review of area and integration from Calculus 1 . . . . . . . . . . . . . 3657.2 Double integrals and volume . . . . . . . . . . . . . . . . . . . . . . . 3707.3 Fubini’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3777.4 Triple integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3887.5 Change of variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3927.6 New ways of computing area and volume . . . . . . . . . . . . . . . . 4037.7 Homework exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409

8 Line integrals 4268.1 Vector fields and flows . . . . . . . . . . . . . . . . . . . . . . . . . . . 4288.2 Operators on vector fields . . . . . . . . . . . . . . . . . . . . . . . . . 435

3

Contents

8.3 Paths and parametrizations . . . . . . . . . . . . . . . . . . . . . . . . 4408.4 Line integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4448.5 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4558.6 Conservativity and path independence . . . . . . . . . . . . . . . . . . 4638.7 Summary of Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . 4698.8 Homework exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4708.9 Review material for Exam 3 . . . . . . . . . . . . . . . . . . . . . . . . 493

A Mathematica reference 497A.1 What is Mathematica? . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497A.2 Important general concepts re: Mathematica syntax . . . . . . . . . . . 499A.3 Entering vectors and matrices in Mathematica . . . . . . . . . . . . . . 501A.4 Mathematica quick reference guides . . . . . . . . . . . . . . . . . . . . 502

Index 512

4

Chapter 1

Preliminary material

1.1 What is this class about?What was Calculus 1 and 2 about?

Let’s review what you learned in the first two semesters of calculus. Based on yourprior calculus experience, think about these three questions with your neighbors(don’t cheat, and look at the answers on the next page):

1. What is calculus?

2. What is the difference between a math problem that is a calculus problemand a math problem that is NOT a calculus problem?

3. What are some typical kinds of problems you learn how to solve / mathe-matical procedures you learn in Calculus I?

5

1.1. What is this class about?

Now for the answers

1. What is calculus?

Calculus is the study of limits. Loosely speaking, a limit is a piece ofmathematical language that allows us to formalize a method of solvingproblems via approximations.

2. What is the difference between a math problem that is a calculus problemand a math problem that is NOT a calculus problem?

A calculus problem is one that contains a limit; a non-calculus problemcontains no limit.

Here is a diagram indicating the three most important classes of limits,and a summary of the material you lern in Calculus 1 and 2:

LIMIT INTEGRAL

DERIVATIVE

SERIES

APPLICATIONS

APPLICATIONS

APPLICATIONS-

-

-

-

@@@R

3. What are some typical kinds of problems you learn how to solve / mathe-matical procedures you learn in Calculus I?

Each of the three main types of limit have significant application:

Applications of derivatives:

• Slopes of tangent lines• Instantaneous rates of change• Analysis of motion along a line (position, velocity and acceleration)• Optimization (finding maxima and/or minima of a function)• Graphical analysis (increasing/decreasing, concavity, etc.)• Tangent line approximation• Related rates problems• Newton’s method (for approximating solutions of equations)• L’Hôpital’s Rule (for evaluating limits of the form 0

0 or ∞∞ )

6


Applications of integrals:

• Areas of regions whose boundaries are described by graphs• Volumes of solids with a known cross-sectional shape (i.e. washer

and shell methods)• Arc length (lengths of curves)• Probability and expected value• Moments and centers of mass• Work• Net change, as an accumulation of a rate of change

Applications of series:

• Applications of geometric series (compound interest, fractal geom-etry, repeating decimals, etc.)

• Approximation of functions and integrals via Taylor polynomials• Evaluation of 0

0 and ∞∞ limits without L’Hôpital’s Rule

So that’s the theory you know. What don’t you know?

What you don’t know yet

In Calculus 1 and 2, you study the calculus of what are called “single-variablefunctions”:

• the domain and range are subsets of the real numbers, i.e.• the input is a single number (usually called x) and the output is a single

number (usually called y) and the function is of the form

y = f(x) i.e. xf7−→ y.

• such a function is specified by writing “f : R→ R”.

In the grand scheme of things, we want to study functions where the inputsand/or outputs are not necessarily single numbers:

EXAMPLE 1The volume V a gas occupies depends on its pressure P , its temperature T and thenumber of moles n of gas there are. This means we need a function

V = f(P, T, n) i.e. (P, T, n) f7−→ V.

Such a function would be described by saying “f : R3 → R”.

7


A natural question in this context: Suppose the pressure and temperature of thegas are changing at a certain rate at a certain instant. How fast is the volume chang-ing at that instant?

EXAMPLE 2If an airplane is flying across the sky, to specify its position at a given time t, youneed to give its longitude x, latitude y and altitude z. This means we need a func-tion

(x, y, z) = f(t) i.e. tf7−→ (x, y, z).

Such a function would be described by saying “f : R→ R3”.

A natural question in this context: What is the acceleration of the airplane at acertain instant?

EXAMPLE 3You finance a car by taking out a loan with principal P and term t which chargesyou interest rate r. You are interested in knowing the monthly payment m and thetotal amount A you will end up paying over the life of the loan. This means weneed a function

(m,A) = f(P, t, r) i.e. (P, t, r) f7−→ (m,A).

Such a function would be described by saying f : R3 → R2”.

A natural question in this context: How much more or less would you expect topay if your interest rate is slightly lowered and/or your term is slightly increased?

Our “big picture” goal

We want to study the calculus of functions f : Rn → Rm; that is, functions whichtake inputs with n coordinates and transform them into outputs with m coordi-nates. As another (purely mathematical) example, here is a function f : R4 → R3:

f(w, x, y, z) = (w3 sin y, x− 2w2 + 5yz, zezy)

Such f are called functions of several variables, and Calculus 3 is often calledseveral-variable calculus or multivariable calculus. The inputs and outputs ofthese functions will be called vectors, so Calculus 3 is also called vector calculus.

8


Main questions of vector calculus

Given a function f : Rn → Rm, i.e. a function (x1, x2, ..., xn) f−→ (y1, y2, ..., ym):

1. What is meant by the “limit” of such a function?

When do limits of such functions exist or fail to exist?

How do you compute a limit of a function like this?

2. What is meant by the “derivative” of such a function? How do you computethe derivative of such a function?

What are the analogues of the Product Rule, Quotient Rule, Chain Rule, etc.?

For single-variable functions, the derivative represents the slope of the tan-gent line, but what does the derivative represent in this setting?

3. What is meant by the “integral” of such a function?

How would you compute an integral in this setting?

For single-variable functions, the integral represents an area computation,but what does it represent in this setting?

4. How do we write the Taylor series of such a function? 1

Single-variable functions can be approximated by their Taylor polynomials.Can you do the same thing in this setting; if so, how do you define the Taylorpolynomials of a function of several variables?

We expect the following applications of our study:

1. Analysis of motion: Given an object whose position (either in a plane or in3-dimensional space or perhaps m-dimensional space) at time t is given bysome function t f−→ (x1, ..., xm), compute its velocity and acceleration at anyinstant, determine how far it travels in a given interval of time, etc.

Intuition: velocity and acceleration should have something to do with “derivatives”,and the distance traveled should have something to do with an “integral”.

2. Approximation: Given a function (x1, x2, ..., xn) f−→ (y1, y2, ..., ym), find apolynomial which approximates the function. (We’re especially interestedin writing down something “linear” which approximates the function, thatplays the role of the tangent line to a single-variable function.)

Intuition: this should have something to do with series whose terms contain “deriva-tives” akin to the Taylor series of a single-variable function.

1We probably won’t have time to answer this.

9

1.2. Limits

3. Optimization: Given a function (x1, ..., xn) f−→ y, determine the inputs whichmake y as large or as small as possible (possibly with some constraints on thexjs).

Intuition: this should have something to do with the “derivative” of the function;for a single-variable function, you do this by setting the derivative equal to zero, etc.

4. Volume: Given a 3-dimensional region whose boundary surfaces are de-scribed by graphs, find the volume of the blob.

Intuition: this should have something to do with the “integral” of a function ofseveral variables.

In fact, we’ll do all this and a lot more.

But first, it is important to review what you should know from Math 220 and230, and we cover that in the next four sections (read these at home)!

1.2 LimitsReviewing the theory of Calculus 1 & 2

Let’s start by reviewing a bit more of the theory from Calculus 1 & 2. Here’s whatyou should have in mind:

1. The three important calculus objects (i.e. classes of limits) are derivatives,integrals and series.

2. Each of the three main classes of limits have a specific problem that motivatestheir discovery.

3. To solve each motivating problem theoretically, you “approximate and take alimit”. This means you approximate the solution, observe how your approx-imation can be made better, and last, decree the answer to the motivatingproblem to be a limit of your approximation.

4. In practice you do not actually use the theory to compute the calculus objects.You use “shortcuts”.

The details are on the next page:

10

1.2. Limits

Calculus 1 & 2 theory in one chart

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atio,C

omparison,etc.)

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f(x)=∞∑n=

0

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0

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11

1.2. Limits

Limits

Recall: Suppose f : R→ R. To say

limx→a

f(x) = L

means that as x gets closer and closer to a, then f(x) gets closer and closer to L.This suggests that the graph of f looks like one of the following three pictures:

() =

()

The graph on the left is “continuous” at a; the other two graphs are not. Moreprecisely,

Definition 1.1 A function f : R → R is continuous at a if limx→a

f(x) = f(a). f iscalled continuous if it is continuous at every point in its domain.

Most “reasonable” functions are continuous, as seen in the following theorem:

Theorem 1.2 Any function which is the sum, difference, product, composition, and/orquotient of functions made up of constants, powers of x, sines, cosines, arcsines, arc-tangents, exponentials and/or logarithms is continuous everywhere except where anydenominator is zero.

This theorem suggests that to evaluate most limits, you should start by plug-ging in a for x. If you get a number, that is usually the answer.

EXAMPLE 4Evaluate these limits:

(a) limx→π/3

(3 sin 2x)

Solution: limx→π/3

(3 sin 2x) = 3 sin 2(π3 ) = 3√

32 .

12

1.2. Limits

(b) limx→4

x+3x−2

Solution: limx→4

x+3x−2 = 4+3

4−2 = 72 .

EXAMPLE 5Evaluate these limits:

(a) limx→2

x−5(x−2)2

Solution: limx→2

x−5(x−2)2 = 2−5

(2−2)2 = −30 .

Since we obtain a fraction of the form nonzero #0 , the answer is either∞ or −∞.

To determine which it is, analyze the signs of the numerator and denominatorcarefully:

Numerator: as x→ 2, x− 5→ −3, so the numerator is negative.

Denominator: (x− 2)2 is something squared, so it must be positive.

Since the fraction is of the form −+ , the answer must be −∞ (if it was −− or +

+ ,the answer would be∞).

(b) limx→5

x2−3x−10x−5

Attempted solution: limx→5

x2−3x−10x−5 = 52−3(5)−10

5−5 = 00 .

When you get 00 in a limit, the answer could be anything. To deal with this

situation, do one of the following things:

• Rewrite the function you’re taking the limit of (usually this means fac-toring and cancelling). In Example 5 (b), this means

limx→5

x2 − 3x− 10x− 5 = lim

x→5

(x− 5)(x+ 2)x− 5 = lim

x→5(x+ 2) = 7 .

• Alternatively, use L’Hôpital’s Rule (see the next page)

13

1.2. Limits

Theorem 1.3 (L’Hôpital’s Rule) Suppose f and g are differentiable functions. Sup-pose also that either

limx→a

f(x) =limx→a

g(x) = 0 or limx→a

f(x) =limx→a

g(x) = ±∞.

Then:limx→a

f(x)g(x)

L=limx→a

f ′(x)g′(x) .

In Example 5 (b), the solution via L’Hôpital’s Rule looks like this:

limx→5

x2 − 3x− 10x− 5 = 0

0L= limx→5

2x− 31 = 2(5)− 1 = 7 .

One-sided limitsEXAMPLE 6

Let f(x) = |x|x

=

1 if x > 0−1 if x < 0 . What is lim

x→0f(x)?

Here is a graph of f :

-4 -3 -2 -1 1 2 3 4

-2

-1

1

2

Definition 1.4 Suppose f(x) is defined for all x near a with x > a. If (whenever xgets closer and closer to a from the right, f(x) approaches L), then we say the limit off(x) as x approaches a from the right is L and we write

limx→a+

f(x) = L.

Suppose f(x) is defined for all x near a with x < a. If (whenever x gets closer andcloser to a from the left, f(x) approaches L), then we say the limit of f(x) as xapproaches a from the left is L and we write

limx→a−

f(x) = L.

These are also called, respectively, left-hand limits and right-hand limits. Collec-tively, left- and right-hand limits are referred to as one-sided limits.

14

1.2. Limits

EXAMPLE 6, REVISITED

In the previous example where f(x) = |x|x

,

limx→0+

f(x) = 1 but limx→0−

f(x) = −1 .

Theorem 1.5 limx→a

f(x) exists only if limx→a+

f(x) and limx→a−

f(x) both exist and areequal. In this situation,

limx→a

f(x) = limx→a+

f(x) = limx→a−

f(x).

So we see that if f(x) = |x|x

,

limx→0

f(x) DNE

because the left- and right-hand limits are unequal.

15

1.3. Derivatives

1.3 DerivativesDefinition 1.6 (Limit definition of the derivative) Let f : R → R be a functionand let x be in the domain of f . If the limit

limh→0

f(x+ h)− f(x)h

exists and is finite, say that f is differentiable at x. In this case, we call the value ofthis limit the derivative of f and denote it by f ′(x) or df

dxor dy

dxor d

dx(y).

Differentiable functions are smooth, i.e. are continuous and do not have sharpcorners, vertical tangencies or cusps.

Assuming it exists, the derivative f ′(x) computes:

1. the slope of the line tangent to f at x (a.k.a. the slope of the graph of f at x);

2. the instantaneous rate of change of y (or f ) with respect to x;

3. the instantaneous velocity at time x (assuming f(x) is the position of theobject at time x).

Derivatives have many applications. The most important (for Calculus 3 pur-poses) is that given a differentiable function f , you can approximate values of fnear a using the tangent line to f at a:

Definition 1.7 Given a differentiable function f and a number a at which f is differ-entiable, the tangent line to f at a is the line whose equation is

y = f(a) + f ′(a)(x− a)

a.k.a.L(x) = f(a) + f ′(a)(x− a).

For values of x near a, f(x) ≈ L(x); approximating f(x) via this procedure is calledlinear approximation.

16

1.3. Derivatives

EXAMPLE 7Estimate sin 1

3 by linear approximation.

Solution: To apply linear approximation, we need to choose a function f , a valueof a, and a value of x. Since we are approximating a value of sine, let’s let f(x) =sin x. It seems natural to let x = 1

3 . That leaves a, which should be a number closeto 1

3 and easy to work with. Let’s try a = 0. So the linear approximation is

sin 13 = f

(13

)≈ L

(13

)= f(0) + f ′(0)

(13 − 0

)= sin 0 + (cos 0)

(13 − 0

)= 0 + (1)

(13

)

= 0 + 13 = 1

3 .

You can also conclude many things about the graph of f from looking at itsderivative and its higher-order derivatives, as defined below:

Definition 1.8 Let f : R→ R be a function.

• The zeroth derivative of f , sometimes denoted f (0), is just the function f itself.

• The first derivative of f , sometimes denoted f (1) or dydx

, is just f ′.

• The second derivative of f , denoted f ′′ or f (2) or d2ydx2 , is the derivative of f ′:

f ′′ = (f ′)′.

• More generally, the nth derivative of f , denoted f (n) or dnydxn

,is the derivative off (n−1):

f (n) = ((((f ′)′) · · ·′)′

The first derivative of a function measures its tone (when f ′ > 0, the functionis increasing; when f ′ < 0, the function is decreasing). The second derivative of afunction measures its concavity (when f ′′ > 0, the function is concave up; whenf ′′ < 0, the function is concave down).

17

1.3. Derivatives

Differentiation rules

We do not compute derivatives using the limit definition given above. We usedifferentiation rules (given on the next two pages), i.e. a list of functions whosederivative we memorize, and a list of rules which tell us how to differentiate morecomplicated functions made up of pieces whose derivatives we remember.

Theorem 1.9 (Derivatives to memorize) .

Constant Functions ddx

(c) = 0

Power Rule So long as n 6= 0, ddx

(xn) = nxn−1

Special cases of the Power Rule include:

d

dx(mx+ b) = m

d

dx(√x) = 1

2√x

d

dx

(1x

)= −1

x2

Trigonometric Functions

d

dx(sin x) = cos x d

dx(cosx) = − sin x d

dx(tan x) = sec2 x

Exponential and Logarithmic Functions

d

dx(ex) = ex

d

dx(ln x) = 1

x

Inverse Trigonometric Functions

d

dx(arctan x) = 1

x2 + 1d

dx(arcsin x) = 1√

1− x2

There may be some other derivatives you know, like these:

d

dx(cotx) = − csc2 x

d

dx(secx) = secx tan x d

dx(cscx) = − cscx cotx

But the really important ones are in Theorem 1.9 above.

18

1.3. Derivatives

Theorem 1.10 (Differentiation Rules) Let f : R→ R and g : R→ R be differen-tiable functions, and let k be a constant. Then:

Sum Rule (f + g)′(x) = f ′(x) + g′(x)

Difference Rule (f − g)′(x) = f ′(x)− g′(x)

Constant Multiple Rule (kf)′(x) = k · f ′(x)

Product Rule (fg)′(x) = f ′(x) g(x) + g′(x) f(x)

Quotient Rule(fg

)′(x) = f ′(x) g(x)−g′(x) f(x)

[g(x)]2

Chain Rule (f g)′ (x) = f ′(g(x)) g′(x)

To take the derivative of a function of the form f(x)g(x), you rewrite it as eg(x) ln f(x)

and use the Chain and Product Rules.

EXAMPLE 8Find the derivative of f(x) = 3x4 cos 2x.

Solution: Use the Product Rule first, and the Chain Rule to find the derivativeof cos 2x:

f ′(x) = [3x4]′ cos 2x+ [cos 2x]′3x4

= [12x3] cos 2x+ [− sin 2x · 2]3x4

= 12x3 cos 2x− 6x4 sin 2x .

EXAMPLE 9Find the slope of the line tangent to f(x) = 6e4x

ex+2 at the point (0, 2).

Solution: We need to compute the derivative. Use the Quotient Rule first, thenthe Chain Rule to compute the derivative of the numerator:

f ′(x) = [6e4x]′(ex + 2)− [ex + 2]′(6e4x)(ex + 2)2 = [24e4x](ex + 2)− [ex](6e4x)

(ex + 2)2

Now plug in x = 0 to get the slope:

f ′(0) = [24e4(0)](e0 + 2)− [e0](6e4(0))(e0 + 2)2 = [24]3− [1](6)

(3)2 = 669 = 22

3 .

19

1.4. Integrals

1.4 IntegralsDefinition 1.11 Given function f : [a, b] → R, the definite integral of f from a tob is ∫ b

af(x) dx = lim

||P||→0

n∑k=1

f(ck)∆xk,

where the expression inside the limit is a Riemann sum for f .

Note: In Math 230, the limit above always exists (but it doesn’t always exist forcrazy functions f ... take Math 430 for more on that).

The definite integral of a function is a number which is supposed to give thesigned area of the region between the graph of f and the x-axis. Area above thex-axis is counted as positive area; area below the x-axis is counted as negative area.

As with derivatives, we do not compute integrals with this definition. We usethe following important circle of ideas:

Definition 1.12 Given function f , an antiderivative of f is a function F such thatF ′ = f .

Ex: F (x) = sin x is an antiderivative of f(x) = cos x.

Every continuous function has an antiderivative (although you may not be ableto write its formula down); any two antiderivatives of the same function must dif-fer by a constant (so if you know one antiderivative, you know them all by addinga +C to the one you know).

Definition 1.13 Given function f , the indefinite integral of f , denoted∫f(x) dx,

is the set of all antiderivatives of f .

Ex:∫

cosx dx = sin x+ C.

20

1.4. Integrals

Theorem 1.14 (Fundamental Theorem of Calculus Part II) Let f be continuouson [a, b]. Suppose F is any antiderivative of f . Then

∫ b

af(x) dx = F (x)|ba = F (b)− F (a).

Despite the similar notation,∫f(x) dx and

∫ ba f(x) dx are very different objects.

The first object is a set of functions; the second object is a number.

Theorem 1.15 (Integration Rules to Memorize)∫0 dx = C∫M dx = Mx+ C∫xn dx = xn+1

n+ 1 + C (so long as n 6= −1)∫x−1 dx =

∫ 1xdx = ln |x|+ C (I don’t care so much about the | |)∫

sin x dx = − cosx+ C∫cosx dx = sin x+ C∫sec2 x dx = tan x+ C∫ex dx = ex + C∫ 1x2 + 1 dx = arctan x+ C∫ 1x2 + a2 dx = 1

aarctan x

a+ C∫ 1√

1− x2dx = arcsin x+ C

As with derivatives, there may be other integrals you know, but those are lessimportant.

21

1.4. Integrals

Theorem 1.16 (Linearity of Integration) Suppose f and g are integrable functions.Then: ∫ b

a[f(x) + g(x)] dx =

∫ b

af(x) dx+

∫ b

ag(x) dx;∫ b

a[f(x)− g(x)] dx =

∫ b

af(x) dx−

∫ b

ag(x) dx;∫ b

a[k · f(x)] dx = k

∫ b

af(x) dx for any constant k;

(and the same rules hold for indefinite integrals).

Here are some other basic properties of integrals:

Definition 1.17 Let a < b and let f : [a, b]→ R be integrable. Then

∫ a

bf(x) dx = −

∫ b

af(x) dx

and ∫ a

af(x) dx = 0.

Theorem 1.18 (Additivity property of integrals) Suppose f is integrable. Thenfor any numbers a, b and c,

∫ b

af(x) dx+

∫ c

bf(x) dx =

∫ c

af(x) dx

EXAMPLE 10Compute the integral:

∫ (sin x

4 + 2x3 − 4x

+ ex − 24√x

)dx

Solution: Rewrite the integrand using exponent rules, then use usual integrationrules:∫ (

sin x4 + 2x3 − 4

x+ ex − 24√

x

)dx =

∫ (14 sin x+ 2x3 − 41

x+ ex − 24x−1/2

)dx

= −14 cosx+ 1

2x4 − 4 ln x+ ex − 48x1/2 + C .

22

1.4. Integrals

EXAMPLE 11Suppose

∫ 72 f(x) dx = 8 and

∫ 74 f(x) dx = 1. Find

∫ 24 f(x) dx.

Solution:∫ 7

2 f(x) dx =∫ 4

2 f(x) dx +∫ 7

4 f(x) dx so we have 8 =∫ 4

2 f(x) dx + 1.Therefore

∫ 42 f(x) dx = 8− 1 = 7 so

∫ 24 f(x) dx = −

∫ 42 f(x) dx = −7 .

Integration by u-substitution

Difficult integrals require more advanced techniques. You should be very profi-cient with u-substitutions, which are useful to integrate functions which are theproduct of two or more related terms:

Theorem 1.19 (Integration by u−substitution - Indefinite Integrals)∫f(g(x)) · g′(x) dx =

∫f(u) du

by setting u = g(x).

Theorem 1.20 (Integration by u−substitution - Definite Integrals)

∫ b

af(g(x)) · g′(x) dx =

∫ g(b)

g(a)f(u) du

by setting u = g(x).

EXAMPLE 12Compute ∫ 3

2

8x(x2 + 2)2 dx

Solution: Let u = x2 + 2 so that du = 2x dx and therefore 4 du = 8x dx. Next,change the limits of integration: when x = 2, u = 22 + 2 = 6 and when x = 3,u = 32 + 2 = 11. So the integral becomes∫ 3

2

8x(x2 + 2)2 dx =

∫ 11

6

4u2 du =

∫ 11

64u−2 du =

[−4u−1

]11

6

= −411 −

−46 = 2

3 −411 = 10

33 .

23

1.4. Integrals

It is helpful to memorize the following idea, which comes from using a u-substitution u = mx+ b:

Theorem 1.21 (Linear Replacement Principle) Suppose you know∫f(x) dx = F (x) + C.

Then for any constants m and b,∫f(mx+ b) dx = 1

mF (mx+ b) + C.

EXAMPLE 13Compute ∫ 1

0ex/3 dx.

Solution: Using the Linear Replacement with m = 13 , we see

∫ 1

0ex/3 dx =

[3ex/3

]10

= 3e1/3 − 3 .

Integration by parts

Integrals whose integrands are products of unrelated terms are often computedusing integration by parts:

Theorem 1.22 (Integration by Parts (IBP) Formula)∫u dv = uv −

∫v du.

EXAMPLE 14Compute

∫x sin x dx.

Solution: Use integration by parts. Let u = x and dv = sin x dx so that v =∫dv =

− cosx and du = dx. Then by the parts formula,∫x sin x dx =

∫u dv = uv −

∫v du = x(− cosx)−

∫(− cosx) dx

= −x cosx+ sin x+ C .

One other technique useful to evaluate some integrals is to find the partial frac-tion decomposition of the integrand; we omit this technique in these notes becauseintegrals using this method are rare in Math 320.

24

1.5. Series

1.5 SeriesAn infinite series is an attempt to add an infinite list of numbers

∞∑n=1

an = a1 + a2 + a3 + ...

If the infinite list of numbers has a finite sum, we say the series converges; other-wise we say the series diverges. More formally:

Definition 1.23 Let∑an be an infinite series. For each N , let SN be the N th partial

sum of the series; this is defined to be the sum of all the an for which n ≤ N . Then:

1. If L is a real number such that limN→∞

SN = L, then we say the infinite seriesa1 + a2 + a3 + ... converges (to L) and write

∑an = L. In this setting L is

called the sum of the series.

2. If limN→∞

SN = ±∞ or if limN→∞

SN DNE, then we say the infinite series∑an

diverges.

Important: There is a big difference between saying “∑an converges” and say-

ing “an converges”. Without the Σ, you aren’t adding the numbers. Therefore, youshould never omit the Σ when describing whether or not an infinite series con-verges.

We divide convergent series into two classes as follows:

Definition 1.24 Let∑an be an infinite series. We say the series is absolutely con-

vergent (or that the series converges absolutely) if∑ |an| converges.

If∑an converges but

∑ |an| diverges, then we say∑an is conditionally con-

vergent (or that the series converges conditionally).

The reason we care whether a series converges absolutely or conditionally isthe following theorem:

Theorem 1.25 (Rearrangement Theorem) Suppose∑an is an infinite series.

1. If∑an converges conditionally, then the terms of that series can be rearranged

so that the rearranged series converges to any number you like! (The series canalso be rearranged so that the rearranged series diverges.)

2. If∑an converges absolutely to L, then no matter how the terms of the series are

regrouped or rearranged, the rearranged series still converges absolutely to L.

25

1.5. Series

The major application of series is to obtain alternate representations of func-tions which can be used for approximations:

Definition 1.26 Suppose f is a function which can be differentiated over and overagain at x = 0. The Taylor series (centered at 0) of f (a.k.a. Maclaurin series off is the power series

∞∑n=0

f (n)(0)n! xn.

If we truncate this series at the N th power term, we obtain a partial sum of the Taylorseries called the N th Taylor polynomial (centered at 0) of f . This polynomial isdenoted PN(x).

General properties of Taylor polynomials:

1. PN(x) is a polynomial of degree ≤ N ;

2. P0(x) is the constant function of height f(0);

3. P1(x) is the tangent line to f when x = 0;

4. PN(x) is the best N th degree polynomial approximation to f near 0.

The point of Taylor series and Taylor polynomials is that if a function f hasN th Taylor polynomial PN , then PN(x) ≈ f(x) (this approximation improves as Nincreases, but is pretty good even for small N ).

Theorem 1.27 (Uniqueness of power series) Suppose f is a function which canbe differentiated over and over again at x = 0. Then if we write f as a power series ofthe form

f(x) =∞∑n=0

anxn,

then the coefficients an must satisfy

an = f (n)(0)n! for all n.

In other words, the only power series of the form∑anx

n which can represent f is itsTaylor series centered at 0.

26

1.5. Series

EXAMPLE 15Suppose f(x) = ex

2 . Find the fifth Taylor polynomial of f , and use it to approxi-mate f(1

2) and∫ 1

0 f(x) dx.

Solution: In Calculus 2, we memorize the Taylor series of ex:

ex =∞∑n=0

xn

n! = 1 + x+ x2

2 + x3

3! + ...

Replacing all the xs with x2s on both sides of the above equation, we get the Taylorseries of f :

f(x) = ex2 = 1 + (x2) + (x2)2

2 + (x2)3

3! + ...

= 1 + x2 + x4

2 + x6

3! + ...

Now the fifth Taylor polynomial of f is obtained by keeping all the terms of theabove Taylor series up to and including the x5 term. This leaves only three nonzeroterms:

P5(x) = 1 + x2 + x4

2Now we can answer the questions:

f(12) ≈ P5(1

2) = 1 +(1

2

)2+

(12

)4

2 = 1 + 14 + 1

32 = 4132 .

∫ 1

0f(x) dx ≈

∫ 1

0P5(x) dx =

∫ 1

0

(1 + x2 + 1

2x4)dx =

[x+ 1

3x3 + 1

10x5]1

0

= 1 + 13 + 1

10

= 4130 .

27

1.6. Homework exercises

1.6 Homework exercisesWARNING: At any time in this (or any) math course, you may be asked to

compute a quantity which does not exist, in which case you should say so, withappropriate justification.

SECOND WARNING: Correct units should always be used in the answer toany problem that has units provided to you.

In each of Problems 1-14, you are given a real world situation which needs tobe modeled by a function f , similar to those of Examples 1-3 in Section 1.1. Youare to write down the domain and codomain of the function f , using appropriatenotation. For example, in the context of Example 1 on p. 7, the answer I’m lookingfor here would be “f : R3 → R”.

1. At each point in a river, you want to determine the water temperature at thatpoint as a function of the location.

2. At each point in a river, you want to determine the velocity of the water as afunction of the location.

3. A bug is crawling around on a table. You want to determine, at each time,the velocity of the bug.

4. A bug is crawling around on a table. You want to determine, at each time,the speed of the bug.

5. A bird is flying through the sky. You want to determine, at each time, theposition of the bird.

6. A bird is flying through the sky. You want to determine, at each time, theacceleration of the bird.

7. You want to determine the wind chill (how cold it “feels”) in terms of thetemperature and the wind speed.

8. You want to determine an amortized loan payment, in terms of the principalof the loan, the interest rate, and the frequency of payments.

9. You want to determine the effective rate of interest on an account, in terms ofthe interest rate and the number of periods.

10. You want to determine the concentration of acid in a solution (held in a jar)at a particular point, in terms of the point and the elapsed time.

28


11. You are an actuary, and you want to determine the probability that a policyholder will have an accident and the expected amount of damage he causes,in terms of the policy holder’s age, the value of the policy holder’s car andthe average number of miles per day the policy holder drives his car.

12. You want to determine consumer demand for a product, in terms of the priceof the product, the price of related goods, and the price of competing goods.

13. You want to determine the (magnitude of the) gravitational force betweentwo objects, in terms of the distance between the objects and the masses ofthe two objects.

14. You want to determine the revenue, profit and costs of a company basedon the number of employees the company hires, the number of factories thecompany uses, the tax rate of the company, and the price of four differentraw materials.

In Problems 15-19, evaluate the given limit.

15. limx→0

3e−2x

16. limx→3

x+4(x−3)2

17. limx→−1

x2−4x−5x2+3x+2

18. (R) limx→0

arctan 2xx

19. limx→π

cosx1+sinx

20. Find limx→3

f(x), where f : R→ R be the piecewise-defined function

f(x) =

x2 − x+ 2 if x < 3

4 if x = 313x

3 − 1 if x > 3.

21. Find limx→π

4

g(x), where g : R→ R be the piecewise-defined function

g(x) =

sin 2x if x < π4

tan 4x if x > π4.

22. Let f(x) = 5 ln x− 3x4 +√x

6 − 2. Find dfdx

.

23. Find g′(x) and g′(1), if g(x) = x3e−3x.

29


24. Find dydx

and d2ydx2 , if y = 3

x−1 .

25. Differentiate f(t) = sin3 4t cos2 2t.

26. (R) Find the derivative of g(x) = x−2+5x−33√8x2+7

27. Let h(x) = 83x−1 . Find h′′(x) and h′′′(x).

28. An object moves back and forth along a horizontal line so that it’s position,in meters, at time t, in minutes, is given by f(t) = 1

3x3 − 4x2 + 12x− 2.

a) Find all times where the velocity of the object is 0.

b) Find the acceleration of the object at time 6.

c) At time 3, is the object speeding up, or slowing down? Explain.

d) At time 5, is the object moving to the left, or moving to the right? Ex-plain.

29. Write the equation of the line tangent to f(x) = 2x2 − 3x+ 4 when x = 5.

30. (R) Estimate arctan 14 using linear approximation.

31. Suppose g is some unknown function such that g(4) = 5 and g′(4) = −2.Estimate g(4.3) using linear approximation.

32. Find the locations of any local maximum value(s) of the function f(x) =2x − 4x3. Explain, using calculus, how you know the value(s) you give aremaxima.

33. Find∫

(12x3 − 2x2 + 6) dx.

34. Compute∫ 3

1 (x+ x−1) dx.

35. Find∫

sin(3x+ π) dx.

36. Compute∫ ( e−x

6 + 3 sec2 x− 8x4

)dx.

37. Find all functions F such that F ′(x) = 3 cosx+ 2 sin x.

38. Evaluate∫ 2

0 xex2dx.

39. (R) Evaluate∫ 2

0 xex dx.

40. Suppose f is some unknown function such that∫ 3

0 f(x) dx = 2 and∫ 1

0 f(x) dx =5. Find

∫ 31 f(x) dx.

41. Suppose an object is moving back and forth along a horizontal line, so thatits velocity at time t is given by 3 sin t+ cos t in/sec. Find the displacement ofthe object between times π

4 and π/2.

30


Selected answers

WARNING: Since I found these (and all) answers throughout this text by hand,it is possible that these contain errors. If you find any, please bring them to myattention.

1. f : R3 → R

2. f : R3 → R3

3. f : R→ R2

4. f : R→ R

8. f : R3 → R

10. f : R4 → R

16. ∞

19. −1

21. DNE

23. g′(x) = 3x2e−3x − 3x3e−3x;

g′(1) = 0.

27. h′′(x) = 144(3x− 1)−3;

h′′′(x) = −1296(3x− 1)−4.

29. y = 17(x− 5) + 39.

30. 14

31. g(4.3) ≈ 4.4.

34. 4 + ln 3

35. −13 cos(3x+ π) + C

37. F (x) = 3 sin x− 2 cosx+ C.

39. 1 + e2

31

Chapter 2

Vectors and matrices

2.1 Some basic set theoryThe fundamental objects of mathematics are sets and functions. In this section,

we establish some notation regarding the first class of fundamental objects (i.e.sets) that will be used not only in this course, but throughout your future mathe-matics coursework. (We’ll get to functions in Chapter 3.)

Definition 2.1 (Basic language associated to sets)

1. A set is any definable collection of objects. Sets are usually denoted by capitalletters.

2. The members of a set are called elements of the set; if x is an element of set Athen we write x ∈ A. If x is not an element of A, we write x /∈ A.

3. If every element of set E is also an element of set F , we say E is a subset of Fand write E ⊆ F or F ⊇ E.

4. Two sets E and F are said to be equal if E ⊆ F and F ⊆ E, in which case wewrite E = F .

5. The empty set, denoted ∅, is the set with no elements.

Remarks:

1. the key word in part (1) of the above definition is “definable”. This basicallymeans that the set can be described without creating any kind of logical con-tradiction. For more on a collection which isn’t definable, Google “Russell’sparadox”.

2. To say two sets are equal means that they contain exactly the same elements.

32

2.1. Some basic set theory

3. Note the difference between “∈” and “⊆”: the first symbol should be pre-ceded by an element; the second symbol should be preceded by a subset.

4. There is only one empty set, so we say “the empty set”, not “an empty set”.

Set-builder notation

We often describe sets with “set-builder” notation. For instance, to say somethinglike

E = x ∈ R : 2 < x ≤ 5

means (in English) that E is the set of real numbers x such that 2 < x ≤ 5 (in otherwords, E is the interval (2, 5]).

A picture of E would look something like this:

EXAMPLE 1Let F = [0, 10] and let G = x ∈ R : 7 < x ≤ 9. Classify the following statementsas true or false:

1. 8 ∈ G

2. 12 ∈ F

3. G ∈ F

4. G ⊆ F

5. F ⊆ G

Last, draw a picture illustrating F and G.

33


Operations on sets

Definition 2.2 Let E be a set. The complement of E, which is denoted Ec, EC , E ′,Ω− E, E (and other ways as well), is the set of objects which do not belong to E.

Remark: The complement of a set depends on what the “universal set” is; i.e.the complement of 0, 1, 2might mean the integers which aren’t 0, 1 or 2; it mightmean the real numbers which aren’t 0, 1 or 2; etc. So exactly what is meant by“complement” depends on the context.

Definition 2.3 Let E and F be sets. The union of E and F , denoted E ∪ F , is theset of objects which are elements of E or elements of F (or elements of both E and F ).

Definition 2.4 Let E and F be sets. The intersection of E and F , denoted E ∩ F ,is the set of objects which are elements of both E and F .

E and F are called disjoint if they have no elements in common, i.e. E ∩ F = ∅.

If E and F are disjoint, then they should have no “overlap”, like this:

34


Definition 2.5 Let E and F be sets.

1. The Cartesian product of E and F , denoted E × F , is the set of ordered pairs(x, y) such that x ∈ E and y ∈ F .

2. Given set E, the Cartesian square of E is E2 = E×E and the nth Cartesianpower of E is En = E × E × · · · × E, the set of ordered n-tuples of elementsof E.

EXAMPLE 2Let A = [0, 4); let B = [1, 6]; and let E = [2,∞). Sketch a picture of each set:

1. A ∩B

2. B ∪ E

3. B ∩ AC

4. B × E

35


EXAMPLE 2, CONTINUED

Recall that A = [0, 4); B = [1, 6]; E = [2,∞). Sketch a picture of each set:

5. E ×B

6. A2

7. H = (B × A) ∩ (x, y) ∈ R2 : 0 ≤ y ≤ x2

36


A description of Rn

We are most interested in the Cartesian powers of R, because these sets serve asthe domain and codomain of the functions we study in Math 320.

Definition 2.6 Rn is called n-dimensional Euclidean space.

• R1 = R = the set of real numbers (for more on what exactly the set of realnumbers is, take Math 430)

– R usually depicted geometrically as a number line or axis; real numberscan be thought of as points on this axis

– R is ordered, i.e. given two real numbers, one is greater than (or equalto) the other

• R2 = (x, y) : x ∈ R, y ∈ R

– R2 usually depicted geometrically as a plane; elements of R2 are de-picted as points in this plane

– R2 is not ordered in any meaningful sense

• R3 = (x, y, z) : x ∈ R, y ∈ R, z ∈ R

– R3 usually depicted geometrically as 3-D space (in mathematics, we usea right-hand rule to draw the “x”, “y” and “z” directions); elements ofR3 are depicted as points in space

37

2.2. Introducing vectors

– R3 is not ordered in any meaningful sense

• Rn = (x1, x2, ..., xn) : xj ∈ R for all j

– Rn for n ≥ 4 is hard to depict. If we had to, we could think of a picturelike this:

– Rn is not ordered in any meaningful sense (unless n = 1)

2.2 Introducing vectorsIf I were to ask you “what is a vector?” you would probably tell me (assuming

you have heard of vectors before) that...

a vector is an object with a and a .

This is not quite right. There are exotic examples of vectors which do not haveand/or , but we won’t deal with those in

this course.

In reality, vectors are rather abstract objects. If someone were to ask you whata vector is, the correct answer is that...

a vector is an element of a vector space.

Of course, that begs the question of what a vector space is. Put simply, a vectorspace is a collection of “things” (called vectors) for which:

• you can add two of the “things”, and the sum is the same type of “thing”;

• you can multiply any of the “things” by a real number, and the result is a“thing”.

The only restriction is that we want the addition and multiplication defined aboveto be “reasonable”, i.e. to have rules that mirror the rules for addition and multi-plication of numbers. More formally, we make the following definition:

38


Definition 2.7 A vector space V is a set, together with two operations:

• addition: + : V × V → V (i.e. (u,v) +7−→ u + v)

• scalar multiplication: R× V → V (i.e. (c,v) ×7−→ cv)

such that the addition and scalar multiplication obey all the usual rules of arithmetic.In particular, there must be a special vector in V , called the zero vector and de-

noted 0, for which v + 0 = v for all v ∈ V .In this context, real numbers are called scalars; elements of the vector space are

called vectors.

Remarks:

1. The “usual rules of arithmetic” mean things like commutativity (i.e. u + v =v + u), associativity (i.e. u + (v + w) = (u + v) + w), distributive laws (likek(u + v) = ku + kv) and the identity laws (including 0v = 0, 1v = v andv + 0 = v). In Math 320, we don’t need to dwell on this (if you need details,consult my Math 322 lecture notes); you just need to know that all the normalrules for addition and scalar multiplication still work.

2. Vectors are usually referred to by boldface letters (like v) when typed, and asletters with arrows over them (like −→v ) when hand-written. However, some-times we get lazy and just refer to a vector with a letter (like v). The zeroscalar is denoted 0; the zero vector is denoted 0 or

−→0 .

3. Warning: Vectors from two different vector spaces cannot be added to oneanother; for example, if V and W are two unrelated vector spaces with u,v ∈V but w ∈ W , then u + v makes sense but u + w is nonsense.

Rn as a vector space

For our purposes, what we really care about is that each Rn is a vector space. Thismeans you can add things in Rn to other things in Rn, and multiply things in Rn

by scalars. You do this coordinate-wise, as indicated below:

Theorem 2.8 The set Rn is a real vector space, where the addition and scalar multi-plication are defined coordinate-wise, i.e.

x + y = (x1, x2, ..., xn) + (y1, y2, ..., yn) = (x1 + y1, x2 + y2, ..., xn + yn)

andcx = c(x1, ..., xn) = (cx1, cx2, ..., cxn).

39


EXAMPLE 3Let v = (2,−1, 3, 0) and w = (−3,−2, 0, 2).

1. To what vector space should one assume v and w belong?

2. Compute 4v.

3. Compute v + w.

4. Compute 2v− 3w.

5. Compute v + (2,−5, 1).

Pictorial representations of vectors in Rn: To get a picture of elements of Rn

(or any vector space, really), we think of them as “arrows”:

vPPPPPq

PPPPPq

In reality, vectors in Rn are more like “floating” arrows, in that two vectorspointing the same direction with the same length are really the same vector (evenif they are drawn in different spots).

Using the idea of vectors as “arrows”, vector addition then corresponds to“head-to-tail” or “parallelogram” addition:

w

PPPPPqv

Scalar multiplication corresponds to “stretching”:

:

v

40


Vectors vs. points

In §2.1, I told you that elements of Rn can be thought of as points in coordinatespace (which have a fixed position), but now I just said that elements of Rn canbe thought of as floating arrows. These concepts seem to contradict one another -which of these ideas is correct?

In reality, there is no difference between a point in a vector space and a vectorin a vector space, because you can do all the same math with points that you cando with vectors: if you take a vector v ∈ Rn and draw it so that it starts at theorigin, the vector will end at a point whose coordinates are specified by v, so wemay as well also call that point v.

In this course, we will use the word “point” to emphasize that we are thinkingabout something with a fixed position, and we will use the word “vector” whenwe are describing something that conceivably we want to “float around”.

Examples of vector arithmetic in Rn

EXAMPLE 4Let u = (3,−1) and v = (2, 1).

-7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

41


EXAMPLE 5Let v = (1, 2, 4) and w = (−6,−2, 5).

Important:

The vector that goes from the end of w to the end of v is v−w .

EXAMPLE 6Let a = (0,−2, 3, 1) and b = (2, 8,−1, 4).

EXAMPLE 7Suppose v1,v2 and v3 are vectors in R3; suppose w1 and w2 are vectors in R4; sup-pose all other letters represent scalars. Determine whether each of the followingexpressions are scalars, vectors, or if the expression is nonsense.

1. v1 + 5cv3

2. c(v1 − v1)

3. c(dv2)

4. v1(v2 + cv3)

5. v2(c+ d)

6. v1 + w1

7. 1cv1

8. 0w2

9. c(d+ c)

42


Standard basis vectors

Notice that any vector v = (v1, v2) ∈ R2 can be written as

v = v1(1, 0) + v2(0, 1)

More generally, we define for each n a special list of vectors in Rn, which playthe same role in Rn that the set (1, 0), (0, 1) does in R2:

Definition 2.9 The standard basis of Rn is the list of vectors e1, e2, ..., en (writ-ten in that order) of Rn where

e1 = (1, 0, 0, ..., 0), e2 = (0, 1, 0, ..., 0), · · · en = (0, 0, ..., 0, 1)

In general, ej is a vector that has a 1 in the jth position and 0s in all other positions.

Theorem 2.10 Let e1, ..., en be the standard basis of Rn. Then for any vector v =(v1, ..., vn) ∈ Rn,

v =n∑j=1

vjej = v1e1 + ...+ vnen.

PROOF This is a direct calculation similar to what was done above for R2:

v = (v1, ..., vn) = (v1, 0, ..., 0) + (0, v2, 0, ..., 0) + ...+ (0, ..., 0, vn)= v1(1, 0, ..., 0) + v2(0, 1, 0, ..., 0) + ...+ vn(0, ..., 0, 1)= v1e1 + v2e2 + ...+ vnen.

Here’s an example of Theorem 2.10 in action:

(7,−5, 3, 0) = 7(1, 0, 0, 0)+(−5)(0, 1, 0, 0)+3(0, 0, 1, 0)+0(0, 0, 0, 1) = 7e1−5e2+3e3+0e4.

Alternate notation for the standard basis vectors in R2 and R3:In R2,

e1 = (1, 0) is also called i;e2 = (0, 1) is also called j.

So to say v = 4i− 2j means v = (4,−2).

43


In R3,

e1 = (1, 0, 0) is also called i;e2 = (0, 1, 0) is also called j;e3 = (0, 0, 1) is also called k.

So to say v = 2i− j + 4k means v = (2,−1, 4).

i

j

1

1

j

k

i1

1

1

I dislike the i, j,k notation because:

1. it doesn’t extend to higher dimensions very well (only 26 letters);

2. i is also used in math for√−1;

3. writing v = 2i + j by itself doesn’t tell you whether v is in R2 (i.e. v = (2, 1))or if v is in R3 (i.e. v = (2, 1, 0)).

That said, this notation has its uses, and we will use it from time to time.

EXAMPLE 8Let v = (4,−2, 3, 5). Compute v− 3e3 + 7e2.

44

2.3. Vector geometry and dot product

2.3 Vector geometry and dot productMuch of this section is adapted from Frank Jones’ notes, which can be found

online at this link.

Goals of this section:

1.

2.

Standard notation: Throughout this section, v is the vector (v1, v2, ..., vn) ∈ Rn

and w is the vector (w1, ..., wn) ∈ Rn.

Distances

Recall that distance in R2 is computed using the Distance Formula, which itselfcomes from the Pythagorean Theorem:

a

bc

x2-x1

(x2 ,y2)

(x1 ,y1)

y2-y1

x1 x2

y2

y1

c2 = a2 + b2

c =√a2 + b2 dist((x1, y1), (x2, y2)) =

45

http://www.owlnet.rice.edu/~fjones/chap1.pdf


Now, let’s try figuring the distance between two points/vectors in R3. Let v =(v1, v2, v3) and w = (w1, w2, w3); this gives us a picture like this:

(v1 ,v2 ,v3)

(w1 ,w2 ,w3)

y

x

z

Although we can’t draw a picture in higher dimensions, we can do the samekind of reasoning to find the distance between points in R4, R5, etc. We wouldobtain the following formula:

46


Definition 2.11 Let v,w ∈ Rn. Then the distance from v to w is

dist(v,w) =√√√√ n∑j=1

(vj − wj)2.

The norm of v, denoted ||v||, is the distance from v to 0:

||v|| = dist(v,0) =√√√√ n∑j=1

v2j .

The norm of a vector is also called its length, its magnitude, and its absolute value.

Theorem 2.12 (Elementary properties of distance and norm) Let v,w ∈ Rn.Then:Distance is symmetric: dist(v,w) = dist(w,v);Distances and norms are positive: dist(v,w) ≥ 0, and ||v|| ≥ 0;Distance and norm are definite: dist(v,w) = 0 only if v = w, and

||v|| = 0 only if v = 0;Distance is the norm of the difference: dist(v,w) = ||v−w||;Norms are multiplicative: for any scalar a, ||av|| = |a| ||v||.

EXAMPLE 9Find ||v||, if v = (−4, 2, 5, 1, 0).

Solution: ||v|| =√

5∑j=1

v2j =

√(−4)2 + 22 + 52 + 12 + 02 =

√46 .

EXAMPLE 10Find the distance between the points (3,−2, 5) and (0,−1,−3).

47


Right angles

The next thing we know about planar geometry that we want to generalize tohigher dimensions is the concept of a right angle (i.e. the concept of perpendicular-ity. To do this, we will again appeal to the Pythagorean theorem, which says:

Theorem 2.13 (Pythagorean Theorem) Let a and b be the lengths of two legs of atriangle, and let c be the length of the hypotenuse. Then a2 + b2 = c2 if and only ifthe triangle is a right triangle.

The “only if” part is usually not emphasized in high school, but it’s importanthere, because it tells us how the formula of the Pythagorean theorem can be usedas a test for whether or not an angle is a right angle.

Take two vectors v and w in Rn, and draw them both so that they start at 0.Those two vectors will make a plane, so if you rotate your persepective so that youlook at this plane head-on, you’ll see something like this:

θ

Question: When is θ a right angle?

Answer: Use the Pythagorean Theorem. The angle is a right angle if

(third side)2 = (first side)2 + (second side)2

which in this case is the vector equation

Let’s work this out and see what this implies:

48


To summarize, there is a right angle at 0 if and only ifn∑j=1

vjwj = 0.

The expressionn∑j=1

vjwj has sort of appeared out of nowhere. It turns out that

this quantity is really, really important, so we give this expression a name:

Definition 2.14 The dot product on Rn is the function Rn × Rn → R (where theoutput associated to inputs v = (v1, ..., vn) and w = (w1, ..., wn) is written v · w orvTw) defined by

v ·w =n∑j=1

vjwj = v1w1 + v2w2 + ...+ vnwn.

Note: v ·w is a scalar, not a vector.

EXAMPLE 11

(3,−2, 5) · (1, 0,−2) =

Theorem 2.15 (Elementary properties of dot products) Let v,w,x ∈ Rn andlet a ∈ R. Then:Dot product is symmetric: v ·w = w · v.Dot product is “bilinear”:

(v + w) · x = v · x + w · xv · (w + x) = v ·w + v · x

a(v ·w) = (av) ·w = v · (aw)

Dot product is positive: v · v ≥ 0.Dot product is definite: v · v = 0 only if v = 0.

Note: The bilinearity property extends to the following fact: if vi,wj ∈ Rn andci, dj ∈ R, then (

m∑i=1

civi)·

n∑j=1

djwj

=m∑i=1

n∑j=1

cidj(vi ·wj).

One other property deserves its own theorem: there is the following connectionbetween dot products and norms:

49


Theorem 2.16 Let v ∈ Rn. Then:

v · v = ||v||2.

Restated, this means||v|| =

√v · v.

PROOF This is a direct calculation:

||v||2 = (dist(v,0))2 =√√√√ n∑

j=1(vj − 0)2

2

=n∑j=1

v2j =

n∑j=1

vjvj = v · v.

Unit vectors and normalization

Definition 2.17 A vector v ∈ Rn is called a unit vector if ||v|| = 1.

EXAMPLE 12Describe all the unit vectors in R2.

i

j

1

1

Theorem 2.18 Given any nonzero vector v ∈ Rn, there is a unit vector in the samedirection as v. This unit vector is called a normalized version of v.

PROOF Given v 6= 0, define u = 1||v||v. This vector is clearly in the same direction

as v; u is a unit vector since

||u|| =∣∣∣∣∣∣∣∣∣∣ 1||v||

v∣∣∣∣∣∣∣∣∣∣ = 1||v||||v|| = 1.

EXAMPLE 13Find a unit vector in the same direction as v = (2,−3, 2, 1). Find a vector w oflength 10 in the same direction as v.

50


Orthogonality

Recall that we originally wanted to extend the notion of “perpendicularity” or“right angle” for R2 to higher-dimensions. Based on what we have done, we de-fine:

Definition 2.19 Two vectors v,w ∈ Rn are called orthogonal (a.k.a. perpendicu-lar) if v ·w = 0, in which case we write v ⊥ w.

The interesting thing here is that we have obtained an algebraic test (via the dotproduct) for a geometric fact (whether the vectors are orthogonal).

Theorem 2.20 0 is orthogonal to every vector in Rn, and 0 is the only vector orthog-onal to every v ∈ Rn.

PROOF For the first statement, let x = (x1, ..., xn) ∈ Rn and observe

0 · x = 0x1 + 0x2 + ...+ 0xn = 0,

so 0 ⊥ x.For the second statement, suppose v = (v1, ..., vn) ∈ Rn is orthogonal to every

vector in Rn. That means in particular that v ⊥ v, i.e. v · v = 0. Thus ||v|| = 0, so vmust be the zero vector.

EXAMPLE 14Determine whether or not the vectors (2,−3, 1,−4) and (0, 1, 7, 1) in R4 are orthog-onal.

Solution: Compute the dot product:

(2,−3, 1,−4) · (0, 1, 7, 1) = 2(0) + (−3)1 + 1(7) +−4(1) = 0− 3 + 7− 4 = 0

Since the dot product of the vectors is 0, the two vectors are orthogonal .

Angles and the Cauchy-Schwarz inequality

Our last goal is to generalize what we know about arbitrary angles in the plane(not just right angles) to angles between vectors in Rn. To get started, supposeyou have two vectors in Rn, and you want to find the angle between them. Drawboth vectors so that they start at zero; then, the two vectors make a plane. Rotateyour perspective so that you’re looking directly at this plane, and as before, yousee something like this:

51


θ

From basic trigonometry, we know the Law of Cosines, which says

In this case, we have c = ||v − w||, b = ||w|| and a = ||v||, so the Law of Cosinesbecomes

||v−w||2 = ||v||2 + ||w||2 − 2||v|| ||w|| cos θ(v−w) · (v−w) = v · v + w ·w− 2||v|| ||w|| cos θ

v · v− v ·w−w · v + w ·w = v · v + w ·w− 2||v|| ||w|| cos θv · v− 2 v ·w + w ·w = v · v + w ·w− 2||v|| ||w|| cos θ

−2 v ·w = −2||v|| ||w|| cos θv ·w = ||v|| ||w|| cos θ

We have proven:

Theorem 2.21 (Angle formula for dot product) Let v,w ∈ Rn. Then

v ·w = ||v|| ||w|| cos θ,

where θ is the angle between the two vectors.

EXAMPLE 15Find the angle between the vectors (5, 2) and (1, 7) in R2. Give both the exact valueand a decimal approximation.

Solution: Let v = (5, 2) and let w = (1, 7). From Theorem 2.21, we have

v ·w = ||v|| ||w|| cos θ(5, 2) · (1, 7) =

√52 + 22

√12 + 72 cos θ

19 =√

29√

50 cos θ

arccos(

19√29 · 50

)= θ

1.048 rad ≈ 60.06 ≈ θ

52


What does dot product “mean”?

The formula v ·w = ||v|| ||w|| cos θ gives some insight to what dot product actuallycomputes. Let’s first think about the sign of v · w. Since ||v|| and ||w|| are bothpositive, the angle formula tells us that

v ·w has the same sign as .

Now:

• θ is acute⇒ θ is in , so cos θ is .

(Also, the closer θ is to 0, the closer cos θ is to 1, so the dot product gets closeto ||v|| ||w|| (i.e. the dot product is larger).

• θ is obtuse⇒ θ is in , so cos θ is .

(Also, the closer θ is to π = 180, the closer cos θ is to −1, so the dot productgets close to −||v|| ||w|| (i.e. the dot product is more negative).

Summarizing:

Theorem 2.22 (Interpretation of dot product) Let v,w ∈ Rn.• v ·w > 0 if and only if the angle between v and w is acute. The closer v ·w is

to ||v|| ||w||, the more the vectors v and w point in the same direction.• v ·w = 0 if and only if v ⊥ w.• v ·w < 0 if and only if the angle between v and w is obtuse. The closer v ·w is

to −||v|| ||w||, the more the vectors v and w point in the opposite direction.

Cauchy-Schwarz and triangle inequalities

We can now prove two important inequalities, which appear in various forms ingeometry, linear algebra, abstract algebra, functional analysis, and probability (aswell as here):

Theorem 2.23 (Cauchy-Schwarz Inequality) For any v,w ∈ Rn,

|v ·w| ≤ ||v|| ||w||.

PROOF Using Theorem 2.21,

|v ·w| = |||v|| ||w|| cos θ| = ||v|| ||w|| | cos θ| ≤ ||v|| ||w|| 1.

This proves the result.

53


Theorem 2.24 (Triangle Inequality) For all v,w ∈ Rn,

||v + w|| ≤ ||v||+ ||w||.

More generally, for all u,v,w ∈ Rn,

dist(v,w) ≤ dist(v,u) + dist(u,w).

Remark: Why is this called the “Triangle Inequality”?

+

PROOF To prove the first statement, start with the Cauchy-Schwarz inequality:

v ·w ≤ ||v|| ||w||⇒ 2v ·w ≤ 2||v|| ||w||

⇒ v · v + 2v ·w + w ·w ≤ ||v||2 + 2||v|| ||w||+ ||w||2

⇒ (v + w) · (v + w) ≤ (||v||+ ||w||)2

⇒ ||v + w||2 ≤ (||v||+ ||w||)2 .

Take the square root of both sides to get the desired result.

To prove the second statement of the Triangle Inequality, let x = v − u andy = u−w. Then

x + y = (v− u) + (u−w) = v−w,

so the statement we are looking to prove is

||x + y|| ≤ ||x||+ ||y||.

This is exactly what was proven in the first statement.

Corollary 2.25 (Generalized Triangle Inequality) For all v1,v2, ...,vn ∈ Rn,∣∣∣∣∣∣∣∣∣∣∣∣n∑j=1

vj

∣∣∣∣∣∣∣∣∣∣∣∣ ≤

n∑j=1||vj||.

54

2.4. Matrices and matrix operations

2.4 Matrices and matrix operationsNotation and definitions

Definition 2.26 Given positive integers m and n, an m×n matrix with entries in Ris an array of numbers aij ∈ R where 1 ≤ i ≤ m and 1 ≤ j ≤ n. We denote matricesby capital letters (usually); a matrix with entries aij is usually denoted A. We arrangethe entries of the matrix in a rectangle as follows:

A =

a11 a12 a13 · · · a1na21 a22 a23 · · · a2n

a31 a32. . . ...

...... . . . ...

am1 am2 · · · · · · amn

The set of matrices of size m × n with entries in R is denoted Mmn(R). If m = n,we call A a square matrix; the set of n × n square matrices with entries in R isdenoted Mn(R). Two matrices are equal if they are the same size and if all theirentries coincide, i.e. A = B if they have the same size and if aij = bij for all i, j.

In particular, aij is the entry of A in the ith row and jth column. m is the numberof rows of A; n is the number of columns of A.

Definition 2.27 The trace of a square n× n matrix, denoted tr(A), is the sum of thediagonal entries of that matrix, i.e. tr(A) = a11 + a22 + ...+ ann.

EXAMPLE 16

A =

7 −2 41 −2 50 1 −1

tr(A) =

Definition 2.28 The transpose of anm×nmatrixA, denotedAT orAt, is the n×mmatrix satisfying (aT )ij = aji for all i, j.

EXAMPLE 17

A =

1 −42 −13 0

AT =

55


Definition 2.29 The n × n identity matrix, denoted I or In, is the diagonal n × nmatrix with all diagonal entries equal to 1.

I =

1 0 0 · · · 00 1 0 · · · 00 0 1 ......

... . . . 00 0 · · · 0 1

EXAMPLE 18

Let B =(

3 1−2 −5

).

1. What is the size of B?

2. What is the (1, 2)−entry of B? How is this entry denoted?

3. Compute the trace of B.

4. Write down the transpose of B.

EXAMPLE 19

Let A =

−4 2 5 67 0 −3 24 2 1 −2

.

1. What is the size of A?

2. To what set does A belong?

3. What is the (4, 4)−entry of A?

4. What is the second row of AT ?

5. Compute a21 + a33 − a24.

56


Associating vectors in Rn to column matrices: We associate vectors in Rn ton× 1 column matrices as follows:

x = (x1, x2, ..., xn) ∈ Rn ←→ x =

x1x2...xn

∈Mn1(R).

In particular, an n× 1 matrix is also called a column vector. A column vector withn entries is the same thing as a vector in Rn. Given a vector x = (x1, ..., xn) ∈ Rn, ifwe want to think of that vector as a row vector, we take the transpose of x:

xT =(x1 x2 x3 · · · xn

)∈M1n(R).

Matrix operations

For each m and n, the set Mmn(R) of m × n matrices is a vector space. This meanswe can add matrices, and multiply matrices by scalars. We do this by decreeingthat addition and scalar multiplication of matrices are performed entry-by-entry. Moreprecisely:

Definition 2.30 Given two matrices A,B ∈ Mmn(R) and a scalar r ∈ R, we definethe matrices A+B ∈Mmn(R) and rA ∈Mmn(R) by

A+B =

a11 + b11 a12 + b12 · · · a1n + b1na21 + b21 a22 + b22 · · · a2n + b2n

...... . . . ...

am1 + bm1 am2 + bm2 · · · amn + bmn

and

rA =

ra11 ra12 · · · ra1nra21 ra22 · · · ra2n

...... . . . ...

ram1 ram2 · · · ramn

(equivalently we set (a+ b)ij = aij + bij and (ra)ij = r(aij) for all i, j).

Note:

You can only add matrices of the same size.

57


EXAMPLE 20

Let A =(

4 1 −12 0 −3

), B =

(1 1 20 1 −1

)and C =

(2 11 5

). Compute:

1. A+B

2. 3A− 2B

3. A+ 2C

Theorem 2.31 The operations defined in Definition 2.30 make each Mmn(R) into areal vector space; the additive identity element of Mmn(R) is the m× n zero matrix

0m×n =

0 0 · · · 00 0 · · · 0...

... . . . ...0 0 · · · 0

.

Matrix multiplication

There is another operation one can perform on matrices. The importance of thisoperation will be seen later in the course; for now we simply define it.

Definition 2.32 Suppose A ∈ Mmn(R) and B ∈ Mpq(R). If n = p, then we definethe product AB, which is an m× q matrix AB defined entry-wise by setting

the (i, j)−entry of AB = (the ith row of A) · (the jth column of B).

In symbols, this means

(ab)ij =n(=p)∑k=1

aikbkj.

(If n 6= p, AB is undefined.)

58


Remarks:

1. If A is a square matrix, we write A2 for AA, A3 for AAA, etc.

2. If A isn’t square, then A2 is undefined.

3. In general matrix multiplication is not commutative: AB 6= BA most of thetime, even if both products are defined and are of the same size.

EXAMPLE 21

Let A =(

1 −1 02 1 −3

), B =

(−2 11 3

)and C =

5 −11 20 1

. Compute:

1. AB

2. BA

3. A2

59



Let A =(

1 −1 02 1 −3

); B =

(−2 11 3

); C =

5 −11 20 1

. Compute:

4. B2

Solution:

(−2 11 3

)(−2 11 3

) (5 11 10

)

5. CBA

EXAMPLE 22Compute Ax, where

A =(

1 −1 02 1 −3

)and x = (1,−2, 1).

60


Important (based on Example 23):

Suppose A ∈Mmn(R), i.e. A is an m× n matrix. Then, if x ∈ Rn is a vectorin Rn, then Ax is a vector in Rm. That means the formula

f(x) = Ax

defines a function f : Rn → Rm. This f is something of a “prototype”function of several variables.

(This remark explains why I have been calling the domain of a function of severalvariables Rn and the range Rm throughout these notes, instead of the other wayaround.)

Theorem 2.33 (Properties of matrix multiplication) Let A,B,C be matrices, letI be the identity matrix of the appropriate size and let k ∈ R. Then, so long aseverything is defined, we have:

1. IA = A and BI = B

2. A(BC) = (AB)C

3. k(AB) = (kA)B = A(kB)

4. A(B + C) = AB + AC and (A+B)C = AC + AC

5. (AT )T = A

6. tr(AT ) = tr(A)

7. (rA)T = rAT

8. (A+B)T = AT +BT

9. (AC)T = CTAT

10. tr(A+B) = tr(A) + tr(B)

11. tr(AB) = tr(BA)

Recall: Earlier, I wrote that the dot product v · w can also be written as vTw.The reason is that if v and w are both vectors in Rn, they are both n × 1 matrices,so vT is 1× n, so

vTw = vT1×nwn×1 = (vTw)1×1

so you get a scalar (and the scalar you get is exactly v ·w).

61


EXAMPLE 23Suppose A ∈ Mmn(R), b ∈ Rn, C ∈ Mnn(R), D ∈ Mnm(R). For each of the fol-lowing expressions, determine if the expression is a matrix (in which you shouldgive its size), a vector in some Rk (in which case you should give the value of k), ascalar, or nonsense.

As a convention, if an answer can be thought of as both a matrix and a vectorin some Rk, let’s call it a vector.

1. Ab

2. bA

3. AC2

4. CAb

5. bbT

6. bTb

7. bTCb

8. (bTb)D

9. DACD

Solution: Dn×mAm×nCn×nDn×m = (DACD)n×m so this is an n×m matrix.

10. DAC2b

Solution: Dn×mAm×nCn×nCn×nbn×1 = (DAC2b)n×1 so this is a vector in Rn.

62

2.5. Determinants

2.5 DeterminantsGoal: Assign, to every square matrix, a real number which tells you something

about the matrix.

Motivation: Consider a 2× 2 square matrix

A =(a bc d

).

Treat the columns ofA as vectors: let v = (a, c) and w = (b, d). Use those vectorsto make a parallelogram:

b a

c

d

What is the area of this parallelogram?

b a a+b

c

d

c+d

A = (a+ b)(c+ d)− 2(1

2bd)− 2

(12ac

)− 2bc

= (ac+ bc+ ad+ bd)− bd− ac− 2bc= ad− bc.

(There is a catch: ad− bc could be negative, in which case we need to take absolutevalue of the answer to get the area.)

This computation motivates the following definition:

63

2.5. Determinants

Definition 2.34 Let A =(a bc d

)∈ M2(R). The determinant of A, denoted

detA or det(A), is the number detA = ad− bc.

The determinant of A is also denoted |A|, but this a poor choice of notation thatshould be avoided.

What we have shown is that if you have a parallelogram in R2, and you treatthe sides of the parallelogram as vectors, and put those vectors in as columns of amatrix A, then

area of the parallelogram = | detA|.EXAMPLE 24

Find the determinant of(

2 33 1

).

What, is the signficance of the fact that this determinant is negative?

-1 1 2 3 4

-1

1

2

3

4

In general, the columns of an n× n matrix can be thought of as n vectors in Rn.Those n vectors can be used to make a n−dimensional box called a parallelepiped.

If you take the same vectors and put them in as columns of a matrix, this makesan n × n matrix A. The determinant of this matrix is either +1 or −1 times thevolume of this parallelepiped1. Whether or not it is positive or negative depends

1There is a more rigorous definition of the determinant of a square matrix, but that requiresmathematics beyond the scope of this course.

64

2.5. Determinants

on the “orientation” of the column vectors, i.e. which order they are written in. InR3, this goes like this:

A =

a11 a12 a13a21 a22 a23a31 a32 a33

⇒ detA = ± volume of

(a12,a22,a32)

(a11,a21,a31)

(a13,a23,a33)

Note: If A is not square, then detA is undefined.

Theorem 2.35 (Properties of determinants) Let A,B ∈Mn(R). Then:1. detA = detAT .2. If B is obtained from A by swapping two columns of A, then detB = − detA.3. If B is obtained from A by swapping two rows of A, then detB = − detA.4. If two rows or columns of A coincide, then detA = 0.5. det I = 1, where I is the identity matrix.6. If B is obtained from A by multiplying one row (or column) of A by a constantr, then detB = r detA.

7. det(rA) = rn detA.8. det(AB) = detA · detB.

PROOF Take linear algebra (Math 322).

Computing the determinant of a 1× 1 matrix

The determinant of a 1× 1 matrix is just the number in the matrix: if A = (a), thendetA = a.


As we have seen, the determinant of a 2 × 2 matrix is given by the following for-mula:

det(a bc d

)= ad− bc.

EXAMPLE 25

Find the determinant of A =(

7 123 5

).

Solution: detA = 7 · 5− 3 · 12 = 35− 36 = −1.

65

2.5. Determinants


Definition 2.36 Let A =

a b cd e fg h i

∈M3(R). Then

detA = aei+ cdh+ bfg − bdi− ceg − afh.

The formula in Definition 2.36 is impossible to remember, but there is a trick,called the Rule of Sarrus. Given a 3 × 3 matrix, copy the first two columns to theright of the matrix.

Then multiply along the diagonals, and add the “upper” and “lower” products.The determinant is the bottom sum minus the top sum.

a b cd e fg h i

a bd eg h

WARNING:

The Rule of Sarrus does not work for 4× 4 and larger matrices.

EXAMPLE 26

Let A =

1 −2 32 4 −13 −1 1

. Find detA.

Solution: Use the Rule of Sarrus:

1 −2 32 4 −13 −1 1

1 −22 43 −1

Computing the determinant of a 4× 4 or larger matrix

The easiest and most efficient way to compute a determinant of a large matrix isto use technology (see the last section of this chapter). However, there is a way tocompute determinants by hand called evaluation by minors (look this up onlineif you are interested; this is taught in Math 322 but will not be needed in Calc 3).

66

2.6. Cross products

2.6 Cross productsMuch of our work in this course will be done in R3 (three-dimensional space).

In this setting, it will be valuable to have a method of finding a nonzero vectorwhich is orthogonal to each of two given vectors. Here is a standard method:

Definition 2.37 Let v,w ∈ R3, and let θ be the angle between them. The crossproduct of v and w, denoted v×w, is the vector in R3 with the following properties:

It is orthogonal to both v and w: (v×w) ⊥ v and (v×w) ⊥ w.

Its length is an area computation: ||v × w|| = ||v|| ||w|| | sin θ|, the area of theparallelogram formed by v and w.

Right-hand rule: If you take your right-hand so that its fingers point toward v whenthe hand is open and point toward w when the hand is closed, the thumb pointsin the direction of v×w.

Note that this definition specifies one and only one vector for v × w. The firstcondition tells you a line along which v × w must lie; the second condition spec-ifies the length of the cross product (which reduces you to two possibilities), andthe third condition tells you how to choose between the two possibilities.

EXAMPLE 27Using the geometric definition of cross product, find (0, 3, 0)× (2, 0, 0).

Remark: There is such a thing as a cross product in Rn when n 6= 3, but it israrely used, so for the purposes of our course, we will say the the cross product ofvectors not in R3 is undefined.

Theorem 2.38 (Elementary properties of cross product) Let v,w,x ∈ R3. Then:Anticommutativity: v×w = −(w× v);Distributivity: (v+w)×x = (v×x)+(w×x) and v× (w+x) = v×w+v×x;Associativity with scalars: for any r ∈ R, rv×w = r(v×w) = v× rw;Test for parallelism: v||w if and only if v×w = 0.

67

2.6. Cross products

EXAMPLE 28Suppose v,w and x are three vectors in R3 such that v×w = (1, 3,−2) and w×x =(0, 2, 5). Compute these:

1. v× 2w

2. w× v

3. 3w× (v + x)

4. x× 4x

5. The area of the parallelogram formed by v and w

Note: Cross product is not associative, i.e. in general,

v× (w× x) 6= (v×w)× x

This means that you should never write something like

v×w× x

because it is ambiguous (the answer depends on how you group). However, thereare some interesting formulas that you can concoct out of three vectors in R3, calledtriple products:

Theorem 2.39 (Triple products) Let v,w,x ∈ R3. Then:Scalar triple product: v · (w× x) = (v×w) · x;Bracket rule (a.k.a. vector triple product): v× (w× x) = (v · x)w− (v ·w)x.

PROOF HW

68

2.6. Cross products

How to compute cross product

METHOD 1: Remember some basic cross products, and use linearity (i.e. FOIL)(easy to remember, and useful for vectors with zeros in them):

Cross products to memorize: let i = e1 = (1, 0, 0); let j = e2 = (0, 1, 0); letk = e3 = (0, 0, 1). Then:

i× i = . j× i = . k× i = .

i× j = j× j = k× j =

i× k = j× k = k× k =

An easy way to memorize this:

i

j

k

+

i

j

k

-

EXAMPLE 29Compute the cross product of (2,−3, 0) and (0, 4,−1).

Now let’s find a general formula for cross product. Let v = (v1, v2, v3) andw = (w1, w2, w3). Then:

v×w = (v1i + v2j + v3k)× (w1i + w2j + w3k)

= v1w1(i× i) + v1w2(i× j) + v1w3(i× k)+ v2w1(j× i) + v2w2(j× j) + v2w3(j× k)

+ v3w1(k× i) + v3w2(k× j) + v3w3(k× k)

=

69

2.6. Cross products

METHOD 2: Memorize the formula we just derived (useful for proving theoremsabout cross product, but hard to remember):

Theorem 2.40 Let v = (v1, v2, v3) and w = (w1, w2, w3) be two vectors in R3. Then

v×w = (v2w3 − v3w2, v3w1 − v1w3, v1w2 − v2w1).

EXAMPLE 30(1, 4,−3)× (2,−1, 3) =

METHOD 3: Use a 3 × 3 determinant (easy to remember, and useful for vectorswith no zeros in them):

Theorem 2.41 (Computing cross product via determinants) Let v = (v1, v2, v3)and w = (w1, w2, w3) be two vectors in R3. Let e1, e2 and e3 be the standard basisvectors of R3. Then:

v×w = det

e1 e2 e3v1 v2 v3w1 w2 w3

= det

i j kv1 v2 v3w1 w2 w3

.PROOF Applying the Rule of Sarrus, we see that

det

e1 e2 e3v1 v2 v3w1 w2 w3

= (v2w3e1 + v3w1e2 + v1w2e3)− (v2w1e3 + v3w2e1 + v1w3e2)

= (v2w3 − v3w2)e1 + (v3w1 − v1w3)e2 + (v1w2 − v2w1)e3

= (v2w3 − v3w2, v3w1 − v1w3, v1w2 − v2w1).

By Theorem 2.40, this last expression is equal to v×w.

70

2.7. Equations of lines, planes and hyperplanes

EXAMPLE 30, REPEATED

(1, 4,−3)× (2,−1, 3) =

2.7 Equations of lines, planes and hyperplanesSymmetric and parametric equations of lines

Review of what we know about lines in R2: Every line in an xy-plane is deter-mined by two things:

1.

2.

From these two bits of information, we can derive three equivalent equationsfor such a line:

• the “slope-intercept equation” y = mx+ b

• the “point-slope equation” y = y0 +m(x− x0)

• the “standard equation” Ax+By = C

71


What about lines in Rn? The first thing we need to know is what “determines”a line in Rn. As above, we need two things to distinguish a particular line:

1.

2.

Given this, what does it mean for a point x = (x1, ..., xn) to lie on this line?

We have obtained the following set of equations for a line in Rn:

Definition 2.42 The symmetric equations of the line in Rn passing through pointp = (p1, p2, ..., pn) with direction vector v = (v1, v2, ..., vn) are

x1 − p1

v1= x2 − p2

v2= ... = xn − pn

vn.

Symmetric equations of a line are the analogue of the “ point-slope” equationyou know and love. However, they aren’t that useful for lines in R3 or Rn wheren ≥ 4, so this is the last time we will see symmetric equations in this course.

72


To get a more useful set of equations that describes a line in Rn, recall thatthe equation of the line on the previous page was x = p + tv. Writing this outcoordinate-wise, we get:

Definition 2.43 A set of parametric equations of the line in Rn passing throughpoint p = (p1, p2, ..., pn) with direction vector v = (v1, v2, ..., vn) are

x = p + tv⇔

x1 = p1 + tv1x2 = p2 + tv2...

...xn = p1 + tvn

Parametric equations of this type represent a function R → Rn, i.e. a functionwhich takes input t and produces output (x1, x2, ..., xn). Parametric equations fora line “lay a t−axis” on that line, i.e. coordinatize the line where

t = 0 ↔ choice of point p on the line .one unit of t ↔ choice of direction vector v

WARNING:

Parametric equations for a line are NEVER unique. More on this in Chapter5.

EXAMPLE 31Write the parametric equations for the line passing through the points (3, 3, 1) and(2, 5, 2).

Solution: A point on the line is p = (3, 3, 1) (using (2, 5, 2) as p is fine also).

A direction vector for the line is v = (2, 5, 2)− (3, 3, 1) = (−1, 2, 1).

So parametric equations (from Definition 2.43) are

x = p + tv, i.e.

x = 3− ty = 3 + 2tz = 1 + t

73


EXAMPLE 32Find the parametric equations for the line passing through the points (1, 2,−1) and(−3, 2, 0).

EXAMPLE 33Find the parametric equations of the line passing through (−1, 3, 2) which is per-pendicular to the vectors (2, 1,−3) and (5, 4,−1).

Solution: We are given that the line passes through p = (−1, 3, 2), but need tofind the direction vector v. What we know is that v ⊥ (2, 1,−3) and v ⊥ (5, 4,−1),so

v = (2, 1,−3)× (5, 4,−1) = det

i j k2 1 −35 4 −1

= (11,−13, 3).

So by applying Theorem 2.43, we get the parametric equations

x = p + tv, i.e.

x = −1 + 11ty = 3− 13tz = 2 + 3t

74


EXAMPLE 34Determine if the following two lines intersect:

x = 4− 2ty = 3tz = −1 + 2t

x = 3 + ty = −2t+ 2z = −4 + 3t

75


Parametric equations of planes

What “determines” a plane in Rn?

1.

2.

Given this, what makes a point x = (x1, ..., xn) lie in such a plane?

Definition 2.44 A set of parametric equations of the plane in Rn passing throughpoint p = (p1, p2, ..., pn) that has nonparallel direction vectors v = (v1, v2, ..., vn) andw = (w1, w2, ..., wn) are

x = p + sv + tw⇔

x1 = p1 + sv1 + tw1x2 = p2 + sv2 + tw2...

...xn = p1 + svn + twn

Parametric equations of this type represent a function R2 → Rn, i.e. a functionwhich takes inputs (s, t) and produces output (x1, x2, ..., xn). Parametric equationsfor a line “lay an st−plane” on that line, i.e. coordinatize the plane where

s = 0, t = 0 ↔ choice of point p on the line .one unit of s ↔ choice of direction vector vone unit of t ↔ choice of direction vector w

WARNING:

Parametric equations for a plane are NEVER unique.

76


Normal equations of hyperplanes (including lines in R2 andplanes in R3)

Definition 2.45 Let n ∈ Rn be a nonzero vector and let d be a scalar. A hyperplanein Rn is a subset H ⊆ Rn consisting of all points x satisfying

x ∈ H ⇐⇒ n · x = d.

n is called a normal vector to the hyperplane, and the equation n ·x = d is called thenormal equation or standard equation of the hyperplane.

EXAMPLE 35Consider the hyperplane in R2 with normal vector n = (−3, 2) and d = −1. Thenormal equation is therefore

More generally, the hyperplane in R2 with normal vector n = (a, b) has normalequation

(a, b) · (x, y) = d

ax+ by = d.

This is, of course, the equation of a line, so hyperplanes in R2 are just lines.

Suppose we take any point on the line −3x + 2y = −1. Call that point p =(x0, y0). Notice that for any other point x = (x, y) on the same line,

−3x0 + 2y0 = −1 and − 3x+ 2y = −1.

so

This idea generalizes:

Theorem 2.46 Let H ⊆ Rn be a hyperplane with normal vector n. Then, for anypoint p ∈ H , the hyperplane has equation

n · (x− p) = 0.

77


EXAMPLE 36Consider the hyperplane in R3 with normal vector (1,−3, 2) passing through (6, 2,−7).The normal equation is therefore

More generally, the hyperplane in R3 with normal vector n = (a, b, c) (assumingn 6= 0) has normal equation

(a, b, c) · (x, y, z) = d (2.1)ax+ by + cz = d. (2.2)

Question: What kind of geometric object do you think this the equation of?

To see this algebraically (assuming c 6= 0; if c = 0 the set is still a plane but youhave to do different algebra) we can solve (2.2) above for z to get z = −a

cx− b

cy+ d

c.

Let x = s and y = t, so thatx = 0 + 1s+ 0ty = 0 + 0s+ 1tz = d

c+ −a

cs+ −b

ct

This is the equation of a plane passing through p =(0, 0, d

c

)with direction vectors

v =(1, 0, −a

c

)and w =

(0, 1, −b

c

).

78


To summarize, we have shown:

Theorem 2.47 (Normal equations of lines and planes) Every hyperplane in R2

is a line, which has a normal equation of the form ax+ by = d.Every hyperplane in R3 is a plane, which has a normal equation of the form

ax+ by + cz = d.

In general, every hyperplane in Rn has “dimension n − 1”. To understand thisstatement more fully, take Math 322 (Linear Algebra).

EXAMPLE 37Write a normal equation of the plane whose parametric equations are

x = 2− 3s+ 4ty = 1− sz = 5s+ 7t

EXAMPLE 38Write parametric equations of the plane 2x− 3y + 7z = 14.

79


EXAMPLE 39Find a normal equation of the plane passing through the three points (2, 5, 6),(1,−1, 2) and (4, 0, 6).

Solution: First, draw a crude picture:

Now, to get the vectors v and w lying in the plane, subtract the points wherethese vectors begin from the points where they end:

v = (2, 5, 6)− (1,−1, 2) = (1, 6, 4)w = (4, 0, 6)− (1,−1, 2) = (3, 1, 4)

To get the normal vector n, since n ⊥ v and n ⊥ w, use cross product:

n = v×w =

i j k1 6 43 1 4

= (20, 8,−17)

Thus the normal equation of the plane is n·(x−p) = 0, which if we let p = (1,−1, 2)is

(20, 8,−17) · (x− 1, y + 1, z − 2) = 020(x− 1) + 8(y + 1)− 17(z − 2) = 0

Simplifying, we get20x+ 8y − 17z = −22

Note: in this problem, we could have subtracted any pairs of the given points toget v and w, and we could have used any of the three given points as p. Ultimatelythis would have produced the same answer.

80


EXAMPLE 40Find a (normal) equation of the plane containing the line whose parametric equa-tions are x = 2t + 1, y = −3t − 1, z = t + 4 which is parallel to the intersection ofthe planes 2x− y + z = 10 and y + z + 1 = 0.

81


Graphing planes in R3

The easiest way to graph a plane in R3 is to find the points where the plane crossesthe x-, y- and z-axes, and make a triangle.

EXAMPLE 41Sketch the graphs of the following planes:

x+ 2y − 3z = 6 2x+ z = 4

In general, the graph of any equation in R3 which is missing one of the variablesx, y and z is called a cylinder (this doesn’t mean it is round) and will be parallel tothe axis of the missing variable.

82


Coordinate planes

In R3, there are three planes that are particularly special. They are called the coor-dinate planes of R3:

the xy-plane the xz-plane the yz-planez = 0 y = 0 x = 0

Any plane of the form x = constant, y = constant or z = constant is parallel toone of these three planes.

EXAMPLE 42Sketch the graphs of the following planes:

x = 3 y = −2 z = 4

83

2.8. Polar, cylindrical and spherical coordinates

2.8 Polar, cylindrical and spherical coordinatesPolar coordinates

So far, we have described vectors in R2 by giving x- and y-coordinates, which rep-resent (respectively) how far east-west and how far north-south the vector points:

x-coordinate↔

y-coordinate↔

The x- and y-coordinates of a point are called its Cartesian coordinates or its rect-angular coordinates.

One way to think about Cartesian coordinates is to imagine the plane R2 beingfilled out with a bunch of vertical and horizontal lines:

-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

However, it is sometimes useful to specify the location of a point in R2 a differ-ent way. To do this, imagine filling out R2 with a bunch of circles centered at theorigin, and a bunch of rays pointing outward from the origin, like this:

0

π

6

π

3

π

4

π

23 π

4

π

5 π

4

3 π

2

3 π

4

7 π

4

1 2 3 4 5

This coordinate system describes any vector in R2 by giving its length r, andthe direction in which it points (by measuring an angle θ in the usual way). Thenumbers (r, θ) are called polar coordinates for the vector.

84


NOTE: Polar coordinates of a point are NEVER unique. Suppose (r, θ) are polarcoordinates for some point x ∈ R2. What are some other polar coordinates thatrepresent x?

We convert between Cartesian and polar coordinates using basic trig:

x

yr

θ

Theorem 2.48 (Cartesian↔ polar coordinates) If x ∈ R2 has polar coordinates(r, θ), then its rectangular coordinates are (x, y), where

x = r cos θ y = r sin θ.

If x ∈ R2 has Cartesian coordinates (x, y), then its polar coordinates are (r, θ), where

r = ||x|| =√x2 + y2

tan θ = y

x

Technically speaking, the polar coordinates of the origin are undefined, but youcan use r = 0 and any value of θ without a problem.

85


EXAMPLE 43Find the Cartesian coordinates of the point whose polar coordinates are (5, 5π

6 ).

Solution:x = r cos θ = 5 cos 5π

6 = 5 · −√

32 = −5

√3

2y = r sin θ = 5 sin 5π

6 = 5 · 12 = 5

2.

So the Cartesian coordinates are(−5√

32 ,

52

).

Graphs of polar functions

Long ago you learned to graph functions of the form y = f(x). You pick x-values,solve for the corresponding y-values, plot those points, and connect them with acurve.

You can do the same thing with polar functions of the form r = f(θ). In thegrand scheme of things, you pick values of θ, and solve for the corresponding val-ues of r. Then you plot the points you get (using polar coordinates, of course), andconnect them with a curve.

Here’s an example of how this works. Let’s take r = 3 cos θ.

θ r point(pick these) (solve for these) (r, θ)

0 r = 3 cos 0 = 3(1) = 3 (3, 0)π6 r = 3 cos π

6 = 3(√

32

)= 3

√3

2

(3√

32 , π6

)π4

3√

22

(3√

22 , π4

)π3

32

(32 ,

π3

)...

......

11π6

3√

32

(3√

32 , 11π

6

)2π 3 (3, 2π)

Now take the points in the last column and plot them, then connect them with acurve (in order of increasing θ) as shown on the next page:

86


(3,0)

(3 3

2,π

6)

-4 -2 2 4

-4

-2

2

4

-4 -2 2 4

-4

-2

2

4

Thus we see that the graph of r = 3 cos θ is a circle which goes through the points(0, 0) and (3, 0) (these are the Cartesian coordinates of the left- and right- edge ofthe circle, respectively).

Graphing polar functions with Mathematica

If you have to graph a lot of polar functions by hand, you will find that this isa very annoying procedure. Fortunately, Mathematica will graph polar equationsrather quickly. To get the graph of r = 3 cos θ, execute the following command:

PolarPlot[3 Cos[θ], θ, -10, 10]

(To get the θ, use the Basic Math Assistant Pallette, although you can use anotherletter in place of the θ (like t) so long as you use it throughout the command.) Youwill get this picture, which is exactly the plot you got above by hand:

0.5 1.0 1.5 2.0 2.5 3.0

-1.5

-1.0

-0.5

0.5

1.0

1.5

In general, to plot the graph of a polar equation in Mathematica, execute

PolarPlot[formula in terms of θ, θ, θmin, θmax]

This will plot r = f(θ) where θ ranges from θmin to θmax. If you care about therange of x and y-values you see on your graph, insert PlotRange into the commandas follows:

PolarPlot[formula in terms of θ, θ, θmin, θ max,PlotRange -> xmin, xmax, ymin, ymax]

87


Polar functions whose graphs you should memorize

CARTESIAN POLARGRAPH EQUATION EQUATIONCircle,

centered at (0, 0)with radius a

a

x2 + y2 = a2 r = a

Circle,with endpoints (0, 0)

and (a, 0)

aa/2

(x− a2)2 + y2 =

(a2

)2

(not important)r = a cos θ

Circle,with endpoints (0, 0)

and (0, a)a

a/2

x2 + (y − a2)2 =

(a2

)2

(not important)r = a sin θ

the x-axisy = 0 θ = 0

(or θ = π)

the y-axisx = 0 θ = π

2(or θ = 3π

2 )

Diagonal line,passing through (0, 0)

θ

y = mx(m = tan θ)

θ = constant(the constant is arctanm)

Vertical lineh

x = h r = h sec θ = hcos θ

Horizontal linek y = k r = k csc θ = k

sin θ

88


Cylindrical coordinates

We will also need three-dimensional coordinate systems akin to polar coordinates.There are two ways of doing this; the first is to use polar coordinates in the xy-planeand to keep the z coordinate “as is”. This generates three-dimensional coordinates(r, θ, z) called cylindrical coordinates.

EXAMPLE 44Sketch the graph of the region in R3 described by the given equations and/or in-equalities:

r = 2 π4 ≤ θ ≤ π

2 , z = −20 ≤ r ≤ 3,0 ≤ z ≤ 4

89


Theorem 2.49 (Cartesian↔ cylindrical coordinates) If x ∈ R3 has cylindricalcoordinates (r, θ, z), then its Cartesian coordinates are (x, y, z), where

x = r cos θ y = r sin θ z = z.

If x ∈ R3 has Cartesian coordinates (x, y, z), then its cylindrical coordinates are(r, θ, z), where

r =√x2 + y2

tan θ = y

xz = z

EXAMPLE 45Find the cylindrical coordinates of the point whose rectangular coordinates are(4,−4,−3).

EXAMPLE 46Find the Cartesian coordinates of the point x, if the cylindrical coordinates of x are(3, π,−1).

Solution: Using the above theorem, we have

x = r cos θ = 3 cos π = 3 · −1 = −3y = r sin θ = 3 sin π = 3 · 0 = 0z = z = −1

So the Cartesian coordinates are (−3, 0,−1) .

90


Graphs of equations in cylindrical coordinates

You should be familiar with the graphs of the following equations in cylindricalcoordinates:

CYLINDRICAL DESCRIPTIONEQUATION OF GRAPH PICTURE

r = a(r = constant)

cylinder about z-axiswith radius a

θ = a(θ = constant)

plane containing z-axisat angle θ frompositive x-axis

a

z = a(z = constant)

plane parallel toxy-plane at height a

a

Helpful note:

In general, given any graph like the three above, or others that have rota-tional symmetry about the z-axis, you should instinctively think that con-verting the graph to cylindrical coordinates might be helpful, in that itsequation in cylindrical coordinates might be much easier to work with.

91


Spherical coordinates

The second version of “three-dimensional polar coordinates” use two angles andone length to represent a point. These are called spherical coordinates:

ρ = “rho”

ϕ = φ = “phi”

Note: the order is important when writing spherical coordinates: the angle ϕswinging down from the positive z-axis always comes before θ, the angle of rota-tion from the positive x-axis.

(That said, some physicists and engineers reverse the θ and ϕ, writing the θfirst. These people should be severely punished.)

EXAMPLE 47Sketch the graph of each region in R3 described by the given equations and/orinequalities:

ρ = 3 ϕ = π6 ρ ≥ 0,

θ = π4 ,

0 ≤ ϕ ≤ π2

92


Theorem 2.50 (Cartesian↔ spherical coordinates) If x ∈ R3 has spherical co-ordinates (ρ, ϕ, θ) are (x, y, z), where

x = ρ sinϕ cos θ y = ρ sinϕ sin θ z = ρ cosϕ.

If x ∈ R3 has Cartesian coordinates (x, y, z), then its spherical coordinates are (ρ, ϕ, θ),where

ρ = ||x|| =√x2 + y2 + z2

tanϕ = r

z=√x2 + y2

z

tan θ = y

x

EXAMPLE 48Find spherical coordinates of the point whose Cartesian coordinates are (0, 4,−4

√3).

EXAMPLE 49Find Cartesian coordinates of the point whose spherical coordinates are (2, π4 ,

2π3 ).

Solution:

x = ρ sinϕ cos θ = 2 sin π

4 cos 2π3 = 2 ·

√2

2 ·−12 = −

√2

2y = ρ sinϕ sin θ = 2 sin π

4 sin 2π3 = 2 ·

√2

2 ·√

32 =

√6

2z = ρ cosϕ = 2 cos π

4 = 2 ·√

22 =

√2

So the Cartesian coordinates are(−√

22 ,

√6

2 ,√

2)

.

93


Graphs of equations in spherical coordinates

SPHERICAL DESCRIPTIONEQUATION OF GRAPH PICTURE

ρ = a(ρ = constant)

sphere of radius acentered at (0, 0, 0)

θ = a(θ = constant)

plane containing z-axisat angle θ frompositive x-axis

(same as incylindrical

coordinates)a

ϕ = a(ϕ = constant)

cone opening aroundthe z-axis

(at angle a fromthe positive z-axis)

a

94

2.9. Some basic topology

2.9 Some basic topologyOpen and closed sets

In precalculus and/or calculus, you learn about subsets of the real numbers calledintervals. As examples:

(a, b) = . . [a,∞) = . . [a, b] =x ∈ R : a < x < b x ∈ R : a ≤ x x ∈ R : a ≤ x ≤ b

The first interval above is called an open interval (essentially because it doesnot contain its endpoints). The second and third intervals above are called closedintervals (because they contain their endpoint(s)). An interval like [a, b) is calledhalf-open, half-closed (it is neither open nor closed).

Goal: We want to extend the ideas expressed above to Rn, and characterizecertain sets as “open”, “closed” (or perhaps neither open nor closed, or both openand closed).

Enrichment: In fact, one can talk about “open” and “closed” sets in sets thatare far more general than Rn; such things arise in the branch of mathematicscalled topology which studies the properties of objects which are preservedunder stretching, twisting, rotating, etc.

Definition 2.51 Let x ∈ Rn and let ε > 0. The open ball of radius ε centered at x,denotedBε(x), is the set of points y ∈ Rn such that dist(x,y) < ε. A neighborhoodof x is a set which contains some open ball centered at x.

Generic pictures of open balls in dimensions 1, 2 and 3:

Note: All open balls have a radius which is positive, but finite.

Note: Open balls do not contain their boundaries.

95


Definition 2.52 LetE be a subset of Rn. A point x ∈ Rn is called a boundary pointof E if every open ball centered at x contains both points in E and points in EC . Theboundary of E, denoted ∂E, is the set of boundary points of E.

EXAMPLE 50Describe the boundary of the set in R2 described as follows:

E = (x, y) : y > 3− 2x

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

EXAMPLE 51Describe the boundary of the set in R2 described as follows:

E = (x, y) : 1 ≤ x2 + y2 < 16

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

EXAMPLE 52Describe the boundary of the interval (3,∞) in R.

-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8

Note: In general,• The boundary of a one-dimensional set is usually a collection of endpoints

(i.e. is zero-dimensional);• The boundary of a two-dimensional set is usually a collection of curves (i.e.

is one-dimensional);• The boundary of a three-dimensional set is usually two-dimensional;• The boundary of a region of an n-dimensional set has dimension n− 1.

96


Definition 2.53 A sphere is a boundary of an open ball in Rn. (If n = 2, we call asphere a circle.)

EXAMPLE 53Write an equation which describes the sphere that is the boundary of the open ballin R2 of radius 4 centered at (−5, 3).

Solution: If a point (x, y) is on the boundary of this open ball, then

-9 -8 -7 -6 -5 -4 -3 -2 -1 1 2

-3

-2

-1

1

2

3

4

5

6

7

8

Example 53 generalizes as follows:

Theorem 2.54 The equation of the sphere centered at h = (h1, h2, ..., hn) ∈ Rn withradius r > 0 is

dist(x,h) = r;

written coordinate-wise, this is

(x1 − h1)2 + (x2 − h2)2 + ...+ (xn − hn)2 = r2.

Theorem 2.54 takes the equation of a circle in R2 you know:

(x− h)2 + (y − k)2 = r2

and generalizes it to higher dimensions.

97


Definition 2.55 Let E be a subset of Rn.

1. E is called an open set (or just open) ifE contains none of its boundary points,i.e. ∂E ∩ E = ∅.

2. E is called a closed set (or just closed) if E contains all of its boundary points,i.e. ∂E ⊆ E.

3. E is called a clopen set (or just clopen) if E is both closed and open. (This isonly possible if ∂E = ∅.)

Generic pictures of open and closed sets:

The next two theorems are usually proven in a real analysis (Math 430) or topol-ogy course. The first theorem gives a technical description of open sets:

Theorem 2.56 A set E ⊆ Rn is open if and only if for every point x ∈ E, there is anumber ε > 0 such that Bε(x) ⊆ E.

OPEN NOT OPEN

E-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

F

-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

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Theorem 2.57 A set E ⊆ Rn is closed if and only if its complement is open.

More enrichment: Definition 2.55 and Theorems 2.56 and 2.57 are “backwards”from the way they “should be”. It is better (i.e. more general) to take Theorem2.56 as the definition of an open set, and to take Theorem 2.57 as the definition ofa closed set, in which case Definition 2.55 becomes a theorem you can prove. Ichose this presentation because I think it is easier to explain.

WARNING:

Sets are not doors!

99


Bounded sets, compact sets and connected sets

Definition 2.58 Let E be a subset of Rn.

1. E is called a bounded set (or just bounded) if there is a number ε and a pointx such that E ⊆ Bε(x).

2. E is called a compact set (or just compact) if E is both closed and bounded.

UNBOUNDED BOUNDED, BUT COMPACT(i.e. not bounded) NOT COMPACT

F

-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

E-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

G-3 -2 -1 1 2 3 4 5 6 7

-5

-4

-3

-2

-1

1

2

3

4

5

More enrichment: This is a “poor” definition of what it means for a set tobe compact, because it doesn’t generalize to situations beyond subsets of Rn.There is a better definition of compactness which turns out to be equivalent to“closed and bounded” for subsets of Rn; for more on this, do a Google searchfor “compactness” or the “Heine-Borel” property.

Definition 2.59 Let E be a subset of Rn. E is called a disconnected set (or justdisconnected) if there are two disjoint open sets A and B such that E ∩ A 6= ∅,E ∩ B 6= ∅ and E ⊆ A ∪ B. E is called a connected set (or just connected) if it isnot disconnected.

A disconnected set is one containing two or more pieces that “don’t touch”.

CONNECTED NOT CONNECTED

E-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

H

-3 -2 -1 1 2 3 4 5 6 7

-5

-4

-3

-2

-1

1

2

3

4

5

100

2.10. Mathematica and calculator commands for vector and matrix operations

2.10 Mathematica and calculator commands for vector and matrixoperations

Vector and matrix operations for large-sized matrices can and should be doneelectronically using either a computer software package or calculator. In this sec-tion, we provide directions for matrix operations using the computer softwarepackage Mathematica.

Typing in vectors

To type in a vector, use a set of squiggly braces to surround the components,and separate the components with commas: for example, to save the vector v =(2,−3, 5, 0, 1), execute

v = 2,-3,5,0,1

Typing in matrices

To type in a matrix, use one of two methods:

1. Use squiggly braces and commas to separate the entries. Each row should besurrounded by a squiggly brace, and the entire matrix should be surroundedby a set of squiggly braces, and everything should be separated by commas.For example, to define A as (

1 23 4

)one could execute

A = 1, 2, 3, 4

Note that if you have a column matrix like

123

, this matrix can be defined

by thinking of B as a vector and typing something like B = 1, 2, 3 (insteadof having to type B = 1, 2,3).

2. On the Basic Math Assistant palette, under Basic Commands, click the matrix.Then type A =, then click the matrix in the palette. To add rows and columns,click AddRow or AddColumn until the matrix is the appropriate size. Then gointo the matrix and type in each entry, moving between the locations usingthe [TAB] key or clicking on the location you want.

101

2.10. Mathematica and calculator commands for vector and matrix operations

If the entries of the matrix are functions, then define the matrix as a function byexecuting A[t_] = ... or A[x_, y_] = ... instead of A = ...

Once you have defined all necessary matrices and/or vectors, Mathematica com-mands for operations on those objects are as follows2:

EXPRESSION MATHEMATICA SYNTAX

VE

CT

OR

OP

ER

AT

ION

S

Vector addition / subtraction:(2, 5,−1) + (5, 0, 2) 2,5,-1 + 5,0,2v−w v - w

Scalar multiplication:3(1, 2,−3,−4) 31,2,-3,-45v− 3w 5v - 3w

Dot product v ·w v.w(3,−4) · (2, 10) 3,-4.2,10

Norm ||v|| Norm[v]Unit vector in same direction as v Normalize[v]

(i.e. v||v|| )

Projection of v onto w Projection[v,w](i.e. πwv)

Angle between two vectors VectorAngle[v,w] (answer is in radians)(to get degrees, click degree measure in the

suggestions bar)Cross product v×w Cross[v,w]To get the number of components of v Length[v]To get the ith component of v v[[i]]

MA

TR

IXO

PE

RA

TIO

NS

Matrix addition / subtractionA+B A + BA−B A - B

Scalar multiplication3A 3AnA n A (space important)−5A+ 1

2B -5A + (1/2)BMatrix product AB A.B (the period is important)

A2 A.A or MatrixPower[A,2] (not Aˆ2)A7 MatrixPower[A,7]

Trace tr(A) Tr[A]Determinant detA Det[A]Transpose AT Transpose[A]To get the entry of matrix A in the ith row A[[i,j]]

and jth columnTo call the n× n identity matrix I IdentityMatrix[n]

To make Mathematica display an answer as a matrix:1. follow your command with // MatrixForm, or2. once you’ve executed the command, choose Display as... matrix from the sug-

gestions bar.

2Other operations on matrices that we won’t need much, but that are very useful in Math 322and Math 330, can be found in the Appendix.

102

2.11. Summary of Chapter 2

2.11 Summary of Chapter 2Sections 2.1 and 2.2

• A set is any definable collection of objects called elements. Vocabulary re-lated to sets includes subset, equal (sets), complement (the set of things notin a set), empty set (the set with no elements), union (the set of things inone set or the other or both), intersection (the set of things in both sets) andCartesian product (the set of ordered pairs of elements in the given sets).

• A vector space is a set of objects which can be added together and multipliedby scalars in a reasonable way. Elements of a vector space are called vectors.

• For each n ∈ 1, 2, 3, ..., Rn, the set of ordered n-tuples of real numbers,is a vector space, where the addition and scalar multiplication are definedcoordinate-wise.

• A good pictorial representation of a vector is a floating arrow. In this context,addition of vectors corresponds to parallelogram (or head-to-tail) addition,and scalar multiplication corresponds to stretching. When you multiply avector by a negative scalar, the vector points in the opposite direction.

• The standard basis vectors e1, ..., en of Rn are defined as follows: ej hasa 1 in the jth position and 0s everywhere else. Every vector v ∈ Rn can bewritten uniquely as a linear combination of the standard basis vectors by the

equation v =n∑j=1

vjej . In R3, the standard basis vectors can also be written as

i, j and k.

Section 2.3

The dot product of two vectors in Rn is the scalar v ·w = vTw =n∑j=1

vjwj .

Dot products are useful because they lead to algebraic solutions of geometric prob-lems involving vectors. In particular, we use dot products to define:

• The distance between two vectors v,w ∈ Rn is the scalar

dist(v,w) = ||v−w|| =√√√√ n∑j=1

(vj − wj)2.

• The norm (length) of a vector v ∈ Rn is the scalar ||v|| =√

v · v =√

n∑j=1

v2j .

(In other words, v · v = ||v||2.)

103


• A unit vector is a vector whose norm is 1.

Given any nonzero v ∈ Rn, there is always a unit vector u ∈ Rn in the samedirection as v, called a normalized version of v, which is computed by set-ting u = 1

||v||v.

• Two vectors v and w are called orthogonal (written v ⊥ w) if v ·w = 0. Or-thogonality generalizes the idea of “perpendicularity” from planar geometry.

• v · w = ||v|| ||w|| cos θ, where θ is the angle between vectors v and w. So ifv ·w > 0, θ is acute, and if v ·w < 0, θ is obtuse.

Sections 2.4 and 2.5

• Anm×nmatrix is an array of numbers arranged intom rows and n columns.A vector x ∈ Rn can be thought of as a n× 1 matrix called a column vector.

• Matrix addition and scalar multiplication of matrices are done term-by-term.

• If A is an m × n matrix and B is an n × p matrix, the matrix product AB is

defined and is a m× p matrix where (ab)ij =n∑k=1

aikbkj (so each entry of AB is

a dot product of a row from A with a column from B).

• The determinant of a 2 × 2 matrix is (plus or minus) the area of a parallelo-gram formed by the columns of the matrix:

det(a bc d

)= ad− bc.

• The determinant of a 3 × 3 matrix is (plus or minus) the volume of a paral-lelepiped formed by the columns of the matrix:

det

a b cd e fg h i

= aei+ cdh+ bfg − bdi− ceg − afh.

Usually, such a determinant is computed using the Rule of Sarrus.

Section 2.6

The cross product of two vectors in R3 is a vector which is orthogonal to both ofthe inputs, has length equal to the area of the parallelogram formed by the two

104


vectors, and which points in a direction determined by the right-hand rule. Crossproducts can be computed by determinants:

v×w = det

e1 e2 e3v1 v2 v3w1 w2 w3

.or by memorizing the cross products of i, j and k and distributing.

Section 2.7

• A line in Rn is determined by a point p on the line and a direction vector vfor the line. Such a line has parametric equations

x = p + tv⇔

x1 = p1 + tv1x2 = p2 + tv2...

...xn = p1 + tvn

and if the line is in R2, it has normal equation ax+ by = c where n = (a, b) isa normal vector to the line.

• A plane in Rn is determined by a point p on the plane and two nonparallelvectors v and w lying in the plane. Such a plane has parametric equations

x = p + sv + tw⇔

x1 = p1 + sv1 + tw1x2 = p2 + sv2 + tw2...

...xn = p1 + svn + twn

and if the plane is in R3, it has normal equation ax + by + cz = d wheren = (a, b, c) is a normal vector to the plane.

• The normal equation of the plane in R3 containing point p with normal vectorn is n · (x− p) = 0.

• The three coordinate planes in R3 are the yz-plane x = 0, the xz-plane y = 0,and the xy-plane z = 0.

105


Section 2.8

• Polar coordinates (r, θ) of a point in R2 record the point’s length and its di-rection from the origin. Polar coordinates of a point are not unique.

Cartesian↔ polar conversion formulas

r = ||x|| =

√x2 + y2

tan θ = yx

x = r cos θy = r sin θ

Circles and lines have nice equations in polar coordinates which should bememorized.

• Cylindrical coordinates (r, θ, z) of a point in R3 record the polar coordinatesin the xy-plane and “leave the z alone”.

Cartesian↔ cylindrical conversion formulas

r =√x2 + y2

tan θ = yx

z = z

x = r cos θy = r sin θz = z

In cylindrical coordinates, the equation r = constant is a cylinder around thez-axis; the equation θ = constant is a plane containing the z-axis; the equationz = constant is a plane parallel to the xy-plane.

• Spherical coordinates (ρ, ϕ, θ) of a point in R3 record the distance from thepoint to the origin and use two angles to measure the direction the point is in.ϕ is measured downward from the positive z-axis, and θ is measured aroundthe z-axis, from the positive x-axis.

Cartesian↔ spherical conversion formulas

ρ = ||x|| =

√x2 + y2 + z2

tanϕ = rz

=√x2+y2

z

tan θ = yx

x = ρ sinϕ cos θy = ρ sinϕ sin θz = ρ cosϕ

In spherical coordinates, the equation ρ = constant is a sphere centered at theorigin; the equation ϕ = constant is a cone opening around the z-axis; theequation θ = constant is a plane containing the z-axis.

106


Section 2.9

• An open ball in Rn is the set of points less than distance ε from some fixedx ∈ Rn. A neighborhood of x is a set in Rn containing an open ball centeredat x.

• The boundary of a set is its “edge”. A sphere is the boundary of an open ballin Rn; every sphere has equation

(x1 − h1)2 + (x2 − h2)2 + ...+ (xn − hn)2 = r2

where (h1, ..., hn) is the center of the sphere and r is the radius of the sphere.

• A subset of Rn is called open if it contains none of its boundary points andclosed if it contains all its boundary points. Sets can be open, closed, clopen(i.e. open and closed), or neither open nor closed.

• A subset of Rn is called bounded if it is contained in some open ball. A subsetof Rn is called compact if it is closed and bounded. A subset of Rn is calledconnected if it does not consist of two or more pieces that “don’t touch”.

107


2.12 Homework exercisesProblems from Section 2.1

1. Draw a rough picture of each set:

a) A = [0, 4]b) A× A, where A is as in part (a)

c) B = (x, y) ∈ R2 : x ≥ 4d) BC , where B is as in part (c)

e) D = (x, y) ∈ R2 : x2 + y2 ≤ 25f) B ∩D, where B and D are as above


a) C = (x, y) ∈ R2 : x+ y < 10b) B ∪ C, where B is as in Problem 1 and C is as above

c) B ∪ ∅, where B is in Problem 1

d) A×D, where A and D are as in Problem 1

e) D × A, where A and D are as in Problem 1


a) E = (x, y, z) ∈ R3 : 0 ≤ z ≤ 3b) F = (x, y, z) ∈ R3 : x2 + y2 + z2 ≤ 16c) G = (x, y, z) ∈ R3 : y ≤ z ≤ y + 1d) H = (x, y, z) ∈ R3 : x+ y + z ≤ 4, x ≥ 0, y ≥ 0, z ≥ 0e) E ∩H , where E and H are as in parts (a) and (d)

f) G ∩GC , where G is as in part (c)

Problems from Section 2.2

4. Plot each of the following points in xyz-space:

A = (0, 4,−1) B = (1, 2, 5) C = (−3, 0, 0) D = (−5,−1, 2)

5. In this problem, let v = (2,−7, 4), let w = (0, 3, 5) and let x = (−2,−1,−5).

a) Compute v + x.

b) Compute 3w− 2x.

108


c) Compute 12w.

d) Compute 2v− (1, 4,−2).

e) If y is a vector such that 2(y + v) = w, find y.

6. Compute the following quantities, if they make sense. If they don’t make anysense, just write “nonsense”.

a) (2,−3, 1) + (−7, 5, 2)b) 3(−1, 0, 4)− 2(1, 3, 0)

c) 2(1, 4,−2, 5) + (2, 3, 0, 1, 0)d) 4(1, 2) + 3(0,−1) + (5, 6)

7. Suppose vector v ∈ R2 can be drawn as an arrow that begins at (3,−2) andends at (−1,−1).

a) What are the coordinates of v?

b) If v is drawn as an arrow that starts at (−5, 4), where does this arrowend?

c) If v is drawn as an arrow that ends at (3, 7), where does this arrow start?

8. Let u = 3i− k, let v = 2i + 3j− 4k, and let w = −i + 5j− 3k.

a) Compute u + 2j− k.

b) Write the answer to part (a) in coordinates (i.e. without using i, j, k-notation).

c) Compute 2u− 3w.

d) If x is a vector such that 13x + u = v + w, find x.

9. a) Write the coordinates of the vector e2 ∈ R3.

b) Write the coordinates of the vector 2e1 − 3e5 ∈ R6.

Here is a picture which indicates three vectors a,b and c in R2:

5 10 15

2

4

6

8

In each part of Problems 10 and 11, you are given some vector which is written interms of a, b and/or c. Write the coordinates of that vector.

109


10. a) −ab) 2b

c) a + bd) a − b

11. a) 2a + e2

b) c− 3a + bc) 2

3a + cd) c− c

The vectors a, b, c, ... i in some vector space are indicated by the picture below:

Use the picture to answer the questions asked in Problems 12-14.

12. a) What vector is h + g?

b) What vector is f + d?

c) What vector is c− g?

d) Write f in terms of d and e.

13. a) Find x if x + g = −f .

b) Find x if a + c + x = b.

c) Write e in terms of d and f .

d) Write g in terms of c and h.

14. a) Write c− d in terms of e, h and i.b) Simplify c + f + i− h.

c) Simplify a + h + g + f − e.

d) Find three vectors in the picture, all of which (assuming the picture is toscale) appear to be scalar multiples of one another.

110



In Problems 15 and 16, let v = (−2, 3, 1), let w = (0, 2, 3) and let x = (5,−1,−1).Compute each given quantity:

15. a) v ·wb) x · 2w

c) x · (v− x)d) dist(v,w)

16. a) ||v||b) ||v + x||

c) ||v||vd) dist(3v, 2x−w)

In Problems 17 and 18, let v,w,x ∈ Rn be such that v ·w = 8, ||v|| = 2, v · x = −7and x · x = 16. Compute each given quantity:

17. a) −x · 2v

b) v · (4x− 3w)c) v · v

d) w · 1||w||2 w

e) ||x||f) || − 3v||

18. a) (v + x) · (v + x)b) 2x ·w− x · 2w

c) v · ||v||vd) dist(w,w + v)

19. Let v = (2, 3,−1, 2).

a) Normalize v (this means find a unit vector in the same direction as v).

b) Find a unit vector in the opposite direction as v.

c) Find a vector of length 7 in the same direction as v.

d) Find the distance between v and 4e2 + e3.

20. (F) Find all real numbers k, such that the vectors (1, 2, k) and (2, 4, 3) aredistance 3 apart.

21. a) Determine whether or not the vectors (4,−7, 2, 0) and (3, 1, 8, 5) are or-thogonal.

b) Determine whether or not the vectors (1,−3,−5) and (7, 4,−1) are or-thogonal.

c) Find all values of a (if any) such that the vectors (a, 1, 1) and (1, a, 4) areorthogonal.

d) (R) Describe all vectors in R2 which are orthogonal to (3, 2).

111


22. Compute the angle between the vectors v and x, where v and x are as de-scribed just before Problem 17.

23. a) Find the cosine of the angle between the vectors (2,−3, 5) and (1, 0,−4).b) (R) Find the cosine of the angle between the vectors (5,−3) and (7,−1).c) Find the angle between the vectors (3, 0) and (−4, 4). Give the exact

answer, in radians.d) Find the angle between the vectors (2, 7, 1) and (4, 3,−4). Give the exact

answer and a decimal approximation, in degrees.

24. Suppose v,w,x ∈ Rn are such that ||v|| = 6 and ||w|| = 10.

a) What is the largest possible value of v ·w?b) What is the smallest possible value of v ·w?c) If the angle between v and w is π

3 , what is v ·w?d) If v · x = 12, is the angle between v and x acute, right or obtuse?e) If v · x = −1, is the angle between v and x acute, right or obtuse?f) What inequality do you know about ||v + w||?

25. (F) A main diagonal of a cube is a line segment connecting two opposite cor-ners of the cube. Find the angle between a main diagonal of a cube and adiagonal of one of its faces.

26. Let v ∈ R2. Let α be the angle between v and the positive x-axis, and let βbe the angle between v and the positive y-axis. Use dot products to simplifycos2 α + cos2 β.

27. a) (F) Let v ∈ R3. Let α be the angle between v and the positive x-axis,let β be the angle between v and the positive y-axis, and let γ be theangle between v and the positive z-axis. Use dot products to simplifycos2 α + cos2 β + cos2 γ.

b) (F) Let v ∈ R3. Suppose the angle between v and the positive x-axis isπ4 and the angle between v and the positive y-axis is π

3 . Find the anglebetween v and the positive z-axis (assume this angle is between 0 andπ2 ).

c) (F) Prove that there is no vector v ∈ R3 such that the angle between vand the positive x-axis and the angle between v and the positive y-axisare both equal to π

6 .

28. Prove that for any real numbers a and b, their geometric mean√ab is always

less than or equal to their arithmetic mean a+b2 .

Hint: Apply an appropriate inequality to the vectors u = (√a,√b) and v =

(√b,√a).

112



In problems 29 through 35, let

A =(

3 2 1−5 0 2

); B =

4 01 −2−1 7

; C =

2 −2 13 1 30 1 −4

;

D =(

5 −21 −1

); E =

−3 −4 1−3 2 14 0 3

; F =(−4 1−3 −2

);

x = (−3, 5,−2); y = (4, 3).Compute each indicated quantity, if it makes sense (if it doesn’t make sense, justwrite “nonsense”):

29. a) the size of Ab) b12

c) b21

d) the second row of A

e) the vector space to which x be-longs

f) the trace of D

30. a) A+B b) C + E c) 2F − 3D + I

31. a) AT b) (2 + 3)B c) 2(3D)

32. a) AB b) BA c) B2

33. a) (R) BF b) (R) FB c) (R) A(C + E)

34. a) AB +D b) DBT c) DAT

35. a) Ax b) xA c) Fy

36. For each function f or f given below, write down a matrix A such that f(x) =Ax:

a) f(x, y, z) = (2x+ y − 5z, x+ 4z)b) f(x) = (4x, 7x, 0,−2x)c) f(x, y) = 3x− 5yd) f(w, x, y, z) = (w, y, z, x)


37. Compute each indicated quantity, if it makes sense (if it doesn’t make sense,just write “nonsense”). The matrices being referred to are the ones givenbefore Problem 30.

113


a) detAb) detC

c) detDd) detF

38. Use determinants to find the area of the parallelogram in R2 whose verticesare (1,−3), (2,−1), (4, 0) and (5, 2).

39. Find the volume of the parallelepiped whose non-parallel edges are the vec-tors (1, 2, 4), (2, 5, 3) and (4, 7, 1).


40. Given each pair of vectors v and w in R3, compute the cross product v×w.

a) v = (−2, 5,−3); w = (1,−3, 7)b) v = (3, 6, 1); w = (8,−1, 0)

c) v = 5i; w = 7kd) v = 3j + k; w = −5i− j

41. (R) Given each pair of vectors v and w in R3, compute the cross product v×w.

a) v = (−8,−8, 4); w = (6, 6,−3)b) v = (0, 0, 6); w = (5, 0,−2)

c) v = 5i; w = 3i + 2kd) v = 6k; w = −2j

42. a) (R) Find a nonzero vector which is orthogonal to both (2,−1, 4) and(−3, 1,−2).

b) Find two unit vectors which are orthogonal to both (1,−5, 0) and (2, 4, 1).

43. (F) Use an appropriate cross product to solve Problem 38. (Yes, there is away to apply cross product to this situation, even though the vectors are inR2.)

44. (R) Find the area of the triangle in R3 whose vertices are (7, 11, 3), (4, 5, 8) and(5, 9, 2).

45. Suppose a, b and c are vectors in R3 such that

a × b = (1, 2, 3) b× c = (0,−1, 5)

Use this given information to evaluate the following:

a) (−b)× cb) a × 3bc) b× a

d) (4a + c)× be) (a × b) · bf) b× b

114


46. Choose one from (a), (b), (c):

a) Prove the scalar triple product formula stated in Theorem 2.39.Hint: Write each vector coordinate-wise, work out both sides in terms oftheir coordinates, and observe that both sides are equal.

b) Prove the bracket rule (i.e. the vector triple product formula) stated inTheorem 2.39.Hint: Write each vector coordinate-wise, work out both sides in terms oftheir coordinates, and observe that both sides are equal.

c) Let v and w be vectors in R3. Prove ||v×w||2 = ||v||2 ||w||2 − (v ·w)2.

47. Let u, v and w be vectors in R3. Suppose that u×w = 0 and u · v = 0. Whatconclusion about v and w can be drawn from this? (Do you know v ·w = 0?Do you know v×w = 0?) Explain.

In Problems 48-54, you are to assume:

• x and y are vectors in R4;• v and w are vectors in R3;• a and b are vectors in R2;• A is a 4× 3 matrix;

• B is a 3× 3 matrix;• f is a function R3 → R3;• g is a function R3 → R2;• h is a function R2 → R3.

In Problems 48-54, determine whether each given quantity is a scalar, a vector (inwhich case you should say whether it belongs to R2, R3 or R4), a matrix (in whichcase you should give its size), or nonsense.

Note: For the purposes of this problem, matrices with a single column shouldbe called vectors.

48. a) x · xb) x · x · x

c) (x · x) · xd) (x · x)x

e) (x + 2y) · yf) Ax

49. a) Avb) ABv

c) BAvd) Bx

e) 4Ax + 2Bvf) v · Ax

50. a) Av · Awb) x · ATy

c) (v ·w)Axd) A((x · y)Aw · y)

e) f(v)×wf) a · g(v)

51. a) dist(x,x)b) ||2y + x||

c) ||y||yd) 1

||w||xe) dist(w, 2v)Af) dist(x,w)y · x

115


52. a) v×wb) v×Bw

c) −4a × bd) ||v||||x||

e) (Ax · v)af) 4v + 3h(x)

53. a) v×w× vb) v× (w× v)

c) v× (w · v)d) v · (w× v)

e) (v ·w)× (w · v)f) v×w · (1, 4,−2)

54. a) h(g(2v) · a)

b) B(v×w)×w

c) f(v)−Bw + 4h(2a)

d) ||f(v)−v||v

e) ||f(v)−Av||||v||

f) ||f(v)−v||||v||


55. Write parametric equations of each given line:

a) the line in R3 passing through (11,−6, 7) parallel to vector 3i + j− 12kb) the line in R3 passing through the points (3, 1, 3) and (2,−4,−2)c) the x-axis (in R2)

56. Write parametric equations of each given line:

a) the line in R4 passing through the points (−1, 0, 3,−12) and (6,−15, 11,−8)b) the line in R2 whose slope-intercept equation is y = −1

3 x−53

c) the x-axis (in R3)

57. Consider the line in R3 passing through (0, 15,−11) and (6, 10,−8). Find threeother points on this line.

58. Determine if the given pair of lines intersect; if so, find the coordinates of theintersection point.

x = −7 + 2ty = −2− 3tz = −3 + t

;

x = 4 + ty = 9 + 4tz = 8 + 2t

59. (R) Determine if the given pair of lines intersect; if so, find the coordinates ofthe intersection point.

x = 2ty = 7− 2tz = 6− t

;

x = 4 + 2ty = 8− tz = −1− 2t

60. Two airplanes fly along straight lines. At time t, plane 1 is at x = 75 + 5t, y =50 + 10t, z = 25 + t and plane 2 is at x = 60 + 10t, y = 80 + 5t, z = 34− t.

116


a) Do the flight paths of these planes intersect? Explain.

b) Do the planes crash into one another? Explain.

61. Write parametric equations of each given plane:

a) The plane in R3 containing the points (1, 3, 0), (−2, 5,−2) and (7, 0,−2).

b) The plane in R5 containing the points (0, 3,−4, 7, 3), (1, 1, 3,−4, 0) and(10,−7, 2, 11, 13).

c) The xy−plane (in R3).

62. Write parametric equations of each given plane:

a) The plane in R3 containing the points (3, 0,−1), (−2, 7, 9) and (8, 5,−10).

b) The plane in R3 whose normal equation is 2x+ 7y − z = 17.Hint: first find three points in this plane and then proceed as in part (a).

c) The plane in R3 containing the point (1,−5, 7) and the line whose para-metric equations are

x = 3− ty = 7 + 2tz = −2t

63. (R) Write parametric equations of each given plane:

a) The plane in R3 with normal vector (3,−5,−2) containing the point(−5,−1,−3).

b) The plane in R3 containing the two linesx = t+ 3y = −5t− 2z = 3t+ 10

x = 4t− 7y = 9t+ 1z = −2t− 5

64. Write a normal equation of each given plane in R3:

a) The plane containing the points (4,−8, 1), (0, 2, 3) and (7,−5,−1).

b) The plane whose parametric equations arex = 3− 6s+ 11ty = 2 + 3s− tz = −2 + 5s

65. Write a normal equation of each given plane in R3:

a) The plane containing the line whose parametric equations are x = −t, y =5t− 2, z = −3t+ 1 and the point (17,−7, 4).

117


b) The plane parallel to the plane 4x+5y−7z = 5, passing through (6,−7, 0).

66. Consider the plane in R3 passing through the points (4, 2, 1), (0,−3, 7) and(−8,−3, 2). Find five other points on this plane.

67. Find parametric equations of the line which is the intersection of the twoplanes 5x+ 3y − z = 9 and −2x+ 3y − 11z = 15.

68. Choose three from (a), (b), (c), (d), (e).

In each part, you are given a line and a plane in R3. Determine whether theline lies entirely within the plane, hits the plane in one point, or lies parallel tothe plane. If the line hits the plane in exactly one point, find the coordinatesof that point.

a) line has parametric equations x = 7t− 5, y = 3t+ 2, x = −5t+ 3;plane has normal equation 3x+ y + 4z = 3

b) line passes through (−4, 6,−1) and has direction vector 5i− 3j− 5k;plane has normal equation 2x− 5y + 3z = −21

c) line has parametric equations x = 2− t, y = 3t+ 5, z = 2t− 3;plane has normal equation −7x+ 3y − 8z = 25

d) line has parametric equations x = (−2, 4, 1) + t(2,−3,−1);plane has normal equation 4x+ y + 5z = 13

e) line passes through (3, 0,−2) and (4,−2, 5);plane passes through (3, 1,−3), (7,−2,−4) and (−2,−5, 5)

69. a) Define what you think it means for two planes in R3 to be perpendicu-lar.

b) Based on your definition, find the value(s) of a for which the two planesax− y + 3z = 9 and 2ax+ ay − 2z = 14 are perpendicular.

70. Give a rough sketch of the graphs of each of the following planes in R3:

a) −x− 2y + z = 4b) x+ 2z = 6c) 3x+ 5y − 5z = 15

d) y = 2

e) x− y + z = 0

71. Give a rough sketch of the graphs of each of the following planes in R3:

a) z = 0b) y − 2z = 2c) x = −1

d) 2x+ 3y = 12

e) x+ 4y + 2z = 14

118



72. a) Give a set of polar coordinates for the point whose Cartesian coordinatesare (2

√3, 2).

b) Give a set of polar coordinates for the point whose Cartesian coordinatesare (0,−3).

c) Give three different sets of polar coordinates for the point whose polarcoordinates are (3, 7π

5 ).

d) If a point has polar coordinates (3, π2 ) and (s, 3π2 ), what is the value of s?

e) Give the Cartesian coordinates for the point whose polar coordinates are(5, 2π

3 ).

f) Give the Cartesian coordinates for the point whose polar coordinates are(4, 3π).

73. Sketch the graph of the planar region described by the following polar coor-dinate inequalities:

a) 0 ≤ θ ≤ π2 , r ≥ 0

b) 0 ≤ θ ≤ π4 , 0 ≤ r ≤ 3

c) π ≤ θ ≤ 2π, 2 ≤ r ≤ 4d) π

2 ≤ θ ≤ 3π2 , 0 ≤ r ≤ 1

74. Write inequalities (using polar coordinates) describing each set graphed be-low:

a)

-4 -2 2 4

-4

-2

2

4

b) -4 -2 2 4

-4

-2

2

4

c)

-4 -2 2 4

-4

-2

2

4

d)

-4 -2 2 4

-4

-2

2

4

75. Sketch a graph of each polar function using Mathematica (make sure to use alarge enough range of θ values to get the entire graph):

119


a) r = cos 2θb) r = 1 + sin 2θc) r = 4

2+cos θ

d) r = 4 cos 3θe) r = 2 sin θ + cos 3θf) r = 3 cos 2

5θ


a) r = 2 + cos θb) r = 2 + 2 cos θc) r = 1

6 − cos 3θ

d) r = 3 sin t cos 2te) r = 3 cos 3

5θ + sin 15θ

f) r = eθ


a) r = 1 + 4 sin θ3

b) r = cos2 t+ sin 2tc) r = esin θ − 2 cos 4θd) r = tan θ − cot θ

78. Write a polar function r = f(θ) which represents each given graph:

a) -5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

b) -5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

c) -5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

d) -5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

79. Write a polar function r = f(θ) which represents each given graph:

a) -5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

b) -5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

c) -5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

120


d) -5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

e) -5-4-3-2-1 1 2 3 4 5

-5-4-3-2-1

12345

f)3

1

80. a) Give cylindrical coordinates for the point whose rectangular coordinatesare (−1, 0, 2).

b) Give cylindrical coordinates for the point whose Cartesian coordinatesare (−1,

√3, 3).

c) Give the Cartesian coordinates of the point whose cylindrical coordi-nates are (4, π2 , 1).

d) Give the Cartesian coordinates of the point whose cylindrical coordi-nates are (1, 3π

4 ,−3).

81. Sketch a rough graph of each of these equations, given in cylindrical coordi-nates:

a) r = 5b) θ = π

2

c) θ = π

d) z = −2e) z = 1f) z = r2

82. a) Give spherical coordinates for the point whose rectangular coordinatesare (−1, 1,

√2).

b) Give Cartesian coordinates of the point whose spherical coordinates are(4, π6 ,

3π4 ).

c) Give Cartesian coordinates of the point whose spherical coordinates are(6, π2 , π).

83. a) Give spherical coordinates for the point whose rectangular coordinatesare (3, 0, 2).

b) Give Cartesian coordinates of the point whose spherical coordinates are(2, π, π6 ).

c) Give Cartesian coordinates of the point whose spherical coordinates are(2, 5π

6 ,π4 ).

84. Sketch a rough graph of each of these equations, given in spherical coordi-nates:

a) ρ = 2b) ϕ = π

4

c) ϕ = π2

d) θ = 2π3

121



85. Write the equation of the sphere with the given properties:

a) The sphere is in R3, is centered at (2,−3, 0) and has radius 4.

b) The sphere is in R2, is centered at (−5, 2) and has radius 6.

c) The sphere is in R3, is centered at (4,−1, 5) and passes through the point(2,−3, 1).

d) The sphere is in R4, and a diameter of the sphere has endpoints (3,−5, 2, 4)and (1, 3, 0,−6).

86. For each given set, describe the boundary of the set, either by drawing apicture of it or by describing it using set-builder notation.

a) A = [0, 4]b) B = (7,∞)c) C = (x, y) ∈ R2 : x ≤ 3d) D = (x, y) ∈ R2 : x > 1, y ≥ 1e) E = (x, y) ∈ R2 : x2 + y2 ≤ 25f) F = (x, y, z) ∈ R3 : x2 + y2 ≤ 16g) G = (x, y, z) ∈ R3 : |y| > 1h) H = (x, y, z) ∈ R3 : x2 + y2 + z2 < 36

87. For each given set, determine if it is open, closed, clopen, or neither open norclosed:

a) [0, 9]b) (x, y) ∈ R2 : x2 ≤ 4

c) (x, y) ∈ R2 : x2 < yd) (x, y, z) ∈ R3 : x2 +y2 +z2 > 10


a) (3,∞)b) (3, 6]

c) (x, y) ∈ R2 : 3 ≤ x < 8d) (x, y, z) ∈ R3 : 4 < x, y ≥ 3


a) B2((3, 8))b) (x, y, z) ∈ R3 : x+ y + z ≤ 8

122


c) (w, x, y, z) ∈ R4 : x ≥ 3d) (w, x, y, z) ∈ R4 : x+ w > 8, x− 2w2 < 4, y + x > 0, z > 3y + xe) ∅

90. a) Determine which one or ones of the setsA-H from Problem 86 are bounded(no proofs necessary).

b) Determine which one or ones of the sets A-H from Problem 86 are com-pact (no proofs necessary).

91. For each given set, determine if the set is connected.

a) The set A from Problem 86.

b) The set B = (−∞,−3) ∪ (0, 6) ∪ (10,∞).

c) The set C from Problem 86.

d) The set D = (x, y) ∈ R2 : x2 ≥ 16e) The set E = (x, y, z) ∈ R2 : x+ y − z > 0f) The set F = (x, y) ∈ R2 : x2 + y2 > 121g) The set G from Problem 86.

123


Selected answers

DISCLAIMER # 1: I did these by hand, often using Mathematica. They may con-tain errors, and the answers may be simplified in an unusual way (so they may bethe same as yours, even if they don’t “look” the same).

DISCLAIMER # 2: In many of these problems, answers may vary (especially whenyou are writing equations of lines and or planes, because those equations are notunique).

1. b) ×

-1 1 2 3 4 5-1

1

2

3

4

5

c)

-1 1 2 3 4 5 6 7

-3

-2

-1

1

2

3

e)

-5 -3 -1 1 3 5

-5

-3

-1

1

3

5

f)⋂

-1 1 3 5

-3

-1

1

3

2. a)

-1 1 3 5 7 9-1

1

3

5

7

9

b)⋃

-1 1 3 5 7 9-1

1

3

5

7

9

c)

⋃∅

-1 1 2 3 4 5 6 7

-3

-2

-1

1

2

3

3. a)

3

E

d)

4

4

4

H

f) ∅

5. a) (0,−8,−1)c)(0, 3

2 ,52

)e)(−2, 17

2 ,−32

)6. a) (−5, 2, 3)

b) (−5,−6, 12)

7. a) (−4, 1)

8. c) 9i− 15j + 7kd) (−6, 24,−18)

9. a) (0, 1, 0)

10. a) (0,−6)

124


c) (5, 4)

11. b) (10,−14)c) (5, 10)

12. b) ed) f = e− d.

13. a) x = i.b) x = −d.

c) e = d + f .

14. c) b

15. a) 9c) −41

16. b)√

13c) (−2

√14, 3√

14,√

14)d)√

489

17. a) 14b) −52e) 4

18. a) 6c) 8

19. a)(√

23 ,√

22 ,−

√2

6 ,√

23

)c)(

7√

23 , 7

√2

2 ,−7√

26 , 7

√2

3

)20. k = 1, k = 5.

21. b) They are orthogonal.

22. arccos(−7

8

)23. a) −9

√2

323

d) arccos(

253√

246

)≈ 57.9.

24. a) 60e) obtuse

f) ||v + w|| ≤ 16.

25. arccos√

23

26. 1

27. a) 1b) π

3

29. a) 2× 3d) (−5, 0, 2)e) R3

f) 4

30. a) DNE

b)

−1 −6 20 3 44 1 −1

31. a)

3 −52 01 2

b)

20 05 −10−5 35

32. a)

(13 3−22 14

)c) DNE

33. a)

−16 42 5−17 −15

34. b)

(20 9 −194 3 −8

)a) DNE

35. a) (−1, 11)

36. a) A =(

2 1 −51 0 4

)

37. a) DNEb) −35

125


38. 3

39. 20

40. a) (26, 11, 1)c) (0,−35, 0)

41. b) (0, 30, 0)c) (0,−10, 0)

42. b)(− 5√

222 ,−1√222 , 7

√2

111

)and(

5√222 ,

1√222 ,−7

√2

111

)44.

√4612

45. a) (0, 1,−5)c) (−1,−2,−3)f) (0, 0, 0)

48. a) scalar

b) nonsense

d) vector in R4

e) scalar

49. a) vector in R4

b) vector in R4

50. b) nonsense

c) nonsense

d) 4× 3 matrix

51. a) scalar

b) scalar

c) vector in R4

d) vector in R4

e) 4× 3 matrix

f) scalar

52. a) vector in R3

d) scalar

e) nonsense

53. b) vector in R3

e) nonsense

f) nonsense

54. a) vector in R3

e) nonsense

55. a)

x = 11 + 3ty = −6 + tz = 7− 12t

56. a)

x1 = −1 + 7tx2 = −15tx3 = 3 + 8tx4 = −12 + 4t

c)

x = ty = 0z = 0

59. (−6, 13, 9)

62. b)

x1 = s+ 10tx2 = 3− 2s− 10tx3 = −4 + 7s+ 6tx4 = 7− 11s+ 4tx5 = 3− 3s+ 10t

c)

x = sy = tz = 0

63. a)

x = −5 + 5sy = −1 + 3s− 2tz = −3 + 5t

64. b) 5x+ 55y − 27z = 179

65. a) 48y + 80z = −16

67.

x = 2− 30ty = −1 + 57tz = −2 + 21t

68. a) intersection point is (2, 5,−2)

126


c) line lies entirely within theplane

d) line is parallel to the plane

e) intersection point is(

24583 ,

883 ,−194

83

)69. b) a = 2, a = −3

2

70. b) 3

6

d)2

e)

71. a)

c)-1

d)4

6

e)

14

7

3.5

72. a) (4, π6 )

d) s = −3.

e) (−52 ,

5√

32 )

f) (−4, 0)

73. a) -4 -2 2 4

-4

-2

2

4

b) -4 -2 2 4

-4

-2

2

4

c) -4 -2 2 4

-4

-2

2

4

74. a) r ≤ 3

b) r ≤ 3, 0 ≤ θ ≤ π4

75. a) -1 1

-1

1

c) -4 -3 -2 -1 1

-2

-1

1

2

127


e)

-1 1

1

2

f) -3 -2 -1 1 2 3

-3

-2

-1

1

2

3

76. b) 1 2 3 4

-2

-1

1

2

c) -1

-1

1

f) -2 -1 1 2

-2

-1

1

2

77. a) -4 -3 -2 -1 1 2 3 4

-5

-4

-3

-2

-1

1

2

3

c)-3 -2 -1 1 2 3

-1

1

2

3

78. a) r = 3 sin θd) r = −2 cos θ

79. a) r = −2 csc θ

c) θ = 3π4

d) θ = π3

e) r = 2 sec θ

80. a) (1, π, 2)d)

(−√

22 ,√

22 ,−3

)82. a) (2, π4 ,

3π4 )

b) (−√

2,√

2, 2√

3)

83. a) (√

13, arctan 23 , 0)

84. a)

b)π

4

d)

2 π3

85. a) (x− 2)2 + (y + 3)2 + z2 = 16.b) (x+ 5)2 + (y − 2)2 = 36.d) (w− 2)2 + (x+ 1)2 + (y− 1)2 +

(z + 1)2 = 43.

86. b) ∂B = 7.c) ∂C = (x, y) ∈ R2 : x = 3.e) ∂E = (x, y) ∈ R2 : x2 + y2 =

25.h) ∂H = (x, y, z) ∈ R3 : x2 +

y2 + z2 = 36.

87. a) closedb) open

128


88. a) openb) neitherd) neither

89. a) openc) closedd) open

90. a) A,E, F and H are bounded.

b) A,E and F are compact.

91. a) A is connected.

d) D is not connected.

e) E is connected.

129

Chapter 3

Functions and limits

3.1 Functions from Rn → Rm

Recall from Chapter 2 that the basic building blocks of mathematics areand . In Chapter 2, we dealt with the first of these buildingblocks (in particular, we learned a lot about the vector spaces Rn). In this chapter,we study the basics of functions of several variables. First, remember the definitionof a function:

Definition 3.1 Given two sets E and F , a function from E to F is a rule (or proce-dure) f which assigns to each element of E at most one element of F . In this case wewrite

f : E → F.

The set of points in E which have an element of F assigned to them is called thedomain of f and is denoted Dom(f); the set of points in F which are assigned tosome element of E is called the range or image of f and is denoted Im(f). F is calledthe codomain of F .

Given x ∈ Dom(f), denote by f(x) the output associated to input x.

Given y ∈ F , define f−1(y) = x ∈ E : f(x) = y; this is called the inverseimage (of y under f ).

Given a set A ⊆ E, let f(A) = f(x) : x ∈ A.Given a set B ⊆ F , let f−1(B) = x ∈ E : f(x) ∈ B.

130

3.1. Functions from Rn → Rm

In this course, we study “functions of several variables”, i.e. functions from Rn

to Rm. First, a word about notation. If the codomain is R, we will usually name thefunction by a regular letter like

f, g, h, F,G, ...

to emphasize that the output is a scalar. If the codomain is Rm where m ≥ 2, inthese notes (barring typos) we will refer to the function by a boldface letter like

f ,g,h,F,T, ...

to emphasize that the output is a vector (in at least two dimensions). When writ-

ing the name of such a function by hand, we might write−→f to represent f , but

often we forget the arrow and just write f .

Keep in mind that when we discuss anything about functions f : Rn → Rm, thematerial we learn always includes the situation f : Rn → R (and includes regularold functions f : R → R that you learned about in calculus 1 and 2, if not before),unless otherwise stated.

Component functions

Let f : Rn → Rm. The output of such a function has m coordinates; the first thingwe want to do is establish some notation for these coordinates:

Definition 3.2 Let f : Rn → Rm be a function. The component functions of f aref1, f2, ..., fm, each taking Rn to R, defined by setting

fj = πj f

where πj(y1, ..., ym) = yj , the jth coordinate of the vector. The functions πj are calledcoordinate projections.

Idea: if f : Rn → Rm, then we can always write

f(x) = (f1(x), f2(x), ..., fm(x)) where each fj : Rn → R.

The big picture: It turns out that studying most aspects of the differential cal-culus of a function of several variables (especially limits) boils down to study-ing each component function separately.

131


EXAMPLE 1Suppose x = (2, 3, 5,−4, 2). Compute:

π2(x) = π4(x) =

EXAMPLE 2π2(x+ 3y2, 5− 2 sin xy) = 5− 2 sin xy.

EXAMPLE 3Suppose f : R2 → R4 is given by

f(x) = f(x, y) =(x2y, x+ 2ey, x

y2 + 1 , cos(x− 2y)).

Compute:

f1(x) = f1(x, y) = f1(3, 2) =

f2(x) = f2(x, y) =

f3(x) = f3(x, y) =

f4(x) = f4(x, y) = f4(π, 0) =

Alternate notation for component functions, if m = 2 or 3

If f : Rn → R2, the component functions are often called f and g, x and y, or u andv, rather than f1 and f2. In this setting, we might denote f “variable-wise” andwrite (if the input variables are (x1, ..., xn))

f : (x1, ..., xn) 7→ (x, y) or f : (x1, ..., xn) 7→ (u, v), etc.

EXAMPLE 4If f : R→ R2 is given by f(t) = (3et cos t, 4e−t sin t), then we might write

f : t 7−→ (x, y)

andf = (x, y) = (x(t), y(t)) where x(t) = 3et cos t, y(t) = 4e−t sin t.

132


If f : Rn → R3, the component functions are often called “f , g, h” or “x, y, z”, or“u, v, w”, rather than f1, f2 and f3.

EXAMPLE 5If f : R3 → R3 is given by f(x) = f(x, y, z) = (x + 2y, 3x − z, z2 cos y + x), then wemight write

f : (x, y, z) 7→ (f, g, h)

f = (f, g, h), where

f(x, y, z) = x+ 2y g(x, y, z) = 3x− z h(x, y, z) = z2 cos y + x.

NOTE: Avoid the use of the same letters for the variables of the domain andrange (if your inputs are x and y, don’t call the component functions x and/ory.

Every function Rn → Rm can be written as a “sum” of its component functionsin the same way that a vector can be written as a linear combination of the standardbasis vectors:

Theorem 3.3 If f : Rn → Rm is a function with component functions f1, f2, ..., fm,then

f(x) =m∑j=1

fj(x)ej.

PROOF If f : Rn → Rm has component functions f1, ..., fm, then

f(x) = (f1(x), f2(x), ..., fm(x))

= (f1(x), 0, ..., 0) + (0, f2(x), 0, ..., 0) + ...+ (0, ..., 0, fm(x))

= f1(x)e1 + f2(x)e2 + ...+ fm(x)em

=m∑j=1

fj(x)ej.

133


Domain issues

Given any function, one of the first things you should consider is what the domainof that function is. Recall from Definition 3.1 that the domain of a function is the setof “allowable inputs”.

For a function of several variables, we can determine its domain by finding thedomain of each component function:

Theorem 3.4 Let f : Rn → Rm have component functions f1, ..., fm. Then

Dom(f) =m⋂j=1

Dom(fj),

i.e. the domain of f is the intersection of the domains of its component functions.

PROOF x ∈ Rn is a valid input to the function f if and only if it is a valid input toeach component function.

134

3.2. Graphs of functions Rn → Rm

3.2 Graphs of functions Rn → Rm

Motivating example: Let f : R → R be f(x) = x2. Let’s think about pictureswhich represent this function, and how those pictures might generalize to func-tions Rn → Rm.

Picture # 1: the graph of f (i.e. the blue curve in the picture below):

-3 -2 -1 1 2 3

1

2

3

4

5

Thinking about this picture: Think of this blue curve as a set, called graph(f).

1. What vector space is graph(f) a subset of?

2. Remembering that the picture above shows graph(f) where f : R→ R, whatvector space do you think graph(g) is a subset of, if g : R2 → R? Why?

3. What vector space do you think the graph of a function f : R7 → R5 shouldbe a subset of?

4. Describe graph(f) in set notation.

5. Suppose h : (x, y) 7→ (u, v, w). Describe graph(h) in set notation.

135


Based on what we came up with on the previous page, we define:

Definition 3.5 Let f : Rn → Rm. The graph of f is the subset of Rm+n defined asfollows:

graph(f) = (x,y) ∈ Rm+n : y = f(x)= (x1, x2, ..., xn, y1, ..., ym) ∈ Rm+n : (y1, ..., ym) = f(x1, ..., xn)

Picture # 2: the image of f

Think of the input variable x as time, and the output variable y as positionalong an axis (useful if the function is describing an object’s position at timex):

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10

NOTE: this isn’t a picture of the graph of f : this is a picture of the image or rangeof f only, with some points “tagged” by their corresponding inputs . The image ofa function f : R → Rm is called a path or curve; properties of curves are discussedin detail in Chapter 5.

Goal of this section

We want to discuss various pictures which describe functions of several variables;the initial idea is to generalize Picture # 1 (the graph of a function) and Picture # 2(a picture of the image of the function) on the previous page.

Remark: If m and/or n is/are too big, then drawing a picture is hopeless, butif m + n ≤ 4 then there are various sketches you can make which may help youget a handle on the behavior of the function. So we’ll stick to these relatively low-dimensional situations.

Other Remark: Mathematica is an awesome tool for producing graphs of func-tions of several variables. So rather than spending too much time on how to createthese graphs (other than by using technology), we will focus on what the graphs meanand how various pictures relate to one another.

136


Parametrized curves: images of functions R→ Rm

EXAMPLE 6Sketch a picture which represents the function f : R→ R2 defined by

f(t) = (t2 − 2, 2t− t2).

Solution: This is a function from to , so the graph(f) ⊆ .

If we tried to sketch the graph of this function (i.e. repeat Picture # 1 fromabove),

• the picture would need three dimensions;

• this would be very difficult to draw correctly; and• you wouldn’t get the most useful information from the picture.

Instead, let’s repeat Picture # 2 from above, sketching only the image of f :• this displays only the outputs of the function (so requires only two dimen-

sions);• we can also include the inputs by labeling points on the picture with appro-

priate values of t;• this picture will show useful information.

Table of values:

t x(t) y(t) f(t) = (x(t), y(t))(pick these) (solve for this) (solve for this)−2 (−2)2 − (2) = 2 2(−2)− (−2)2 = −8 (2,−8)−1 −1 −3 (−1,−3)0 −2 0 (−2, 0)1 −1 1 (−1, 1)2 2 0 (2, 0)

137


Plot the points obtained in the table: Connect to get a picture of the image of f :

t=-2

t=-1

t=0t=1

t=2-5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

t=-2

t=-1

t=0t=1

t=2-5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

Thinking of this function as giving the position (x, y) of an object at time t, thisgraph should have an arrow on it indicating the direction of motion (from nega-tive/smaller t to positive/larger t).

Easier solution: have Mathematica sketch this graph

Here are the commands to define a function which takes R to R2, and then tosketch a picture of its image:

f[t_] = tˆ2 -2, 2t-tˆ2ParametricPlot[f[t], t, -20, 20]

Mathematica uses the command ParametricPlot because we are thinking of the curveas being given by what are called parametric equations (in this case, x = t2 − 2 andy = 2t− t2). More on this in Chapter 5.

If you want to add various things to the second command above, you can. Forexample, to plot the graph for x-values ranging from xmin to xmax and y-valuesfrom ymin to ymax, execute something like this (all in one cell):

ParametricPlot[f[t], t,tmin,tmax,PlotRange -> xmin,xmax, ymin,ymax]

You can also include AspectRatio -> Automatic to make the scales on the x- andy-axes match, and by using PlotStyle you can make the graph dashed, or in a cer-tain color, or thick, etc.

ParametricPlot is used to sketch the images of functions R → R2. To sketch theimage of a function R→ R3, use ParametricPlot3D (see the example coming up twopages from now). Unfortunately, Mathematica does not sketch the image of func-tions from R to Rm where m ≥ 4.

138


Drawback: Mathematica does not indicate the direction of motion.

EXAMPLE 7Sketch the image of of f(t) =

(t cos t+ t

2 , sin(3t)− t):

Mathematica input:

f[t_] = t Cos[t] + t/2, Sin[3t]-t

ParametricPlot[f[t], t,-20,20,PlotRange -> -20,20,-18,18,PlotStyle -> Thick, Red]

Output:

-20 -10 10 20

-15

-10

-5

5

10

15

What is the direction of motion? To figure this out, just figure some points onthe graph using Mathematica:

Execute f[0] to get output 0,0. This tells you f(0) = (0, 0).

Execute f[11] (and get a numerical approximation) to get output 5.54868, -10.0001. This tells you f(11) ≈ (5.5,−10).

Since f(11) is below and to the right of f(0), we see that the direction of motionis generally downward (and can then add an arrow to the graph Mathematica pro-duces).

Note: There’s nothing special about the 0 and 11 we chose; you just play aroundwith the inputs until you figure out the direction of motion.

139


EXAMPLE 8f(t) =

(2 cos t, 2 sin t, t2

).

Partial table of values:

(pick these) (solve for these)t x = f1(t) y = f2(t) z = f3(t) f(t) = (x, y, z)0 2 cos 0 = 2 2 sin 0 = 0 0

2 = 0 (2, 0, 0)π2 0 2 π

4

(0, 2, π4

)π −2 0 π

2

(−2, 0, π2

)2π 2 0 π (2, 0, π)

A bunch of points plotted: The image of f :

Mathematica code that generates something similar to the graph shown above:

f[t_] = 2 Cos[t], 2 Sin[t], t/2

ParametricPlot3D[f[t], t, -20,20]

140


Surfaces: graphs of functions R2 → REXAMPLE 9

Sketch the graph of f : R2 → R defined by

f(x, y) = y + x2.

Solution: We make an analogue of Picture # 1 from earlier. This time, we pick xand y, and solve for z = f(x, y). Then we plot the points (x, y, z) in R3 and make apicture connecting all the points (as best we can):

Table of values: The points coming from this table:

x y z = f(x, y)(pick (pick (solve forthese) these) these) (x, y, z)

0 0 0 + 02 = 0 (0, 0, 0)1 0 0 + 12 = 1 (1, 0, 1)−1 0 1 (−1, 0, 1)0 1 1 (0, 1, 1)0 −1 −1 (0,−1,−1)2 0 4 (2, 0, 4)−2 0 4 (−2, 0, 4)2 1 5 (2, 1, 5)2 −1 3 (2,−1, 3)2 2 6 (2, 2, 6)

A lot more points: The graph of f(x, y) = y + x2:

To get an idea of what the graph of this function looks like without having toresort to plotting so many points, we consider what are called traces of the function:

141


Definition 3.6 A trace of a function f : Rn → R is a function R → R obtained bysetting all but one of the input variables of f equal to a constant.

Suppose f : R2 → R2. Then graph of the y = k trace is the intersection of thegraph of f with the plane y = k, and the graph of the x = h trace is the intersectionof the graph of f with the plane x = h.

EXAMPLE 9, REPEATED

Let f(x, y) = y + x2. Here is a hand-drawn picture of the graph of f :

Let’s look at some traces of this graph. First, for the y = constant traces:

y = 0 trace, y = 1 trace: y = 2 trace: y = k trace:a.k.a. xz-trace: f(x, 1) = 1 + x2 f(x, 2) = 2 + x2 f(x, k) = k + x2

f(x, 0) = .

Conclusion: curves on the graph of f parallel to the x-axis are parabolas.

Now for the x = constant traces:

x = 0 trace, x = 1 trace: x = 2 trace: x = h trace:a.k.a. yz-trace: f(1, y) = y + 1 f(2, y) = y + 4 f(h, y) = y + h2

f(0, y) = .

Conclusion: curves on the graph of f parallel to the y-axis are lines of slope 1.

Notice how the traces tell you information about the graph of f .

142


EXAMPLE 10Sketch some traces of the function f(x, y) = y2 sin x, and use those traces to sketchthe graph of f .

Traces parallel to x-axis:

y = 0 trace: y = 1 trace:f(x, 0) = 0 f(x, 1) = sin x

y = 2 trace: y = −1 trace:f(x, 2) = 4 sin x f(x,−1) = sin x

Traces parallel to y-axis:

x = 0 trace: x = π2 trace:

f(0, y) = 0 f(π2 , y) = y2

x = π trace: x = 3π2 trace:

f(π, y) = 0 f(3π2 , y) = −y2

The entire graph:

143


Alternatively, we can use Mathematica to sketch plots of surfaces. The appropri-ate command to plot a function z = f(x, y) where x ranges from xmin to xmax andy ranges from ymin to ymax is:

Plot3D[function,x,xmin,xmax, y,ymin,ymax]Here is code for the graph we constructed by hand in Example 10:

Plot3D[yˆ2 Sin[x], x,-4 Pi, 4 Pi,y,-12, 12, AxesLabel -> "x","y","z"]

The AxesLabel part of the above command is useful to make sure you know whichaxis is which.

To specify a range of z-values, include PlotRange as follows:

Plot3D[function,x,xmin,xmax, y,ymin,ymax, PlotRange -> zmin,zmax]

There are a lot of bells and whistles which can (and sometimes should) beadded to the Plot3D command; for more on this, consult the Mathematica file “Graphsof functions of several variables” which is available on my Math 320 web page.

Vertical Line Test

The Vertical Line Test, which determines which sets in R2 are graphs of functionsy = f(x), still holds in this setting:

Theorem 3.7 (Vertical Line Test) A surface is the graph of a function z = f(x, y)if and only if every vertical line (i.e. every line parallel to the z-axis) hits the surfacein at most one point.

144


Cylinders

As we saw in Chapter 2, if a function f : R2 → R is missing one of the variables inits rule, then we expect the function’s graph to be parallel to the axis of the missingvariable:

EXAMPLE 11Sketch crude graphs of these functions by hand:

f(x, y) = 3− x2 f(x, y) = 4 cos y

Contour plots of functions R2 → REarlier in this section we saw that for a function z = f(x, y), it might be useful toset x or y equal to a constant (to obtain traces of the function). It might also beuseful to set z to be constant and look for the set of x and y that produce that z asoutput.

Definition 3.8 Let f : R2 → R and let c be a constant. The level curve (at heightc) of f is the set f−1(c), i.e. the set of points (x, y) ∈ R2 such that f(x, y) = c. Acollection of level curves drawn on the same (x, y)-plane is called a contour plot forf .

EXAMPLE 12Let f(x, y) = y2 − x. Sketch the level curve at height 2 for this function.

145



Again, let f(x, y) = y2 − x. Sketch a contour plot for this function.

Solution:z = 0: 0 = y2 − x⇒ x = y2

z = 1: 1 = y2 − x⇒ x = y2 − 1

z = 2: 2 = y2 − x⇒ x = y2 − 2

z = 3: 3 = y2 − x⇒ x = y2 − 3

z = −1: −1 = y2 − x⇒ x = y2 + 1

z = −2: −2 = y2 − x⇒ x = y2 + 2

arbitrary z:

How does this contour plot connect with the graph of f(x, y) = y2 − x?

146


EXAMPLE 13The contour plot of f(x, y) = 1

4(y3−2y−x2) is given below, at left. Sketch a possiblegraph of f .

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

To obtain a contour plot in Mathematica, use the ContourPlot command. Forexample, to sketch contour plot for the function f(x, y) = x2− y+xy2 for values ofz ∈ −30,−29,−28, ..., 29, 30 where x ranges from −3 to 3 and y ranges from −3to 3, use the command (all in one cell)

ContourPlot[xˆ2 - y + x yˆ2, x, -3, 3, y, -3, 3,Contours -> Range[-30,30], ContourLabels -> True, ContourShading -> False]

The Contours part of the command tells Mathematica which z-values to use; theContourLabels -> True part of the command labels each contour with its z-value.If you remove the ContourShading -> False part of the command, Mathematica pro-duces a contour plot with shading; the lighter the region, the greater the value of z.

As another example, code similar to the following was used to generate thepicture in Example 14 on the next page:

ContourPlot[Sin[x - y] Cos[x + 2 y], x, -3, 3, y, -3, 3, ContourLabels -> True]

147


EXAMPLE 14Here is a contour plot for the function f(x, y) = sin(x− y) cos(x+ 2y):

-0.8

-0.8

-0.8

-0.8

-0.8 -0.8

-0.8

-0.6

-0.6

-0.6-0.6-0.6

-0.6-0.6

-0.4

-0.4

-0.4

-0.4

-0.4

-0.4

-0.4

-0.4

-0.2

-0.2

-0.2

-0.2

-0.2

-0.2

-0.2

-0.2

-0.2

0

0

0

0

0

0

00

0

00

0.2

0.2

0.2

0.2

0.2

0.2

0.2

0.2

0.2

0.4

0.4

0.40.4 0.4

0.4

0.4

0.4

0.60.6

0.6

0.6

0.60.6

0.6

0.80.8

0.8

0.8 0.8

0.8

0.8

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

Based on this picture, answer the following questions:

1. Estimate f(−1,−2).

2. Which is greater, f(1, −12 ) or f(0, −3

2 )?

3. If you are at point (−2, 0) and move in direction (0,−1) (not towards the point(0,−1)), will the value of f(x, y) increase or decrease?

4. If you are at the point(

32 ,

12

), in what direction (given as a vector) should you

move if you want f(x, y) to increase as rapidly as possible?

148


Level surfaces of functions R3 → RSuppose f : R3 → R. It is impossible to draw the graph of f , because you wouldneed 3 + 1 = 4 dimensions. However, if you think of the function as w = f(x, y, z),then by setting w equal to a constant you will get something similar to a levelcurve:

Definition 3.9 Let f : R3 → R and let c be a constant. The level surface (at heightc) of f is the set f−1(c), i.e. the set of points (x, y, z) ∈ R3 such that f(x, y, z) = c.

More generally:

Definition 3.10 Let f : Rn → R and let c be a constant. The level set (at height c)of f is the set f−1(c), i.e. the set of points x ∈ Rn such that f(x) = c.

It doesn’t make any sense to draw more than one level surface on the same xyz-axes because things would get too crowded, but you can still think of a collectionof level surfaces for a function f : R3 → R as comprising a “contour plot” for thefunction f .

EXAMPLE 15Sketch level surfaces for the function f(x, y, z) = x2 +y2−z2 at heights−1, 0, 1 and3.

Solution: We sketch these using the ContourPlot3D command in Mathematica:

Levelsurface Graph Mathematica command

w = −1 ContourPlot3D[xˆ2 + yˆ2 - zˆ2 == -1,x, -3, 3, y, -3, 3,z, -3, 3]

w = 0 ContourPlot3D[xˆ2 + yˆ2 - zˆ2 == 0,x, -3, 3, y, -3, 3, z, -3, 3]

149


Levelsurface Graph Mathematica command

w = 1 ContourPlot3D[xˆ2 + yˆ2 - zˆ == 1,x, -3, 3, y, -3, 3,z, -3, 3]

w = 3 ContourPlot3D[xˆ2 + yˆ2 - zˆ2 == 3,x, -3, 3, y, -3, 3, z, -3, 3]

Vector fields: functions Rn → Rn

Definition 3.11 A vector field on Rn is a function f : Rn → Rn.

A good way to think about a vector field is to think of each input as being apoint in the sky, and the output as being the direction and magnitude of the windat that point. Vector fields are also used to describe ocean currents, magnetic andelectric fields, etc. They are also of fundamental importance in differential equa-tions (MATH 330). We will discuss vector fields in greater detail in Chapter 8.

The best picture of a vector field is to draw a bunch of vectors f(x), each start-ing at their value of x. (Usually we shrink the length of f(x) by a common factorto emphasize the direction of the f(x)s.)

EXAMPLE 16Sketch a picture of the vector field f : R2 → R2 defined by f(x, y) = (x+ y, x− y).

Solution: Make a table of values, and draw the vector f(x, y) starting at eachpoint (x, y):

(choose these) (solve for this)x y f(x, y)0 0 (0 + 0, 0− 0) = (0, 0)0 1 (0 + 1, 0− 1) = (1,−1)1 0 (1 + 0, 1− 0) = (1, 1)3 −2 (1, 5)2 2 (2, 0)−3 1 (−2,−4)

-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

150


Mathematica sketches pictures of vector fields (with rescaled arrows) with theVectorPlot (for a function f : R2 → R2) and VectorPlot3D (for a function f : R3 → R3)commands. For the example above, f(x, y) = (x+ y, x− y), execute

VectorPlot[x + y, x - y, x, -4, 4, y, -6, 6]to get the picture below at left:

-4 -2 0 2 4

-6

-4

-2

0

2

4

6

-4 -2 0 2 4

-6

-4

-2

0

2

4

6

For the picture above at right, which has arrows all of the same length, add thesebells and whistles:

VectorPlot[x + y, x - y, x, -4, 4, y, -6, 6,VectorScale -> .05, Automatic, None, VectorStyle -> Red, Arrowheads[.03]]

For a picture of the vector field f : R3 → R3 defined by f(x, y, z) = (x+y+z, x−y, y + z), execute

VectorPlot3D[x+y+z, x-y, y+z, x,-4,4, y,-4,4, z,-4,4]

You will get this picture, which is very hard to get any useful information from:

151


Summary of Mathematica commands for functions of severalvariables

Here are the basic Mathematica commands discussed in this section which can beused to produce useful pictures of functions of several variables. By inserting op-tional statements into these commands, you can adjust the pictures in several ways(see the files on my web page for examples of this):

Class of function Basic Mathematica command to produce pictureCalculus I/II function Plot[f(x), x, xmin, xmax]

f : R→ Ry = f(x)

Polar function r = f(θ) PolarPlot[f(θ), θ, θmin, θmax]image (in a plane) of ParametricPlot[f(t), t, tmin, tmax]

f : R→ R2 ParametricPlot[x(t), y(t), t, tmin, tmax]i.e. (x, y) = f(t)

image (in 3D space) of ParametricPlot3D[f(t), t, tmin, tmax]f : R→ R3 ParametricPlot3D[x(t), y(t), z(t),

i.e. (x, y, z) = f(t) t, tmin, tmax]

surfacef : R2 → R

i.e. z = f(x, y)

graph: Plot3D[f(x, y), x, xmin, xmax,y, ymin, ymax]

level curve at height z = c:ContourPlot[f(x, y) == c,x, xmin, xmax, y, ymin, ymax]

basic contour plot (shaded, without labels):ContourPlot[f(x, y),x, xmin, xmax, y, ymin, ymax]

contour plot with z-values from a to b (unshaded,with labels):

ContourPlot[f(x, y),x, xmin, xmax, y, ymin, ymax,Contours -> Range[a,b],ContourShading -> None,ContourLabels -> True]

f : R3 → R level surface for w = c:i.e. w = f(x, y, z) ContourPlot3D[f(x, y, z) == c, x, xmin, xmax,

y, ymin, ymax, z, zmin, zmax]planar vector field VectorPlot[f(x, y), x, xmin, xmax,

f : R2 → R2 y, ymin, ymax]i.e. (u, v) = f(x, y)

3D vector field VectorPlot3D[f(x, y, z), x, xmin, xmax,f : R3 → R3 y, ymin, ymax, z, zmin, zmax]

i.e. (u, v, w) = f(x, y, z)

152


A last word on graphs

We started this section by looking at this picture:

-3 -2 -1 1 2 3

1

2

3

4

5

Based on what we learned, we can think of this blue shape as coming from threedifferent places:

1. It is the graph of

2. It is the image of

3. It is a level curve of

Thinking of the same set in different ways is key to being able to use vector calculusefficiently.

153


EXAMPLE 17Let E be a circle of radius 4 in the xy-plane, centered at (0, 0).

1. Describe E as the graph of one or more functions f : R→ R.

2. Describe E as the image of a function f : R→ R2.

3. Describe E as a level set of some function f : R2 → R.

154

3.3. Conic sections

3.3 Conic sectionsMotivation: We need a bank of shapes to use as examples for some of our

forthcoming calculus computations. These shapes need the following attributes:

1. they need to be realistic models for real-world applications;

2. they need to be relatively easy to work with; but

3. they need to be just hard enough so that they lead to an interesting variety ofproblems.

The right class of planar curves we want to study are called conic sections, and theright class of surfaces we care about are called quadric surfaces. In this section, wediscuss conics.

Definition 3.12 A conic section (or just conic) is any subset of a plane which canbe described in any of these three equivalent ways:

1. It is the intersection of a plane with a cone.

Technicality: when a mathematician says “cone”, they really mean what a nor-mal person would think of as two cones... so each picture below is really thepicture of one cone):

155

3.3. Conic sections

2. It the set of points (x, y) ∈ R2 such that the distance from (x, y) to a fixed pointis some fixed multiple of the distance from (x, y) to a line;

3. It is the set of (x, y) ∈ R2 satisfying an algebraic equation of the form

Ax2 +Bxy + Cy2 +Dx+ Ey + F = 0

where A,B,C,D,E, F are constants, not all equal to zero.

We will not prove that the three conditions given above are equivalent (i.e. thatthey describe the same class of objects). Also, there are many other equivalentcharacterizations of conics... do a Google search if you are interested in this.

Conic sections have been continually studied since ancient Greece, becausethey describe the motion of objects which are acted on by relatively simple forces(like gravity, or electrical or magnetic fields).

EXAMPLE 18By the third part of the definition of conic, the set

E = (x, y) ∈ R2 : 2x2 − 5xy + 4y2 − 9x+ 7y − 6 = 0

is a conic section, meaning that it must be the intersection of a plane with a cone.How might you get a picture of this set, using Mathematica? Equivalently, howwould you think about this set, given the types of graphs we studied in the previ-ous section?

E

-3 -1 1 3 5 7 9 11 13x

-5

-3

-1

1

3

5

7

9

11

y

156

3.3. Conic sections

Simplifying and classifying conics

Bad news: The equation

Ax2 +Bxy + Cy2 +Dx+ Ey + F = 0

has six constants in it, which makes it hard to work with.

Good news: If you are willing to change what you call “x” and “y”, you can greatlysimplify things. For example, here is how we can change the equation that de-scribes the set E from Example 16:

E

-3 -1 1 3 5 7 9 11 13x

-5

-3

-1

1

3

5

7

9

11

y

E

-3 -1 1 3 5 7 9 11 13

-5

-3

-1

1

3

5

7

9

11

E

-10 -8 -6 -4 -2 2 4 6 8 10new x

-4

-2

2

4

new y

Relative to the “new” x and y, the graph at right is centered at the origin, andsymmetric about both the x- and y-axes. That will make its equation much simpler(in fact, as we’ll see later, the equation is

x2

a2 + y2

b2 = 1

for suitable constants a and b).

In general, if you start with a conic section whose equation is

Ax2 +Bxy + Cy2 +Dx+ Ey + F = 0, (3.1)

you can:• get rid of the B term by rotating the x- and y-axes;• get rid of the D term (so long as A 6= 0) by a translation;• get rid of the E term (so long as C 6= 0) by a translation;• make F = 1 (so long as F 6= 0) by dividing through the equation by F ;• and move F to the other side of the equation.

This is not a course in analytic geometry, so you don’t actually have to be able todo the above stuff, but what you should be aware of is that after doing this, youcan reduce all conic sections to one of the following forms which can be easilyclassified:

157

3.3. Conic sections

Ellipses and circles

Definition 3.13 An ellipse is a conic section which can be described (after a rotationand/or translation of axes) as the set of points (x, y) ∈ R2 satisfying an equation of theform

x2

a2 + y2

b2 = 1.

Ellipses describe (among other things) the orbits of planets and are used to modelcertain types of random data applied in actuarial science.

Remark: If a = b, then the ellipse is

x2

a2 + y2

a2 = 1⇒

so circles are specific types of ellipses.

EXAMPLE 19Sketch the graph of the ellipse x2

36 + y2

9 = 1.

Recall: we want to think of curves in three different ways, if possible:

• as the graph of a function y = f(x) (not feasible here, since ellipses fail theVertical Line Test);

• as a level curve to a function f : R2 → R;

• as the image of a function f : R→ R2.

158

3.3. Conic sections

Theorem 3.14 (Parametric equations of an ellipse) The ellipse

x2

a2 + y2

b2 = 1

can be described as the image of the function f : R→ R2 where

f(t) = (x, y) = (a cos t, b sin t).

One complete revolution of the ellipse corresponds to the interval 0 ≤ t ≤ 2π.

PROOF Suppose x = a cos t and y = b sin t. Then

x2

a2 + y2

b2 = a2 cos2 t

a2 + b2 sin2 t

b2 =

In this setting, t refers to the angle at which the point (x, y) on the ellipse sits:

t

(x,y)

a-a

b

-b

x

y

Special case: a circle centered at the origin with radius r is the image of f : R→R2 given by f(t) = (r cos t, r sin t) for 0 ≤ t ≤ 2π.

EXAMPLE 20Find parametric equations, and sketch the graph of, the ellipse 64x2 + y2 = 16.

159

3.3. Conic sections

Hyperbolas

Definition 3.15 A hyperbola is a conic section which can be described (after a rota-tion and/or translation of axes) as the set of points (x, y) ∈ R2 satisfying an equationof the forms

x2

a2 −y2

b2 = 1 ory2

b2 −x2

a2 = 1.

The graph of a hyperbola is disconnected; it consists of two pieces which openeither to the left and right, or up and down.

If a hyperbola has equation x2

a2 − y2

b2 = 1, then the two pieces open to the left andright, like the picture below at left.

If a hyperbola has equation y2

b2 − x2

a2 = 1, then the two pieces open upward anddownward, like the picture below at right.

a-a x

y

b

-b

x

y

Theorem 3.16 (Parametric equations of hyperbolas) The hyperbola

x2

a2 −y2

b2 = 1


f(t) = (x, y) = (a sec t, b tan t).

The hyperbolay2

b2 −x2

a2 = 1


f(t) = (x, y) = (a tan t, b sec t).

PROOF HW (as a hint, remember that sec2 θ − tan2 θ = 1).

160

3.3. Conic sections

EXAMPLE 21Find the Cartesian equation of the hyperbola which is the image of the curve f :R→ R2 where f(t) = (3 tan t, 2 sec t). Then sketch the graph of this hyperbola.

EXAMPLE 22Find parametric equations for the left half of the hyperbola whose Cartesian equa-tion is

x2

25 −y2

9 = 1.

Solution: From the theorem on the previous page, we know the entire hyperbola isthe image of f : R→ R2 where

f(t) = (5 sec t, 3 tan t).

To get only the left-half, we restrict to t-values which produce points on the left-half of the Cartesian plane. From basic trig, we know those t-values go from π

2 to3π2 . So the answer, which includes both the function and the range of t-values, is

f(t) = (5 sec t, 3 tan t) whereπ

2 < t <3π2 .

161

3.3. Conic sections

Parabolas

Definition 3.17 A parabola is a conic section which can be described (after a rotationand/or translation of axes) as the set of points (x, y) ∈ R2 satisfying an equation of theforms

y = ax2 or x = ay2

where a 6= 0.

After translating axes, parabolas of both types can be assumed to have their vertexat the origin, and open either up (y = ax2, a > 0), down (y = ax2, a < 0), left(x = ay2, a < 0) or right (x = ay2, a > 0).

Theorem 3.18 (Parametric equations of a parabola) The parabola y = ax2 is theimage of the function f : R→ R2 where

f(t) = (x, y) = (t, at2).

The parabola x = ay2 is the image of the function f : R→ R2 where

f(t) = (x, y) = (at2, t).

In both cases, to get the entire parabola, you need to consider all t-values in (−∞,∞).

EXAMPLE 23Sketch the graph of the parabola x = 2y2.

EXAMPLE 24Sketch the graph of the parabola which is the image of f(t) = (t,−1

2t2).

162

3.3. Conic sections

Summary of nondegenerate conics

The phrase nondegenerate conic refers to any conic that is an ellipse, a hyperbola, ora parabola. Having studied each of these, we can now summarize their key facts:

Class

ofnon-degenerate

conicC

artesianequation

Parametric

equationsG

raph

Ellipse(circle,if

a=b)

x2

a2 +

y2b 2

=1

x

=acos

ty

=bsin

t

0≤t≤

2πa

-a

b

-b

x

y

Hyperbola

(openingleftand

right)x

2

a2 −

y2b 2

=1

x

=asec

ty

=btan

t

0≤t≤

2πa

-a

x

y

Hyperbola

(openingup

anddow

n)y

2b 2−

x2

a2

=1

x

=atan

ty

=bsec

t0≤t≤

2π

b

-b

x

y

Parabola(opening

upor

down)

y=ax

2

x

=t

y=at 2

−∞

<t<∞

x

y

Parabola(opening

leftorright)

x=ay

2

x

=at 2

y=t

−∞

<t<∞

x

y

163

3.3. Conic sections

Degenerate conics

There are some situations where a conic section is not an ellipse, hyperbola orparabola. These are called degenerate cases, and should be thought of as “freak” or“flukish” situations.

Definition 3.19 A conic section that is not an ellipse, hyperbola or parabola is calleda degenerate conic.

Class ofdegenerate

conic

General formof equation

Specificexample

Graph of specificexample

Empty set Ax2 + Cy2 = 1where A < 0, C < 0 −x2 − y2 = 1

Point Ax2 + Cy2 = 0where AC > 0 x2 + 2y2 = 0 x

y

Line Dx+ Ey = F 3x+ 2y = 6 x

y

Parallellines

Ax2 = 1 or Cy2 = 1where A > 0, C > 0 x2 = 9 x

y

Intersectinglines

Ax2 + Cy2 = 1where AC < 0

4x2 − y2 = 0(a.k.a. y = ±2x)

x

y

164

3.4. Quadric surfaces

3.4 Quadric surfacesRecall that all conics, which are subsets of R2, satisfy an equation of the form

Ax2 +Bxy + Cy2 +Dx+ Ey + F = 0.

If we add a third variable to this equation (i.e. adding an extra dimension), we geta class of objects living in R3, called quadric surfaces:

Definition 3.20 A quadric surface (or just quadric) is a set of points (x, y, z) ∈ R3

satisfying some equation of the form

Ax2 +By2 + Cz2 +Dxy + Exz + Fyz +Gx+Hy + Iz + J = 0

for constants A,B, ..., J , not all zero.

Similarly to conic sections, we can get rid of some of these terms by rotation ofaxes, translations, etc. so we can restrict our study to the situation where:

• D = E = F = 0 (rotate axes to get rid of these terms)

• A or G is zero (complete square on x to get rid of G)

• B or H is zero (complete square on y)

• C or I is zero (complete square on z)

• J is on the other side of the equation and J = 1 or 0 (divide through by Jotherwise)

This leaves us with some classes of quadrics which we run through in the next fewpages (we’ll ignore the degenerate cases).

Initially, we will learn the shape of each class of quadric by graphing the tracesand/or level curves of their associated equations, but with practice I hope you willbecome more familiar with what equation goes with what type of quadric.

165


Ellipsoids and spheres

Definition 3.21 An ellipsoid (centered at the origin) is a quadric section whoseequation is

x2

a2 + y2

b2 + z2

c2 = 1.

If a = b = c, then by setting r = a = b = c, we can rewrite the equation of theellipsoid as

x2 + y2 + z2 = r2,

the equation of a sphere of radius r centered at the origin given in Chapter 2. So ifa = b = c, the ellipsoid is a sphere.

EXAMPLE 25Sketch the graph of 2x2 + 4y2 + z2 = 16.

x = 0 trace y = 0 trace z = 0 trace4y2 + z2 = 16 2x2 + z2 = 16 2x2 + 4y2 = 16

166


Hyperboloids

Definition 3.22 A hyperboloid (centered at the origin) is a quadric section whoseequation is

Ax2 +By2 + Cz2 = 1

where A,B,C 6= 0 and A,B and C do not all have the same sign.If exactly two of A,B,C are positive, and one is negative, we say the hyperboloid

is a hyperboloid of one sheet, and if exactly two of A,B,C are negative, and oneis positive, we say the hyperoloid is a hyperboloid of two sheets.

EXAMPLE 26Sketch the graph of x2 − y2 − z2 = 1.

x = 0 trace y = 0 trace z = 0 trace−y2 − z2 = 1 x2 − z2 = 1 x2 − y2 = 1

167


EXAMPLE 27Sketch the graph of y2 = x2 + z2

4 − 1.

x = 0 trace y = 0 trace z = 0 tracey2 = z2

4 − 1 0 = x2 + z2

4 − 1 y2 = x2 − 1z2

4 − y2 = 1 x2 + z2

4 = 1 x2 − y2 = 1

168


Cones

Definition 3.23 A cone (centered at the origin) is a quadric section whose equationis

Ax2 +By2 + Cz2 = 0

where exactly two of A,B,C are positive, and one is negative.

Note: If one of A,B,C is positive and two are negative, the equation is stilla cone: you can multiply through the entire equation by (−1) to satisfy the abovedefinition.

EXAMPLE 28Sketch the graph of x2

16 = y2 + z2.

x = 0 trace y = 0 trace z = 0 trace0 = y2 + z2 0 = z2 − x2

16 0 = y2 − x2

16

169


Paraboloids

Definition 3.24 A paraboloid (centered at the origin) is a quadric section whoseequation is one of the following three forms:

z = Ax2 +By2 y = Ax2 + Cz2 x = By2 + Cz2

where A,B and/or C represent nonzero constants. If these two constants have thesame sign, we call the paraboloid an elliptic paraboloid and if they have oppositesigns, we call the paraboloid a hyperbolic paraboloid.

EXAMPLE 29Sketch the graph of y − 3x2 = z2.

x = 0 trace y = 0 trace z = 0 tracey = z2 z2 + 3x2 = 0 y = 3x2

170


EXAMPLE 30Sketch the graph of z = y2 − x2.

x = 0 trace y = 0 trace level curve z = 0z = y2 z = −x2 y2 − x2 = 0

y2 = x2

171


Summary of quadric surfaces

Classification of ±Ax2 ±By2 ± Cz2 = 1

SIGN ONx2 TERM

SIGN ONy2 TERM

SIGN ONz2 TERM

CLASS OFQUADRIC

EXAMPLEGRAPH

+ + + ellipsoid (sphere if A = B = C)

+ + − hyperboloid of one sheet,opening around z-axis

+ − + hyperboloid of one sheet,opening around y-axis

− + + hyperboloid of one sheet,opening around x-axis

+ − − hyperboloid of two sheets,opening around x-axis

− + − hyperboloid of two sheets,opening around y-axis

− − + hyperboloid of two sheets,opening around z-axis

− − − empty set

Classification of ±Ax2 ±By2 ± Cz2 = 0

SIGN ONx2 TERM

SIGN ONy2 TERM

SIGN ONz2 TERM

CLASS OFQUADRIC

EXAMPLEGRAPH

+ + + single point (0, 0, 0)− − − single point (0, 0, 0)+ + − cone opening around z-axis+ − + cone opening around y-axis− + + cone opening around x-axis+ − − cone opening around x-axis− + − cone opening around y-axis− − + cone opening around z-axis

Classification of z = ±Ax2 ±By2

(y = ±Ax2 ±Bz2; x = ±Ay2 ±Bz2 similar,but open around y and x axes respectively)

SIGN ONx2 TERM

SIGN ONy2 TERM

CLASS OFQUADRIC

EXAMPLEGRAPH

+−

+−

paraboloid opening to positive z-axisparaboloid opening to negative z-axis

+−

−+ hyperbolic paraboloid (saddle)

172

3.5. Limits of functions Rn → R

3.5 Limits of functions Rn → R

The idea of the limit

Recall the following facts about limits of functions f : R→ R:

• Suppose f : R→ R. To saylimx→a

f(x) = L

means that as x gets closer and closer to a (but ignoring any value of f(a)),the corresponding values f(x) get closer and closer to L.

• To evaluate a limit of the form limx→a

f(x), first try plugging in a for x.

– If you get an answer, that is probably the answer to the limit.– If you get 0

0 , then rewrite the limit by1. using L’Hôpital’s Rule, or2. factoring and cancelling, or3. conjugating square roots, etc.

Then you plug in again.– If you get nonzero

0 , then the answer is ±∞.

Most of these ideas carry through when we talk about limits of functions Rn →R.

Definition 3.25 Let f : Rn → R and let a ∈ Rn. To say

limx→a

f(x) = L

means that as x gets closer and closer to a (without actually getting to a), the corre-sponding values f(x) get closer and closer to L.

Remark: This is an imprecise definition of limit; there is a more precise defini-tion of limit involving εs and δs which is covered in real analysis (Math 430).

EXAMPLE 31

lim(x,y)→(0,0)

x2y + cosx+ e−y

x2 + y2 + 1

Alternate notation: the limit in Example 29 could also be written as

limx→0

x2y + cosx+ e−y

x2 + y2 + 1 .

173


Note: recall that derivatives and integrals are the two most important exam-ples of limits. Their definitions are

f ′(x) = limh→0

f(x+ h)− f(x)h

and∫ b

af(x) dx = lim

||P||→0

n∑j=1

f(cj)∆xj

Note that both limits are taken as the variable tends to zero. Thus limits as thevariable approaches zero are of particular importance, and we focus on thosetypes of limits in this course.

Basic computational techniques

As with limits from Calculus 1, the easiest computational method is “plugging andchugging”:

EXAMPLE 32

lim(x,y,z)→(0,0,0)

4exyz + 32(x+ 1)(y − 1)(z − 2)

You identify infinite limits the same way you did in Calculus 1:

EXAMPLE 33

limx→0

cos(xy) + 1x2y2

Factoring and canceling is still a valid technique:

EXAMPLE 34

limx→0

x2 − y2

x+ y

Not every limit is where the variable tends to 0:

EXAMPLE 35

limx→(3,−2)

x− 2yx+ y

174


Squeeze Theorem

In Calculus 1, you may have heard of something called the Squeeze Theorem. Thisfact also holds for functions Rn → R, and is more useful here than it was in Calcu-lus 1:

Theorem 3.26 (Squeeze Theorem) Suppose f, g and h are functions from Rn to Rsuch that for all x 6= a ∈ Rn, f(x) ≤ g(x) ≤ h(x). If

limx→a

f(x) = limx→a

h(x) = L,

then limx→a

g(x) = L as well.

A picture to explain: Suppose f, g, h : R → R satisfy f(x) ≤ g(x) ≤ h(x). Then, iflimx→a

f(x) =limx→a

h(x) = L, then we have the following picture:

h

g

fa

L

It should be clear from the picture that limx→a

g(x) = L as well.

EXAMPLE 36

limx→0

x4

x2 + y6

175


Determining when limits fail to existEXAMPLE 37

limx→0

x+ y

x− y

Recall the following facts about limits of functions f : R→ R:

• To say limx→a

f(x) = L means that f(x) must approach L as x approaches a nomatter which direction you approach a from.

• For a function f : R→ R, you can only approach a from two directions:

from the right: limx→a+

f(x)

from the left: limx→a−

f(x)

• limx→a

f(x) exists only if limx→a+


f(x) both exist and are equal, in

which case limx→a+


f(x) is equal to the common value of the two

one-sided limits.

• So if limx→a+

f(x) 6= limx→a−

f(x), we can conclude limx→a

f(x) DNE.

The same principle holds for limits of functions Rn → R. To say limx→a

f(x) = L

means that f(x) must approach L as x approaches a no matter which direction youapproach a from.

The big difference is...

176


Here are some various paths that you can approach (0, 0) from:


limx→0

x+ y

x− y

EXAMPLE 38

lim(x,y)→(0,0)

xy

x2 + y2

177


EXAMPLE 39

limx→0

xy + xz

x2 + y2 + z2

Solution: Along the x-axis, we have

lim(x,0,0)→(0,0,0)

xy + xz

x2 + y2 + z2 = limx→0

0 + 0x2 + 02 + 02 = lim

x→0

0x2 = 0.

Along the line x = y = z, we have

lim(x,x,x)→(0,0,0)

xy + xz

x2 + y2 + z2 = limx→0

x2 + x2

x2 + x2 + x2 = limx→0

2x2

3x2 = 23 .

Since these limits are unequal, we can conclude

limx→0

xy + xz

x2 + y2 + z2 DNE .

EXAMPLE 40

lim(x,y)→(0,0)

x3 + y4

x2 + y2

WARNING

You can never conclude that a limit does exist, based on argument usingpaths alone. Evaluating limits along paths is a technique to show that alimit does not exist.

178


The polar coordinates trick

Suppose f : R2 → R. We have seen that in order for

limx→0

f(x) = L,

f(x) must approach L no matter the direction in which x approaches 0.

What do all these paths have in common?

Theorem 3.27 (Polar coordinates trick) Suppose that the rule for a function f :R2 → R is rewritten in terms of polar coordinates.

1. If limr→0

f(r, θ) = L no matter what θ is, then limx→0

f(x) = L.

2. If there are values of θ which produce two or more different values of limr→0

f(r, θ),then lim

x→0f(x) DNE.


lim(x,y)→(0,0)

x3 + y4

x2 + y2

179


EXAMPLE 41

lim(x,y)→(0,0)

x2

x2 + y2

Note: Example 41 could also be done using paths:

• Along x = 0: lim(0,y)→(0,0)

x2

x2+y2 =limy→0

00+y2 =lim

y→00 = 0, but

• Along y = 0: lim(x,0)→(0,0)

x2

x2+y2 =limx→0

x2

x2+0 =limy→0

1 = 1.

The polar coordinates trick works only for limits of functions R2 → R. Whattrick do you think works for limits of functions f : R3 → R?

EXAMPLE 42

lim(x,y,z)→(0,0,0)

xy + yz + xz

x2 + y2 + z2

180


Theorem 3.28 (Spherical coordinates trick) Suppose that the rule for a functionf : R3 → R is rewritten in terms of spherical coordinates.

1. If limρ→0

f(ρ, ϕ, θ) = L no matter what ϕ and θ are, then limx→0

f(x) = L.

2. If there are different values of ϕ and/or θ which produce two or more differentvalues of lim

ρ→0f(ρ, ϕ, θ), then lim

x→0f(x) DNE.

EXAMPLE 42, REVISTED

lim(x,y,z)→(0,0,0)

xy + yz + xz

x2 + y2 + z2

= limρ→0

ρ2 sin2 ϕ cos θ sin θ + ρ2 sinϕ sin θ cosϕ+ ρ2 sinϕ cos θ cosϕρ2

= limρ→0

(sinϕ)[sinϕ cos θ sin θ + sin θ cosϕ+ cos θ cosϕ]

181

3.6. Limits and continuity of functions Rn → Rm

3.6 Limits and continuity of functions Rn → Rm

Having discussed the limits of functions taking values in R, we can now studythe limits taking values in any Rm. All you do is evaluate the limit coordinate-wise:

Theorem 3.29 Let f : Rn → Rm and let a ∈ Rn. Then, if the coordinate functions off are denoted (f1, f2, ..., fm) (where each fj takes Rn to R):

limx→a

f(x) exists ⇐⇒ limx→a

fj(x) exists for all j,

in which case

limx→a

f(x) =(

limx→a

f1(x), limx→a

f2(x), ..., limx→a

fm(x)).

The definition of continuity you learn in Calculus I extends in the natural way:

Definition 3.30 Let f : Rn → Rm and let a ∈ Dom(f). We say f is continuous(cts) at a if

limx→a

f(x) = f(a).

A function f : Rn → Rm is called continuous (cts) if it is continuous at every pointa in its domain.

Since limits of functions are computed coordinate-wise, we therefore know:

Theorem 3.31 Let f : Rn → Rm. Then

1. For any a ∈ Rn, f is continuous at a if and only if each of the componentfunctions of f are continuous at a; and

2. f is continuous if and only if each of the component functions of f are continuous.

From Calculus I, we know that any function R → R comprised of usual mathoperations is continuous at all points except where the denominator of any fractionis zero. This generalizes:

Theorem 3.32 Any function f : Rn → Rm whose component functions are com-prised of addition, subtraction, multiplication, division, powers and roots of the com-ponents of x, sines and cosines, arctangents, arcsines, exponential and logarithmicterms of the components of x is continuous at any x ∈ Dom(f) except at those xwhich make the denominator of some fraction in any of its component functions zero.

182

3.6. Limits and continuity of functions Rn → Rm

EXAMPLE 43Let f(t) = (3 cos t, 4 sin t, 2et, t+ 2). Then

limt→0

f(t) =

EXAMPLE 44Let f(x) =

(sinxx, x+2x−1

). Then

limx→0

f(x) =

EXAMPLE 45Let f(x, y) =

(x3+2y3

x2+y2 , 4 cos y, x+ 2y − 3). Then

lim(x,y)→(0,0)

f(x, y) =

183



1. Suppose f(x, y, z) = (x+ 3y, 2− 7y, x+ y + z, z − 2).

a) What is f2(x, y, z)?

b) What is f3(4, 7,−3)?

c) What is the codomain of f?

d) What is the codomain of f1?

2. a) Suppose f(t) = (3t, 2t2, 1− 2t). Find z(4).

b) Let x = (3,−7, 2). Find 5π2(x) + 8π1(x).

c) Suppose f : (x, y) 7→ (u, v) where u = x2 − y2 and v = 2xy. Find f(3, 1).


3. If f : R3 → R8, what vector space is graph(f) a subset of?

In Problems 4-9, sketch a picture of the image of these functions by hand, mak-ing a table of values, plotting the points you get in your table, and connecting themappropriately:

4. f(t) = (t− 1)2 − 4

5. f(t) = (3− t, 1 + 2t)

6. f(t) = (0, 0,−t)

7. f(t) = (0, t−1, 2t+1)

8. f(t) = (2, t, t2)

9. f(t) = (3 cos t, sin t)

In Problems 10-15, sketch a picture of the image of these functions using Mathe-matica. Make sure you use a range of t-values and an appropriate viewing windowso that you can see the relevant characteristics of the graph, and add an arrow (byhand) on the graph that indicates the direction of motion for increasing t.

10. f(t) = (6 cos t, 3 sin t)

11. f(t) = (3 cos 3t, 2 sin t)

12. f(t) = (ln(t2 + 1), sin t)

13. f(t) = (2 cos t, 2 sin t, t/4)

14. f(t) = (t cos 2t, t sin t, t)

15. f(t) = (cos 3t, sin 2t, sin t)

In Problems 16-21, sketch a picture of the graph of each function using Math-ematica. Make sure to use an appropriate viewing window, so you can see therelevant characteristics of the graph.

184


16. f(x, y) = cos x sin y

17. f(x, y) = sin(4− x2 − y2)

18. f(x, y) = ye−12y(x+y)

19. f(x, y) = ln(|x2 − y2|)

20. f(x, y) = 2− x sin y

21. f(x, y) = x+3yx−2y

22. Let h(x, y) = y + ex.

a) Sketch traces of h corresponding to x = −2,−1, 0, 1, 2, each on its ownyz-plane.

b) Sketch the traces you drew in part (a) as they would appear in xyz-spaceas part of the graph of h.

c) Sketch traces of h corresponding to y = −2,−1, 0, 1, 2, each on its ownxz-plane.

d) Sketch the traces you drew in part (c) as they would appear in in xyz-space as part of the graph of h.

e) Using the pictures you drew in parts (b) and (d), sketch a crude graphof h.

23. Let g(x, y) = (4− x2)y2.

a) Sketch traces of g corresponding to x = −3,−2,−1, 0, 1, 2, 3, each on itsown yz-plane.

b) Sketch the traces you drew in part (a) as they would appear in xyz-spaceas part of the graph of g.

c) Sketch traces of g corresponding to y = −2,−1, 0, 1, 2, each on its ownxz-plane.

d) Sketch the traces you drew in part (c) as they would appear in in xyz-space as part of the graph of g.

e) Using the pictures you drew in parts (b) and (d), sketch a crude graphof g.

24. Let f(x, y) = x2 + y2.

a) Sketch a contour plot for f , including labelled level curves at heights9, 4, 1, 0 and −1.

b) Sketch the level curves you drew in part (a) as they would appear inxyz-space as part of the graph of f .

c) Based on the picture you drew in part (b), sketch a crude graph of f .

25. (R) Let h(x, y) = yx.

185


a) Sketch a contour plot for h, including labelled level curves at heights−2, 1, 0, 1 and 2.

b) Sketch the level curves you drew in part (a) as they would appear inxyz-space as part of the graph of f .

c) Based on the picture you drew in part (b), sketch a crude graph of h.

26. Let g(x, y, z) = 3− x+ 3y − z.

a) Sketch the level surface for the function g at height 0.

b) Sketch the level surface for the function g at height 9.

27. Let g(x, y, z) = x2 + y2 + z2.

a) Sketch the level surface for the function g at height 9.

b) Sketch the level surface for the function g at height 4.

c) Sketch the level surface for the function g at height 0.

d) Sketch the level surface for the function g at height −4.

In Problems 28-33, use Mathematica to sketch the indicated level curves, contourplots or level surfaces:

28. The level curve to the function f(x, y) = x3 − 2xy3 + 4y − 8x at height −7.

29. A contour plot for the function g(x, y) = 4x sin(y−x) cos(x+y), showing levelcurves at heights −8,−4, 0, 4 and 8.

30. The set of points (x, y) ∈ R2 satisfying x3 − 8xy + y3 = 4.

31. A contour plot for the function f(x, y) = ecos(x+y) + esinx sin(x−y).

32. The set of points (x, y, z) ∈ R3 satisfying z2 = 8x2

1+y2 + 8y2

1+x2 .

33. The level surface of g(x, y, z) = xy3 − zy + xz2 + 4z at height 3.

34. The contour plot for some unknown function f : R2 → R is given on the nextpage. Use that contour plot to answer these questions:

a) Estimate the value of f(−2,−3).

b) Estimate the value of f(1, 2).

c) Which value is greater: f(3,−2) or f(3, 0)?

d) If you move in a straight line from (−2,−2) to (−2, 0), would you expectthe value of f to increase or decrease? By how much does the value of fincrease or decrease?

186


e) If you were at the point (0, 2), in what compass direction would youwant to move if you wanted to head towards a smaller value of f asquickly as possible?

f) For what value of y is f(3, y) minimized? Estimate the minimum valueof f(3, y).

g) Sketch the level curve for f corresponding to height 1.

h) Sketch a graph of the y = 0 trace of f .

-8-8 -6

-6

-4-4

-2-2

0

0

2

2

4

4

4

4

6

6

6 6

6 66-4 -3 -2 -1 0 1 2 3 4

-4

-3

-2

-1

0

1

2

3

4

-4 -3 -2 -1 0 1 2 3 4

-4

-3

-2

-1

0

1

2

3

4

x

y

35. A 12′′ × 12′′ metal plate occupies the region [0, 12] × [0, 12] in the xy-plane.Suppose that this metal plate is heated unevenly, so that its temperature (inF) at point (x, y) on the plate is T (x, y) = 170 + 100 cos x

2 sin x+y3 . Use Mathe-

matica to obtain a contour plot of T , and use that contour plot to answer thesequestions:

a) What is the temperature at point (9, 4)?

b) What is the coldest point on the plate?

c) What is the temperature at the coldest point on the plate?

d) If a heat-seeking particle, which at all times moves in the direction thatT most greatly increases, is placed on the plate at position (2, 11), sketchthe path along which the particle will move.

e) What will be the final position of the particle in part (d)?

f) What will be the temperature at the final location the particle in part (d)reaches?

187


g) Suppose an object moves along this plate in a straight line from point(6, 1) to point (10, 9). What is the net change in temperature the objectexperiences along its trip?

h) What is the average rate of change in temperature (per inch travelled)that the object in part (g) experiences along its trip?

Here are pictures of nine surfaces in R3, each of which is the graph of a functionf : R2 → R:

A. B. C.

D. E. F.

G. H. I.

In each of Problems 36-41, you are given a contour plot of some function f :R2 → R. Determine which of the pictures A-I above, if any, is a possible graph forthe function whose contour plot is given.

Note: It is conceivable that more than one graph matches the given contour plot,since the contour plots are unlabelled.

188


36.

-4 -2 0 2 4

-4

-2

0

2

4

37.

-4 -2 0 2 4

-4

-2

0

2

4

38.

-4 -2 0 2 4

-4

-2

0

2

4

39.

-4 -2 0 2 4

-4

-2

0

2

4

40.

-4 -2 0 2 4

-4

-2

0

2

4

41.

-4 -2 0 2 4

-4

-2

0

2

4

In Problems 42-47, sketch a crude graph of each function by hand:

42. f(x, y) = ex

43. f(x, y) = y2 − 4y

44. f(x, y) = 12x

45. f(x, y) = √y

46. (R) f(x, y) = |y|

47. (R) f(x, y) = x2 − 5

In Problems 48-57, you are given an equation or a formula for some function.Determine which Mathematica command is most appropriate for generating a pic-ture associated to that equation or formula. Your choices are Plot, PolarPlot, Plot3D,ParametricPlot, ParametricPlot3D, ContourPlot, ContourPlot3D, VectorPlot, and Vector-Plot3D.

48. r = cos 3θ

49. x2 + 3y2 = 24

50. f(x, y) = (3x+ y2, sin(x− y))

51. x2 − 3xz + 4xyz − 5z + y = 17

52. f(x, y) = 3x2 cos exy

53. y = 17x− 4

189


54. f(t) = (et, e2t, e3t−1)

55. f(t) = (4 sin t, 3 cos 2t− 7 cos t)

56. f(x) = 4x3 − 3ex +√x2 + 1

57. f(x, y) = x− 6

In Problems 58-60, you are given a picture which can be produced by one ormore Mathematica commands. In each part of the problem, fill in the blank in theincomplete Mathematica command, so that the picture given in the problem is pro-duced by that command.

58. -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

a) Plot[ , x, -3, 3]b) ContourPlot[ , x, -3, 3, y, -3, 3]

59. -3 -2 -1 1 2 3

-3

-2

-1

1

2

3

a) Plot[ , x, -3, 3]b) ParametricPlot[ , t, -3, 3]c) ContourPlot[ , x, -3, 3, y, -3, 3]

60. -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

a) Plot[ , x, -7, 7]b) ParametricPlot[ , t, 0, 2π]c) PolarPlot[ , θ, 0, 2π]d) ContourPlot[ , x, -7, 7, y, -7, 7]

In Problems 61-66, you are given either a description or a picture of some subsetE ⊆ R2. For each problem:

(a) if possible, describe the set as the graph of one or more functions from R toR;

(b) if possible, describe the set as the image of a function from R to R2; and(c) if possible, describe the set as a level set of some function from R2 to R.

61. E = (x, y) ∈ R2 : x2 + y2 = 36

62. E = (x, y) ∈ R2 : 5x+ 4y = 20

190


63.

E

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

64. E =the y-axis

65. E = (x, y) ∈ R2 : y = ex

66.

E

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4


67. (F) Prove Theorem 3.16 from the notes, in which parametric equations for ahyperbola are established.

In Problems 68-73, sketch (by hand) the graph of the given conic section, andfind a set of parametric equations describing the conic.

68. x2

25 −y2

4 = 1

69. x2

16 + y2

4 = 1

70. y = −2x2

71. x = 14y

2

72. y2 = x2 + 1

73. x2 + y2 = 14

In Problems 74-77, find a set of Cartesian equations describing the conic. Thensketch the graph of the conic.

74.x = 1

3 cos ty = 2

3 sin t

75.x = 4 sec ty = tan t

76.x = 2t2y = t

77.x = 8 cos ty = 5 sin t

In Problems 78-81, use Mathematica to graph the given conic section. Use thegraph you produce to classify the conic as an ellipse, a hyperbola, a parabola, aline, parallel lines, intersecting lines, a point, or the empty set.

78. xy = 1

79. 8x2 − 5xy + y2 − 16x+ 7y = 17

80. 3xy − y2 + 4x− y + 7 = 0

81. y2 − xy − 6x2 = 0


In Problems 82-91, sketch (by hand) a crude graph of the given quadric surface.

191


82. z2 − x2 − y2 = 4

83. 2x2 = 4y2 + z2

84. (R) x = y2 + z2

85. y2 − x2

4 + z2

9 = 1

86. z = x2 − y2

87. (R) z = −2x2 − y2

88. z2 = y2 − x2

89. (R) z2 = y2 − x2 + 1

90. (R) z2 = y2 − x2 − 1

91. x2 + y2

25 + z2

9 = 1

In Problems 92-95, use Mathematica to sketch the graph of the given quadricsurface. Use the picture you produce to classify the quadric as as an ellipsoid, a hy-perboloid of one sheet, a hyperboloid of two sheets, a cone, an elliptic paraboloid,or a hyperbolic paraboloid.

92. x2 + 2xy − 3xz + z2 = 1

93. x2 + 2y2 + z2 − 5xy − 3xz + 4− 2z + 1 = 0

94. 5yz + 3xz − 4x− z = 1

95. 5− 10x+ 5x2 − 4y + 4xy − 2y2 + 6z − 6xz − 2yz = 0

Problems from Sections 3.5 and 3.6

In Problems 96-127, evaluate the given limit (or explain why the limit does notexist).

96. lim(x,y)→(1,3)

4x+yx−2y

97. limx→0

x+1y−2

98. lim(x,y)→(0,0)

x2

x2+y4

99. limx→0

xx2+y2

100. limx→0

4x2+y4

101. lim(x,y)→(0,0)

y8−x6

y4+x3

102. limx→0

x−2yx2−4y2

103. limx→0

x−2yx2+4y2

104. lim(x,y)→(0,0)

ex

x+y−1

105. lim(x,y,z)→(0,0,0)

x2+2xzy2+z

106. limx→0

2x6

3x2+4y6

107. lim(x,y)→(0,0)

x12/5

x2+y2

108. lim(x,y)→(0,0)

y5/3

x2+y2

109. limx→0

xx2−y2

110. lim(x,y)→(0,0)

2x+yx−2y

111. (R) limx→0

x2+y2

4xy

112. limx→0

(x+y)2

x2+y2

113. lim(x,y)→(0,0)

x+y√x2+y2

192


114. lim(x,y)→(0,0)

3xy√x2+y2

115. limx→0

x3+y4

x3+y4

116. lim(x,y)→(0,0)

x2+y8

x2+y2

117. (R) limx→0

2x2

x2+y2

118. (R) lim(x,y)→(0,0)

x2+y2

sin(x2+y2)

119. lim(x,y,z)→(0,0,0)

e(x2+y2+z2)−1x2+y2+z2

120. limx→0

xzx2+y2+z2

121. lim(x,y,z)→(0,0,0)

x2+z2√x2+y2+z2

122. limx→0

xy+xz+yzx2+y4+z2

123. (F) lim(w,x,y,z)→0

wy+wz+xy+xzw2+x2+y2+z2

124. (F) lim(w,x,y,z)→0

w2y+wz2+x2y+xz2

w2+x2+y2+z2

125. limx→0

(exy, e−xy, 4)

126. limx→0

(x2−16x−4 ,

xtanx , 5 +

√x+ 4, x− 3

)127. lim

(x,y)→(0,0)

(x−y√x2+y2

, x+ y − 2)

193


Selected answers

1. a) 2− 7yb) 8c) R4

d) R

2. a) −7b) −27

3. graph(f) ⊆ R11

4. -10-9-8-7-6-5-4-3-2-1 1 2 3 4 5 6 7 8 9 10

The direction of motion is initiallyfrom right to left, then turns at x =−4 (when t = 0) and goes left toright.

5.

-4 -3 -2 -1 1 2 3 4 5 6 7 8

-4

-3

-2

-1

1

2

3

4

5

6

7

8

7.

8.

12.-1 1 2 3 4 5 6 7 8 9

-2

-1

1

2

13.

14.

16.

19.

194


20.

22. a)

x = −2 x = 0 x = 2

-2 -1 1 2y

-1

1

2

3

4

5

6

7

8

9

10z

-2 -1 1 2y

-1

1

2

3

4

5

6

7

8

9

10z

-2 -1 1 2y

-1

1

2

3

4

5

6

7

8

9

10z

b)

c)

y = −2 y = −1 y = 0

-2 -1 1 2x

-2

-1

1

2

3

4

5

6

7

8

9

10z

-2 -1 1 2x

-2

-1

1

2

3

4

5

6

7

8

9

10z

-2 -1 1 2x

-2

-1

1

2

3

4

5

6

7

8

9

10z

d)

e)

23. e)

24. Let f(x, y) = x2 + y2

a)1

4

9

0

-4 -2 0 2 4-4

-2

0

2

4

(The level curve at height −1is the empty set.)

b)

195


c)

26. a)

3

-1

3

b)

-6

2

-6

27. a)

d) ∅

28.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-5

-4

-3

-2

-1

0

1

2

3

4

5

-5 -4 -3 -2 -1 0 1 2 3 4 5

-5

-4

-3

-2

-1

0

1

2

3

4

5

x

y

30.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-5

-4

-3

-2

-1

0

1

2

3

4

5

-5 -4 -3 -2 -1 0 1 2 3 4 5

-5

-4

-3

-2

-1

0

1

2

3

4

5

xy

31.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-5

-4

-3

-2

-1

0

1

2

3

4

5

-5 -4 -3 -2 -1 0 1 2 3 4 5

-5

-4

-3

-2

-1

0

1

2

3

4

5

x

y

32.

34. a) f(−2,−3) ≈ 6.

d) Decrease, by about 11 units.

f) The minimum value occurs atabout y = −1

2 , and the mini-mum value is about −6.

196


g)

-4 -3 -2 -1 0 1 2 3 4

-4

-3

-2

-1

0

1

2

3

4

-4 -3 -2 -1 0 1 2 3 4

-4

-3

-2

-1

0

1

2

3

4

x

y

35. a) T (9, 4) ≈ 190F.f) ≈ 270F.

38. A

39. D

h) ≈ 52

√5F/in.

43.

45.

46.

49. ContourPlot

50. VectorPlot

53. Plot

55. ParametricPlot

56. Plot

58. a) (2/5)x-2

b) 2x - 5y == 10

59. c) y + xˆ2 == 2

60. a) Sqrt[25-xˆ2], -Sqrt[25-xˆ2]

b) 5 Cos[t], 5 Sin[t]

d) xˆ2 + yˆ == 25

61. a) E = graph(f)⋃ graph(g), where f(x) =√

36− x2 and g(x) = −√

36− x2.

b) E = image(f), where f(t) = (6 cos t, 6 sin t).

c) E is the level curve to f(x, y) = x2 + y2 at height 36.

63. a) E = graph(f)⋃ graph(g), where f(x) =√

1− x and g(x) = −√

1− x.

b) E = image(f), where f(t) = (t2 − 1, t).

64. a) Not possible (E fails the Vertical Line Test).

65. a) E = graph(f), where f(x) = ex.

b) E = image(f), where f(t) = (t, et).

197


66. b) E = image(f), where f(t) = (2 cos t+ 2, 2 sin t).

c) E is the level curve to f(x, y) = (x− 2)2 + y2 at height 4.

68. -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

x = 5 sec ty = 2 tan t

69. -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

x = 4 cos ty = 2 sin t

70. -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

x = ty = −2t2

72. -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

73.x =√

14 cos ty =√

14 sin t

74.- 43

-1 - 23

- 13

1

3

2

31 4

3

- 43

-1

- 23

- 13

1

3

2

3

1

4

3

76. x = 2y2

77. -10-9-8-7-6-5-4-3-2-1 1 2 3 4 5 6 7 8 9 10

-10-9-8-7-6-5-4-3-2-1

12345678910

78. hyperbola

82.

84.

198


85.

86.

91.

92. hyperboloid of one sheet

95. cone

96. −75

97. −12

98. DNE

102. DNE

103. DNE

104. −1

105. DNE

106. 0

109. DNE

112. DNE

113. DNE

116. DNE

119. 1

120. DNE

121. 0

125. (1, 1, 4)

127. DNE

199

3.8. Review material for Exam 1

3.8 Review material for Exam 1Typical tasks on this exam

Vectors and matrices: Perform basic operations with vectors and matrices (addi-tion, scalar multiplication, dot and cross product, norms, distances, matrixmultiplication, determinants); determine whether a complicated-looking ex-pression represents a vector, scalar or nonsense

Lines and planes: Solve problems involving lines and planes in 3D (write equa-tions of lines and/or planes, solve intersection problems, sketch graphs ofplanes)

Coordinate systems: Convert between Cartesian and polar coordinates in R2; con-vert between Cartesian, cylindrical and spherical coordinates in R3; sketchcrude graphs of basic equations in Cartesian, polar, cylindrical and/or spher-ical coordinates

Graphs: Answer questions related to graphs associated to functions of severalvariables, especially how traces and level curves associate to the graph ofa surface; interpret contour plots; sketch crude graphs of planes, cylindersand/or quadric surfaces; translate between Cartesian and parametric equa-tions of conics; represent a curve as a graph, image and/or level curve ofappropriate functions; write Mathematica commands to create appropriatepictures

Limits: Compute a limit of a function of several variables (or state with justifica-tion why the limit fails to exist)

Extra practice problems

1. Let v = (8,−2, 5) and w = (3, 7,−1).

a) Find v · (3w− 2v).

b) Find w× v.

c) Find a vector of length 19 in the opposite direction as w.

d) Find the distance from v to w.

2. Let M =

3 −1 40 2 52 1 −3

.

200


a) Find Mv, where v is as in Prob-lem 1.

b) Find detM .

c) Find 5M .

d) Find M2.

e) Find the transpose of M .

3. Suppose that in the given expressions, all (non-bold) lowercase letters arescalars, all boldface lowercase letters are vectors in R3, and all capital lettersare 3 × 3 matrices. Determine if each given quantity is nonsense, a scalar, avector (in R3), or a (3× 3) matrix:

a) (a · b)ab) (ra ·Ma)× b

c) ||||a||a × c||

d) ||a − sb||c + 5Qae) (bTm)XT

f)(

a||a|| × c

)M

4. Find parametric equations of the line in R3 passing through (10, 0,−4) and(7, 3,−1).

5. Write the normal equation of the plane in R3 containing the point (−2, 3,−6)and the line whose symmetric equations are

x− 34 = y + 2

7 = z − 5

6. Find parametric equations of the plane in R3 containing the points (0, 5, 0),(2, 3,−4) and (7,−8, 6).

7. Find the normal equation of the plane described in Problem 6.

8. Find the normal equation of the hyperplane in R4 whose normal vector is(5,−2, 3, 0) and which contains the point (2,−3, 7, 2).

9. Find the intersection of the line in R3 whose parametric equations arex = 3t− 4y = 5t+ 3z = −8t+ 9

and the plane in R3 whose normal equation is 7x − y − 4z = −11. If there isno intersection, explain why.

10. Find the distance from point (2,−5, 4) to the plane whose normal equation is7x− y + 4z = 19.

11. Sketch the graph of each given plane in xyz-space:

3x+ 2y + z = 12 − x− 4y + 2z = 8 x+ 3y = 9 y = 5

201


12. Find the polar coordinates of each point whose Cartesian coordinates aregiven:

(−6, 0) (4, 0) (−8, 8)

13. Find the Cartesian coordinates of each point whose polar coordinates aregiven: (

3, 3π2

) (8, 2π

3

)14. Give a crude sketch of the following polar equations:

r = 3 r = 2 sin θ r = 5 cos θ θ = π

3

15. Find the cylindrical coordinates of each point whose Cartesian coordinatesare given:

(0, 5, 0) (3,−3, 7)

16. Find the Cartesian coordinates of each point whose cylindrical coordinatesare given: (

6, 3π2 ,−4

) (4√

2, π4 , 3)

17. Find the spherical coordinates of each point whose Cartesian coordinates aregiven:

(4, 0, 4) (0, 0, 7) (5, 5√

3, 10√

3) (−3, 3, 0)

18. Find the Cartesian coordinates of each point whose spherical coordinates aregiven: (

8, π2 ,π

4

)(5, π, 0)

19. Sketch crude graphs in R3 of the following equations given in cylindricalcoordinates:

z = 4 r = 3 θ = π

2

20. Describe the shape in R3 described by each of the following equations givenin spherical coordinates:

ρ = 1 ϕ = π

6 ϕ = π

2

21. Sketch crude graphs in R3 of the following equations given in Cartesian co-ordinates:

z = y2 y = cosx x2 + y2 + z2 = 16

202


22. Sketch crude graphs in R3 of the following equations given in Cartesian co-ordinates:

x2 = y2 + z2 x2 − y2 + z2 = 1 z = y2 − x2 y = −x2 − z2

23. Let f(x, y) = x2 − y2. Sketch the graph of each given trace:

x = 0 y = 0 x = 2 y = −1

24. Below is a contour plot for some unknown function f : R2 → R:

-3

-2

-1

0

12 2 3

3

4

4

56

7

8

9

10

11

12

13

14

-4 -2 0 2 4

-4

-2

0

2

4

Use this contour plot to estimate the answer to these questions:

a) Which is greater, f(−2,−1) or f(−2, 3)?

b) Estimate f(1, 2).

c) Estimate a value of y such that f(4, y) = 1.

d) From the point (4, 3), which compass direction corresponds to the direc-tion in which f most rapidly increases?

e) Sketch a rough graph of the x = 0 trace of f .

25. Here are a bunch of pictures, labelled A through L:

203


A.

-4 -2 2 4

-4

-2

2

4

B.

-4 -2 2 4

-6

-4

-2

2

4

6

C.

-4 -2 2 4

-4

-2

2

4

D.

-6 -4 -2 2 4 6

-6

-4

-2

2

4

6

E. F. G. H.

I. J. K. L.

For each given type of object below, write the letter of all the possible picturesA-L (there may be more than one possible picture, and there may be none)which could be an example of that kind of object:

a) the graph of a function f : R→ R;b) the graph of a function f : R2 → R;c) the graph of a function f : R2 → R2;d) a curve defined parametrically by a function f : R→ R2;e) a curve defined parametrically by a function f : R→ R3;f) the x = 0 trace of a function f : R2 → R;g) a level curve for a function f : R2 → R;h) a level curve for a function f : R3 → R;i) a level surface for a function f : R3 → R;j) the graph of an equation of the form f(x, y) = c where c is a constant;

k) the graph of an equation of the form f(x, y, z) = c where c is a constant;l) a vector field on R2;

m) a vector field on R3.

26. Find each of the following limits. If the limit does not exist, explain why itdoes not exist. Appropriate justification is required on all limits.

a) limx→0

x+y+2x2+y2+1

b) lim(x,y)→(0,0)

x2−4y2

x−2y

c) limx→0

x+yx2+y2

204


d) lim(x,y)→(0,0)

x2−y2

x2+y2

e) lim(x,y)→(0,0)

x2y2

x2+y2

f) limx→0

xy−xz+yzx2+y2+z2

g) lim(x,y,z)→(0,0,0)

x3 cos( yz )x2+y2+z2

h) limx→0

(x2 − y + z − 4, x−z+8

yz−4

)

205


My answers

1. a) −171b) (33,−23,−62)

c)(−57√

59 ,−133√

59 ,19√59

)d)√

142

2. a) (46, 21,−1)b) −59

c)

15 −5 200 10 2510 5 −15

d)

17 −1 −510 9 −50 −3 22

e)

3 0 2−1 2 14 5 −3

3. a) vector

b) nonsense

c) scalar

d) vector

e) matrix

f) nonsense

4.

x = 10− 3ty = 3tz = −4 + 3t

(answers may vary)

5. 82x− 39y − 55z = 49

6.

x = 2s+ 7ty = 5− 2s− 13tz = −4s+ 6t

(answers may vary)

7. 16x+ 10y + 3z = 50

8. 5w − 2x+ 3y = 37

9. The intersection is at the point(−12 ,

536 ,−13

).

10. 16√66

11. Here are descriptions of the graphs:

a) plane that passes through (4, 0, 0), (0, 6, 0) and (0, 0, 12)b) plane that passes through (−8, 0, 0), (0,−2, 0) and (0, 0, 4)c) plane passing through (9, 0, 0) and (0, 3, 0) which is parallel to the z-axisd) plane passing through (0, 5, 0) parallel to x- and z-axes

12. a) (6, π) b) (4, 0) c)(8√

2, 3π4

)13. a) (0,−3) b) (−4, 4

√3)


a) circle of radius 3 centered at (0, 0)

206


b) circle of radius 1 centered at (0, 1) (passing through (0, 0) and (0, 2)c) circle of radius 2.5 centered at (2.5, 0) (passing through (0, 0) and (5, 0)d) line passing through origin with slope

√3 (at 60 angle with x−axis)

15. a)(5, π2 , 0

)b)

(3√

2, −π4 , 7)

16. a) (0,−6,−4) b) (4, 4, 3)

17. a)(4√

2, π4 , 0)

b) (7, 0, 0) c)(20, π6 ,

π3

)d)

(3√

2, π2 ,5π4

)18. a)

(4√

2, 4√

2, 0)

b) (0, 0,−5)


a) z = 4 is a horizontal plane at height 4b) r = 3 is a cylinder around the z-axis of radius 3c) θ = π

2 is the plane x = 0 (sitting above/below the y-axis)


a) ρ = 1 is a sphere of radius 1, centered at the originb) ϕ = π

6 is a cone opening around the positive z-axisc) ϕ = π

2 is the xy-plane, a.k.a. z = 0


a) z = y2 is a cylinder; draw a parabola in the yz plane and extend it paral-lel to the x-axis

b) y = cosx is a cylinder; draw the curve y = cosx in the xy plane andextend it vertically, parallel to the z-axis

c) x2 + y2 + z2 = 16 is a sphere of radius 4 centered at the origin


a) cone opening around the positive and negative x-axisb) hyperboloid of one sheet, opening around the y-axisc) hyperbolic paraboloid (saddle)d) paraboloid, opening around the negative y-axis


a) the x = 0 trace is z = −y2, a parabola opening downward with vertex at(0, 0)

b) the y = 0 trace is z = x2, a parabola opening upward with vertex at (0, 0)c) the x = 2 trace is z = 4− y2, a parabola opening downward with vertex

at (0, 4)d) the y = −1 trace is z = x2 − 1, a parabola opening upward with vertex

at (0,−1)

207


24. a) f(−2, 3) is greater

b) f(1, 2) ≈ −1c) y ≈ −3d) southwest

e) the graph should be a smooth curve connecting the points (−5, 1), (0, 0),and (5, 1).

25. a) Bb) E,F,G,Jc) Dd) A,Be) L

f) Bg) A,B,Ch) none of thesei) E,F,G,H,J,Kj) A,B,C

k) E,F,G,H,J,K

l) D

m) I

26. Justification (not given here) is required in these answers:

a) 2b) 0

c) DNE

d) DNE

e) 0f) DNE

g) 0h) (−4,−2)

208

Chapter 4

Differentiation

4.1 The total derivativeSuppose f : R → R. In Math 220, you learn what it means for a function to be

differentiable (i.e. for the derivative f ′(x) to exist. In particular, f is differentiableat x if some limit exists:

If this limit exists, we call its value f ′(x), the derivative of f at x.

To define what it means for a function f : Rn → Rm to be differentiable at somex ∈ Rn, and to define its derivative, the first thing we might try to do is to copy thedefinition above and just change the fs, xs and hs into vectors as appropriate:

First attempt at a definition of differentiability (WRONG)

A function f : Rn → Rm is differentiable at x ∈ Rn if there is a number f ′(x) ∈ Rsuch that

limh→0

f(x + h)− f(x)h

= f ′(x).

Question: In this definition, to what vector space would each of these items be-long?

x ∈ h ∈ x + h ∈ f(x) ∈ f(x + h) ∈

209

4.1. The total derivative

Let’s try to get around the problem of dividing by a vector, by first rearrangingthe limit from the previous page that defines the derivative of f : R→ R by usingsome algebra:

limh→0

f(x+ h)− f(x)h

= f ′(x)

⇐⇒ limh→0

f(x+ h)− f(x)h

− f ′(x) = 0

⇐⇒ limh→0

[f(x+ h)− f(x)

h− f ′(x)

]= 0

⇐⇒ limh→0

∣∣∣∣∣f(x+ h)− f(x)h

− f ′(x)∣∣∣∣∣ = 0

⇐⇒ limh→0

|f(x+ h)− f(x)− f ′(x)h||h|

= 0.

This algebra proves:

Theorem 4.1 (Differentiability of function f : R→ R, restated) A function f :R→ R is differentiable at x ∈ R if there is a number f ′(x) ∈ R such that

limh→0

|f(x+ h)− f(x)− f ′(x)h||h|

= 0.

Given this theorem, we can try again to extend the definition to functions f :Rn → Rm, by changing the f , x and h to vectors and changing the absolute values| · | to

Second attempt at definition of differentiability (ALSO WRONG)

A function f : Rn → Rm is differentiable at x ∈ Rn if there is a number f ′(x) ∈ Rsuch that

limh→0

||f(x + h)− f(x)− f ′(x)h||||h||

= 0.

210


Definition 4.2 (Differentiability of function f : Rn → Rm) A function f : Rn →Rm is said to be differentiable at x ∈ Rn if there exists an m × n matrix Df(x) ∈Mmn(R), such that

limh→0

||f(x + h)− f(x)−Df(x)h||||h||

= 0.

In this situation we call the matrix Df(x) the (total) derivative of f at x. If f isdifferentiable at every point in its domain, we say that f is differentiable, and callthe functionDf : Rn →Mmn(R) which takes x Df7−→ Df(x) the derivative mappingof f .

You should be able to precisely state the entire first sentence of this definition.

Leibniz notation: The total derivative Df is also called dfdx .

EXAMPLE 1

1. Suppose f : R2 → R3 is some function. Which one or ones of these expres-sions conceivably could be Df(1, 5)?

(7,−5) (3,−5,−2)(

3 7−2 5

) 3 7−2 51 4

(3 7 −34 −2 5

) 3 7 −34 −2 52 8 3

2. Suppose g : R→ R4 is some function. Which one or ones of these expressions

conceivably could be Dg(−2)?

8 (7,−5, 3, 0)(

3 −7 −2 −15)

3−7−2−15

3. Suppose f : R3 → R2. Which expression makes sense? For the expression

that makes sense, what type of object should it be (and to what vector spacedoes it belong)?

Df(3, 4) Df(2,−4, 3)

211


EXAMPLE 2Suppose f : Rn → Rm is a constant function, i.e. f(x) = c where c ∈ Rm. GuessDf(x), and verify using the definition that your guess is correct.

Guess:

Verification: We need to verify the limit definition of derivative, i.e.

limh→0

||f(x + h)− f(x)−Df(x)h||||h||

= 0.

For our proposed Df(x), we see

limh→0

||f(x + h)− f(x)−Df(x)h||||h||

=

In Example 2, we have proven our first differentiation rule for functions Rn →Rm:

Theorem 4.3 If f : Rn → Rm is a constant function, then it is differentiable and itstotal derivative is the m× n zero matrix (i.e. Df(x) = 0m×n).

EXAMPLE 3Suppose f : R3 → R2 is given by f(x) = f(x, y, z) = (2x + 3y − z, x + 4z). GuessDf(x), and verify using the definition that your guess is correct.

Hint: This function can also be described by saying f(x) = Ax where A is the

2× 3 matrix(

2 3 −11 0 4

).

Guess:

Verification: For our proposed Df(x), we have

limh→0

||f(x + h)− f(x)−Df(x)h||||h||

=

212


Example 3 generalizes to any function given by multiplication by a matrix:

Theorem 4.4 Let A ∈ Mmn(R) be an m × n matrix of numbers. Then the functionf : Rn → Rm given by f(x) = Ax is differentiable, and its total derivative satisfies,for all x ∈ Rn, Df(x) = A.

Differentiation rules

Many of the same properties of derivatives you know for functions R → R carryover to total derivatives. It turns out, however, that these rules are not as use-ful as they look (in that they don’t usually help us compute total derivatives offunctions). But they are needed to justify some important formulas and theoreticalitems, so I’ll list them here for reference:

Theorem 4.5 (Linearity of total derivative) Suppose f : Rn → Rm and g : Rn →Rm are differentiable functions at x ∈ Rn. Then:

1. (Sum Rule) f + g : Rn → Rm is differentiable at x and

D(f + g)(x) = Df(x) +Dg(x).

2. (Difference Rule) f − g : Rn → Rm is differentiable at x and

D(f − g)(x) = Df(x)−Dg(x).

3. (Constant Multiple Rule) For any constant c ∈ R, cf : Rn → Rm is differen-tiable at x and

D(cf)(x) = cDf(x).

There is no direct analogue of the product rule for functions R → R, but thesefacts hold for dot products, scalar products and cross products of functions:

Theorem 4.6 (Product Rules) Suppose f : R → Rm and g : R → Rm are differen-tiable functions at x ∈ R. Then:

1. (Dot Product Rule) f · g : R→ R is differentiable at x and

D(f · g)(x) = Df(x) · g(x) + f(x) ·Dg(x).

2. (Scalar Product Rule) If h : R→ R is differentiable at x, then hf : R→ Rm isdifferentiable at x and

D(hf)(x) = f(x)Dh(x) + h(x)Df(x).

213


3. (Cross Product Rule) If m = 3, then f × g : R→ R3 is differentiable at x and

D(f × g)(x) = Df(x)× g(x) + f(x)×Dg(x).

Note: The order of the cross products is important in the Cross Product Rule,because cross product is not commutative.

Since you can’t divide vectors, it makes no sense to talk about a quotient rulefor functions of several variables if m ≥ 2, but if the functions of several variablesare real-valued we have:

Theorem 4.7 (Quotient Rule) Suppose f : Rn → R and g : Rn → R are differen-tiable functions at x ∈ Rn. If g(x) 6= 0, the quotient f

g: Rn → R is differentiable at x

and

D

(f

g

)(x) = g(x)Df(x)− f(x)Dg(x)

[g(x)]2 .

As with functions R→ R, differentiability implies continuity:

Theorem 4.8 (Differentiability implies continuity) Suppose f : Rn → Rm isdifferentiable at x ∈ Rn. Then f is continuous at x.

EXAMPLE 4Suppose f : R → R3 and g : R → R3 are functions such that f(3) = (3,−1, 2),g(3) = (0, 4,−1),

Df(3) =

3−51

and Dg(3) =

−232

.Compute each quantity:

1. D(3f − g)(3)

2. D(f · g)(3)

214

4.2. Partial derivatives

4.2 Partial derivativesMOTIVATING EXAMPLE

Let f(x, y) = (x2 + y2, x2y). Compute Df(x, y).

Remark: Recall that f(x, y) = (f1(x, y), f2(x, y)) where f1(x, y) = x2 + y2 andf2(x, y) = x2y.

What we know at this point: Df(x, y) should be a

What we have no clue about as of now:

Solution: For now, write

Df(x) =(A BC D

);

let’s try to figure out what A, B, C and D should be.

From the definition of total derivative, we have

limh→0

||f(x + h)− f(x)−Df(x)h||||h||

= 0. (4.1)

The limit statement in line (4.1) must hold no matter what direction h approaches0 from.

It turns out that some clever choices of direction will tell us something aboutthe entries A,B,C and D. We’ll make two clever choices of direction on the nexttwo pages:

215


Let h = (h1, h2) approach 0 along the positive h1 axis (i.e. the “x”-direction).

(Recall that in this motivating example, f(x, y) = (x2 + y2, x2y).)

Rewriting equation (4.1) in this setting, we get

limh1→0

∣∣∣∣∣∣∣∣∣∣f(x + (h1, 0))− f(x)−Df(x)

(h10

)∣∣∣∣∣∣∣∣∣∣

||(h1, 0)|| = 0

⇒ limh1→0

∣∣∣∣∣∣∣∣∣∣f(x+ h1, y)− f(x, y)−

(A BC D

)(h10

)∣∣∣∣∣∣∣∣∣∣

|h1|= 0

216


Now let h = (h1, h2) approach 0 along the positive h2 axis (i.e. the “y”-direction).

(Recall that in this motivating example, f(x, y) = (x2 + y2, x2y).)

Again rewriting equation (4.1), we get

limh2→0

∣∣∣∣∣∣∣∣∣∣f(x + (0, h2))− f(x)−Df(x)

(0h2

)∣∣∣∣∣∣∣∣∣∣

||(0, h2)|| = 0

⇒ limh2→0

∣∣∣∣∣∣∣∣∣∣f(x, y + h2)− f(x, y)−

(A BC D

)(0h2

)∣∣∣∣∣∣∣∣∣∣

|h2|= 0

⇒ limh2→0

1|h2|

∣∣∣∣∣∣∣∣∣∣(x2 + (y + h2)2, x2(y + h1))− (x2 + y2, x2y)−

(Bh2Dh2

)∣∣∣∣∣∣∣∣∣∣ = 0

⇒ limh2→0

1|h2|

∣∣∣∣∣∣(x2 + (y + h2)2 − (x2 + y2)−Bh2, x2(y + h1)− x2y −Dh2)

∣∣∣∣∣∣ = 0

⇒ limh2→0

1|h2|

∣∣∣∣∣∣(x2 + y2 + 2h2y + h22 − x2 − y2 −Bh2, x

2y + x2h1 − x2y −Dh2)∣∣∣∣∣∣ = 0

⇒ limh2→0

1|h2|

∣∣∣∣∣∣(2h2y + h22 −Bh2, x

2h1 −Dh2)∣∣∣∣∣∣ = 0

⇒ limh2→0

1|h2|

∣∣∣∣∣∣h2(2y + h2 −B, x2 −D)∣∣∣∣∣∣ = 0

⇒ limh2→0

1|h2||h2|

∣∣∣∣∣∣(2y + h2 −B, x2 −D)∣∣∣∣∣∣ = 0

⇒ limh2→0

∣∣∣∣∣∣(2y + h2 −B, x2 −D)∣∣∣∣∣∣ = 0

⇒ ||(2y −B, x2 −D)|| = 0⇒ (2y −B, x2 −D) = 0

Therefore B = 2y and D = x2.

217


Putting together the work on the previous two pages, we see that for the func-tion

f(x) = (x2 + y2, x2y),

if the total derivative of f at x exists, then it must be equal to(A BC D

)=(

2x 2y2xy x2

).

(In fact, this matrix is the total derivative of f ; this can be proven by verifyingthe limit in the definition of total derivative, but we won’t do that here.)

How might we have figured the entries of this matrix out without the longcomputations? Observe that

• the first column of Df(x), i.e. A and C, came from

• the second column of Df(x), i.e. B and D, came from

• the first row of Df(x), i.e. A and B, came from

• the second row of Df(x), i.e. C and D, came from

More generally: If Df(x) is the total derivative of some function f : Rn → Rm,then

• the entries in the ith row of Df(x) have to do only with the ith componentfunction fi;

• the entries in the jth column of Df(x) have to do only with the behavior withrespect to xj , the jth component of the input variable x;

• so the (i, j)-entry of Df(x) should be the derivative of fi with respect to xj .

For instance, if f : (w, x, y, z) 7→ (u, v), we would expect Df(x) to be a matrixlike this:

218


Definition 4.9 Let f : Rn → Rm be a function denoted by

f(x) = f(x1, ..., xn) = (f1(x1, ..., xn), f2(x1, ..., xn), ..., fm(x1, ..., xn)).

Then for each i ∈ 1, ..., n and each j ∈ 1, ...,m, we define the partial derivative(of fj with respect to xi) by

∂fj∂xi

(x) = limh→0

fj(x + hei)− fj(x)h

(assuming this limit exists).

The ∂fj∂xi

notation is pronounced “del fj del xi”, similar to how dydx

is pronounced“dee y dee x”.

Question: Why do we use ∂ rather than d? Why can’t we just use primes?

Answer: Coming later (there’s a subtle difference between ∂y∂x

and dydx

).

Partial derivatives are computed by looking at the jth component function anddifferentiating that expression with respect to variable xi and treating all the othervariables as constants.

EXAMPLE 5Suppose f(x, y, z) = (3x2y3, z sin y+xe2z). Then this function has six partial deriva-tives:

∂f1∂x

= ∂f2∂x

=

∂f1∂y

= ∂f2∂y

=

∂f1∂z

= ∂f2∂z

=

The next two theorems (which I won’t prove) make clear the connection be-tween partial derivatives and the total derivative:

219


Theorem 4.10 If f : Rn → Rm is differentiable at x, then for every i ∈ 1, ..., n andevery j ∈ 1, ...,m, the partial derivative ∂fj

∂fiexists at x. Furthermore,

Df(x) =

∂f1∂x1

∂f1∂x2

· · · ∂f1∂xn

∂f2∂x1

∂f2∂x2

· · · ∂f2∂xn

...... . . . ...

∂fm∂x1

∂fm∂x2

· · · ∂fm∂xn

.

WARNING: There are functions of several variables where all partial deriva-tives exist at x, but the function fails to be differentiable at x. An example of this isf : R2 → R defined by

f(x, y) = 3√xy

(you’ll see in the homework why this function isn’t differentiable at (0, 0), despitethe partial derivatives existing there.)

However: under an extra assumption, the partial derivatives of a function doguarantee the existence of the total derivative:

Theorem 4.11 Suppose f : Rn → Rm is a function. If for every i ∈ 1, ..., n andevery j ∈ 1, ...,m, the partial derivative ∂fj

∂xiexists and is a continuous function on

a neighborhood of x, then f is differentiable at x and its total derivative satisfies

Df(x) =

∂f1∂x1

∂f1∂x2

· · · ∂f1∂xn

∂f2∂x1

∂f2∂x2

· · · ∂f2∂xn

...... . . . ...

∂fm∂x1

∂fm∂x2

· · · ∂fm∂xn

.

Based on this theorem, we make the following definition:

Definition 4.12 A function f : Rn → Rm is called C1 (or C1-smooth) if for ev-ery i ∈ 1, ..., n and every j ∈ 1, ...,m, the partial derivative ∂fj

∂xiexists and is

continuous at all x in the domain of f .

220


More examples of partial derivative computations

Alternate notation for partial derivatives: Suppose f : Rn → R is a real-valuedfunction of variables x1, ..., xn (if n = 3, these variables might be x, y, z, etc.)

Then fxj means the same thing as ∂f∂xj

.

Suppose that f : Rn → R2 is such that its component functions are written uand v instead of f1 and f2. Then ux means ∂u

∂x, etc.

EXAMPLE 6Suppose f(x, y) = x2 − 3x4y + y3. Find fx and fy, and find Df(x, y).

EXAMPLE 7Suppose f(x, y, z) = sin(xyz) + x2

y− 3 cos z. Find the partial derivatives of f .

Solution: Use the Chain Rule on the first term:

fx(x, y, z) = cos(xyz) · yz + 2xy

fy(x, y, z) = cos(xyz) · xz − x2

y2

fz(x, y, z) = cos(xyz) · xy + 3 sin z

EXAMPLE 8Suppose f : (x, y) 7→ (u, v) where u = ex+2y and v = e3y−x. Find Df(x, y).

Higher-order partial derivatives

Analogues of the second derivative, third derivative, etc. for functions of severalvariables are most useful for functions which are real-valued (i.e. f : Rn → R) sowe’ll restrict to those here.

221


Definition 4.13 Suppose f : Rn → R is a real-valued function of variables x1, ..., xn.We define the second-order partial derivatives of f as follows: for each i, j ∈1, ..., n, set

fxixj = fxi,xj = ∂2f

∂xj∂xi= ∂

∂xj

(∂

∂xi(f)

).

Higher-order partial derivatives are defined similarly: as an example, supposef : (x, y, z) 7→ w. Then

fxzzy = ∂4f

∂y∂z2∂x= ∂

∂y

[∂

∂z

(∂

∂z

(∂

∂x(f)

))].

Some comments on the notation:

Note: If f : Rn → Rm has component functions f1, ..., fm, then for each k ∈1, ...,m, we can define

∂2fk∂xj∂xi

using the above definition, but the following is nonsense:

∂fk∂fl∂xj∂xi

EXAMPLE 9Let f(x, y, z) = ln(xy + z) + 2xy2z. Find ∂2f

∂y∂xand fzx(1, 0, e).

222


EXAMPLE 10Let f(x, y) = x10y8. Compute ∂2f

∂x∂yand ∂2f

∂y∂x.

The fact that the answers in the preceding example are the same isn’t a coinci-dence:

Theorem 4.14 (Clairaut’s Theorem (Equality of mixed partials)) Let f : Rn →R and suppose that the mixed second-order partials fxy and fyx are both continuouson a neighborhood of x ∈ Rn. Then fxy(x) = fyx(x).

More generally, if f : Rn → R is such that all mixed rth-order partials of f existand are continuous on a neighborhood of x ∈ Rn, then the order in which thosepartials are taken at x doesn’t matter.

So we make the following definitions, to indicate the functions where the orderdoesn’t matter when taking partial derivatives:

Definition 4.15 A function f : Rn → Rm is called C2 (or C2-smooth) if for everyk ∈ 1, ..,m and every i ∈ 1, ..., n and every j ∈ 1, ...,m, the partial derivative∂2fk∂fj∂fi

exists and is continuous at all x in the domain of f .Let r ∈ 1, 2, 3, .... f : Rn → Rm is called Cr (or Cr-smooth) if for every

k ∈ 1, ...,m and every i1, i2, ..., ir ∈ 1, ..., n, the partial derivative ∂rfk∂fxir ···∂fi2∂fi1

exists and is continuous at all x in the domain of f .f is called C∞ (or C∞-smooth) if it is Cr for every r ∈ 1, 2, 3, ....

223


Geometric interpretation of partial derivativesEXAMPLE 11

Let f : R2 → R be f(x, y) =√y − x2. Consider the point x = (x, y) = (1, 5).

a) Compute f(1, 5), fx(1, 5) and fy(1, 5).

Solution: By direct computation, we have:

f(1, 5) =√

5− 12 =√

4 = 2

fx(x, y) =

fx(1, 5) = −2(1)2√

5− 12= −2

2 · 2 = −12

fy(x, y) =

fy(1, 5) = 12√

5− 12= 1

2√

4= 1

4 .

b) Interpret the value of fx(1, 5) you found in part (a) geometrically.

Solution: Let’s look at the graph of f and the point we are considering:

We will interpret the partial derivative fx(1, 5) as a slope. Remember that thispartial derivative is obtained by approaching (1, 5) in the x-direction (i.e. inthe direction (1, 0)). Let’s mark the line through (1, 5) in this direction with ared dashed line in the xy-plane, and then see what the trace of the graph of flooks like above that dashed line:

224


Extracting only the red curve from this picture, we get the following:

(1,5,2)

(1,5,0)

- 5 1 5x

25

z

The partial derivative fx(1, 5) measures the slope on the above picture wheny = 5. In other words, fx(1, 5) = −1

2 is the

c) Interpret the value of fy(1, 5) you found in part (a) geometrically.

Solution: This partial derivative is obtained by approaching (1, 5) in the y-direction (i.e. the direction (0, 1)). Let’s mark this direction with a bluedashed line in the xy-plane, and then see what the trace of the graph of flooks like above that dashed line:

225


Extracting only the blue curve from this picture, we get:

(1,5,2)

(1,5,0)

1 5y

2

z

The partial derivative fy(1, 5) measures the slope on the above picture whenx = 1. In other words, fy(1, 5) = 1

4 is the

Previewing the next section: Suppose you superimposed the pictures we drew forthe partial derivatives fx and fy on the same image:

What do you think might be a useful thing to do with the dark red and dark bluetangent lines?

226


EXAMPLE 12Suppose f(x, y) =

√x2 + y2. Calculate ∂f

∂y(0, 2) and interpret this partial derivative

geometrically.

EXAMPLE 13The contour plot for some unknown function f : R2 → R is shown below. Use thecontour plot to estimate fx(−2, 1) and ∂f

∂y(2,−2).

-7

-7-6

-6

-5

-5

-4

-4-3

-3

-2

-2

-1

-1

0

0

11

2

2

3

3

4

4

5

5 6

6

7

7

8

8

9

9

10

10

11

121314

-3 -2 -1 0 1 2 3-3

-2

-1

0

1

2

3-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

x

y

fx(−2, 1) ≈

∂f∂y

(2,−2) ≈

.

227

4.3. Linear approximation, tangent lines and tangent planes

4.3 Linear approximation, tangent lines and tangent planesMotivation: We have seen that the total derivative Df of a function f : Rn →Rm is an

We know that the entries of this matrix are the partial derivatives, which fromthe previous section can be interpreted geometrically as

But how do you interpret the total derivative Df?

To answer this, let’s think back to something you did in Calculus 1: linearapproximation. Recall that if f : R → R is a differentiable function, then values off(x) for x-values near a can be approximated by L(x), where L is the tangent lineto f at a:

f

L

a

f(a)

f

L

a

f(a)

a

f(a)

Calculus 1 Pop Quiz: What was the formula for L?

This idea of approximating a function by something “linear” generalizes com-pletely to functions f : Rn → Rm. However, in this case L is sometimes an objectthat is more abstract than a tangent line, so we give the function L, which is ob-tained by essentially the same formula you wrote down in the pop quiz, a differentname:

228


Definition 4.16 Let f : Rn → Rm be differentiable at a ∈ Rn. The linearization off at a (also called the linear approximation to f at a) is the function L : Rn → Rm

given byL(x) = f(a) +Df(a)(x− a).

The point of this is that for x values near a, f(x) ≈ L(x).

Observe: L(a) = f(a) +Df(a)(a− a) = f(a), so the graph of L always contains(a, f(a)), and if you sketch the image of L, it will contain f(a).

EXAMPLE 14Let f : R2 → R2 be defined by f(x, y) = (

√x cos y, exy). Use linearization to estimate

f(3.8, .1).

229


Question: How do you interpret linearization geometrically?

Short answer: It depends on the dimension of the domain and codomain.

Case 1: f : R→ R (i.e. y = f(x), i.e. x f7−→ y, i.e. n = 1, m = 1)

Here, Df(a) means f ′(a) so the linearization is

L(x) = f(a) + f ′(a)(x− a),

i.e. the linearization is the same as the tangent line to f at a.

Case 2: f : R→ R2 (i.e. (x, y) = f(t), i.e. t f7−→ (x, y), i.e. n = 1, m = 2)

Here, Df(a) =(x′(a)y′(a)

)= (x′(a), y′(a)) is a vector in R2, so

L(t) = f(a) +Df(a)(t− a)= [f(a)−Df(a)a] + tDf(a)

t = a

f

-3 -2 -1 1 2 3x

-3

-2

-1

1

2

y

Case 3: f : R→ R3 (i.e. t f7−→ (x, y, z), i.e. n = 1, m = 3)

Similar to Case 2, Df(a) =

x′(a)y′(a)z′(a)

= (x′(a), y′(a), z′(a)) is a vector so

L(t) = f(a) +Df(a)(t− a)= [f(a)−Df(a)a] +Df(a) t

so the image of L is a line in R3 passing through f(a) with direction vectorDf(a) = (x′(a), y′(a), z′(a)).

230


Case 4: f : R→ Rm (i.e. t f7−→ (x1, ..., xm), i.e. n = 1)

Generalizing Cases 2 and 3, L(x) gives parametric equations for a line in Rm

passing through f(a) with direction vector Df(a).

Definition 4.17 If f : R → Rm is differentiable at a ∈ R, we call the linearizationof f at a the tangent line to f at a. This line passes through f(a) and has directionvector Df(a).

EXAMPLE 15Find parametric equations for the tangent line to f(t) = (3 cos t, 2 sin 2t, 3t) whent = π

3 .

231


Case 5: f : R2 → R (i.e. z = f(x, y) i.e. (x, y) f7−→ z)

This time, Df(a) =(

∂f∂x

(a) ∂f∂y

(a))

=(fx(a) fy(a)

)is a row vector. So

writing a = (a, b), we have

L(x) = f(a) +Df(a)(x− a)

= f(a, b) +(fx(a, b) fy(a, b)

)( x− ay − b

)= f(a, b) + fx(a, b)(x− a) + fy(a, b)(y − b)= f(a, b) + fx(a, b)x+ fy(a, b)y − fx(a, b)a− fy(a, b)b= (const) + fx(a, b)x+ fy(a, b)y.

We can rearrange the equation above by thinking of L(x) as z:

L(x) = (const) + fx(a, b)x+ fy(a, b)yz = (const) + fx(a, b)x+ fy(a, b)y

⇒ fx(a, b)x+ fy(a, b)y − z = (const)⇒ (fx(a, b), fy(a, b),−1) · (x, y, z) = (const)

Definition 4.18 If f : R2 → R is differentiable at a ∈ R2, we call the linearization off at a the tangent plane to f at a. The line in R3 passing through (a, f(a)) which isnormal to the tangent plane is called the normal line to the graph of f at a.

Here are some pictures of a tangent plane to some function f : R2 → R at (1,−1)(this is the same picture, viewed from different angles):

232


Theorem 4.19 (Tangent planes of functions R2 → R) Let f : R2 → R be differ-entiable at a = (a, b) ∈ R2. Let P be the tangent plane to f at (a, b). Then:

1. P goes through the point p = (a, b, f(a, b));

2. P has normal vector n = (fx(a, b), fy(a, b),−1);

3. The normal equation n · (x− p) = 0 of P is

fx(a, b)(x− a) + fy(a, b)(y − b)− (z − f(a, b)) = 0

which can be rewritten as

L(x, y) = z = f(a, b) + fx(a, b)(x− a) + fy(a, b)(y − b).

Information needed to write equations of normal lines to a function is below:

Theorem 4.20 (Normal lines of functions R2 → R) Let f : R2 → R be differen-tiable at a = (a, b) ∈ R2. Let N be the normal line to f at (a, b). Then:

1. N passes through p = (a, b, f(a, b));

2. N has direction vector n = (fx(a, b), fy(a, b),−1).


f(x, y) =√y − x2; x = (x, y) = (1, 5).

Recall: f(1, 5) = 2; fx(1, 5) = −12 ; fy(1, 5) = 1

4 .

233


EXAMPLE 16Find the equation of the tangent plane to f when x = 1, y = 0, and parametricequations of the normal line to f at the same point, if f(x, y) = xe−2y.

Solution: We are given a = 1, b = 0. f(a, b) = 1e−2(0) = 1.

Therefore, the tangent plane and normal line both pass through

p = (a, b, f(a, b)) = (1, 0, 1).

fx(x, y) = e−2y ⇒ fx(a, b) = fx(1, 0) = e−2(0) = 1fy(x, y) = −2xe−2y ⇒ fy(a, b) = fy(1, 0) = −2(1)e0 = −2

Therefore n = (fx(a, b), fy(a, b),−1) = (1,−2,−1).

EXAMPLE 17Find a point on the graph of f(x, y) = 2x2 + 3y2 where the tangent plane to f atthat point is parallel to the plane 8x− 3y + z = 0.

234

4.4. Higher-dimensional Chain Rule

4.4 Higher-dimensional Chain RuleA diagram that describes the Chain Rule from Calculus 1

Recall from Calculus 1: the Chain Rule for functions of one variable is used to findderivatives of compositions:

EXAMPLE FROM CALC 1Suppose y = sin2 x. Then, we can write y = (f g)(x) where

Therefore, by the Chain Rule,

y′(x) = dy

dx=

Two diagrams illustrating what’s going on here:

Functional diagram Derivative diagram

R g //

fg

##R f // R . .

Rdudx //

dydx

= dydu

dudx

77Rdydu // R

x g // u f // y

Question: How might these pictures (especially the “derivative diagram”) gener-alize to functions Rn → Rm?

Functional diagram Derivative diagram

Rn g //

fg

%%Rp f // Rm . .

Rn Dg(x) // 66Rp Df(u)// Rm

x g // u f // y

235


Theorem 4.21 (Higher-dimensional Chain Rule) Let f : Rp → Rm and g :Rn → Rp. Suppose g is differentiable at x ∈ Rn and f is differentiable at g(x) ∈ Rp.Then f g : Rn → Rm is differentiable at x and

D(f g)(x) = Df(g(x))Dg(x),

where the multiplication on the right is the matrix multiplication of a m × p matrixand a p× n matrix. Equivalently, if u = g(x), we can write

D(f g)(x) = Df(u)Dg(x).

Remark: This Chain Rule is a major reason why matrix multiplication has theweird definition that it does.

EXAMPLE 18Suppose z = x2ey, x = t2 − 1 and y = sin t. Compute dz

dttwo different ways.

First method: Use the higher-dimensional Chain Rule:

Second method: Substitute to get a formula for z in terms of t, then differentiate:

Example 18 suggests that the Chain Rule isn’t always that useful to actuallycompute derivatives (because you can compute them by substituting and then dif-ferentiating directly). However, the Chain Rule is very useful in proving theoremsand deriving ideas we need later. Also, this Chain Rule can be used to revisita differentiation technique originally introduced in Calculus 1 (that many Calc 1students struggle with):

236


Implicit differentiation using the Chain RuleEXAMPLE 19

Suppose x2 + y2 = 25. What is dydx

?

A solution you “should” know from Calculus 1: Differentiate implicitly to get

A second way of doing this problem:

Think of the graph of x2 + y2 = 25 as the image of a set of parametricequations:

The (Calculus 1) Chain Rule says

dy

dt= dy

dx· dxdt

sody

dx=

dydtdxdt

=

Pros of this method: You get a correct answer, without having to remem-ber how to differentiate implicitly.

Cons of this method:

237


EXAMPLE 19, RESTATED

Suppose x2 + y2 = 25. What is dydx

?

A third way of doing this problem:

Define the function g : R→ R2 by g(x) = (x, y) = (x, y(x)).

Note Dg(x) is a 2× 1 matrix:

Next, define F : R2 → R by F (x) = F (x, y) = x2 + y2.

The total derivative of F is a 1× 2 matrix:

The trick with this method is to think about the composition F g : R→ R.The formula for this composition is

and from the given equation in the problem, this is always equal to 25. SoF g = 25, and we therefore know D(F g)(x) = (because that’s whatthe derivative of the constant 25 is).

But by the Chain Rule,

0 = D(F g)(x) = DF (g(x))Dg(x)

=(

2x 2y)( 1

dydx

)

= 2x+ 2y dydx

← Remember this from the first method?

At first, this third method may seem unnecessarily hard (or a stupid wayof redoing the first method). But what is great about this new method is that itgeneralizes.

238


General method for implicit differentiation

Suppose you have any equation of the form f(x1, x2, ..., xn) = constant. Supposeyou want to find dxj

dxi, the rate of change of variable xj with respect to xi. Let’s do

the same procedure as in the third method we used in the previous example. Hereare the details:

Define g : R → Rn by g(xi) = (0, 0, 0, ..., 0, xi, 0, ..., 0, xj, 0, ..., 0). Note thatDg(xi) is a n× 1 matrix:

Dg(xi) =

00...010...0dxjdxi

0...0

00...0

← ith position0...

← jth positionab

0...0

Now define F : Rn → R by F (x1, ..., xn) = f(x1, ..., xn). The total derivative of F isa 1× n matrix:

DF (x1, ..., xn) =(fx1 fx2 · · · fxn

).

But by assumption, we know F g = f is constant, so D(F g)(x) = 0. So by theChain Rule,

0 = D(F g)(x) = DF (g(x))Dg(x)

=(fx1 fx2 · · · fxn

)

0...1...dxjdxi...0

= fxi(1) + fxj

dxjdxi

.

239


From the previous page, we have

0 = fxi + fxjdxjdxi

.

Solving for dxjdxi

, we obtaindxjdxi

= −fxifxj

.

We have now proven this general formula:

Theorem 4.22 (Implicit Differentiation Formula) Suppose f : Rn → R is con-stant. Then for all i, j ∈ 1, ..., n,

dxjdxi

= −fxi(x1, ..., xn)fxj(x1, ..., xn) .

EXAMPLE 20Suppose 3xy2 cos z − 2zx+ 4x2y2 = 8. Find dy

dx.

EXAMPLE 21Suppose z = 3− 1

2x2y. What is the difference between ∂z

∂xand dz

dx?

∂z∂x

= 0− 12(2x)y = −xy

= rate of change of z with respect to xas y is held constant

dzdx

= ddx

(z) = ddx

(3− 1

2x2y)

= rate of change of z with respect to xas you move along a curve in thexy-plane whose derivative is dy

dx

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4.5. Directional derivatives and gradients

4.5 Directional derivatives and gradientsDefinition of the directional derivative

Recall: Suppose f : Rn → R. The partial derivatives of f represent the rate ofchange (or slope) of f as x changes in a direction parallel to one of the coordinateaxes (see the pictures in Section 4.2 for a good reference).

Question: Suppose you change x in a direction other than one parallel to a coor-dinate axis. Is there a way to measure the rate of change of f in this context? Wewant to call this rate of change the “directional derivative of f in the direction v atpoint x”.

First attempt (wrong): Let’s try defining the directional derivative to be

limh→0

f(x + hv)− f(x)h

. (4.2)

EXAMPLE 22Compute the expression in line (4.2) above for the function f(x, y) = 3x + y at thepoint x = (1, 1) in the direction v = (−1, 1).

limh→0

f(x + hv)− f(x)h

= limh→0

f((1, 1) + h(−1, 1))− f(1, 1)h

= limh→0

f(1− h, 1 + h)− 4h

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EXAMPLE 23Compute the expression in (4.2) above for the function f(x, y) = 3x+y at the pointx = (1, 1) in the direction v = (−2, 2).

Since we are using the same function as Example 22, with the same x as Exam-ple 22, and a vector which is pointing in the same direction as the one in Example22, the answer should be the same. But, if we work it out:

limh→0

f(x + hv)− f(x)h

= limh→0

f((1, 1) + h(−2, 2))− f(1, 1)h

= limh→0

f(1− 2h, 1 + 2h)− 4h

= limh→0

3− 6h+ 1 + 2h− 4h

= limh→0

−4hh

= −4.

The problem with our first attempt at the definition is that it depends not onlyon the direction v points in, but on the length of v. (In Example 23, the vector wastwice as long as the vector in Example 22, so the answer was twice as big.)

What we really want to measure is the rate of change of f per unit length inthe direction of v, so to fix this we only allow unit vectors to count as “directions”:

Definition 4.23 Let f : Rn → R be C1; let x ∈ Rn and let u be a unit vector in Rn.The directional derivative of f in the direction u at x is the scalar

Duf(x) = limh→0

f(x + hu)− f(x)h

.

Remarks:

1. If f : Rn → R is C1, then the directional derivative exists for every choice ofunit vector u.

2. Dejf(x) = fxj(x), the partial derivative of f with respect to xj .

3. If f : Rn → Rm where m ≥ 2, then the expression Duf(x) makes no sense.

4. Without the x, we think of Duf as a function Rn → R (the function is x Duf7→Duf(x)).

5. This definition is not that useful to actually compute directional derivatives(see the HW). Is there a way to compute Duf(x) without using the limit def-inition?

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Gradients

Definition 4.24 Let f : Rn → R be C1. The gradient of f , denoted ∇f , is thefunction∇f : Rn → Rn defined by

∇f(x) = (fx1(x), fx2(x), ..., fxn(x)) .

EXAMPLE 24Suppose f(x, y) = x2y + 4y3 − x. Find∇f(x, y).

Remarks:

1. If f : Rn → Rm where m ≥ 2, then∇f is not defined (yet).

2. The gradient∇f of a function f : Rn → R is a vector field on Rn.

(For instance, in Example 24, your answer gives a formula for a function∇f : R2 → R2.)

3. For each choice of x ∈ Rn, the output ∇f(x) of the gradient function is avector in Rn.

Question: Why are we bothering to define this?

EXAMPLE 25Suppose f(x, y, z) = ex sin y + 2x4 cos z.

1. Find the gradient of f . Write the answer as a matrix (usually you don’t writea gradient as a matrix).

2. Compute the total derivative of f .

3. What is the connection between the previous two answers? (Think in termsof matrices.)

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Theorem 4.25 Let f : Rn → R be C1. Then for any x ∈ Rn,

∇f(x) = [Df(x)]T .

This theorem gives the first indication that gradients might be important (be-cause they are related to total derivatives). It turns out that gradients have lots ofapplications.

Applications of gradients

Application # 1: Gradients can be used to compute directional derivatives

Theorem 4.26 Let f : Rn → R be C1. Then for any x ∈ Rn and any unit vectoru ∈ Rn, the directional derivative of f in direction u satisfies

Duf(x) = ∇f(x) · u.

PROOF Define the function k : R→ Rn by setting k(t) = x + u t.

Now define g : R→ R by setting g = f k. Then:

Duf(x) =limh→0

f(x+hu)−f(x)h

=limh→0

f(k(h))−f(k(0))h

=limh→0

g(h)−g(0)h−0

= g′(0) (by the Calc 1 definition of derivative)

= Dg(0)= Df(k(0))Dk(0) (by the Chain Rule applied to g = f k)

= Df(x)Dk(0) (since k(0) = x + u 0 = x)= Df(x) u (since Dk(t) = u for every t, including t = 0)= [∇f(x)]Tu (by Theorem 4.25 above)= ∇f(x) · u (since a · b = aTb).

This proves the theorem.

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EXAMPLE 26Find the directional derivative of f(x, y) = 4x2y3− 2xy in the direction of (1,−1) atthe point (2, 1).

Application # 2: Gradients can be used to quickly obtain vectors which arenormal to level sets of a function

Theorem 4.27 Let f : Rn → R be C1. Let c ∈ R. For any point x0 ∈ Rn on the levelset f(x) = c, the gradient∇f(x) is normal to the level set f(x) = c.

PROOF Let u be a unit vector tangent to the level curve at x0. Then since movingalong u corresponds to moving along the level set and because moving along alevel curve doesn’t change the value of f (i.e. the height of the graph), we have

Duf(x0) = 0.

By Theorem 4.26, this means∇f(x0) · u = 0

so∇f(x0) ⊥ u as required.

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EXAMPLE 27Find a vector normal to the curve given by the equation x2 + y4 − 3y2 = 7 at thepoint (3, 1).

-4 -2 0 2 4

-4

-2

0

2

4

x

y

EXAMPLE 28Let f(x, y) = x

x2+y2 . Find a normal vector to the level curve f(x, y) = .2 at the point(1, 2).

Solution: First, compute the gradient (use the Quotient Rule):

∇f(x, y) = (fx, fy) =(x2 + y2 − 2x2

(x2 + y2)2 ,−2y

(x2 + y2)2

).

Then plug in (1, 2):

∇f(1, 2) =(

1 + 4− 2(1 + 4)2 ,

−4(1 + 4)2

)=( 3

25 ,−425

).

So a normal vector to f(x, y) = .2 at (1, 2) is( 3

25 ,−425

).

Remark: Any nonzero multiple of ∇f(1, 2) also is a solution to this problem, so(for instance) you could take 25 times the previous answer to get (3,−4), which isan easier-to-work-with normal vector to the level curve f(x, y) = .2.

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Application # 3: Gradients can be used to quickly write equations of tangentplanes to level surfaces and tangent lines to level curves

EXAMPLE 29Find the equation of the tangent plane to the ellipsoid x2 + 2y2 + 3z2 = 6 at thepoint (1, 1, 1).

Old solution:(coming from §4.3)

First, solve for z to get

z =√

13(6− x2 − 2y2) =

√2− 1

3x2 − 2

3y2.

Next, let f(x, y) =√

2− 13x

2 − 23y

2.

By Theorem 4.20, the tangent plane has equation

z = f(1, 1) + fx(1, 1)(x− 1) + fy(1, 1)(y − 1)= 1 + (const)(x− 1) + (const)(y − 1)

Problem: Working out these constants is hard.

New solution: Recognize the ellipsoid as a level surface for the functionw = f(x, y, z) defined by f(x, y, z) = x2 + 2y2 + 3z2.

(The ellipsoid is the level surface at w = 6).

By Theorem 4.27 above,∇f is normal to the level surface,hence normal to the tangent plane. So we compute∇f :

Since the tangent plane has normal vector (2, 4, 6) and goesthrough (1, 1, 1), its equation is

This new method of solution generalizes:

Theorem 4.28 (Tangent plane to a level surface) Let f : R3 → R be C1. Theequation of the plane tangent to the level surface f(x, y, z) = c at point (x0, y0, z0) is

∇f(x0, y0, z0) · (x− x0, y − y0, z − z0) = 0.

And the same procedure works in other dimensions:

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Theorem 4.29 (Tangent hyperplane to a level set) Let f : Rn → R be C1. Theequation of the hyperplane tangent to the level set f(x) = c at point x0 ∈ Rn is

∇f(x) · (x− x0) = 0.

Application # 4: Gradients indicate the direction in the domain correspondingto the greatest increase and decrease of the output of the function

Theorem 4.30 Let f : Rn → R be C1, and let x ∈ Rn. Then:

• the maximum value of Duf(x) is ||∇f(x)||, occurring when u is in the samedirection as∇f(x); and

• the minimum value of Duf(x) is −||∇f(x)||, occurring when u is in the oppo-site direction as∇f(x);.

PROOF f increases (decreases) most rapidly in the direction of unit vector u whichmaximizes (minimizes) the directional derivative Duf(x). Now

Duf(x) = ∇f(x) · u= ||∇f(x)|| ||u|| cos θ (where θ is the angle between∇f(x) and u)= ||∇f(x)|| cos θ (since u is a unit vector)

Since ||∇f(x)|| is fixed, the only way this expression changes is by changing θ. Soto maximize it, maximize cos θ by setting θ = . This makes ∇f(x) and upoint in the same direction; in this case Duf(x) = ||∇f(x)||.

The same quantity is minimized when cos θ is minimized, i.e. when cos θ = −1,i.e. when θ = π, i.e. when ∇f(x) and u point in the opposite direction; in this caseDuf(x) = ||∇f(x)|| (−1) = −||∇f(x)||.

EXAMPLE 30The contour plot for some unknown function f : R2 → R is given below. Use thecontour plot to estimate the direction in which∇f(1, 3) points.

-12

-10

-8

-6

-4 -2

0

2

2

4

6

8

10

12

0 1 2 3 4 50

1

2

3

4

5

x

y

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EXAMPLE 31The temperature of a point (x, y) on a metal plate is given by T (x, y) = 100−2x2+y2.If a heat-seeking particle is dropped on the plate at the point (2, 1), what directiondoes it start to move in?

Enrichment: In a situation like this example, we can not only figure out whatdirection the particle starts in, but we can determine the entire path the particlewill travel. To do this, let x(t) = (x(t), y(t)) be the position of the particle at time t.We know

x(0) = 2 y(0) = 1

and that for all times t,(dx

dt,dy

dt

)= ∇T (x(t), y(t)) = (−4x, 2y).

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Gradients as vector fieldsEXAMPLE 30, REVISITED

-12

-10

-8

-6

-4 -2

0

2

2

4

6

8

10

12

∇f(1,3)

0 1 2 3 4 50

1

2

3

4

5

x

y

Suppose we sketched ∇f(x) (with rescaled arrow lengths) at a lot of different val-ues of x. In this example, we get the following picture:

0 1 2 3 4 5

0

1

2

3

4

5

x

y

This is a picture of the vector field ∇f : R2 → R2. (In general, if f : Rn → R, then∇f : Rn → Rn, so ∇f is a vector field on Rn.)

Suppose we superimpose the picture of the vector field ∇f and the contourplot of f . We get this:

-12

-10

-8

-6

-4 -2

0

2

2

4

6

8

10

12

0 1 2 3 4 5

0

1

2

3

4

5

x

y

250

4.6. Mathematica commands for derivatives

4.6 Mathematica commands for derivatives


Partial derivatives of f : Rn → Rfx = ∂f

∂xD[f[x,y], x] or D[f[x,y,z], x], etc.

fy = ∂f∂y

D[f[x,y], y] or D[f[x,y,z], y], etc.

Partial derivatives of f : Rn → Rm

∂f1∂x

D[f[x,y][[1]],x] or D[f[x,y,z][[1]], x], etc.∂f2∂x

D[f[x,y][[2]],x] or D[f[x,y,z][[2]], x], etc.

(alternatively, compute total derivativeusing the command given below, andread off the answer)

Higher-order partial derivativesof f : Rn → Rfxx = ∂2f

∂x2 D[f[x,y], x,x] or D[f[x,y], x,2]fyyyyy = ∂5f

∂y5 D[f[x,y,z], y,5]fxy = ∂2f

∂x∂yD[f[x,y], x,y]

∂10f∂x3∂y2∂z5 D[f[x,y,z], x,3, y,2, z,5]

Total derivative of f : Rn → Rm D[f[x,y], x,y] or D[f[x,y,z], x,y,z], etc.(to get the answer as matrix, click

MatrixForm)

Total derivative of f : R→ Rm f’[x] or f’[t], etc.(f”[x] and f"[x] work as well)

Gradient∇f Grad[f[x,y], x,y] or Grad[f[x,y,z], x,y,z], etc.(or use total derivative command)

Directional derivative Duf(x) Grad[f[x,y], x,y].Normalize[u]or Grad[f[x,y,z], x,y,z].Normalize[u]

To substitute numerical values for x, y and z, do one of two things:

1. Define the derivative as a function of x, y and z, then ask Mathematica to plugin the values of x, y and z to your newly-defined function:

Example: Suppose you wanted to compute fxy(3, 2,−5). You could executethese commands, one at a time:h[x_,y_,z_] = D[f[x,y,z], x, y]h[3,2,-5]

251


2. Follow any of the commands above with some syntax that causes Mathemat-ica to substitute in numbers for the variables:

Example: Suppose you wanted to compute fxy(3, 2,−5). You could executethis single command:D[f[x,y,z], x, y] /.x -> 3 /.y -> 2 /.z -> -5

In general, you follow the command with a series of /.var->number com-mands; this plugs in number to variable var in the preceding expression.

4.7 Summary of Chapter 4Terminology

• Let f : Rn → Rm. We say f is differentiable at x ∈ Rn if there is an m × nmatrix Df(x), called the total derivative of f , such that

limh→0

||f(x + h)− f(x)−Df(x)h||||h||

= 0.

• Let f : Rn → Rm. The partial derivative of component function fj withrespect to input variable xi is

∂fj∂xi

(x) = (fj)xi = limh→0

fj(x + hei)− fj(x)h

.

This expression gives the slope of the line tangent to the graph of fj at x inthe direction parallel to the xi-axis.

• f : Rn → Rm is called Cr-smooth if all of its partial derivatives (up to orderr) exist and are continuous.

• Let f : Rn → R and let u ∈ Rn be a unit vector. The directional derivative offunction f in the direction u at x is

Duf(x) = limh→0

f(x + hu)− f(x)h

.

This expression gives the slope of the line tangent to f at x in the direction u,and the rate of change in f per unit change of x in the direction of u.

• Let f : Rn → R be C1. The gradient of f is ∇f(x) = [Df(x)]T = (fx1 , ..., fxn).

252


Linearization

The linearization of a differentiable function f : Rn → Rm at a ∈ Rn is the function

L(x) = f(a) +Df(a)(x− a).

For values of x near a, f(x) ≈ L(x).

• When n = 1, L gives parametric equations for the tangent line to f at a.

• When n = 2 and m = 1, L gives an equation for the tangent plane to f at a.

Properties of gradients

• Duf(x) = ∇f(x) · u.

• ∇f(x) is normal to the level set f(x) = c.

• ∇f(x) points in the direction of greatest increase of f .

• The maximum value of Duf(x) is ||∇f(x)||, occurring when u is in the samedirection as∇f(x).

• The minimum value of Duf(x) is −||∇f(x)||, occurring when u is in the op-posite direction as∇f(x).

• The normal equation of the hyperplane tangent to the level set f(x) = c at x0is

∇f(x) · (x− x0) = 0.

Important theoretical concepts

Entries of the total derivative are partial derivatives: If a function f : Rn → Rm isdifferentiable, then all of its partial derivatives must exist, and the (i, j)−entryof its total derivative Df(x) must be the partial derivative ∂fi

∂xj.

Continuity of partial derivatives implies differentiability: If the partial deriva-tives of a function f : Rn → Rm all exist and are all continuous, then f isdifferentiable.

Clairaut’s Theorem: If mixed partials (such as fxy and fyx) are continuous, thenthey are equal.

Chain Rule: If f : Rp → Rm and g : Rn → Rp are differentiable, then so is f g, andD(f g)(x) = Df(g(x))Dg(x).

Implicit differentiation formula: If f(x1, ..., xn) is constant, then dxjdxi

= − fxifxj

.

253



1. Suppose f : R4 → R2 is a differentiable function.

a) Which of these quantities exist, Df(1, 2) or Df(1, 2, 3, 4)?

b) For the quantity in part (a) that exists, how many rows does that quan-tity have?

c) For the quantity in part (a) that exists, how many columns does thatquantity have?

2. Suppose f is some differentiable function such that for some vector x in the

domain of f , Df(x) =(

3 2 7 −1 05 3 −2 4 −2

).

a) To what vector space do the inputs of f belong?

b) To what vector space do the outputs of f belong?

In Problems 3-7, you are to assume that f ,g, h and k are differentiable functionswhere f : R → R3, g : R2 → R3, h : R3 → R2 and k : R2 → R. Assume that p is ascalar, a and b are vectors in R2, and that x and y are vectors in R3.

In each part of each problem, determine whether the given expression is ascalar, a matrix (in which case you should give its size), a vector (in which case youshould determine the vector space to which it belongs), or nonsense. (As usual, acolumn matrix should be thought of as a vector, rather than a matrix.)

3. a) f(x) b) h(x) c) Dh(x)

4. a) Dg(a)Dh(x) b) Dk(a) c) Dg(b)a

5. a) k(pa) b) [Dk(b− a)]T c) Dh(x) + 4Dh(y)

6. a) f(3) + 2g(5a) b) Df(3) + 2Dg(5a) c) k(2a + 7b)

7. a) aT [Dh(x)]x b) 2pDf(a) c) D(f + g)(x + a)

8. Let f : R4 → R2 be defined by f(x) = (3, 5). Compute Df(x).

9. Let f : R2 → R3 be defined by f(x, y) = (3x− 2y + 4, x− 5y, x− 1). Computethe total derivative of f .

10. Suppose f(t) = (2t,−5, 7t+ 3,−4t+ 1). Compute Df(t).

254


11. Suppose f(u, v, w) = 5u− 7w. Compute Df(u, v, w).

In Problems 12-14, assume that f : R2 → R3 and g : R2 → R3 are functions suchthat:

f(1, 3) = (1, 2,−1); g(1, 3) = (0, 5, 1); Df(1, 3) =

−1 30 42 5

; Dg(1, 3) =

−2 01 7−3 −1

.Use this information to compute each given quantity:

12. D(f + g)(1, 3) 13. D(4g)(1, 3) 14. D(3f − g)(1, 3)

In Problems 15-19, assume f : R → R3 and g : R → R3 are such that f(5) =(2, 1,−3), g(5) = (1,−4,−1), Df(5) = (−2, 0, 7) and Dg(5) = (3,−3, 2). Supposealso that h : R → R is a function such that h(5) = 10 and Dh(5) = h′(5) = 2. Usethis information to compute each given quantity:

15. D(f · 2g)(5)

16. D(f × g)(5)

17. D(hf)(5)

18. (F) Dk(5), where

k(x) = 2g(x) + (5,−2, 1)

19. (R) D((f × g) · f)(5)


20. Let f(x, y, z) = ln(xyz). Find all first-order partial derivatives of f .

21. Let f : (x, y) 7→ (u, v), where u = x cos 2y and v = xy2+yx3. Find all first-orderpartial derivatives of f .

22. (R) Find all first-order partial derivatives of F (x, y, z) = x−zx+y .

23. Let f(t) =(t sin t, t2−1

t2+1

). Find all first-order partial derivatives of f .

24. Let f : (x, y, z) 7→ (u, v), where u = 3x4y2z − 2xy3z2 and v = x3yz2 + 2xy2z.Compute the following quantities:

a) ux(1,−1, 1) b) ∂u∂y

(1, 2, 1) c) vz(2, 1,−1) d) ∂v∂x

(1, 3,−1)

25. (R) Suppose f(x, y) = (2x, x+ 3y, xy, x2 + y2). Compute:

a) ∂f2∂x

(8,−3) b) ∂f1∂y

(−4,−1) c) ∂f3∂x

(1, 2) d) ∂f4∂y

(13, 11)

26. Let g : R2 → R2 be g(x, y) = (e2x−y, ex+3y). Find Dg(x, y) and Dg(0, 0).

255


27. (F) Let f : R3 → R3 be f(x) = x× (3,−1, 2). Find the total derivative Df(x).

Hint: Find a formula for f(x, y, z) in terms of x, y and z. Then proceed as inProblem 26.

28. (R) Find Df(t) and Df(3), if f(t) = (t3 − 2,√t+ 1, 2 + ln t, 8t2).

29. (R) Suppose f : (x, y, z) 7→ (u, v) where u = x cos y and v = y sin 2z. Find thetotal derivative of f .

30. (F) In Chapter 2, we learned that to convert from polar coordinates (r, θ)to Cartesian coordinates (x, y) in R2, we use the formulas x = r cos θ andy = r sin θ. These formulas can be thought of as a function f : R2 → R2 wheref : (r, θ) 7→ (x, y) is given by

f(r, θ) = (r cos θ, r sin θ).

Find the total derivative of this f . Then compute detDf(r, θ).

31. (F) In Chapter 2, we also learned formulas for converting from cylindricalcoordinates to Cartesian coordinates. Find the total derivative of the functionf : R3 → R3 that converts cylindrical coordinates to Cartesian coordinates.Then compute detDf(r, θ, z).

32. (F) Find the total derivative of the function f : R3 → R3 that converts spheri-cal coordinates to Cartesian coordinates in R3. Then compute detDf(ρ, ϕ, θ).

33. Throughout this problem, let f(x, y) = 3√xy. We will show that f is an ex-

ample of a function where although its first-order partial derivatives exist at(0, 0), it is not differentiable there.

a) Compute fx(x, y) for any x 6= 0 and y 6= 0 (using usual differentiationrules).

b) Compute fx(0, 0) by using the definition of partial derivative given inDefinition 4.9 (using usual differentiation rules won’t work).

c) Compute fy(0, 0) by using the definition of partial derivative given inDefinition 4.9.

d) Based on your answers to parts (b) and (c), what is the only possiblematrix which Df(0, 0) could equal?

e) Take the matrix you wrote in part (d), and plug it into the limit in thedefinition of differentiability given in Definition 4.2. Show that the limitdoes not exist.

f) Since the limit you found in part (e) doesn’t exist, f cannot be differen-tiable at (0, 0). Why doesn’t this contradict Theorem 4.11, which seemsto give a formula for Df in terms of its partial derivatives?

256


34. Compute all second-order partial derivatives of the function f : R2 → Rdefined by f(x, y) = ln(x2 + y4 + 2).

35. (R) Compute all second-order partial derivatives of the function f : R2 → Rdefined by f(x) = x2 cos y + y2 sin 2x.

36. Compute all second-order partial derivatives of the function f : R3 → Rdefined by f(x, y, z) = exyz.

37. Suppose f(u, v) = arctan vu

. Compute:

a) fuv(u, v) b) fuv(3, 3) c) fuu(u, v)

38. Let f : R2 → R be defined by f(x, y) = 3x3y2 − 2xy2 + 4y4 − 2x2. Compute:

a) fxy(1, 2) b) fxx(−1, 4) c) fyx(3,−1)

39. For the same f as in Problem 38, compute:

a) fxyx(5, 1) b) fxxy(5, 1) c) fyxyx(0, 4)

40. (R) Suppose f(x, y, z) = ex cos y + ey sin z + ez cosx. Compute:

a) fxx(x, y, z) b) fyx(x, y, z) c) fzy(x, y, z)

41. Suppose f : R3 → R is a function such that fx(x, y, z) = xz − 2y2z + 4x.Compute fxz(x, y, z) and fzx(x, y, z).

42. Suppose f : R2 → R is a function such that ∂3f∂x2∂y

= y3 + 3x2y2 + 2x. Computefxyxy.

43. Suppose f(x, y, z) = y sin√x2 + 3+cos(zy+ez−2y)−ln(x+tan2(x2+z4−2xz)).

What is fxyz?

Note: This problem does not require any sophisticated symbol-crunching.

44. (F) Suppose f : Rn → R is a C2 function.

a) Assuming all the first-order partial derivatives of f exist, how manyfirst-order partial derivatives does f have?

b) Assuming all the second-order partial derivatives of f exist, how manysecond-order partial derivatives does f have?

c) If f is C2, what is the most number of different second-order partialderivatives that f could have?

257


45. Find a function f : R2 → R such that fx(x, y) = 6x2y + 2x and fy(x, y) =2x3 + 4.

46. Explain why there is no function f : R2 → R such that fx(x, y) = 4xy + 2xand fy(x, y) = x2 + 3y.

Hint: Clairaut’s Theorem may be useful here.

47. (F) Throughout this problem, let f : R2 → R be defined by

f(x, y) =xy(x2−y2

x2+y2

)if (x, y) 6= (0, 0)

0 if (x, y) = (0, 0)

a) Compute fx(x, y) for any (x, y) 6= (0, 0) (either by using usual differenti-ation rules or by using Mathematica).

b) Use your answer to part (a) to find a (simplified) expression for fx(0, y).

c) Use the limit definition of partial derivative, applied to your answer to(b), to find fxy(0, 0).Hint: The limit you need to compute here is

fxy(0, 0) = (fx)y(0, 0) = limh→0

fx(0, y)− fx(0, 0)h

.

d) Compute fy(x, y) for any (x, y) 6= (0, 0) (either by using usual differenti-ation rules or by using Mathematica).

e) Use your answer to part (d) to find a (simplified) expression for fx(0, y).

f) Use an appropriate limit definition of partial derivative, applied to youranswer to (e), to find fyx(0, 0).

g) Do your answers for fxy(0, 0) and fyx(0, 0) match?

h) Clairaut’s Theorem says that mixed second-order partials are supposedto coincide, but in this example, they don’t. Reconcile this apparentcontradiction.

258


48. The graph of the function f(x, y) = ln(y2 − x+ 2) is shown below.

a) Find the slope of the line tangent to the blue curve at the point (−1,−1, ln 4).

b) Find a direction vector pointing in the direction of the tangent line de-scribed in part (a).

c) Write the parametric equations of the tangent line described in part (a).

49. Find the slope of the line which lies in a plane parallel to the x-axis and istangent to the graph of f(x, y) = cos x sin y at the point

(π4 ,

π2 ,√

22

).

50. A metal plate is heated, so that the temperature at point (x, y) is T (x, y) =36√

4x2+y2. An ant is on the plate at position (1, 2).

a) If the ant starts crawling directly north (parallel to the y-axis), at whatrate (per unit of distance travelled) will the temperature at the ant’s po-sition increase or decrease?

b) If the ant starts crawling directly south from (1, 2), at what rate (per unitof distance travelled) will the temperature at the ant’s position increaseor decrease?

51. Suppose you take out a loan of amount P , at annual interest rate r, com-pounded monthly. If the term of the loan is n months, then the amount Ayou have to pay each month to satisfy the loan is

A(P, r, n) = Pr

1− (1 + r12)−n .

a) In the context of this problem, how should the quantity ∂A∂r

be inter-preted? (I’m looking for a description of what this quantity means inwords, not a math computation.)

259


b) Do you expect ∂A∂r

to be positive, or negative? Why?

c) In the context of this problem, how should the quantity ∂A∂P

be inter-preted?

d) Do you expect ∂A∂P


e) Do you expect ∂A∂n


f) If you borrow $100, 000 at interest rate .3% with a 30-year term, what arethe values of (P, r, n) in this problem?

g) Compute A(P, r, n) for the values of P, r and n you specified in part (f).

h) Compute ∂A∂r

at the values of P, r and n you specified in part (f).

i) Compute ∂A∂P

at the values of P, r and n you specified in part (f).

52. (F) The function (r, y, i) 7→ n given by n = 1+(1−r)y1+i − 1 gives the net gain or

loss of money invested in terms of the annual tax rate r, the annual yield oninvestment y, and the annual rate of inflation i.

a) Find the rate of change of gain (or loss) of money with respect to theannual tax rate when the effective yield is 8%, the annual tax rate is 18%and the inflation rate is 6%.

b) What is the calculus notation for the quantity you computed in part (a)?

c) Find the rate of change of gain (or loss) of money with respect to theeffective yield when the effective yield is 8%, the annual tax rate is 18%and the inflation rate is 6%.

d) What is the calculus notation for the quantity you computed in part (c)?

53. (F) Suppose n moles of a gas is kept in a container of volume V (measuredin cubic meters) at temperature T (measured in Kelvin) and pressure P (mea-sured in pascals). The Ideal Gas Law says that there is a constant R so that atall times, PV = nRT .

a) Starting with the Ideal Gas Law, write V as a function of the other fourvariables, and compute ∂V

∂T.

b) Write T as a function of the other four variables, and compute ∂T∂P

.

c) Write P as a function of the other four variables, and compute ∂P∂V

.

d) Use your answers to parts (a), (b) and (c) to simplify the expression∂V∂T

∂T∂P

∂P∂V

.

e) In the context of the Ideal Gas Law, if the rate of change of volume withrespect to temperature is 3 m3/K and the rate of change of temperaturewith respect to pressure is −1

4 K/p, what must the rate of change of pres-sure with respect to volume be? Write your answer with correct units.Hint: Use your answer to part (d).

260


54. A contour plot for some unknown function g : R2 → R is given below, at left.Use the contour plot to estimate the values of gx(1, 1), gy(3, 0), gy(−2, 1) andgx(2,−2).

-20

-16

-12

-12

-10

-10

-8

-8

-6

-6

-5

-5

-4

-4

-3-2

-1

-1

0

0

1

12

23

4

56 810

12

1620

-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4-4 -3 -2 -1 0 1 2 3 4

-4

-3

-2

-1

0

1

2

3

4

x

y

x→ −1 1 3 5 7y

−1 6 7 9 12 141 3 5 6 9 113 1 4 4 8 125 −5 1 5 11 167 −3 2 9 15 23

55. A table of values for some unknown function h : R2 → R is given above, atright (for example, the second-to-last entry in the bottom row is representingthe fact that h(5, 7) = 15). Use the table to estimate the values of hx(2, 3),hy(5, 6), hy(−1, 2) and hx(5, 6).

56. The graph of some unknown function f : R2 → R is shown below. Use thegraph to determine whether each given quantity is positive, negative, zero,or undefined:

a) ∂f∂x

(4, 3)

b) ∂f∂y

(2, 2)

c) fx(−4, 4)

d) (R) fy(−4, 4)

e) (F) fxy(2, 0)


57. Compute the linearization of the function f(x, y) = (x2y2, 2xy) at (1, 2).

58. Compute the linearization of the function f(x, y, z) = e2x+3y−z at (0, 0, 0), anduse that linearization to estimate f(−1

3 ,14 ,

15).

59. Compute the linear approximation to f(t) = (5 cos 3t, 2 sin 5t) at t = π4 .

261


60. (R) Compute the linearization of the function f(x, y) = (arctan(x2y), ln(xy +1)) at (0, 0), and use that linearization to estimate f(.1, .2).

61. Suppose f : (x, y) 7→ (u, v) is a function such that f(8, 3) = (6, 7), ux(8, 3) = 2,uy(8, 3) = −3, vx(8, 3) = 1 and vy(8, 3) = −4. Use linearization to estimatef(7.8, 3.1).

62. Use linearization of the function f(x, y) = √x cos y to estimate√

65 cos 14 .

63. The tension T in the string holding up a yo-yo is given by

T = 30R2r2 +R2 ,

where R is the radius of the yo-yo and r is the radius of the spool in the mid-dle of the yo-yo that the string wraps around. Use linearization to estimatethe change in tension as R changes from 5 cm to 5.1 cm and r changes from 1cm to 1.2 cm.

64. (R) Write parametric equations of the line tangent to the image of f whent = 2, if f(t) = (2t3 − 4t+ 1, t2 + 5t− 8, t3 − 4t).

65. Write parametric equations for the line tangent to the image of g(t) =(t+1t−1 ,

t2+1t2−1

)at the point

(2, 5

4

).

66. Let C be a curve in R3 whose parametric equations are x = 4 cos 2t, y =3 sin 3t, z = 5t+ 2. Find a vector of length 6 which is tangent to C at the pointwhere t = 0.

67. (R) Write parametric equations for the line tangent to the curve C at the point(2, 4, 1), where C is the image of f(t) =

(3t2 − t, 4

√t, et− 1

).

68. (F) Find all points on the image of f(t) = (t2 − 2t, t3 − 4t2, t2 + 8t − 1) suchthat the tangent line to f at that point is parallel to the line whose parametricequations are x = t− 7, y = 6t+ 1, z = 2t.

69. Write an equation of the plane tangent to the graph of f(x, y) = 3x2 + 2x −4y2 + 2y − 5 at the point (1,−1,−2).

70. Write an equation of the plane tangent to z = ln(x2 + y4) at the point wherex = 0 and y = 1.

71. Write an equation of the plane tangent to z = ex cos y + ey sin x at the point(0, 0, 1).

262


72. (F) Find a point on the graph of the paraboloid y = 2x2 + 5z2 such that thetangent plane to the paraboloid at that point is parallel to the plane−x+4y−3z = 8.

73. a) Write the equation of the tangent plane to the ellipsoid 4x2 + 6y2 + 3z2 =118 at the point (5, 1, 2).

b) Use the equation of the tangent plane you wrote in part (a) to estimatethe (positive) value of y such that (5.1, y, 1.8) is on the ellipsoid.

74. Write parametric equations of the normal line to the graph of f(x, y) = y cosx+x2 sin y at the point (0, π, π).


75. Suppose u, v and w are each functions of x and y, such that u(3, 1) = 4,v(3, 1) = −9, w(3, 1) = 2, ux(3, 1) = 6, uy(3, 1) = −2, vx(3, 1) = −1, vy(3,−1) =5, wx(3, 1) = 2 and wy(3, 1) = −1. Suppose further that s is a function of u, vand w, such that su(4,−9, 2) = 7, sv(4,−9, 2) = 0, and sw(4,−9, 2) = 1. Iff : (x, y) 7→ s, find Df(3, 1).

76. (F) Suppose u and v are functions of t and that x is a function of u and v.Suppose that when t = 6, u = 7 and v = 0, and that x = 9. If du

dt

∣∣∣t=6

= −3,dvdt

∣∣∣t=6

= 4, ∂x∂u

∣∣∣(u,v)=(7,0)

= 5 and dxdt

∣∣∣t=6

= 1, find ∂x∂v

∣∣∣(u,v)=(7,0)

.

77. (F) Suppose f : (x, y) 7→ z, where x and y are the usual Cartesian coordinatesin R2. Suppose we were interested in partial derivatives relative to polarcoordinates, i.e. fr and fθ, rather than fx and fy.

a) Describe in your own words what you think the quantities fr and fθrepresent.

b) Compute fr and fθ, in terms of fx and fy.Hint: Use the Chain Rule in the context of (r, θ) 7→ (x, y) 7→ z; think ofthe f in this situation as the composition that sends (r, θ) to z. The resultof Problem 30 may be useful here.

c) Suppose f(x, y) = y2− x2. Compute fr and fθ at the point (x, y) = (1, 1).

d) Show that for any f : R2 → R, the following identity always holds:

(fx)2 + (fy)2 = (fr)2 + 1r2 (fθ)2.

78. Suppose x3y2 + xy − 2x2y = 14. Find dydx

∣∣∣(x,y)=(2,−1)

.

79. Suppose 2xe3y − 4ye−z + 8ze2x = 7. Find dydx

and dzdx

.

263


80. (R) Suppose y cos 2x = x2 sin y. Find dydx

.

81. Find the slope of the line tangent to the curve x2/3 + y2/3 = 1, when x = −827

and y = 5√

527 .

82. a) Suppose F (x1, ..., xn) = 0. What is the relationship between dxjdxi

and dxidxj

?

b) Give a short proof of your answer, using Theorem 4.22.

c) Suppose F (x, y, z) = 0 and dxdz

= x2−3yz2xz+y2z4 . What is dz

dx?

83. We saw in class that if f : R2 → R is constant, then dydx

= −fxfy

. With somework, one can also derive the following formula for the second derivative ofy with respect to x. If f : R2 → R is constant, then it turns out that

d2y

dx2 = −(fy)2fxx − 2fxfyfxy + (fx)2fyy(fy)3 .

Use this formula to find d2ydx2 if x2y − y3x = 4.

84. (F) Derive the formula for d2ydx2 presented in Problem 83.

85. Suppose z = x2y + 2y2 − x.

a) Compute ∂z∂x

.

b) Compute dzdx

.

c) Suppose y = x2. Compute dzdx

∣∣∣x=2

.


In each of Problems 86-90, find the gradient of the indicated function.

86. f(x, y) = cos (3xy)

87. f(x, y) = xy2ex+3y

88. g(x, y, z) = ln(x2 + 2y2 + 8z4)

89. k(x, y) = (x+ 2y)7 − (2x− y)5

90. f(w, x, y, z) = ewz + eyx − ewy − exz

91. Let f(x, y, z) = 3x2z + 2x√y − z2√x+ 1. Compute∇f(3, 4,−1).

92. Let f(x, y) = x2 − 3xy + y2 + x+ 2y + 8. Find all x ∈ R2 such that∇f(x) = 0.

93. Let f(x, y) = y2 + cos(x− y). Find all (x, y) ∈ R2 such∇f(x, y) = 0.

264


94. Let f(x, y, z) = x2y+ 2xz − 2yz + 4z − 2x+ 3y+ 9. Find all (x, y, z) ∈ R3 such∇f(x, y, z) = 0.

95. Compute Duf(3, 1) if f(x, y) = 2xy + 3y2 and u is in the direction parallel to(2, 1).

96. (R) Compute Duf(3,−2), where f(x, y) = ex2−y2 ; and u is in the direction

parallel to (−5, 12).

97. Compute the directional derivative of f(x, y, z) = x2y − y2z at the point(2, 1, 4), in the direction parallel to (1,−3, 2).

98. Compute Duf(3, 0,−4), if f(x) = ||x|| and u is the unit vector(

13 ,

23 ,−23

).

99. Suppose that f : R2 → R is such that fx(1, 5) = −3 and fy(1, 5) = 2. FindDuf(1, 5), where u is the unit vector

(1√5 ,−√

2√5

).

In Problems 100-103, you are to assume that f : R → R3, g : R2 → R, andh : R3 → R2 are differentiable functions. Assume also that u is a unit vector in R2,and that t ∈ R, x ∈ R2 and y ∈ R3.

In each part of each problem, determine whether the given expression is ascalar, a matrix (in which case you should give its size), a vector (in which case youshould determine the vector space to which it belongs), or nonsense. (As usual, acolumn matrix should be thought of as a vector, rather than a matrix.)

100. a) Dg(x) b) ∂h1∂y1

(x) c) Dug(x)

101. a) Duh(y) b) gx2(x) c) ∇g(x)

102. a) ∇h b) ∂3g∂x2

1∂x2(x) c) ||∇g(x)||

103. a) x · ∇g(x) b) ∂h1∂h2∂y2

1(x) c) [∇g(x)][Dg(x)]

104. (F) Consider the curve C in R2, which is the graph of the equation x2 − y2 −3exy = 1.

a) Use the implicit differentiation formula to find the slope of the line tan-gent to C at the point (2, 0).

b) Use the slope you found in part (a) to write the equation of the linetangent to C at (2, 0).

c) Use gradients to find a normal vector to C at the point (2, 0).

d) Use the normal vector you found in part (c) to write the normal equationof the line tangent to C at (2, 3).

265


e) Verify that your answers to parts (b) and (d) are equations of the sameline.

105. Consider the surface S in R3 which is described by the equation xz2 + zy −x2y = 0.

a) Solve this equation for z to obtain a function f : (x, y) 7→ z which givesthe part of S containing the point (2, 4, 2).

b) For the function you found in part (a), find fx and fy.

c) Use your answers to part (b), together with the formula for tangentplanes to functions z = f(x, y) in Theorem 4.20, to write the equationof the tangent plane to S at (2, 4, 2).

d) S can be thought of as a level surface to some function F : R3 → R. Whatfunction should F be, and at what height is the level surface?

e) Find the gradient of the function F you found in part (d) at (2, 4, 2).

f) Use the gradient you found in part (e) to write the equation of the planetangent to S at (2, 4, 2).

g) Verify that your answers to parts (c) and (f) are equations of the sameplane.

106. Find an equation of the plane tangent to sin(x+z)+sin(y+z) = 1 at the point(0, π2 ,

π2

).

107. Find an equation of the plane tangent to x2 + 2y2 − 5z2 = 29 at the point(4, 3, 1).

108. Find a vector of length 5 which is normal to the ellipsoid 3x2 + 6y2 + 2z2 = 24at the point (2, 1,

√3).

109. Find parametric equations of the normal line to√x + √y +

√z = 6 at the

point (1, 9, 4).

110. (F) The graphs of f(x, y) = −x2

2 − y2 and g(x, y) = x4

100 −y2

2 − 2 are shown inthe figure below (f is the orange surface; g is the blue surface). They intersectin the red curve. Write parametric equations for the line tangent to the redcurve at the indicated point

(√52 ,√

118 ,−

218

).

266


Hint: It is not a good idea to try to figure out the equation of the red curve.

111. (F)

a) Make a good definition of what you think it means for two surfaces inR3 to be tangent to one another at some point p ∈ R3.

b) Use the definition you wrote in part (a) to show that the ellipsoid 3221x

2 +1621y

2 + 821z

2 = 1 is tangent to the graph of f(x, y) = (x + y)2 − 7x + 112 at

the point(

34 ,−

14 ,

12

).

112. (F) Show that for any sphere in R3 centered at the origin, the normal line toevery point on sphere passes through the origin.

Hint: Write the equation of a sphere centered at the origin (of arbitrary ra-dius), and let (a, b, c) be any point on the sphere. Figure out (using gradients)the direction vector for the normal line passing through (a, b, c). Use the point(a, b, c) and the direction vector to write parametric or symmetric equationsof the normal line, and last, show that (0, 0, 0) is on that line.

113. Find the normal equation of the tangent hyperplane to the hypersurfacew2x2+3wy − 2xyz2 + 4wy2z = 6 at the point (1, 1, 1, 1).

114. Let f(x, y, z) = x arctan zy.

a) Find the direction u for which Duf(3, 2√

3, 2) is maximized.

b) Find the maximum value of Duf(3, 2√

3, 2).

115. A fisherman’s boat sinks at point (e2, 6) in a body of water, where the depthat point (x, y) is f(x, y) = 2 ln y + yx2.

267


a) In what direction should the fisherman swim, if he wants to get to shal-lower water as quickly as possible?

b) If he swims in that direction, what will the rate of change of the depthper unit he swims be?

116. The elevation at point (x, y) in a ski resort is given by h(x, y) = 8000 − (x −10)2(y − 8)2 ft above ski level. If a skier is at point (5, 7) and wants to skidownhill in the steepest direction possible, in what direction should she ski?

117. A basic fact of electromagnetism is that when exposed to an electrical field,charged particles move in the direction in which the electric potential (volt-age) decreases most greatly. If a charged particle at the point (−2, 2, 4) isexposed to an electric potential of V = e−xyz volts, in what direction will theparticle initially move?

118. The temperature at position (x, y) on a metal plate is given by T (x, y) = 120−8x2 − y2. A heat-seeking particle is place on the plate at position (1, 3).

a) In what direction will the particle initially move?

b) What is the Cartesian equation of the curve along which the particle willmove?

268


In Problems 119-122, use the following twelve pictures of vector fields on R2,labelled A through L:

A.

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

x

y

B.

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

x

y

C.

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

x

y

D.

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

x

y

E.

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

x

y

F.

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

x

y

G.

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

x

y

H.

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

x

y

I.

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

x

y

J.

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

x

y

K.

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

x

y

L.

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

x

y

269


119. In each part of this problem, you are given the picture of a contour plot of afunction f : R2 → R. Determine, for each function, which of the pictures A-Labove is the most reasonable picture of the vector field∇f :

a)-4-4 -3-3 -2 -2-1-1 00

1

1

2

2

3

3

4

45

5

6

6

7

7

8

8

9

9

10

10

-4 -2 0 2 4

-4

-2

0

2

4

x

y

c)012

3

4

5

6

7

8

9

9

10

10

11

1112

12

-4 -2 0 2 4

-4

-2

0

2

4

xy

e)

-6

-6

-5

-5

-4

-4

-3

-3

-2-1

0

1

2

3

456

-4 -2 0 2 4

-4

-2

0

2

4

x

y

b)

0

1

2

3

4

5

6

7

8

9

1010

1111

12 12

-4 -2 0 2 4

-4

-2

0

2

4

x

y

d)23 45

6

6

7

7

8 8

8 8

-4 -2 0 2 4

-4

-2

0

2

4

x

y

f)0 1 2

3

4 5 6 7 8

9

9

10

10

11

11

12

12

-4 -2 0 2 4

-4

-2

0

2

4

x

y120. Same directions as the previous problem:

a)

-6

-6

-5

-5

-4

-4 -3

-2

-1

0

1

2

3

4

5

6

-4 -2 0 2 4

-4

-2

0

2

4

x

y

c)

22

22

3

3

4

4

5

67 8

-4 -2 0 2 4

-4

-2

0

2

4

x

y

e)

-6-6

-5 -5

-4 -4

-3 -3-2

-1

0

1

2

3

4

5

6

-4 -2 0 2 4

-4

-2

0

2

4

x

y

b)

-9

-9

-8

-8

-7

-7

-6

-6

-5

-5

-4

-4

-3

-3

-2

-2

-1

-1

0

0

1 1 22 3

3

4

4

55

-4 -2 0 2 4

-4

-2

0

2

4

x

y

d)

0

1

2

3

45

6

7

8

9

9

10 10

1111

12 12-4 -2 0 2 4

-4

-2

0

2

4

x

y

f)

-6

-6

-5

-5

-4

-4

-3

-3

-2 -1

0

1

2

3

4 5 6

-4 -2 0 2 4

-4

-2

0

2

4

x

y

270


121. In each part of this problem, you are given a picture of the graph of a functionf : R2 → R. Determine, for each function, which of the pictures A-L frombefore Problem 119 is the most reasonable picture of vector field∇f :

a) c) e)

b) d) f)

122. Same directions as the previous problem:

a) c) e)

b) d) f)

271


Selected answers

1. a) Df(1, 2, 3, 4) exists.

b) 2c) 4

2. a) R2

b) R5

3. a) nonsense

c) 2× 3 matrix

4. a) 3× 3 matrix

5. a) scalar

c) 2× 3 matrix

6. a) vector in R3

7. a) scalar

9.

3 −21 −51 0

11.

(5 0 7

)

12.

−3 31 11−1 4

16. (21,−8, 1)

17. (−16, 2, 64)

21. ux = cos 2y, uy = −2x sin 2y,

vx = y2 + 3yx2, vy = 2xy + x3

22. Fx = y+z(x+y)2 , Fy = −(x−z)

(x+y)2 , Fz = −1x+y

24. a) 14c) −12d) −9

25. a) 1b) 0

26. Dg(x, y) =(

2e2x−y −e2x−y

ex+3y 3ex+3y

);

Dg(0, 0) =(

2 −11 3

).

27.

0 2 1−2 0 3−1 −3 0

28. Df(t) =

(3t2, 1

2√t+1 ,

1t, 16t

);

Df(3) = (27, 14 ,

13 , 48).

30.(

cos θ −r sin θsin θ r cos θ

); detDf(r, θ) =

r.

32. detDf(ρ, ϕ, θ) = ρ2 sinϕ.

33. a) fx(x, y) = 3√y3x2/3 .

b) 0c) 0

d)(

0 0).

35. fxx(x, y) = 2 cos y − 4y2 sin 2x;

fxy(x, y) = fyx(x, y) = 4y cos 2x− 2x sin y;

fyy(x, y) = −x2 cos y + 2 sin 2x.

36. fxx(x, y, z) = x2y2exyz;

fxy(x, y, z) = fyx(x, y, z) = (z + xyz2)exyz.

272


37. a) v2−u2

(u2+v2)2

b) 0

38. a) 28

39. a) 180c) 0

40. b) −ex sin y

41. fxz(x, y, z) = x− 2y2

42. 0

43. a) n

b) n2

c) 12(n2 + n)

47. a) y(x4+4x2y2−y4)(x2+y2)2

c) −1

d) x(x4−4x2y2−y4)(x2+y2)2

g) No

48. a) −12 .

c)

x = −1y = t+ 1z = ln 4− 1

2t

49. −√

22 .

50. a) The temperature decreases at a rate of 9√8 degrees per unit.

51. a) ∂A∂r

is the instantaneous rate of change in the monthly payment with re-spect to the monthly interest rate, if the principal and loan term are heldfixed.

b) ∂A∂r

should be positive, since a larger interest rate should result in higherpayments.

d) ∂A∂P

should be positive, because if the loan is for a larger amount, youshould have to pay more per month.

f) P = 100000; r = .03; n = 360.h) 99977.3

52. a) −.07547

53. a) V (P,R, T ) = nRTP

; ∂V∂T

(P,R, T ) = nRP

.d) −1b) 4

3 p/m3.

54. gx(1, 1) ≈ −2; gy(3, 0) ≈ −3.5.

55. hx(2, 3) ≈ 0; hy(5, 6) ≈ 2.

56. c) negatived) positive

57. Ł(x, y) = (−12 + 8x+ 4y,−4 + 4x+ 2y)

58. f(−13 ,

14 ,

15) ≈ 53

60 .

273


60. f(.1, .2) ≈ (0, 0).

62. 12916

63. −727

64.

x = 9 + 20(t− 2)y = 6 + 9(t− 2)z = 8(t− 2)

66.(

0,

27√

2√53 , 1515

√2√

53

)69. z = 8x+ 10y

70. z = 4y − 4

72.(

116 ,

23640 ,

340

)74.

x = 0y = π + tz = π − t

76. −11

77. b) fr = fx cos θ + fy sin θ;fθ = fyr cos θ − fxr sin θ.

c) fr = 0 and fθ = 4

79. dydx

= −2e3y−16ze2x

−4e−z+6xe3y ;dzdx

= −2e3y−16ze2x

8e2x+4ye−z .

82. a) They are reciprocals.

c) 2xz+y2z4

x2−3yz

83. 2y(3x3−2x2y2+6xy4−6y6)x2(x−3y2)3

85. c) 95

86. (−3y sin 3xy,−3x sin 3xy)

88.(

2xx2+2y2+8z4 ,

4yx2+2y2+8z4 ,

32z3

x2+2y2+8z4

)

90. (−yewy + zewz, yexy − zexz,xexy − wewy, wewz − xexz)

91.(−57

4 , 32 , 31

)93. (0, 0)

94. (−1, 1, 2) and(−13 ,

53 ,

149

)97.√

14

98. 1115

100. a) 1× 2 matrix

c) scalar

101. a) nonsense

102. b) scalar

c) scalar

104. a) 23

b) y = 23(x− 2)

c) (4,−6)d) 4(x− 2)− 6(y − 3) = 0

105. a) f(x, y) = −y+√

4x3y+y3/2

2x

c) z = (x− 2) + 16(y − 4) + 2

106. −(y − π2 )− (z − π

2 ) = 0

107. 8(x− 4) + 12(y− 3)− 10(z− 1) = 0

108.(

5√

3√7 ,

5√

3√7 ,

5√7

)

109.

x = 1 + 1

2ty = 9 + 1

6tz = 4 + 1

4t

110.

x =

√52 −

√118 t

y =√

118 + 11

2√

10t

z = −218 −

3√

112√

5 t

274


113. 9(w − 1) + 9(y − 1) = 0

114. a)(π6 ,−38 ,

3√

38

)b)

√916 + π2

36

116. In the direction (−1,−5) (or any positive multiple of this)

117. In the direction (−2, 2, 1) (or any positive multiple of this)

118. a) In the direction (−8,−3) (or any positive multiple of this)

b) x = 16561y

8

119. a) C

b) E

c) J

120. a) H

121. a) E

b) K

c) L

122. a) I

b) F

275

Chapter 5

Analysis of motion

In Calculus 1, you learn how to use calculus to analyze “1-dimensional” motion,i.e. motion of an object moving back and forth along a line. Let’s review someprinciples from this setting:

MOTIVATING EXAMPLE

You drive 50 mph due east for 3 hours, then instantaneously do a U-turn and drive50 mph due west for 1 hour.

1. Let x = x(t) be the function that describes your position at time t (measuredin hours after your trip starts). Assume you start at position 0, and that yourposition is measured along an east-west axis. Sketch a graph of x, sketch theimage of x, and write a rule for the function x.

1 2 3 4time t

50

100

150

200

E-W position

50 100 150 200 E-W position

2. What is your velocity, and your speed, at times 1.5 and 3.5?

v(1.5) = speed(1.5) = v(3.5) = speed(3.5) =

276

3. Let v be the function that describes your velocity at time t. Write a rule for v,sketch the graph of v, and describe the relationship between v and x.

1 2 3 4time t

-50

-25

25

50

velocity

4. Let speed be the function that describes your speed at time t (let’s not uses for speed, because s will be something else later). Write a rule for speed,sketch the graph of speed, and describe the relationship between speed and vand/or x.

1 2 3 4time t

-50

-25

25

50

speed

5. What is your displacement (i.e. the net change in your position) from time 0to time 4? Describe one or more formulas in terms of x, v and/or speed thatcould be used to compute this displacement.

6. What is your total distance travelled from time 0 to time 4? Describe one ormore calculus formulas in terms of x, v and/or speed that could be used tocompute this distance travelled.

277

What this chapter is about

In this chapter we look more deeply at the calculus of functions

x : R→ Rm

i.e.t

x7−→ (x1, x2, ..., xm).As mentioned before, such a function can (and maybe “should”) be thought of asgiving the position x(t) = (x1, ..., xm) of an object at time t, so that the functiondescribes the motion of an object at time passes.

If x : R → R3, we’re thinking of a picture like this, where an object is movingalong the red path:

Questions we want to explore:

1. What is the object’s velocity? Its speed? Its acceleration? Is the object speed-ing up? Slowing down?

2. How far does the object travel over a certain time period? What is its dis-placement?

3. How sharply is the object changing direction? Is the path it travels straightor “curvy”?

4. What “linear” objects (like lines and planes) best capture the motion of theobject at a given instant?

278

5.1. Old ideas in higher dimensions

5.1 Old ideas in higher dimensionsVelocity, speed and acceleration

Based on the motivating example we looked at earlier, we can define lots of quan-tities. The crux of what follows is that all the ideas you know from Calculus 1 and2 still work in this context; you just do everything coordinate-wise, and instead ofabsolute values, you use norms:

Definition 5.1 Let x : R→ Rm be C2 and suppose an object is moving in Rm so thatits position at time t is x(t) = (x1(t), x2(t), ..., xm(t)).

1. The velocity of the object at time t is the vector

v(t) = Dx(t) = x′(t) = d

dtx = dx

dt= (x′1(t), x′2(t), ..., x′m(t)).

2. The speed of the object at time t is the scalar

||v(t)|| = ||x′(t)|| =√

[x′1(t)]2 + ...+ [x′m(t)]2.

3. The acceleration of the object at time t is the vector

a(t) = D[Dx(t)] = d2

dt2x = d2x

dt2= x′′(t) = (x′′1(t), ..., x′′m(t)).

Note: The velocity is a vector that indicates both the speed of the object and thedirection in which the object is moving. It should be drawn such that the tail of thevector is at the point x(t) = (x1(t), x2(t), ..., xm(t)). Since the velocity vector is thetotal derivative of x, it is the direction vector for the linearization of x, meaningthat will always point in a direction tangent to the position curve, and point in thesame direction the curve is being traced out.

The speed is the length of the velocity vector; so the longer the velocity vectoris, the faster the object is moving.

(See the figure on the next page.)

279


The acceleration of an object is also a vector which should be drawn such thatthe tail of the vector is at the point x(t). The acceleration vector will always pointto the inside of the position curve, looking forward in time from x(t). (Anotherway of saying this is that the curve will “bend” towards the acceleration vector;this makes sense because by Newton’s second law, acceleration is proportional toforce, and the motion of the object should bend toward the direction of the forceacting on it.)

a(0)x(0)

a(-5)

x(-5)

a(2)

x(2)

Theoretical remark: We ask that x be C2 so that we can differentiate x twice,ensuring all the above stuff exists. For instance, if x was only C1, then you couldcompute the velocity, but maybe not the acceleration.

280


EXAMPLE 1Suppose the position of an object at time t is (t3, et, sin t). Find the velocity, speedand acceleration of the object at t = 0, and sketch a picture of what the path thisobject travels looks like near t = 0.

Solution: The object’s position is x(t) = (t3, et, sin t).

Therefore its position at time 0 is x(0) = (03, e0, sin 0) = (0, 1, 0) .

Differentiate to get its velocity: v(t) = x′(t) = (3t2, et, cos t).

Therefore its velocity at time 0 is v(0) = (3(0)2, e0, cos 0) = (0, 1, 1) .

Next, the object’s speed at time 0 is ||v(0)|| = ||(0, 1, 1)|| =√

02 + 12 + 12 =√

2 .

Furthermore, the object’s acceleration is a(t) = x′′(t) = (6t, et,− sin t).

Therefore its acceleration at time 0 is a(0) = (6(0), e0,− sin 0) = (0, 1, 0) .

Last, let’s sketch a picture of what the path the object travels looks like neart = 0:

1

1

1

281


Displacement and arc length; integration

Definition 5.2 Let x : R→ Rm be a C1 function and suppose an object is moving inRm so that its position at time t is x(t). The displacement of the object from time ato time b is the vector

x(b)− x(a) =∫ b

ax′(t) dt.

Note: Displacement is a vector quantity, which gives the net change in an ob-ject’s position from time a to time b. As an example, if the object moves around acircle from time a to time b, its displacement between those times would be 0.

Definition 5.3 The (arc) length of a curve which is the image of a C1 function x :R→ Rm from t = a to t = b is

∫ b

a||x′(t)|| dt =

∫ b

a

√[x′1(t)]2 + [x′2(t)]2 + ...+ [x′m(t)]2 dt.

Note: the length of a curve is a scalar quantity. You should think of it this way:given a curve in Rm, imagine laying a string on the curve from x(a) to x(b). To findthe length of the curve, pull the string until it is taut, and then measure its length.That length measurement is given by the integral of the speed, as described in Def-inition 5.3.

Theoretical remark: If x is C1, then the integrals in Definitions 5.2 and 5.3 al-ways exist (a proof is beyond the scope of this class).

EXAMPLE 2Write an integral which computes the length of the curve x(t) = (t, et − e−t, e2t)between t = 0 and t = 1.

Ground rule: in a question like this, your integral can only contain the inde-pendent variable of the problem (in this case, t). No x, v, a, etc. are allowed inyour answer.

Solution: The arc length is the integral of the speed, so we need to compute thevelocity, then the speed:

v(t) = x′(t) = (1, et + e−t, 2e2t)⇒ ||v(t)|| = ||x′(t)|| =√

12 + (et + e−t)2 + (2e2t)2.

Therefore the arc length is given by

∫ b

a||x′(t)|| dt =

∫ 1

0

√12 + (et + e−t)2 + (2e2t)2 dt .

282


Remark: A decimal approximation to this integral can be obtained in Mathemat-ica using the NIntegrate command as follows:

Input: NIntegrate[Sqrt[1 + (Eˆt + Eˆ(-t))ˆ2 + 4Eˆ(4t)], t, 0, 1]Output: 6.97019

EXAMPLE 3A bird flies around in R3, so that its position at time t is x(t) = (3 cos t, 3 sin t, 4t).Find the distance the bird travels between the points (3, 0, 0) and (−3, 0, 4π).

EXAMPLE 4Verify (“prove”) that the circumference of a circle of radius r is in fact 2πr.

Solution:

Remark: In the situation above, what is the displacement from t = 0 to t = 2π?

283


Suppose we are given the velocity or acceleration of an object instead of theposition. In this situation, we need to integrate to recover the position function.Toward that end, we make the following definitions (all of which should resemblethings you learned in Calculus 1; you just do everything coordinate-wise):

Definition 5.4 Let f : R→ Rm.

1. A function F : R→ Rm is called an antiderivative of f if DF(t) = f(t) for allt.

2. The set of all antiderivatives of f is called the indefinite integral of f and isdenoted by

∫f(t) dt.

3. Given real numbers a and b, f is said to be integrable on [a, b] if∫ ba fj(t) dt

exists for all j, in which case we define the definite integral of f from a to b bethe vector ∫ b

af(t) dt =

(∫ b

af1(t) dt,

∫ b

af2(t) dt, ...,

∫ b

afm(t) dt

).

Theorem 5.5 Let f : R→ Rm. Then:

1. (Antiderivative theorem) All antiderivatives of f differ by a constant vector.

2. (Coordinate-wise integration) If F is an antiderivative of f , then∫fj(t) dt =

Fj(t) for all j, i.e.∫f(t) dt =

(∫f1(t) dt,

∫f2(t) dt, ...

∫fm(t) dt

).

3. (Fundamental Theorem of Calculus) If f is continuous on [a, b], then f isintegrable on [a, b] and given any antiderivative F of f ,

∫ b

af(t) dt = F(b)− F(a).

WARNING:

Integrals like∫

f(t) dt and∫ ba f(t) dt do not directly solve any problems re-

lated to area or volume. For this, you need more complicated ideas thatwe will study starting in Chapter 7.

284


EXAMPLE 5Suppose an object’s position at time 0 is (0, 0, 0) and its velocity at time 0 is (0, 0, 2).If the acceleration of the object at time t is (−4t2, 12t+ 20t3, 6t− 1) m/sec2, find theposition of the object at time t = 3.

Solution: First, integrate the acceleration to find the velocity:

v(t) =∫

a(t) dt =∫ (−4t2, 12t+ 20t3, 6t− 1

)dt =

(−4t3

3 , 6t2 + 5t4, 3t2 − t)

+ C

To solve for C, use the fact that v(0) = (0, 0, 2):

(0, 0, 2) =(−4(0)3

3 , 6(0)2 + 5(0)4, 3(0)2 − 0)

+ C = (0, 0, 0) + C ⇒ C = (0, 0, 2).

Therefore the velocity is

v(t) =(−4t3

3 , 6t2 + 5t4, 3t2 − t+ 2)

m/sec.

Now integrate again to get the position:

x(t) =∫

v(t) dt =∫ (−4t3

3 , 6t2 + 5t4, 3t2 − t+ 2)dt

=

To solve for C, use the fact that x(0) = (0, 0, 0):

(0, 0, 0) =(−04

3 , 2(0)3 + 05, 03 − 02

2 + 2(0))

+ C = (0, 0, 0) + C ⇒ C = (0, 0, 0).

Last, find the position at t = 3:

x(3) =(−33

3 , 2(3)3 + 35, 33 − 32

2 + 2(3))

=(−9, 297, 57

2

)m .

285


Projectile motion

Suppose that an object is flying through the air so that the only force acting on theobject is gravity. In this setting, the acceleration due to the gravitational force isconstant (by Newton’s law of universal gravitation), so

||a(t)|| = 32 ft/sec2 = 9.8 m/sec2,

and the direction the acceleration acts is towards the Earth’s surface, i.e. down-ward. That means

From this, you can recover the particulars of an object’s motion at any instant,so long as you know the object’s initial position and initial velocity.

EXAMPLE 6A cannon fires a cannonball at an angle of π

6 to the ground, with initial speed 160ft/sec. Assume that only gravity acts on the cannonball.

1. Find the cannonball’s maximum altitude.

2. Find how far away from the cannon the cannonball lands.

3. Find the speed of the cannonball at impact.

Picture (not always needed):

x

y

Strategy: Identify what we are given:

From the given information, figure out v and x and then answer the particularquestions being asked.

286


EXAMPLE 6 (REPEATED FOR CONVENIENCE)A cannon fires a cannonball at an angle of π

6 to the ground, with initial speed 160ft/sec. Assume that only gravity acts on the cannonball.

1. Find the cannonball’s maximum altitude.

2. Find how far away from the cannon the cannonball lands.

3. Find the speed of the cannonball at impact.

Given information (from previous page):

a(t) = (0,−32) v(0) = (80√

3, 80) x(0) = (0, 0)

Computations:

287


EXAMPLE 7 (MORE CHALLENGING)You fire a projectile from an elevation of 640 ft to the north with initial speed 48

√2

ft/sec and launch angle π4 to the horizontal. A crosswind blows to the east, apply-

ing an acceleration of 4 ft/sec2 to the projectile. Where does the projectile land?

v(0)

640

x↔E

y↔N

Given information:

Solution: We compute v and x as in the preceding example. First,

v(t) =∫

a(t) dt =∫

(4, 0,−32) dt = (4t, 0,−32t) + C;

since (0, 48, 48) = v(0) = 0 + C, C = (0, 48, 48) so v(t) = (4t, 48, 48− 32t).

x(t) =∫

v(t) dt =∫

(4t, 48, 48− 16t) dt = (2t2, 48t, 48t− 16t2) + C;

since (0, 0, 640) = x(0) = 0 + C, C = (0, 0, 640) so x(t) = (2t2, 48t, 48t− 16t2 + 640).

Finally, the projectile’s position upon landing is x(8) = (2(8)2, 48(8), 0) = (128, 384, 0) .

288

5.2. New ideas

5.2 New ideasA motivating question: speeding up vs. slowing down

So far, we’ve seen how formulas for velocity, speed, acceleration, displacementand arc length generalize from one-dimensional motion to higher dimensions. But,there are some unresolved issues. Here’s one:

Question: Suppose an object’s position at time t is given by x(t), where x : R→Rm. At time t, is the object speeding up or slowing down?

Answer for 1-dimensional motion: Compute the . Assum-ing your velocity is positive,

Why this doesn’t work in higher dimensions: Ifm ≥ 2, acceleration is a vectorquantity and vectors are not ordered (they can’t be described as positive or nega-tive, or being greater than or less than other vectors).

Two proposed solutions to this problem:

1. Compute the rate of change of the speed (with respect to time). Using propernotation from our course, this would be denoted

If this rate of change is positive, you’re speeding up; if it’s negative, you’reslowing down. If it’s zero, your speed isn’t changing at that instant.

Good news: this works (theoretically).

Bad news: in practice, this is challenging to compute (the norm in the speedcomputation leads to unwieldy square roots).

289

5.2. New ideas

2. Do some thinking coming from physics. Suppose a bird flies due northeast(so its velocity is something like v = (1, 1). If its experiences a force (causingacceleration a) due to some wind, will that wind cause the bird’s speed toincrease or decrease?

To measure whether the vectors v and a have an acute or obtuse angle be-tween them, compute the following quantity:

If this quantity is positive, then the bird is speeding up; it it’s negative, thebird is slowing down.

Summary: We’ve found two quantities, both of which are positive when you’respeeding up and negative when you’re slowing down:

d

dt||v(t)|| and v(t) · a(t) = x′(t) · x′′(t).

Question: What’s the connection between v(t) · a(t) and ddt||v(t)||? We know they

should have the same sign. Do they actually have the same sign? Are they equal?If not, what’s the difference?

290

5.2. New ideas

A short detour: functions R→ Rm of constant normEXAMPLE 8

Suppose f : R→ R2 is given by f(t) = (cos t, sin t).

1. Sketch the image of f .

2. Compute f(0) and f ′(0), and draw them on the sketch you made in part (a).

3. Compute f(

2π3

)and f ′

(2π3

), and draw them on the sketch you made in part

(a).

4. What property seems to hold between f(t) and f ′(t) for all t?

It turns out that what you observed in part (4) of the preceding example is not acoincidence: if f is any function whose image consists of points with constant norm(i.e. the image of f lies entirely on a circle or sphere centered at the origin), thenf(t) and f ′(t) must be orthogonal. Here’s a theorem that makes this more precise:

291

5.2. New ideas

Theorem 5.6 Let f : R→ Rm be C1. Then

||f(t)|| is constant ⇐⇒ for all t, f(t) ⊥ f ′(t).

PROOF (⇒) Suppose ||f(t)|| = c for all t. Then f(t) · f(t) = c2 for all t, so

0 = d

dtc2 = d

dt(f(t) · f(t))

= f ′(t) · f(t) + f(t) · f ′(t) (by the Dot Product Rule)= 2f(t) · f ′(t).

Therefore f(t) · f ′(t) = 0, so f(t) ⊥ f ′(t) for all t.

(⇐) Suppose f(t) ⊥ f ′(t) for all t. Then by reversing the calculations doneabove,

d

dt(f(t) · f(t)) = 0 (5.1)

for all t. Integrate both sides of (5.1) to see that for all t,

f(t) · f(t) = c

where c is a constant. Thus ||f(t)|| =√c for all t.

Unit tangent and principal unit normal vectors

To see what this has to do with velocity, speed and acceleration, we’re going todefine a couple of functions which, by their very definitions, have constant norm:

Definition 5.7 Let x : R→ Rm be C2.

1. The unit tangent vector to x at time t is the normalized version of the velocityat time t, i.e.

T(t) = x′(t)||x′(t)|| = v(t)

||v(t)|| .

2. The principal unit normal vector to x at time t, denoted N(t), is defined by

N(t) = T′(t)||T′(t)|| .

292

5.2. New ideas

EXAMPLE 9Suppose x(t) = (cos 2t, t, sin 2t). Find the unit tangent vector and principal unitnormal vector to x at time 0.

Solution: x′(t) = (−2 sin 2t, 1, 2 cos 2t) so the velocity at t = 0 is x′(0) = (0, 1, 2).Therefore the unit tangent vector is

T(0) = (0, 1, 2)||(0, 1, 2)|| =

(0, 1√

5,

2√5

).

To compute the principal unit normal vector, we need to first compute T(t) for ageneral t.

T(t) = x′(t)||x′(t)|| = (−2 sin 2t, 1, 2 cos 2t)

||(−2 sin 2t, 1, 2 cos 2t)||

= (−2 sin 2t, 1, 2 cos 2t)√4 sin2 2t+ 1 + 4 cos2 2t

= (−2 sin 2t, 1, 2 cos 2t)√4 + 1

=(−2√

5sin 2t, 1√

5,

2√5

cos 2t).

That means T′(t) =(−4√

5 cos 2t, 0, −4√5 sin 2t

), so T′(0) =

(−4√

5 , 0, 0). Finally,

N(0) = T′(0)||T′(0)|| =

(−4√

5 , 0, 0)

||(−4√

5 , 0, 0)||

= (−1, 0, 0) .

Lesson from this example: Computing principal unit normal vectors by hand is apain. Here’s the general flowchart for computing these things:

x(t)ddt // x′(t)

||·||

divide// T(t)

ddt // T′(t)

||·||

divide//N(t)

||x′(t)||

::

||T′(t)||

::

(Fortunately, you don’t have to do this much.)

More complex flowcharts, incorporating some other quantities, will appear atthe end of this section.

293

5.2. New ideas

Remarks on unit tangent and principal unit normal vectors:

• ||T(t)|| = 1 for all t.

• Therefore, by Theorem 5.6, T′(t) ⊥ T(t) for all t (i.e. N(t) ⊥ T(t)).

• Since T(t) and v(t) are parallel, T′(t) ⊥ v(t) for all t.

• This means that T′(t) (and also N(t)) must lie normal to the image of x atx(t).

• ||N(t)|| = 1 for all t.

So we have to have a picture like this:

Let’s bring acceleration into the picture:

a(t) = d

dtv(t)

= d

dt

(||v(t)|| v(t)

||v(t)||

)

= d

dt( ||v(t)||T(t) )

=(d

dt||v(t)||

)T(t) + T′(t)||v(t)|| (by the Scalar Product Rule)

=(d

dt||v(t)||

)T(t) + ||v(t)|| ||T′(t)|| T′(t)

||T′(t)||

=(d

dt||v(t)||

)T(t) + ||v(t)|| ||T′(t)||N(t)

=(

rate of changeof the speed

)T(t) +

(nonnegative

scalar

)N(t).

294

5.2. New ideas

We have just proven this theorem:

Theorem 5.8 (Tangential and normal components of acceleration, version 1)Let x : R→ Rm be C2. The acceleration a(t) = x′′(t) can be decomposed as follows:

a(t) = aTT(t) + aNN(t)

where the scalars aT and aN satisfy

aT (t) = d

dt||v(t)|| and aN(t) = ||T′(t)|| ||v(t)||

aT (t) is called the tangential component of the acceleration, and aN(t) is called thenormal component of the acceleration.

Remarks:

1. aT and aN are scalar-valued functions of t. In other words,

aT : R→ R and aN : R→ R

2. Notice that aT (t) is the derivative of the speed. So:

• if aT (t) > 0, the object is speeding up;• if aT (t) < 0, the object is slowing down;• if aT (t) = 0, the object’s speed is not changing at that instant.

3. aN(t) is always non-negative (since it is the product of two norms).

Corollary 5.9 (Tangential and normal components of acceleration, version 2)Let x : R → Rm be C2. Then the tangential and normal components of accelerationsatisfy the following formulas:

1. aT (t) = a(t) ·T(t);

2. aN(t) = a(t) ·N(t);

3. (Pythagorean Theorem for acceleration) [aT (t)]2 + [aN(t)]2 = ||a(t)||2.

PROOF The first statement is is a direct calculation:

a(t) ·T(t) = (aTT(t) + aNN(t)) ·T(t) (from Theorem 5.8)= aT (T(t) ·T(t) + aN(N(t) ·T(t))= aT ||T(t)||2 + an(0) (since T(t) ⊥ N(t))= aT (1)= aT .

295

5.2. New ideas

The second statement is left as a HW exercise.

For the third statement, notice

||a(t)||2 = ||aT (t)T(t) + aN(t)N(t)||2

= (aT (t)T(t) + aNN(t)) · (aT (t)T(t) + aN(t)N(t))

= [aT (t)]2||T(t)||2 + 2aT (t)aN(t)(T(t) ·N(t)) + [aN(t)]2||N(t)||2

= [aT (t)]2(1) + 2aT (t)aN(t)(0) + [aN(t)]2(1)

= [aT (t)]2 + [aN(t)]2.

EXAMPLE 10Suppose, as in Example 9, that x(t) = (cos 2t, t, sin 2t). Find the tangential and nor-mal components of acceleration at time 0.

What we computed in Example 9:

x′(t) = (−2 sin 2t, 1, 2 cos 2t) x′(0) = (0, 1, 2)

T(0) =(

0, 1√5,

2√5

)N(0) = (−1, 0, 0)

New work: Let’s compute the acceleration at time 0:

a(t) = x′′(t) = (−4 cos 2t, 0,−4 sin 2t)⇒ a(0) = (−4, 0, 0).

Having computed all that stuff, we see from Corollary 5.9 that

aT (0) = a(0) ·T(0) = (−4, 0, 0) ·(

0, 1√5,

2√5

)= 0 ;

aN(0) = a(0) ·N(0) = (−4, 0, 0) · (−1, 0, 0) = 4 .

BUT: We remarked earlier that it is hard to compute N(t) in general. How mightyou compute aN without computing N(t)?

296

5.2. New ideas

Theoretical consequences of this stuff

1. Corollary 5.9 justifies our earlier hypothesis that a(t) · v(t) and aT = ddt||v(t)||

have the same sign. But these are not the same quantity, they are off by afactor of ||v(t)|| (the speed):

d

dt||v(t)|| = aT = a(t) ·T(t) = a(t) · v(t)

||v(t)|| .

2. Since a is a scalar times T plus a scalar times N(t), we see that all of thesevectors must be in the same plane:

v(t) T(t) T′(t) N(t) a(t)More generally, we always have a picture like this:

This picture explains why we call [aT (t)]2 +[aN(t)]2 = ||a(t)||2 the “Pythagoreantheorem for acceleration”.

Summary of Sections 5.1-5.2

Flowchart of computational methods developed so far

x(t)ddt //

x(b)−x(a)

v(t)||·|| %%

ddt //

∫ ba

a(t)dot

22

||·|| // ||a(t)||

$$||v(t)|| divide //

ddt

66

∫ ba

T(t)""

ddt $$

aT (t)a2T+a2

N=||a||2 %%aN(t)

displacement T′(t)||·||

divide//N(t)

dot::

arc length ||T′(t)||

99

297

5.2. New ideas

List of quantities we have defined so far

Quantity Vector orscalar? Description

x(t) vector position at time tv(t) = x′(t) vector velocity at time t

(tangent to image of x at x(t);points in direction of increasing t)

||v(t)|| scalar speed at time ta(t) = x′′(t) vector acceleration at time t

(points to inside of image of x at x(t))T(t) vector unit tangent vector at time t

(normalized velocity vector)N(t) vector principal unit normal vector at time t

(orthogonal to v;points directly to inside of image of x at x(t))

aT (t) scalar tangential component of acceleration(rate of change of speed)

aN(t) scalar normal component of acceleration(what does this describe?)

To get more understanding of aN(t), we will need some other ideas that willlead, among other places, to formulas for aN(t) that we derive in the next two sec-tions.

Some intuition:

Enrichment: If you think there are a lot of letters here, you should be awarethat I am actually leaving some letters out. (Also, as a heads-up, there are acouple of other letters still to come.) A couple of the letters I’m not discussingin the notes will be mentioned in the homework. Others will not be mentionedat all; look up “Frenet-Serret formulas” on the internet if you are interested inlearning more.

298

5.3. Arc length parametrization

5.3 Arc length parametrizationNon-uniqueness of parametrizations

So far, we have been starting with a function x : R→ Rm and studying the motionof an object whose position at time t is x(t).

Now, in part to get a handle on what aN(t) is measuring, we want to think aboutthe path such an object travels solely as a geometric object (ignoring the impact ofthe passage of time), and study the purely geometric properties of this path (forexample, how “curvy” is it?):

Definition 5.10 A path (a.k.a. curve) γ (“gamma”) is a subset of Rm which is theimage of a function x : R→ Rm.

Given a path γ ⊆ Rm, a function x : R → Rm whose path is γ is called aparametrization of γ.

WARNING:

In Definition 5.10, I am lying about what a “path” is. I will “un-lie” to youin Chapter 8.

t=2

t=1 t=0

t=-1

t=-2 γ

EXAMPLE 11Let γ be the circle of radius 2 centered at the origin in R2. Find a parametrizationof γ.

A natural question is to ask is how many parametrizations a certain path has.By definition, a path has at least one parametrization.

299


EXAMPLE 12Let γ be the path associated to the function x : R→ R2 defined by x(t) = (t2− 4, t).Here is a table of values for this function, and a picture of γ:

t x(t)−3 (5,−3)−2 (0,−2)−1 (−3,−1)0 (−4, 0)1 (−3, 1)2 (0, 2)3 (5, 3)

γ

t=-3t=-2

t=-1t=0

t=1t=2

t=3

-6 -4 -2 2 4 6

-4

-2

2

4

Continuing with this example, suppose we were also given the function y :R→ R2 defined by y(u) = (4u2 − 4, 2u). Here is a table of values for this function,and a picture of its path:

u y(u)−32 (5,−3)−1 (0,−2)−12 (−3,−1)0 (−4, 0)12 (−3, 1)2 (0, 2)32 (5, 3)

γ

u=-3/2u=-1

u=-1/2u=0

u=1/2u=1

u=3/2

-6 -4 -2 2 4 6

-4

-2

2

4

Suppose x and y describe the movement of two particles. What are the similar-ities and differences between the ways these particles move?

300


Recall from the previous page that x(t) = (t2 − 4, t) and y(u) = (4u2 − 4, 2u).To explore the connection between these formulas, let φ(u) = 2u. Then when t =φ(u) = 2u, objects whose respective motions are described by x and y are at thesame spot, because

Now let’s analyze the relative speed of the objects whose positions are x(t) andy(u), where t = 2u:

speed of object whose position speed of object whose positionis given by x at time t = 2u is given by y at time u

Note that by the Chain Rule,

y′(u) = Dx(φ(u)) = Dx(φ(u))Dφ(u) = Dx(t)Dφ(u) = x′(t)φ′(u),

so the ratio of the two objects’ speeds is

speed of second objectspeed of first object

= ||y′(u)||

||x′(t)|| = ||x′(t)|| |φ′(u)|||x′(t)|| = |φ′(u)| =

This means the second object goes twice as fast as the first.

301


Continuing with Example 11 where x(t) = (t2 − 4, t), let’s consider z(u) =x(−u) = (u2 − 4,−u):

u x(−u)−3 (5, 3)−2 (0, 2)−1 (−3, 1)0 (−4, 0)1 (−3,−1)2 (0,−2)3 (5,−3)

-γ

u=3u=2

u=1u=0

u=-1u=-2

u=-3

-6 -4 -2 2 4 6

-4

-2

2

4

Note that in this setting, if we let φ(u) = −u, then z(u) = x(−u) = x(φ(u)),so this is similar to the setup on the previous pages. This time, φ′(u) = −1 so|φ′(u)| = 1 so the objects travel at the same speed. But since φ′(u) < 0, the di-rection of motion is reversed. We therefore call this an “orientation-reversingreparametrization” of γ, and technically, the path isn’t γ any more, its called “−γ”(more on this in Chapter 8, if we get that far).

For something more complicated (still building on Example 11, where x(t) =(t2 − 4, t), let’s consider w(u) = x(u3

12 ) =(u6

144 − 4, u3

12

):

u w(u)−3 (1.0625,−2.25)−2 (−3.555,−.6666)−1 (−3.993,−0.8833)0 (−4, 0)1 (−3.993, 0.8833)2 (−3.555, .666)3 (1.0625, 2.25)

γu=3

u=2

u=0u=-2

u=-3

-6 -4 -2 2 4 6

-4

-2

2

4

In this setting, if we let φ(u) = u3

12 , then w(u) = x(φ(u)). But this time, the ratioof the speeds of the objects is |φ′(u)| =

∣∣∣3u2

12

∣∣∣ = u2

4 . This means:

• when u is close to zero,

• when u is far from zero,

302


Theorem 5.11 (Non-uniqueness of parametrizations) Suppose γ is the path ofsome function x : R → Rm. Let φ : R → R be any function which is which isone-to-one and onto. Then:

1. the function x φ : R→ Rm is also a parametrization of γ.

• If φ is increasing, then x φ traces γ in the same direction as x;• if φ is decreasing, then x φ traces γ in the opposite direction as x.

2. If two objects are moving so that their respective positions are x(t) and (xφ)(t),then the objects travel along the same path. Furthermore, when the objects areat the same position,

speed of second object = |φ′(t)|(speed of first object).

Consequence: a curve never has only one function which describes it.

Special cases: if γ is the path of x(t), then for any k 6= 0, x(kt) also has path γ:• if k > 0, x(kt) traces γ out in the same direction as x(t);• if k < 0, x(kt) traces γ out in the opposite direction as x(t);• x(t) and x(kt) are the same point when t = 0;• if |k| > 1, x(kt) traces γ out k times more quickly than x(t);• if |k| > 1, x( t

k) traces γ out 1

ktimes as fast as x(t) (i.e. more slowly);

• if |k| = 1, x(kt) traces γ out at the same speed as x(t).

EXAMPLE 13Find a parametrization of the circle x2 + y2 = 25 which traces the circle out coun-terclockwise with speed 7.

303


Arc length parametrization

Recall: We are thinking about properties of a path (like its “curviness” or “straight-ness”) that depend only on the geometry of the path, and not on the choice ofparametrization of the path. Let’s distinguish these two types of properties:

Definition 5.12 Given path γ, a property of that path is called intrinsic if it is thesame regardless of the choice of parametrization of the path. A property is called ex-trinsic if it might be different, depending on the choice of parametrization.

Morally speaking, intrinsic properties are “geometric” and extrinsic propertiesare “physical”.

Examples of intrinsic properties: Examples of extrinsic properties:tangent lines to γ position x(t)

unit tangent vectors T(t) velocity v(t)at each point of γ

principal unit normal vectors N(t) speed ||v(t)||at each point of γ

acceleration; aT (t) and aN(t)

arc length between two points on γ arc length between two times

displacement between two points on γ displacement between two times

Our next task is define a parametrization of a path γ which is itself somewhat“intrinsic”. The way we do this is to define a parametrization so that if the positionof a point is given by this parametrization, then the speed at which that pointmoves is always equal to 1.

Definition 5.13 Let γ ⊆ Rm. An arc length parametrization of γ is any parametriza-tion x : R→ Rm of γ such that ||x′(s)|| = 1 for all s. The independent variable of thisparametrization is called the arc length parameter of γ, and is usually denoted s.

Suppose you have a path γ parametrized by x(t), where the speed is constantand always equal to k. Then, it is easy to find an arc length parametrization:

304


EXAMPLE 14Let γ be the helix parametrized by x(t) = (5 cos t, 5 sin t, 12t). Find an arc lengthparametrization of γ.

Solution: Notice the speed, at all times t, is

||x′(t)|| =√

(−5 sin t)2 + (5 cos t)2 + 122 =√

25 + 144 = 13.

Therefore, to make the speed 1 instead of 13, we need to parametrize γ by

Let’s check this solution: To be a valid arc length parametrization, the speed mustbe 1 at all s. For our solution x(s), we have

||x′(s)|| =∣∣∣∣∣∣∣∣(− 5

13 sin s

13 ,513 cos s

13 ,1213

)∣∣∣∣∣∣∣∣=√

25169 sin2 s

13 + 25169 cos2 s

13 + 144169

=√

25169 + 144

169=√

1= 1.

But what if the speed isn’t constant? Does an arbitrary path γ have an arc lengthparametrization?

305


Theorem 5.14 Let γ ⊆ Rm be a path, parametrized by the C1 function x : R→ Rm,where x′(t) 6= 0 for all t. Let p be any point on γ, and let t0 ∈ R be such thatx(t0) = p. Given all this, we define the function s : R→ R given by

s(t) =∫ t

t0||x′(u)|| du.

Then, the function s 7→ x(t(s)) is an arc length parametrization of γ.

If you’re wondering what the heck is going on here, don’t panic (this idea is reallydeep). There will be a couple of pictures coming up to help explain this. For now,I’m going to run through some theoretical properties of this machinery (frankly,it’s more important to understand the properties than it is to understand exactlywhat the parametrization is).

Quick remark: The u is just a dummy variable used to define the function s(we can’t use t there, since t is the upper limit of the integral). Don’t worry aboutit too much.

Theorem 5.15 (Properties of arc length parametrization) Let γ ⊆ Rm be a path,parametrized by the C1 function x : R → Rm, and let s = s(t) be the arc lengthparameter as in Theorem 5.14. Then:

1. The derivative of s(t) is the speed associated to parametrization x, i.e.

ds

dt= ||x′(t)|| = ||v(t)||.

2. s(t) is a strictly increasing function of t, i.e. s(t1) < s(t2) whenever t1 < t2.

3. s is an invertible function, i.e. it is theoretically possible to solve for t in termsof s.

PROOF Statement (1) follows from the Fundamental Theorem of Calculus (FTC):

ds

dt= d

dt(s(t)) = d

dt

(∫ t

t0||x′(u)|| du

)FTC= ||x′(t)||.

Since by statement (1) the derivative of s(t) is the speed, and by hypothesis thespeed is never zero. Therefore ds

dt> 0 for all t, so s must be an increasing function

of t. This proves (2).

306


s(t)

t

s

Since s is a strictly increasing function of t, for every s there is exactly one t suchthat s(t) = s. Thus you can theoretically find t in terms of s, proving (3).

PROOF OF THEOREM 5.14 We have to show that the parametrization s 7→ x(t(s)),where s(t) =

∫ tt0||x′(u)|| du, has speed 1. By (3) of Theorem 5.15, we can think of t

as a function of s. Therefore

x′(t) = dxdt

Chain Rule= dxds

ds

dt= x′(s)||x′(t)||.

Divide both sides of the preceding statement by ||x′(t)|| to get

x′(s) = x′(t)||x′(t)||

so x′(s) has length 1, since it is a normalized vector. Thus s 7→ x(t(s)) is an arclength parametrization, as wanted.

There is a connection between arc length parametrization and the tangentialcomponent of acceleration:

Corollary 5.16 (Tangential component of acceleration, version 3) Let γ ⊆ Rm

be a path, parametrized by the C1 function x : R → Rm, and let s = s(t) be the arclength parameter as in Theorem 5.14. Then the tangential component of accelerationsatisfies aT (t) = d2s

dt2.

PROOF Apply two previous results:

aT (t) = d

dt(||v(t)||) (from Theorem 5.8)

= d

dt

(ds

dt

)(from (1) of Theorem 5.15)

= d2s

dt2.

307


What is arc length parametrization, exactly?

1

t=3

t=2t=1 t=0

t=-1

t=-2 γ

1

s=4

s=3s=2

s=1 s=0s=-1

s=-2s=-3

s=-4 γ

s inverse of s(t)

//

x(t(s))

speed 1

::t

xspeed ||x′(t)||

//

s(t)=∫ tt0||x′(u)|| du

oo x(t)↔ γ

Bad news/good news: arc length parametrizations are usually impossible toactually compute. (That’s sort of good news, since that means I can’t ask you to dovery many of them!) Here’s an example showing you why:

EXAMPLE 14Find an arc length parametrization of the path which is the image of x(t) = (t, t2, 2

3t3).

Solution: Let’s choose p = (0, 0, 0) so that t0 = 0. Then:

x′(t) = (1, 2t, 2t2)⇒ ||x′(t)|| =√

1 + 4t2 + 4t4 =√

(1 + 2t2)2 = 1 + 2t2

⇒ s(t) =∫ t

t0||x′(u)|| du =

∫ t

0(1 + 2u2) du = t+ 2

3t3.

Next, we would have to find the inverse of s(t) = t+ 23t

3, i.e. we have to back-solves = t+ 2

3t3 for t.

Unfortunately, this happens in general: usually, you can’t explicitly computet as a nice function of s, so we can’t explicitly write down a formula for the arclength parametrization x(t(s)).

The good thing about arc length parametrizations are that they exist for anyreasonable path, and they have the theoretical properties listed in Theorems 5.15and have the connection with aT (t) given in Corollary 5.16. This stuff will be ap-plied in the next section.

308

5.4. Curvature

5.4 CurvatureGoal: Define an intrinsic quantity of a path γ which measures how “straight”

or “curvy” γ is. This quantity will be called κ,the Greek letter “kappa”.

Initial thoughts:

x(t)

y(t)

z(t)

How might we measure this?

Idea 1 (bad): Use aN(t), the normal component of acceleration.

Problem:

Idea 2 (bad): Look at the rate of change of with respect to .

Problem:

Idea 3 (better): Look at the rate of change of with

respect to .

Problem:

Fix the problem of idea 3 (perfect): .

Definition 5.17 Given a path γ, the curvature κ of γ is the norm of the rate of changeof the unit tangent vector T, with respect to unit length along the curve:

κ =∣∣∣∣∣∣∣∣∣∣dTds

∣∣∣∣∣∣∣∣∣∣ .

309

5.4. Curvature

Remarks on this definition:

1. As wanted, curvature is a non-negative scalar.

2. As wanted, curvature is an intrinsic property.

3. In general, the curvature is non-constant (it depends on the point on thecurve you are at). So κ is really something like κ(s) or κ(t).

4. The definition above does not give a useful method for actually computingcurvature. (We’ll address how you actually compute κ shortly.)

Connections with normal component of acceleration

Recall that our intuition about aN(t) is that it had something to do with how muchthe path bends at x(t) (i.e. it should be related somehow to curvature). Formula(2) in the theorem below makes that connection precise. More importantly, we canuse this connection to get a nice formula for aN(t) in dimension 3:

Theorem 5.18 (Normal component of acceleration, version 3) Let x : R→ Rm

be C2. Then the normal component aN(t) of the acceleration satisfies:

1. aN(t) = ||T′(t)|| ||v(t)||;

2. aN(t) =(dsdt

)2κ(t) = ||v(t)||2κ(t)

(i.e. aN(t) is the curvature, times the speed squared);

3. if m = 3, then aN(t) = ||T(t)× a(t)|| = ||v(t)×a(t)||||v(t)|| .

PROOF Statement (1) was derived in Theorem 5.8.

To establish statement (2), rewrite statement (1) as follows:

aN(t) = ||T′(t)|| ||v(t)||

=∣∣∣∣∣∣∣∣∣∣dTdt

∣∣∣∣∣∣∣∣∣∣ dsdt

=∣∣∣∣∣∣∣∣∣∣dTds ds

dt

∣∣∣∣∣∣∣∣∣∣ dsdt (by the Chain Rule)

=∣∣∣∣∣∣∣∣∣∣dTds

∣∣∣∣∣∣∣∣∣∣ dsdt dsdt

= κ(t)(ds

dt

)2

= κ(t)||v(t)||2.

310

5.4. Curvature

To prove statement (3), note that from Corollary 5.16 and statement (2) of thistheorem, we have

a(t) = aT (t)T(t) + aN(t)N(t)

= d2s

dt2T(t) + κ(t)

(ds

dt

)2

N(t).

Take the cross product of both sides of this with T(t) (on the left) to get

T(t)× a(t) = T(t)×d2s

dt2T(t) + κ(t)

(ds

dt

)2

N(t)

= d2s

dt2[T(t)×T(t)] + κ(t)

(ds

dt

)2

[T(t)×N(t)]

= κ(t)(ds

dt

)2

[T(t)×N(t)] .

Take norms of each side of the above to get

||T(t)× a(t)|| =

∣∣∣∣∣∣∣∣∣∣∣∣κ(t)

(ds

dt

)2

[T(t)×N(t)

∣∣∣∣∣∣∣∣∣∣∣∣

= |κ(t)|

∣∣∣∣∣∣(ds

dt

)2∣∣∣∣∣∣ ||T(t)×N(t)||

= κ(t)(ds

dt

)2

||T(t)|| ||N(t)||∣∣∣∣sin π2

∣∣∣∣= κ(t)

(ds

dt

)2

(1)(1)(1)

= κ(t)(ds

dt

)2

which is equal to aN(t) by the second statement of this theorem.

311

5.4. Curvature

EXAMPLE 15Find the tangential and normal components of acceleration for the function x(t) =(e−t,√

2t, et).

Solution: Differentiate to get

v(t) = x′(t) =(−e−t,

√2, et

)||v(t)|| =

√e−2t + 2 + e2t FACTOR=

√(e−t + et)2 = e−t + et

a(t) = x′′(t) =(e−t, 0, et

)Then:

aT (t) =

aN(t) =

EXAMPLE 16Find the tangential and normal components of acceleration for the function x(t) =(t2 − 4t, 2t− 1, t+ 3) when t = 1.

Solution: Differentiate to get

v(t) = x′(t) = (2t− 4, 2, 1)⇒ v(1) = (−2, 2, 1)

||v(t)|| =√

(2t− 4)2 + 4 + 1 =√

(2t− 4)2 + 5⇒ ||v(1)|| =

√4 + 4 + 1 = 3

a(t) = x′′(t) = (2, 0, 0)Then:

aT (t) = d

dt||v(t)|| = d

dt

(√(2t− 4)2 + 5

)= 1

2√

(2t− 4)2 + 5(2(2t− 4)2)

⇒ aT (1) = 12√

(2(1)− 4)2 + 5(2(2(1)− 4)2) = 1

2(3)(2(−2)2) = −43 .

aN(1) = ||v(1)× a(1)||||v(1)|| = ||(−2, 2, 1)× (2, 0, 0)||

3 = ||(0, 2,−4)||3 =

√203 .

312

5.4. Curvature

Computing curvature

Theorem 5.19 (Curvature formulas) Given a C2 path γ parametrized by x : R→Rm, the curvature κ =

∣∣∣∣∣∣dTds

∣∣∣∣∣∣ satisfies:

1. κ(t) = ||T′(t)||||v(t)|| ;

2. if m = 3, then

κ(t) = ||v(t)× a(t)||||v(t)||3 ;

3. if γ is the graph of y = f(x) in R2, then the curvature of γ at (x, f(x)) is

κ(x) = |f ′′(x)|(1 + [f ′(x)]2)3/2 .

PROOF For statement (1), notice that statement (2) of Theorem 5.18 says

aN(t) = ||v(t)||2κ(t) ⇒ κ(t) = aN(t)||v(t)||2 .

Now statement (1) of Theorem 5.18 says aN(t) = ||T′(t)|| ||v(t)||; plugging this into the above statement gives

κ(t) = ||T′(t)|| ||v(t)||||v(t)||2 = ||T

′(t)||||v(t)|| .

Next, we prove statement (2). From Theorem 5.18, we have

κ(t) ||v(t)||2 = κ(t)(ds

dt

)2

= aN(t) (by statement (2) of Theorem 5.18)

= ||v(t)× a(t)||||v(t)|| (by statement (3) of Theorem 5.18).

Dividing through both sides by ||v(t)||2 gives statement (2).

To prove the last statement, we can represent the path in 3-D space by artifi-cially inserting a z-coordinate which is always zero: let x : R → R3 be defined byx(t) = (t, f(t), 0). Then the image of x is the graph of f , in the xy-plane. Nowv(t) = (1, f ′(t), 0) and a(x) = (0, f ′′(t), 0) so by statement (2), the curvature is

κ(t) = ||v(t)× a(t)||||v(t)||3 .

Substituting in the expressions for v and a and working this out, we get the for-mula of statement (3) (actually working this out is a HW problem).

313

5.4. Curvature

EXAMPLE 17Compute the curvature of the image of x(t) = (e−2t, 2t, 4), at the point (1, 0, 4).

Solution: Notice first that that the point (1, 0, 4) occurs when t = 0. Differentiateto find the velocity and the acceleration at t = 0:

v(t) = x′(t) = (−2e−2t, 2, 0)⇒ v(0) = (−2, 2, 0)a(t) = x′′(t) = (4e−2t, 0, 0)⇒ a(0) = (4, 0, 0)

Then

κ(0) =

EXAMPLE 18Compute the curvature of a circle of radius r.

EXAMPLE 19Compute the curvature of f(x) = ex + e−x at (0, 2).

Solution: Use formula (3) of Theorem 5.19, with x = 0:

κ(0) = |f ′′(0)|(1 + [f ′(0)]2)3/2 =

314


5.5 Summary of Chapter 5• Given a function x : R → Rm, the image of x is called a path; x(t) should be

thought of as the position of some object at time t.

• Given a path γ ⊆ Rm, any function x : R → Rm whose path is γ is called aparametrization of γ.

• An arc length parametrization of γ is a parametrization of γ that has constantspeed 1. One such parametrization is given by s 7→ x(t(s)), where x is anyparametrization with x(t0) = p ∈ γ and

s(t) =∫ t

t0||x′(u)|| du.

Given this setup, we have s′(t) = dsdt

= ||x′(t)|| = ||v(t)||.

• Properties of a path which depend on the parametrization are called extrinsicproperties; properties of a path which do not depend on the parametrizationare called intrinsic properties.

• Key formulas for all properties of a path γ parametrized by x are given below(any formulas with cross products are only valid if m = 3).

Extrinsic properties

position x(t)

velocity v(t) = x′(t)

speed ||x′(t)||

acceleration a(t) = x′′(t) = aT (t)T(t) + aN(t)N(t), where

aT (t) = d

dt||v(t)|| = d2s

dt2= a(t) ·T(t)

aN(t) = ||T′(t)|| ||v(t)|| =(ds

dt

)2

κ(t) = ||v(t)× a(t)||||v(t)||

||a(t)||2 = [aT (t)]2 + [aN(t)]2

arc length from t = a to t = b:∫ ba ||v(t)|| dt

displacement from t = a to t = b: x(b)− x(a) =∫ ba v(t) dt

315


Intrinsic properties

unit tangent vector T(t) = v(t)||v(t)||

principal unit normal vector N(t) = T′(t)||T′(t)||

Note: N(t) ⊥ T(t)

curvature κ(t) =∣∣∣∣∣∣dTds

∣∣∣∣∣∣ = aN (t)||v(t)||2 = ||T′(t)||

||v(t)|| = ||v(t)×a(t)||||v(t)||3

Note: The curvature of y = f(x) at (x, f(x)) is |f ′′(x)|(1+[f ′(x)]2)3/2 .

Generic picture

316



1. Suppose that an object is moving in a plane, so that its position at time t isgiven by x(t) = (cos2 t,− sin2 t).

a) Find the velocity and acceleration of the object at time t.

b) Find the velocity and acceleration of the object at time π3 .

c) Find the speed of the object at time π4 .

2. A bug is flying in the air so that its position at time t is given by x(t) =(t− t3, t, t+ t3).

a) Find the bug’s acceleration at time 1.

b) How fast is the bug flying at time 2?

c) At what time(s), if any, is the bug’s velocity parallel to (22,−2,−26)?

d) Find the displacement of the bug from time 1 to time 2.

3. Suppose an object is moving in R3 so that its position at time t is given byx(t) = (2e−t, 2e2t − e−t, 3et). Find the position, velocity and acceleration ofthe object when t = 0, and sketch a picture of what the path the object travelslooks like near t = 0 (similar to what was done in Example 1).

4. Repeat the directions of the previous problem at t = π2 , if the object’s position

is given by x(t) = (sin 2t, cos 3t, sin t).

5. In each part of this problem, you are given a picture of the image of somefunction x : R → R3. At each the indicated points A, B and C, sketch x′ andx′′ at that point.

a)

x(t)A

B

C

b)

x(t)

A

B

C

6. Find the length of the curve x(t) =(t√

2, et, e−t)

from t = 0 to t = 2.

7. An object is moving in a plane so that its position at time t is (t4 + 2t−2, 8t).

317


a) Find the displacement of the object from time 1 to time 4.

b) Find the distance the object travels from time 1 to time 4.

8. An object is moving in 3D space so that its position at time t is x(t). Supposeyou know that x(0) = (0, 0, 0) and x(1) = (4, 3, 6). What are the possiblevalues for the arc length of x from t = 0 to t = 1?

9. Suppose that an object is moving in R3 so that its speed at time t is 4t + 12t2.Find the distance travelled by the object from time −2 to time 1.

10. a) Use the arc length formula presented in Definition 5.3 to show that for afunction f : R→ R, the length of the graph of f from x = a to x = b is∫ b

a

√1 + [f ′(x)]2 dx.

b) Use the formula given in part (a) to find the length of the graph of y =16x

3 + 12x−1 from x = 1 to x = 4.

11. a) Write an integral which gives the arc length of x(t) = (cos 2t, sin t −cos t, sin 3t) from t = 0 to t = π

4 .

b) Use Mathematica to obtain a decimal approximation to the integral youwrote down in part (a).

12. An object is moving in R3 so that its acceleration at time t is a(t) = (8e2t, 2e−t, et−et/2). If its position at time 0 is (6, 2,−1) and its velocity at time 0 is (4, 2,−1),find the object’s position at time 1.

13. Compute each integral:

a)∫ (

2t3, t−2, 4t−1)dt b)

∫(8 sin 4t, 12 cos 2t) dt

14. Compute each integral:

a)∫ 2

1

(t(t+ 1), t+ 1

t

)dt b)

∫ π/2

0(12 cos 3t, 6 sin 2t) dt

15. Compute this integral:∫ 1

0

(et + e−t, e2t − e−t/2, et/2 − 2e−t

)dt

16. A boy holds a garden hose 4 feet above the ground. He points the nozzle ofthe hose at an angle of 60 to the horizontal. Suppose that water exists thehose at an initial velocity of 48 ft/sec.

318


a) Find the velocity of a water molecule 1 second after it leaves the hose.

b) Find the speed of a water molecule 1 second after it leaves the hose.

c) What is the maximum height achieved by the stream of water comingout of the hose?

d) Find the time it takes for a water molecule to return to a height of 4 feetabove ground.

e) If the boy’s sister doesn’t want to get hit by the water, how far awayfrom her brother would she have to stand?

17. A frisbee is thrown in the air (from ground level) with an initial velocity of21 ft/sec, in the direction (1, 3, 2). Assume that the spin put on the frisbeewhen it is thrown causes the frisbee to experience acceleration (−1,−2, 0)throughout its flight. Assume also that gravity acts on the frisbee. At whatpoint does the frisbee land?


18. In each part of this problem, you are given a picture of the image of somefunction x : R → R3. At each indicated point, sketch the unit tangent vectorT and principal unit normal vector N to x at that point.

a)1

x(t)

BA

b)1

x(t)

B

A

19. Let x(t) = (et cos 4t, et sin 4t).

a) Compute the unit tangent vector to x (at arbitrary t).

b) Compute the principal unit normal vector to x.

c) Compute the tangential component of acceleration to x.

d) Compute the normal component of acceleration to x.

20. Let x(t) = (t2, t3, 4t).

a) Compute T(2), aT (2) and aN(2).

319


b) If an object has position x(t) at time t, is the object speeding up or slow-ing down when t = 2? Explain your answer.

21. Compute the tangential and normal components of the acceleration of x(t) =(t, 2√

2t, ln t).

22. Suppose x : R→ R2 is such that T(0) =(√

64 ,−34 ,

14

). If at time 0, the speed of

an object whose position at time t is x(t) is 8, find x′(0).

23. Suppose x : R→ R3 is such that T(2) =(

13 ,

23 ,−23

), N(2) =

(23 ,−23 ,

13

), aT (2) =

−4 and aN(2) = 2. Find a(2).

24. Suppose x : R → R3 is such that x′(1) = (3,−2, 5), T′(1) = (−1, 4,−2),aT (1) = 2 and aN(1) = 3. Find x′′(1).

25. Suppose x : R→ R3 is such that x′(−3) = (3, 4, 0) and x′′(−3) = (4, 7, 3). FindaT (−3) and aN(−3).

26. Suppose x : R → Rm is such that ||x′(t)|| = 4t2 + 9. Find the tangentialcomponent of acceleration when t = 5.

27. Prove the second statement of Corollary 5.9 , which says that for any C2 func-tion x : R→ Rm, aN(t) = a(t) ·N(t).

28. (R) Given x : R → R3, the binormal vector to x at time t is defined to beB(t) = T(t)×N(t). Compute the binormal vector to x(t) = (4t, 2 sin 3t, 2 cos 3t).

29. (F) Given x : R → R3, the osculating plane to x at time t is the plane in R3

that best contains the image of x at time t. This plane is defined as the planecontaining the point x(t) which has direction vectors T(t) and N(t).

a) In light of the previous problem, what is the relationship of the binormalvector to the osculating plane of the curve?

b) Find a normal equation of the osculating plane to the function x de-scribed in Problem 28 when t = π

4 .

c) Recall that the velocity vector x′(t) and acceleration vector x′′(t) are bothlinear combinations of T(t) and N(t). In light of this, it seems like itmight be easier to define the osculating plane as containing the pointx(t) and having direction vectors x′(t) and x′′(t) (since these are easier tocompute than T(t) and N(t)). But this is potentially problematic. Why?

320



30. Suppose that γ ⊆ Rm is some curve parametrized by x : R → Rm. Supposethat at time t = 8, the speed of the parametrization x is 1

3 . Let φ(t) = 2t3 + 6t.This information is sufficient to find the speed of the parametrization x φ ofγ at a certain value of t.

a) At what value of t do we know the speed of x φ?

b) What is the speed of x φ at that time?

31. Find a parametrization of the circle x2 +y2 = 36, so that the circle is traversedcounterclockwise with speed 4.

32. Find a parametrization of the circle x2 +y2 = 16, so that the circle is traversedclockwise with speed 3.

33. Find a parametrization of the line in R3 passing through (5,−2,−2) and(1, 4, 2), so that the speed of the parametrization is 2.

34. Let γ be the line in R3 passing through x1 = (1, 3,−2) and x2 = (4, 0, 1).

a) Find a parametrization f of γ for which f(0) = x1 and f(1) = x2.

b) Find a parametrization g of γ for which g(0) = x2 and g(1) = x1.

c) Compute the arc length parametrization of γ corresponding to p = x1,starting with the parametrization f you wrote in part (a).

d) Compute the arc length parametrization of γ corresponding to p = x1,starting with the parametrization g you wrote in part (b).

35. Compute the arc length parametrization of the circle x2 + y2 = 64, startingwith the standard parametrization x(t) = (8 cos t, 8 sin t) and choosing p =x(0) = (8, 0).

36. Compute the arc length parameteization of x(t) = (e2t sin t, e2t, e2t cos t), withp = x(0) = (0, 1, 1).


37. Consider the ellipse x2

25 + y2

4 = 1.

a) Sketch a crude graph of this ellipse.

b) Based on the graph you drew in part (a), at what point(s) on the ellipsewould you say the curvature of the ellipse is maximized? Explain youranswer.

321


c) Based on the graph you drew in part (a), at what point(s) on the ellipsewould you say the curvature of the ellipse is minimized? Explain youranswer.

38. Finish the proof of statement (3) of Theorem 5.19 by showing that if x(t) =(t, f(t), 0), then

κ(t) = |f ′′(t)|(1 + [f ′(t)]2)3/2 .

39. Find the normal component of acceleration of x(t) =(t, 2√

2t, ln t)

using for-mula (3) of Theorem 5.18. Verify that you get the same answer you got whenyou computed this quantity in Problem 21.

40. Let x(t) = (t, 2t2, 4t3). Find aT (1) and aN(1).

41. Let x(t) =(

12 cos t, sin t,

√3

2 (1− sin t)). Find aT (t) and aN(t).

42. Suppose x : R → Rm is such that ||x′(3)|| = 8 and aN(3) = 5. Find thecurvature of x when t = 3.

43. Suppose y : R→ Rm is such that ||y′(3)|| = 10 and κ(3) = 2. Find the normalcomponent of acceleration of y when t = 3.

44. Suppose z : R → Rm is such that z′(−2) = (3, 1,−5) and z′′(−2) = (1, 0, 7).Find the normal component of acceleration of z when t = −2.

45. Prove that the curvature of any line in R3 is zero.

Hint: The parametric equations of any line in R3 are x(t) = p + tv. Start withthis, and compute the curvature using the usual formulas.

46. Compute the curvature of x(t) = (cos3 t, 1, sin3 t).

47. Let x(t) =(3t,√

4− 3t2, 2t). Compute κ(0) and κ(1).

48. Compute the curvature of x(t) = (3t2, 3t− t3).

49. Compute the curvature of the graph of f(x) = ex at (0, 1).

50. a) (R) Show that the point on the parabola y = x2 at which the curvature ismaximized is (0, 0).

b) (F) Based on the computation you did in part (a), make a conjecture asto the point on an arbitrary parabola where its curvature is maximized.

c) (F) Prove the conjecture you wrote down in part (b).

51. Use Mathematica to find a decimal approximation to the curvature of x(t) =(t arctan t, t2 arctan 2t, t3 arctan 3t) at the point where t = 1.

322


52. Suppose x : R → Rn gives an object’s position at time t, where the units ofdistance (or position) are meters, and the units of time are seconds. What isthe unit in which curvature is measured? Explain.

53. Suppose f : R → R is C2. If a is the x−coordinate of an inflection point of f ,what is the curvature of f at a? Explain.

54. (F) Let x : R→ R3. The osculating circle to x at time t is the circle in R3 thatbest contains the image of x for points near time t. See the picture below:

osculating circlex(t)

a) What point do you think should be on the osculating circle at time t?

b) In what plane do you think the osculating circle of a curve should lie?

c) If you draw a vector from x(t) to the center of the osculating circle, thisvector should be parallel to what unit vector?

d) If κ(t) is the curvature of x, what do you think the radius of the osculat-ing circle at time t should equal? Why?Hint: Look back at Example 18.

e) Based on your answers to parts (c) and (d), what is the center of theosculating circle to x at time t?

323


Selected answers

1. a) v(t) = (−2 cos t sin t,−2 cos t sin t);a(t) = (−2 cos2 t+ 2 sin2 t,−2 cos2 t+ 2 sin2 t)

c)√

2.

2. a) (−6, 0, 6)b)√

291d) (−6, 1, 8)

3. x(0) = (2, 1, 3); v(0) = (−2, 5, 3); a(0) = (2, 7, 3).

4. x(π2

)= (0, 0, 1); v

(π2

)= (−2, 3, 0); a

(π2

)= (0, 0,−1).

5. In this picture, the first derivatives are shown with solid arrows, and thesecond derivatives are shown with dashed green arrows:

a)

6. e2 − e−2

9. 30

10. b) 878

11. b) 2.08598

12. (4 + 2e2, 4 + 2e−1, 2− 4e1/2 + e)

13. a)(

12t

4, −1t, 4 ln t

)+ C

14. a)(

236 , 1 + ln 2

)16. a) (24,−32 + 24

√3) ft/sec

b) 16√

13− 6√

3 ft/sec

d) 32

√3 sec

17.(

945256 ,

1449128 , 0

)

324


18. In this picture, the unit tangent vectors are shown with solid arrows, and theprincipal unit normal vectors are shown with dashed green arrows:

b)1

19. b) 14√

17 (−16 cos 4t− 4 sin 4t, 4 cos 4t− 16 sin 4t)

c)√

17et

20. a) T(2) = 1√11(1, 3, 1)

b) Since aT (2) > 0, the object is speeding up.

21. aT (t) = −1t2

; aN(t) = 1√2|t|3

.

Note: The absolute value sign is important, since aN must be positive.

22. (2√

6,−6, 2)

26. 40

28. 136√

13 (−108 cos2 3t− 108 sin2 3t, 2 cos 3t,−2 sin 3t)

29. a) The binormal vector is a normal vector to the osculating plane.

c) x′(t) and/or x′′(t) might be parallel or zero, meaning they cannot serveas the two direction vectors for any plane.

30. a) t = 1b) 96

32. x(t) = (4 cos 34t,−4 sin 3

4t)

33. x(t) =(5− 4√

17t,−2 + 6√17t,−2 + 4√

17t)

34. a) f(t) = (1 + 3t, 3− 3t,−2 + 3t)c) f(s) =

(1 + s√

3 , 3−s√3 ,−2 + s√

3

)36. x(s) =

((1 + 2

3s) sin(

12 ln(1 + 2

3s)), 1 + 2

3s, (1 + 23s) cos

(12 ln(1 + 2

3s)))

40. aT (1) = 304√161 ; aN(1) = 4

√181√161 .

325


41. aT (t) = −3 sin t cos t√7 cos2 t+sin2 t

42. 52

44. 11√

6√35

46. 13| sin t cos t|

Note: The absolute value signs are important, since curvature is always non-negative.

48. 23(t2+1)3/2

50. b) The curvature of a parabola is maximized at its vertex.

51. .112229

53. Since f ′′(x) = 0, the curvature κ(a) must be zero.

54. a) x(t)b) The osculating plane to x at t.

c) N(t)d) 1

κ(t)

e) x(t) + 1κ(t)N(t)

326

Chapter 6

Optimization

6.1 Finding and classifying local extremaThis chapter is about solving optimization problems:

Definition 6.1 An optimization problem is any problem in which you are askedto find maximum and/or minimum value(s) of a function called a utility (perhaps onsome fixed domain, and perhaps subject to some constraint on the variables).

Suppose you are tutoring a Calculus 1 student who is having trouble remem-bering how to find the maximum or minimum of a function f : R → R. In onesentence (with at most 10 words), what advice (or what directions) would yougive this student, to jog his/her memory?

Keep this advice in mind as we think about optimization problems in higherdimensions. First, here’s an observation which simplifies things somewhat:

FIRST OBSERVATION:

It makes no sense to ask optimization problems about functions f : Rn →Rm if m > 1, the outputs of f are , and those objects aren’tordered (so there’s no such thing as a “largest” or “smallest” output of f ).

So we only have to learn how to optimize functions f : Rn → R.

327

6.1. Finding and classifying local extrema

EXAMPLE 1Here is the graph of some unknown function f : R2 → R:

This function has a local maximum A and a local minimum B. To think about howto find these points, let’s consider two other pictures associated to f :

Contour plot of f Gradient field∇f

-3 -2 -1 0 1 2 3-3

-2

-1

0

1

2

3-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

x

y

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

x

y

1. Locate A and B on both the contour plot and gradient field above.

2. In general, if a function g : R2 → R has an extremum (i.e. a maximum or aminimum) at x = a and y = b, what do you think the contour plot of g shouldlook like near (a, b)?

328


3. In general, if a function g : R2 → R has a local maximum at (a, b), what doyou think the gradient field of g should look like near (a, b)?

4. In general, if a function g : R2 → R has a local minimum at (a, b), what doyou think the gradient field of g should look like near (a, b)?

5. If (a, b) is the location of a local extrema of g : R2 → R, what do you think∇g(a, b) should be equal to?

EXAMPLE 2The contour plot for some function h : R2 → R is given below, at left. The gradientfield for some other function k : R2 → R is given below at right. Use these picturesto find the locations of the relative extrema of h and k, and classify those extremaas local maxima or local minima.

-1

-1

0

0

1

2

3

3

445

6

-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4-4 -3 -2 -1 0 1 2 3 4

-4

-3

-2

-1

0

1

2

3

4

x

y

-4 -2 0 2 4

-4

-2

0

2

4

x

y

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Critical points of functions f : Rn → R

Definition 6.2 Let f : Rn → R.

1. We say f has a local maximum (a.k.a. relative maximum) at x = c iff(x) ≤ f(c) for all x in a neighborhood of c. In this case f(c) is called a local(relative) maximum value of f .

2. We say f has a local minimum (a.k.a. relative minimum) at x = c iff(x) ≥ f(c) for all x in a neighborhood of c. In this case f(c) is called alocal (relative) minimum value of f .

3. Collectively, all local maxima and local minima of f are called local extrema off .

4. A point c ∈ Rn is called a saddle (a.k.a. saddle point) of f if c is a criticalpoint of f which is neither the location of a local maximum nor the location of alocal minimum.

A note regarding the language: Suppose f : R2 → R. If one says “f has a lo-cal maximum of 5 at (1, 3)”, then one means that 5 is the output (in this case, thez−value) and (1, 3) is the input (i.e. x = 1 and y = 3). So the point on the graph off corresponding to the maximum is (1, 3, 5).

Definition 6.3 A critical point (a.k.a. CP) of a function f : Rn → R is a vectorc ∈ Rn such that either∇f(c) = 0 or ∇f(c) does not exist.

Notice that a critical point is an element of the domain of f .

The observation we made in Question 5 of Example 1 on the previous page sug-gests that to optimize a function, we need to find its critical points:

Theorem 6.4 (Critical Point Theorem (for local extrema)) If f : Rn → R is C1,then all local extrema of f must occur at critical points.

330


EXAMPLE 3Find all critical points of each function:

a) f(x, y) = 2x2 + y2 + 8x− 6y + 20

b) f(x, y) = −x3 + 4xy − 2y2 + 1

Solution: First, compute the gradient:

∇f(x, y) = (−3x2 + 4y, 4x− 4y).

Then, set the gradient equal to 0:

∇f(x, y) = 0⇒−3x2 + 4y = 0

4x− 4y = 0 ⇒ y = x.

Plugging into the first equation, we get −3x2 + 4x = 0, i.e. x(−3x+ 4) = 0.

So x = 0 or x = 43 . Since y = x, the two critical points of f are

(0, 0) and(4

3 ,43

).

c) f(x, y) = x2y + 8x2 − 4y − 32

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Classification of critical points

Question: Suppose f : Rn → R. We know that all local extrema of f occur at crit-ical points, and we know that to find critical points you set ∇f = 0. But how doyou tell if a critical point is a local min, a local max, or a saddle?

To gain intuition for how you do this, let’s again pretend we are tutoring aCalculus 1 student. He/she has correctly found the critical points of some functionf : R → R by setting f ′(x) = 0 and solving for x. Now this student needs todetermine if those critical points are the locations of local maxima or local minima,and he/she is stuck. What should you tell him/her to do?

Hint: Don’t tell them to do a sign chart.

The reason this principle works is because the second derivative measures of afunction f : R→ R the of that function.

• If f ′(c) = 0 and f ′′(c) > 0, then the tangent line to f at c is horizontal but thefunction is at c. Thus the function must look like

f

c

so c has to be the location of a local minimum.

• If f ′(c) = 0 and f ′′(c) < 0, then the tangent line to f at c is horizontal but thefunction is at c. Thus the graph of the function must looklike

f

c

so c must be the location of a local maximum.

332


• In the rare case where f ′(c) = 0 and f ′′(c) = 0, sometimes c is the location ofa local maximum or minimum, but it might be the location of a saddle:

f

c

f

c

f

c

To classify critical points of functions Rn → R, we can use the “second deriva-tive”, but what exactly is meant by the “second derivative” if f : Rn → R? (All weknow about are second-order partials of f : Rn → R; how do they work together?)

Before we do that, however, I want to show you how you can sometimes clas-sify critical points without doing any further calculus:

Method 1 of classifying critical points: reasoningEXAMPLE 4

Find all critical points of the function f(x, y, z) = (x− 2)2 + (y + 3)2 + (z − 5)2 + 6;classify each critical point as a local maximum, local minimum, or a saddle.

Solution: To find the critical points, set∇f(x) = (2(x− 2), 2(y+ 3), 2(z− 5)) = 0and solve for (x, y, z):

∇f(x) = 0⇒

2(x− 2) = 02(y + 3) = 02(z − 5) = 0

⇒

x = 2y = −3z = 5

So the only critical point is (2,−3, 5).

Next, we classify the critical point:

333


EXAMPLE 5Find all critical points of the function f(x, y) = y2 − x2; classify each critical pointas a local maximum, local minimum, or a saddle.

Solution: To find the critical points, set ∇f(x) = (−2x, 2y) = 0 and solve for(x, y); hopefully it is clear that the only critical point is (0, 0).

334


Method 2 of classifying critical points: the Second Derivative Test

Recall: We wanted to understand what it means for the “second derivative” of afunction f : Rn → R to be “positive” or “negative”. Positive “second derivatives”should go with local minima, and negative “second derivatives” should go withlocal maxima.

But first, what “is” the “second derivative”? Suppose f : Rn → R. How manysecond-order partial derivatives does f have?

What do you think is a reasonable way to arrange that number of second-orderpartial derivatives?

This leads to the following definition:

Definition 6.5 Let f : Rn → R be C2. The Hessian of f is the function Hf : Rn →Mnn(R) defined by

Hf(x) =

fx1x1(x) fx1x2(x) · · · fx1xn(x)fx2x1(x) fx2x2(x) · · · fx2xn(x)

...... . . . ...

fxnx1(x) fxnx2(x) · · · fxnxn(x)

n×n

.

EXAMPLE 5Compute the Hessian of f(x, y) = x3 + 3x2y − 2y3 + y2.

Solution: First, compute the first- and second-order partials of f :

fx(x, y) = 3x2 + 6xy fy(x, y) = 3x2 − 6y2 + 2y

fxx(x, y) = 6x+ 6y fxy(x, y) = fyx(x, y) = 6x fyy(x, y) = −12y + 2

335


Since the mixed partials of a C2 function are equal, we know:

Theorem 6.6 Let f : Rn → R be C2. Then, for all x ∈ Rn, the Hessian Hf(x) is asymmetric matrix (meaning (Hf(x))T = Hf(x)).

The Hessian matrix Hf of a function f : Rn → R is the “right” object to use, ifyou need something to serve as the “second derivative” of f . So that takes care of“what” the second derivative is.

Now, we need to understand what makes a symmetric matrix (like Hf ) “posi-tive” or “negative”. First, here’s a complicated way of expressing that a number ispositive:

This leads to the following definition for matrices:

Definition 6.7 Let A be a symmetric n× n matrix.

• A is called positive definite (in which case we write “A > 0”) if for anynonzero vector x ∈ Rn, xTAx > 0.

• A is called negative definite (in which case we write “A < 0”) if for anynonzero vector x ∈ Rn, xTAx < 0.

EXAMPLE 6Determine whether each matrix is positive definite, negative definite, or neither:

A =(

1 33 11

)B =

(3 00 −2

)

336


In addition to the kind of reasoning used in Example 6, there are three otherways to tell if a matrix is positive or negative definite. The first, if you know linearalgebra (Math 322), is to compute the eigenvalues of the matrix:

Theorem 6.8 A symmetric matrix is positive definite if and only if all its eigenvalueshave positive real part. A symmetric matrix is negative definite if and only if all itseigenvalues have negative real part.

There is a sometimes-useful shortcut connected with this: in Math 322, youlearn that the eigenvalues of a matrix add to its trace, and multiply to its determi-nant. So for 2× 2 matrices, it’s not hard to discover this test:

Corollary 6.9 (Definiteness test for 2× 2 matrices) Let A be a symmetric 2 × 2matrix.

1. If detA > 0 and tr(A) > 0, then A is positive definite.

(Both eigenvalues of A have positive real part.)

2. If detA > 0 and tr(A) < 0, then A is negative definite.

(Both eigenvalues of A have negative real part.)

3. If detA ≤ 0, then A is neither positive definite nor negative definite.

(One eigenvalue of A is positive, but one is negative; or at least one eigenvalueis zero.)

The second method to classify a symmetric matrix as positive or negative defi-nite (or neither) is to use what are called “principal minors”:

Definition 6.10 Let

A =

a11 a12 · · · an1a21 a22 · · · an2

...... . . . ...

an1 an2 · · · ann

n×n

.

The sequence of principal minors of A is the sequence d1, d2, ..., dn of numberswhere for each j,

dj = det

a11 a12 · · · aj1a21 a22 · · · aj2

...... . . . ...

aj1 aj2 · · · ajj

j×j

.

337


Theorem 6.11 (Positive/negative definiteness test using minors) LetA be a sym-metric n× n matrix with principal minors d1, d2, d3, ..., dn.

1. if dj > 0 for all j, then A is positive definite.

2. if d1, d3, d5, ... are all negative but d2, d4, d6, ... are all positive, thenA is negativedefinite.

3. if neither (1) nor (2) hold, but none of the dj are zero, then A is neither positivedefinite nor negative definite.

If any of the dj equal 0, then this test is worthless.

A third method is to use the Mathematica command PositiveDefiniteMatrixQ[A],which gives output True if A is positive definite and False if A is not positive defi-nite (NegativeDefiniteMatrixQworks the same way).

Having figured out what the “second derivative” of a function f : Rn → Ris, and having figured out how to classify that object as “positive” or “negative”,we are now able to generalize the Second Derivative Test of Calculus 1 to higherdimensions:

Theorem 6.12 (Second Derivative Test in higher-dimensions) Suppose that f :Rn → R is a C2 function. Suppose that c is a critical point of f with∇f(c) = 0.

1. If Hf(c) is positive definite, then f has a local minimum at c.

2. If Hf(c) is negative definite, then f has a local maximum at c.

3. If Hf(c) is neither positive definite nor negative definite and detHf(c) 6= 0,then f has a saddle at c.

4. If detHf(c) = 0, then this test is inconclusive (use reasoning to classify thecritical point).

338


EXAMPLE 7Let f(x, y) = −x3 +4xy−2y2 +1. In Example 3 (b), we found that f has two criticalpoints (0, 0) and

(43 ,

43

). Classify each critical point as a local maximum, a local

minimum, or a saddle.

Solution: Compute the Hessian:

Hf(x, y) =(fxx fxyfyx fyy

)=(−6x 4

4 −4

)

Then plug each critical point into the Hessian:

Hf(0, 0) =(

0 44 −4

)Hf

(43 ,

43

)=(−8 44 −4

)

Test these matrices for positive/negative definiteness:

339


EXAMPLE 8Find and classify all critical points of each given function:

a) f(x, y) = x2 + xy + y2 + 2x− 2y + 5

Solution: Start by finding critical points. ∇f(x, y) = (2x + y + 2, x + 2y − 2);setting this equal to 0 gives

2x+ y + 2 = 0x+ 2y − 2 = 0 ⇒ (−2, 2) is the only critical point

To classify the critical point, find the Hessian:

Hf(x, y) =(fxx fxyfyx fyy

)=(

2 11 2

).

Now, we determine the “sign” of Hf(−2, 2) =(

2 11 2

):

d1 = det(2) = 2 > 0

d2 = det(

2 11 2

)= 2(2)− 1(1) = 3 > 0

Thus Hf(−2, 2) > 0, meaning (−2, 2) is the location of a local minimum .

3.1

3.5

4

5

68

1214

20

30

(-2,2)

-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4-4 -3 -2 -1 0 1 2 3 4

-4

-3

-2

-1

0

1

2

3

4

x

y

(-2,2)

-4 -2 0 2 4

-4

-2

0

2

4

x

y

340


b) f(x, y, z) = x3 + xy2 + x2 + y2 + 3z2

341

6.2. Absolute extrema

c) f(x, y) = xy2 − 6x2 − 3y2

Solution: ∇f = (y2 − 12x, 2xy − 6y) = 0 givesy2 − 12x = 02xy − 6y = 0 ⇒ 2y(x− 3) = 0

So there are three critical points: (0, 0), (3, 6) and (3,−6). Classify them withthe Hessian:

Hf(x, y) =(−12 2y2y −6

)Now plug in the critical points and find the “sign” of each Hessian:

Hf(0, 0) =(−12 0

0 −6

)Hf(3, 6) =

(−12 1212 −6

)Hf(3,−6) =

(−12 −12−12 −6

)

6.2 Absolute extremaIn the previous section we discussed methods for finding local extrema (i.e.

points c at which f(c) is either the largest or smallest value obtained by f for pointsnear c).

Now we turn to a second kind of optimization problem, where all we care aboutis finding the single largest (or smallest) value of the function on some fixed do-main.

Definition 6.13 Let f : Rn → R, and let D ⊆ Rn.

1. We say f has a absolute maximum on D (a.k.a. global maximum) at x =c ∈ D if f(x) ≤ f(c) for all x ∈ D. In this case f(c) is called a absolute(global) maximum value of f .

2. We say f has a absolute minimum on D (a.k.a. global minimum) at x =c ∈ D if f(x) ≥ f(c) for all x ∈ D. In this case f(c) is called a absolute(global) minimum value of f .

Similar to Calculus 1, we have a theorem that guarantees that under reasonablehypotheses, there is always an absolute minimum and absolute maximum:

342


Theorem 6.14 (Max-Min Existence Theorem) If f : Rn → R is continuous andD ⊆ Rn is a compact set (i.e. is closed and bounded), then f has a absolute maximumvalue and an absolute minimum value on D.

How do you solve problems where you’re looking for absolute extrema?Think back to Calculus 1. If you had to find the absolute extrema of a functionf : R→ R on some interval [a, b], you would first find the critical points of f . Thenwhat? (Let’s pretend to be a Calc 1 tutor again.)

CONCEPT: To find the absolute max and/or min of a function f : Rn → R ona compact region D, you do something similar. The idea is that absolute extremamust occur at critical points, or along the boundary of D (the boundary of D playsthe role of the endpoints of [a, b] from Calculus 1 problems). More precisely:

How to find absolute extrema of f : Rn → R on compact region D ⊆ Rn:

1. Draw a picture of D, if you aren’t sure what D looks like.

2. Find the critical points of f (by setting ∇f = 0). Keep only the criticalpoints which are in D.

3. For each smooth piece of the boundary ∂D:

a) Parameterize it in terms of some function t 7→ (x, y);b) Plug the parameterization into f to obtain a function of t;c) Find the critical points of the resulting function using Calculus 1

methods. Keep only the critical points that are actually on the bound-ary; call these boundary critical points.

NOTE: If the boundary consists of one smooth piece (like a circle or el-lipse), you can do something else in place of this step called Lagrange’sMethod, which will be described in Section 6.3.

4. Take the critical points from Step 2, together with the boundary criticalpoints from Step 3, together with any points ofD which are sharp cornersof ∂D, and plug all these into f . The biggest value you get is the absolutemax, and the smallest value you get is the absolute min.

343


EXAMPLE 9Find the absolute maximum and absolute minimum values of f(x, y) = x2 − y2 onthe region D = (x, y) : x2 + y2 ≤ 4.

344


EXAMPLE 10Find the absolute maximum and absolute minimum values of f(x, y) = 4x2 + y2 −4xy − 2x on the triangular compact set D bounded by the lines y = x, y = −3 andx = 3.

Finally, we test all the points we’ve found by plugging them into the utility f :

CP noneBCP

(−54 ,−3

)f(−5

4 ,−3) = 114

BCP (1, 1) f(1, 1) = −1 ←− abs min is − 1, at (1, 1)CORNER (−3,−3) f(−3,−3) = 15CORNER (3,−3) f(3,−3) = 75 ←− abs max is 75, at (3,−3)CORNER (3, 3) f(3, 3) = 3

345

6.3. Lagrange’s method

6.3 Lagrange’s methodEXAMPLE 11 (FROM CALCULUS 1)

Find the largest possible area of a rectangle with perimeter 24 meters.

Calculus I solution: Let x be the length of the rectangle and let y be the width:

x

y

Then, since the perimeter is 24, we know

2x+ 2y = 24.

Solving this equation for y, we get y = 24−2x2 = 12− x. The goal, therefore, is to

maximize the area which is

A = xy = x(12− x) = 12x− x2.

To do this, differentiate the area function and set the derivative equal to zero:

A′(x) = 12− 2x0 = 12− 2x

⇒ 6 = x.

Since A′′(x) = −2, by the Second Derivative Test x = 6 is the location of amaximum, so the area is maximized when x = 6, i.e. when y = 12 − x =12− 6 = 6 so the maximum area is A = xy = 6(6) = 36 m2 .

Problems like these are called constrained optimization problems. The idea isthat (in Calculus 1) you have two variables x and y and a formula f(x, y) whichyou want to optimize, but only over choices of x and y which satisfy some equationof the form g(x, y) = c where c is a constant. For instance, in the above example,we have

f(x, y) =

g(x, y) =

c =

346


In a setup like this the formula you want to optimize, i.e. f(x, y), is called the util-ity and the equation g(x, y) = c which x and y must satisfy is called the constraint.

In Calculus 1, you learn the following approach to solving constrained opti-mization problems (which is essentially what we did on the previous page):

1. Solve the constraint for one variable in terms of another.

2. Substitute the solution from step 1 into the utility, so that the utility is interms of one variable.

3. Find the critical points of the utility from step 2 by finding critical points,etc.

4. Classify the critical points as maxima or minima.

However: this method has drawbacks:

a)

b)

EXAMPLE 12Find the maximum value of f(x, y) = x− y, given that x2

9 + y2

4 = 1.

First attempt: try the Calculus 1 method outlined above:

x2

9 + y2

4 = 1

347


Second attempt at a solution: do something similar to how I just told you to findextrema along the boundary of some region: write parametric equations forthe constraint, and rewrite the utility in terms of the parameter:

x2

9+y2

4= 1

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3

Drawbacks:

a) If you don’t know how to parameterize the constraint (i.e. if it isn’t a circleor ellipse or line), this method isn’t usable.

b) This method gives you an answer in terms of t, not in terms of x and y.

c) This method also sometimes requires some unpleasant trigonometry or alge-bra.

In this section we will introduce a new method which solves constrained opti-mization problems without the problems of the previous attempts.

348


The idea behind Lagrange’s methodEXAMPLE 12, AGAIN

Find the maximum value of f(x, y) = x− y, given that x2

9 + y2

4 = 1.

Solution: As with our previous attempt, let’s draw a picture of the constraint inan xy-plane:

x2

9+y2

4= 1

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3

What’s different about this method is that we will think of this constraint asa level curve to the function g(x, y) = x2

9 + y2

4 at height 1. By doing this, we knowthat at each point on the constraint, ∇g is a vector which is orthogonal to the levelcurve.

Now, let’s think about a contour plot of the utility f(x, y) = x − y (shown be-low). Based on our knowledge of gradients, we know that at every point on thexy-plane,∇f is a vector orthogonal to these level curves.

-7

-6

-5

-4

-3

-2

-1

0

1

2

3 4

5

6

7

-4 -3 -2 -1 0 1 2 3 4-4

-3

-2

-1

0

1

2

3

4-4 -3 -2 -1 0 1 2 3 4

-4

-3

-2

-1

0

1

2

3

4

x

y

349


The key to understanding our new method is to understand what happens whenwe superimpose the two pictures on the previous page (the level curve g(x, y) = 1and the contour plot of f ):

g(x,y) = 1-6

-4

-3

-1

0

1

3 4

6

-4 -3 -2 -1 1 2 3 4

-3

-2

-1

1

2

3

From this picture, we can answer the following questions:

1. Estimate the point (x, y) on the constraint g(x, y) = 1 at which f(x, y) = x− yis maximized.

2. What is true about the level curves to f at the point you found in Question1? (What is their relationship to the level curve g(x, y) = 1?)

3. What is true about ∇f and ∇g at the point you found in Question 1? De-scribe this answer in words, and give a symbolic formula representing therelationship.

In fact, the same logic described in the example above works for constrainedoptimization problems, even if there are more than two variables. The implemen-tation of this idea is called Lagrange’s method:

350


Theorem 6.15 (Lagrange’s method) Let f : Rn → R and g : Rn → R be C1.Then, the absolute maximum and/or absolute minimum of utility f , subject to con-straint g(x) = c (where c is a constant), occurs at a point x where

∇f(x) = λ∇g(x)

for some constant λ 6= 0 (the constant is called a Lagrange multiplier).

EXAMPLE 12 (SOLUTION USING LAGRANGE’S METHOD)Find the maximum value of f(x, y) = x− y, given that x2

9 + y2

4 = 1.

Solution: To set up Lagrange’s method, let g(x, y) = x2

9 + y2

4 so that the constrainthas the form g(x, y) = c (in this case, c = 1). Then:

∇f = (1,−1) ∇g =(2x

9 ,y

2

)so by Lagrange’s method, we have

∇f(x) = λ∇g(x) ⇒ (1,−1) = λ(2x

9 ,y

2

)⇒

1 = λ2x

9−1 = λy2

351


EXAMPLE 13Find the maximum and minimum of f(x, y, z) = 3x + 2y + z + 5, subject to theconstraint z = 9x2 + 4y2.

352


EXAMPLE 14Find the maximum volume of a box in the first octant (i.e. where x, y and z arepositive) with faces parallel to the coordinate planes, a vertex at the origin, and theopposite vertex on the plane 3x+ 4y + 6z = 12.

353


EXAMPLE 15Find the point (x, y) in the first quadrant at which x2 + y2 is minimized, given thatxy3 = 1.

Solution: Let f(x, y) = x2 + y2 and g(x, y) = xy3, so that we can use Lagrange’smethod. Then,

∇f = (2x, 2y) and ∇g = (y3, 3xy2).

So

∇f = λ∇g ⇒

2x = λy3

2y = λ3xy2

constraint ⇒ xy3 = 1The rest of this is just algebra, to solve the above equations for x and y (and λ,

if necessary). There are a lot of different ways to proceed; here is one method:

From the constraint, we get x = 1y3 . Plugging this into the first two equations,

we get 2y3 = λy3

2y = 3λy

⇒

2 = λy6

2y2 = 3λ ⇒λ = 2

y6

λ = 2y2

3

Setting the two above expressions for λ equal, we obtain

2y6 = 2y2

3 ⇒ 2y8 = 6⇒ y8 = 3⇒ y = ± 8√

3.

Since the question asks for a point in the first quadrant, y > 0, so y = 8√

3.

Finally, we can find x by substituting back into the constraint:

xy3 = 1⇒ x = 1y3 = 1

( 8√

3)3= 3−3/8.

To summarize, the point we want is (x, y), which is(3−3/8,

8√

3)

.

Remark: How do we know this is a minimum (as opposed to a maximum)? Noticethat

f(3−3/8,

8√

3)

= (3−3/8)2 + ( 8√

3)2 = 3−3/4 + 31/4 = 3−3/4(1 + 3) = 4 · 3−3/4 ≈ 1.75.

Taking another point on the constraint like (1, 1) gives

f(1, 1) = 12 + 12 = 2,

which is greater than 4 · 3−3/4.

354



1. The contour plot for some unknown function f : R2 → R is shown below:

a

-9 -4

-2

-0.75

-0.75

-0.5

-0.5

-0.5

-0.5-0.5

-0.25

-0.25

0

2

4 9

-2 -1 0 1 2 3 4 5 60

1

2

3

4-2 -1 0 1 2 3 4 5 6

0

1

2

3

4

Use this contour plot to find the locations of all local extrema in the win-dow [−2, 6] × [−0, 4]; classify each extremum as a local maximum or a localminimum.

2. The gradient field for some unknown function g : R2 → R is shown below,at left. Use the gradient field to find the locations of all local extrema in thewindow [−8, 8] × [−8, 8]; classify each extremum as a local maximum or alocal minimum.

-8 -6 -4 -2 0 2 4 6 8

-8

-6

-4

-2

0

2

4

6

8

-8 -6 -4 -2 0 2 4 6 8

-8

-6

-4

-2

0

2

4

6

8

x

y

-5 -4 -3 -2 -1 0 1 2 3 4 5

-5

-4

-3

-2

-1

0

1

2

3

4

5

-5 -4 -3 -2 -1 0 1 2 3 4 5

-5

-4

-3

-2

-1

0

1

2

3

4

5

x

y

3. The gradient field for some unknown function h : R2 → R is shown above,at right. Use the gradient field to find the locations of all local extrema in thewindow [−5, 5] × [−5, 5]; classify each extremum as a local maximum or alocal minimum.

355


4. Let f(x, y) = 8x3y2 − 4x4y3. Compute the Hessian of f .

5. Let f(x, y) = y cosx− x sin y. Compute Hf(x, y).

6. Let f(x, y, z) = 3x2y3z−2xy2+4yz−5x. Compute the Hessian of f at (1, 2,−1).

7. Determine whether each given matrix is positive definite, negative definiteor neither. (While you can check your answers with Mathematica, you shouldbe able to test these matrices by hand.)

a)(

2 0−3 −1

)b)

(−4 2−4 1

)c)(−3 23 −5

)d)

(7 −3−1 1

)

8. Determine whether the following matrix is positive definite, negative definiteor neither. (While you can check your answer with Mathematica, you shouldbe able to test this matrix by hand.) −2 13 −62

0 23 −1200 6 −31

9. Use Mathematica to determine whether each of the following matrices are

positive definite, negative definite, or neither:

a)

141 4 118 96−12 127 34 7821 14 278 66−51 −34 77 399

b)

−121 89 16 26−33 29 24 633 −155 −126 18−33 115 36 −68

In Problems 10-17, find all critical points of the given function, and classify eachcritical point as a local maximum, local minimum, or saddle.

10. f(x, y) = 12xy + 1

11. f(x, y) = 8− x2 + 4x− 2y2

12. g(x, y) = 4x3 + y3 − 12x− 3y − 5

13. F (x, y) = x4 − 8x2 + 2y2

14. z(x, y) = x3 + y3 − 12xy + 4

15. (R) f(x, y) = x3y3

16. (R) f(x, y) = −2x4y2

17. f(x, y, z) = y3 + yz2 − 3y2 + x2 + 2z2

356


18. In this question, let h be the function whose gradient field is described inProblem 3.

a) Find the coordinates of all saddle points of h in the window [−5, 5] ×[−5, 5].

b) Find the coordinates of a critical point x of h where you would expectHf(x) to be positive definite (if there is no such critical point, say so).

c) Find the coordinates of a critical point x of h where you would expectHf(x) to be negative definite (if there is no such critical point, say so).

d) Find the coordinates of a critical point x of h where Hf(x) is neitherpositive definite nor negative definite (if there is no such critical point,say so).


19. Find the absolute extrema of f(x, y) = x2 − y2, given that (x, y) lies in theelliptical region D = (x, y) : 4x2 + 9y2 ≤ 36.

20. Find a point (x, y) at which f(x, y) = x2 + xy + y2 is maximized, given that(x, y) lies in the set of points which are distance at most 2 from the origin.

21. Find the absolute maximum value and absolute minimum value obtained byf(x, y) = x2 − 6xy + 9y2 on the rectangular region D = [0, 5]× [0, 3].

22. Find the absolute maximum value and absolute minimum value obtained byf(x, y) = xy+6x−3y+4, where D is the triangular subset of R2 with vertices(0, 0), (0, 3) and (6, 3). Also find the points (x, y) at which these absolutemaximum and minimum values occur.

23. The temperature of a metal plate which occupies the set (x, y) : x2 + y2 ≤ 1is T (x, y) = x2 − x+ 4y2. Find the hottest and coldest points on the plate.


24. Consider the problem of finding the maximum and minimum values of f(x, y) =y2 − x2 subject to the constraint x2 + y2 = 9.

a) Solve this problem by parameterizing the constraint, plugging that pa-rameterization into the utility and optimizing via the methods of Calcu-lus 1 (like the second attempt of Example 12).

b) Solve this problem using Lagrange multipliers. Verify that you get thesame answer you got in part (a).

25. Find the minimum value of x2 + 8xy + y2, subject to the constraint x− y = 4.

357


26. Find the maximum value of xy, subject to the constraint 3x2 + 4y2 = 24.

27. Find the locations (x, y) where f(x, y) = xy3 is minimized and maximized,subject to the constraint

√x+√y = 1.

28. Find the maximum and minimum values of f(x, y) = 8x3 + y2, subject to theconstraint 4x2 + y2 = 2.

29. Find the maximum value of xyz, given that (x, y, z) lies on the sphere of ra-dius 2 centered at the origin.

30. Find the maximum value of v + 2w + 3x + 4y + 5z, subject to the constraintv2 + w2 + x2 + y2 + z2 = 1.

31. Find the absolute maximum and absolute minimum value of the functionf(x, y) = 4x2 + y2 on the region D = (x, y) : x2 + y2 ≤ 16.

32. (R) Find the point on the cone z2 = x2 + y2 which is closest to the pointp = (2, 1, 1).

Hint: Instead of minimizing the distance from p to a point (x, y, z) on thecone, minimize the square of that distance. That way, you don’t have to dealwith square roots in your utility function.

33. Find the largest possible product of three non-negative numbers, given thatthe sum of the three numbers is 12.

34. (F) Use Lagrange’s method to prove that of all rectangular (three-dimensional)boxes with fixed surface area A, the one with the largest volume is a cube.

35. (F) Find the volume of the largest rectangular (three-dimensional) box whichfits entirely inside a (three-dimensional) sphere of radius r.

36. In order to carry a piece of luggage onto a commercial airline flight, its length,width and height must sum to less than 48 inches. Find the dimensions of thepiece of luggage with greatest volume that can be carried on such a flight.

37. A soybean farmer determines that his annual yield can be modeled by theformula Y (w, f) = 800 + 6w2 + f 2, where w is the amount of water (in thou-sands of gallons) he uses to fertilize his field, and f is the amount of fertilizer(in pounds) he uses on his crops. Fertilizer costs the farmer $20 per pound,and it costs him $50 to irrigate his field per 1000 gallons of water used. Ifthe farmer has $600 available to spend on water and fertilizer, how should heallocate his money to maximize his soybean yield?

358


Selected answers

1. local maximum at (2, 3);

local minimum at (4, 2)

4.(

48xy2 − 48x2y3 48x2y − 48x3y2

48x2y − 48x3y2 16x3 − 24x4y

)

6.

−48 −80 48−80 −40 4048 40 0

7. a) neither

d) positive definite

10. (0, 0): saddle

12. (−1,−1): local maximum

(1,−1): saddle

(−1, 1): saddle

(1, 1): local minimum

14. (0, 0): saddle

(4, 4): local minimum

15. (0, 0): saddle

17. (0, 0, 0): saddle

(0, 2, 0): local minimum

18. b) (−3.5, 3.5) or (3.5,−1.5)c) (−3.5,−1.5) or (3.5, 3.5)

19. absolute minimum is −4;

absolute maximum is 9.

23. The hottest point is(−16 ,√

356

);

the coldest point is(−12 , 0

).

24. absolute minimum is −9;


25. −24

28. absolute minimum is 5027 ;


30.√

55

31. absolute minimum is 0;


32.(1 + 1√

5 ,12 +

√5

10 ,12 +

√5

2

)33. 64

36. 16′′ × 16′′ × 16′′

359



Definition of differentiability: Precisely write the definition of what it means fora function f : Rn → Rm to be differentiable.

Compute derivatives: Given a function f : Rn → Rm, compute partial derivatives,the total derivative, directional derivatives, gradient, etc.

Applications of derivatives: Write equations of tangent lines and tangent planes;compute derivatives implicitly; find maximum/minimum rate of change anddirection of maximum/minimum rate of change; estimate values of func-tions using linear approximation.

Particle motion: Given information describing the motion of some object, com-pute the object’s position, velocity, speed, acceleration, the tangential andnormal components of its acceleration, unit tangent vectors, principal unitnormal vectors, curvature, distance travelled, displacement, arc length pa-rameterization of its path, the curvature of the path it travels, etc.; solve pro-jectile motion problems

Optimization: Given a function f : Rn → R, find the critical points of f , findand classify the local extrema of f , find the absolute max and/or min of fon a closed and bounded region, find the absolute max/min of f subject to aconstraint g(x) = c.


1. Precisely write the definition of what it means for a function f : Rn → Rm tobe differentiable.

2. What does it mean for a function f to be “C2”?

3. Find the total derivative of the function f(x, y) = (5x3y2+3y, 4(x−y)2, x4y3 sin x).

4. Find the total derivative of the function f(w, x, y, z) =(w ln(x+ z), ewy−x4 cos z

).

5. Let f(x, y, z) = x4z+ y2z3 + xy3. Estimate f(1.01, 1.99, 1) using linear approx-imation.

6. Let f(x, y) = ex+3y + 5e2x−y.

a) Compute ∂2f∂x2 .

b) Compute ∂2f∂x∂y

.

360


c) Compute ∂2f∂y∂x

.

d) In what you are asked to compute in part (c), strictly speaking, doesthat notation mean that you differentiate with respect to x first, or withrespect to y first? (In this example, the order doesn’t matter since f isC2, but for a function that wasn’t C2, the order might matter.)

e) Compute fxy(0, 0).

f) In what you are asked to compute in part (e), strictly speaking, doesthat notation mean that you differentiate with respect to x first, or withrespect to y first?

g) Compute fxxxy.

h) Compute∇f .

i) Find the direction in which f increases most rapidly at the origin.

j) Find the equation of the plane tangent to f at the origin.

k) Find the directional derivative of f at the origin, in the direction (2, 3).

7. Write an equation of the line tangent to the curve f(t) = (3 cos2 t, 4 sin 2t, 3 cos 2t sin t)when t = π

4 .

8. Find an equation of the plane tangent to the ellipsoid 3x2 + y2 + 2z2 = 25 atthe point (1, 2,−3).

9. Suppose x2y3 + x4 cos y − 3y6 = 15. Find dydx

.

10. An object’s position at time t is (5 cos t, 12t,−5 sin t).

a) Find the object’s velocity at time π3 .

b) Show that the object is moving with constant speed.

c) Find the arc length parameterization of the path the object travels.

d) Find the distance the object travels from t = 0 to t = π.

e) Find the displacement of the object from t = 0 to t = π.

f) Find the acceleration of the object (at time t).

g) Find the tangential component of the object’s acceleration at time π4 .

h) Find the normal equation of the osculating plane of the object’s path attime t = π

2 .

i) Find the curvature of the object’s path at the point (5, 0, 0).

j) Find the principal unit normal vector to the curve at time t = π2 .

11. Find the curvature of the graph of y = x3 at the point (2, 8).

361


12. A small rock is launched in the air by a slingshot. If the initial velocity of therock is v(0) = (2, 5, 12) and the rock is fired from an elevation of 28 feet, howlong will it take the rock to hit the ground? What will its speed be on impact?(Assume the only force acting on the rock is gravity, and that you are on theEarth.)

13. Find and classify the critical points of f(x, y) = 2x2 + y3 − x2y − 3y.

14. Find and classify the critical points of f(x, y) = x3 − 3xy + y3.

15. Find the absolute maximum and absolute minimum values of f(x, y) = (2x−y)2−(x−1)2 on the triangular region in the xy-plane whose vertices are (0, 0),(4, 0) and (0, 4).

16. A business makes three products. When they make x, y and z units of thesethree products, their profit is f(x, y, z) = 3xz + 6y. Unfortunately, manufac-turing constraints force x2 + 2y2 + z2 ≤ 6. Find the maximum profit for thecompany.

17. A box is resting on the xy-plane with one vertex at the origin. The oppositevertex lies on the plane 6x+ 4y + 3z = 24. Find the dimensions of the box ofthis type that has the largest volume.

My answers

1. f : Rn → Rm is differentiable at x ∈ Rn if there is an m × n matrix Df(x),called the total derivative of f at x, such that

limh→0

||f(x + h)− f(x)−Df(x)h||||h||

= 0.

2. f is C2 if all its second-order partial derivatives exist and are continuous.

3. Df(x, y) =

15x2y2 10x3y + 38(x− y) −8(x− y)

4x3y3 sin x+ x4y3 cosx 3x4y2 sin x

.

4. Df(w, x, y, z) =(ln(x+ z) w

x+z 0 wx+z

yewy−x4 cos z −ewy−x4 cos z4x3 cos z wewy−x

4 cos z ewy−x4 cos zx4 sin z

).

5. Let a = (1, 2, 1). Notice f(a) = 1 + 4 + 8 = 13 and

Df(a) =(

4x3z + y3 2yz3 + 3xy2 x4 + 3y2z2)

x=a=(

12 16 13).

362


Thus

f(1.01, 1.99, 1) ≈ L(1.01, 1.99, 1) = f(a) +Df(a)(x− a)

= 13 +(

12 16 13) .01−.01

0

= 13 + 12(.01) + 16(−.01)= 12.96.

6. a) ∂2f∂x2 = 20e2x−y + ex+3y.

b) ∂2f∂x∂y

= −10e2x−y + 3ex+3y.

c) ∂2f∂y∂x

= −10e2x−y + 3ex+3y.

d) x first, then y (in (b), you’d do y first, then x).

e) fxy(0, 0) = −7.

f) x first, then y (the same as (c)).

g) fxxxy = −40e2x−y + 3ex+3y.

h) ∇f = (10e2x−y + ex+3y,−5e2x−y + 3ex+3y).

i) (11,−2) (any positive multiple of this vector is also correct).

j) 11x− 2y − z = −6.

k) 16√13

7.

x = 3

2 − 3ty = 4z = −3

√2 t

.

8. 6x+ 4y − 12z = 50.

9. dydx

= −2xy3−4x3 cos y3x2y2−18y5−x4 sin y .

10. a) v(π3

)=(−5√

32 , 12, −5

2

).

b) ||v(t)|| = 13 for all t.

c)(5 cos s

13 ,1213s,−5 sin s

13

).

d) 13πe) (−10, 12Pi, 0)

f) a(t) = (−5 cos t, 0, 5 sin t).

g) 0h) 60x+ 25y = 150π.

i) 5169

j) (0, 0, 1)

11. 12145√

145

363


12. It takes 74 seconds for the rock to hit the ground, at which time its speed is√

1965 feet/sec.

13. (0, 1) is a local min; (0,−1) is a saddle; (−3, 2) is a saddle, (3, 2) is a saddle

14. (0, 0) is a saddle; (1, 1) is a local min

15. Absolute minimum is −1 at (0, 0); absolute maximum is 55 at (4, 0).

16. The maximum profit is 6√

3, occuring at (0,√

3, 0).

17. The dimensions of the largest such box are 43 × 2× 8

3 .

364

Chapter 7

Multiple integration

7.1 Review of area and integration from Calculus 1In Math 220, you learn how to compute areas whose boundaries are defined by

curves which are of the form y = f(x). Suppose you wanted to compute the area ofthe region R pictured below. How do you proceed? (Don’t look at the next page.)

R

y = 4

y =x2

9

y = x +1

y =(x - 1)2

2+ 2

-1 1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

365

7.1. Review of area and integration from Calculus 1

To compute the area of a region like the R shown on the previous page:

1. Break the region into pieces, where each piece can be described as (x, y) : l ≤x ≤ r, b(x) ≤ y ≤ t(x) for nice functions b and t (b is the “bottom” functionand t is the “top” function). In this case, you partition R as follows:

y = 4

y =x2

9

y = x +1

y =(x - 1)2

2+ 2

-1 1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

and you get these three regions, whose union is R:

R1

t(x)=x+1

b(x) =x2

9

l=0 r=1

1

2

3

4

R2

b(x) =x2

9

t(x) =(x - 1)2

2+2

l=1 2 r=3

1

2

3

4

R3

t(x) = 4

b(x) =x2

9

l=3 4 5 r=6

1

2

3

4

Notice that each region Rj has the desired form (x, y) : l ≤ x ≤ r, b(x) ≤ y ≤t(x). For example,

R1 =

(x, y) : 0 ≤ x ≤ 1, x2

9 ≤ y ≤ x+ 1

366


REGION l r bottom function b(x) top function t(x)R1 0 1 x2

9 x+ 1R2 1 3 x2

9(x−1)2

2 + 2R3 3 6 x2

9 4

2. Find the area of each region Rj by computing

∫ r

l[t(x)− b(x)] dx =

∫ right

left[top− bottom] dx.

In this example, you get

∫ 1

0

[(x+ 1)− x2

9

]dx+

∫ 3

1

[((x− 1)2

2 + 2)− x2

9

]dx+

∫ 4

3

[4− x2

9

]dx

3. Evaluate these integrals using the Fundamental Theorem of Calculus, whichsays that you can evaluate definite integrals via antiderivatives:

Theorem 7.1 (Fundamental Theorem of Calculus (FTC)) Let f be a con-tinuous function on [a, b]. Then, if F is any antiderivative of f (i.e. if F ′ = f )on [a, b], then ∫ b

af(x) dx = F (x)|ba = F (b)− F (a).

In this problem, the total area ends up being

∫ 1

0

[(x+ 1)− x2

9

]dx+

∫ 3

1

[((x− 1)2

2 + 2)− x2

9

]dx+

∫ 4

3

[4− x2

9

]dx

=[

12(x+ 1)2 − x3

27

]1

0+[

(x− 1)3

6 + 2x− x3

27

]3

1+[4x− x3

27

]4

3

= 7954 + 118

27 + 7127

= 45754 .

What this discussion doesn’t review is why integrals compute area.

367


Why integrals compute area

You compute definite integrals by using the FTC (usually). But that’s not howdefinite integrals are defined. Definite integrals are defined as

of .

Definition 7.2 Given an interval [a, b], a partition P is a (finite) list of numbersx0, x1, x2, ..., xn such that a = x0 < x1 < ... < xn−1 < xn = b. Such a partitiondivides [a, b] into n subintervals; the kth subinterval is [xk−1, xk]. For each k, set∆xk = xk−xk−1; ∆xk is called the width of the kth subinterval. Call the largest ∆xkthe norm of the partition; denote the norm by ||P||.

EXAMPLE 1Let P be the partition 0, 1, 2, 5 (this is a partition of [0, 5]). A picture of P , withappropriate quantities indicated, is as follows:

x0=0 x1=1 x2=2 x3=5

Δx2=1Δx1=1 Δx3=3

Definition 7.3 Given function f : [a, b] → R and given partition P = x0, ..., xnof [a, b], a Riemann sum associated to P for f is any expression of the form

n∑k=1

f(ck)∆xk

where for all k, ck belongs to the kth subinterval of P . The points c1, c2, ..., cn are calledtest points for the Riemann sum.

A Riemann sum approximates the area under function f from a to b by addingup the area of some rectangles:

f (c3)

f

c1 c3 c5

Δx3

x0=a x1 x2 x3 x4 x5=b

368


Notice that if you take a partition with small norm, that means the width ofevery subinterval is small. This means you have lots and lots of skinny rectangles,which will approximate the area under the curve better and better as the norm getssmaller and smaller. With that in mind, we define:

Definition 7.4 (Limit definition of the integral) Given function f : [a, b] → R,the definite integral of f from a to b is

∫ b

af(x) dx = lim

||P||→0

n∑k=1

f(ck)∆xk

if this limit exists (in Math 220 and Math 230, it always will). If the limit exists, wesay f is integrable on [a, b].)

This theoretical definition explains why integrals compute area. But there’ssomething we haven’t reviewed yet: exactly why the FTC works. A thorough ex-planation of this is given in MATH 220 (several things need to be shown to deriveit). Here is some intuition: suppose f is a velocity function (like a speedometer).Thus any antiderivative F of f measures the object’s position (like an odometer).To find the displacement between times a and b, there are two things you can do:

1.

2.

Since these two methods calculate the same thing, it stands to reason that

∫ b

af(x) dx = F (b)− F (a),

which is the statement in the FTC.

369

7.2. Double integrals and volume

7.2 Double integrals and volumeDefinition of the double integral

Now we want to transition to volume problems. We start with the following goal:let f : R2 → R be some function whose graph is a surface, and let E be some subsetof R2. We want to find the volume of the 3-D solid consisting of the points aboveE (thought of as a subset of the xy-plane) and below the surface z = f(x, y):

Looking ahead: This volume will eventually be denoted∫∫E

f(x, y) dA.

and this expression is called a “double integral”.

Turns out, the problem described above is too hard to start with. So let’s sim-plify things (for now) by assuming E is a rectangle [a1, b1]× [a2, b2]:

x

y

E

a1 b1

a2

b2

First, we’ll define precisely what we mean by a double integral. Similar to theRiemann integrals you learned about in Calculus 1, a double integral will be alimit of Riemann sums.

370


Definition 7.5 Given a rectangle E = [a1, b1] × [a2, b2] ⊆ R2, a partition P ofE is two (finite) lists of numbers x0, x1, x2, ..., xm and y0, y1, ..., yn such thata1 = x0 < x1 < ... < xm−1 < xm = b1 and a2 = y0 < y1 < ... < yn−1 < yn =b2. Such a partition divides E into mn subrectangles; the (i, j)th subrectangle is[xi−1, xi]×[yj−1, yj]. For each i and j, setAi,j = (xi−xi−1)(yj−yj−1); Ai,j is the areaof the (i, j)th subrectangle. Call the maximum of all the (xi−xi−1) and the (yi− yi−1)the norm of the partition; denote the norm by ||P||.

EXAMPLE 2Let P be the partition of E = [0, 12]× [1, 6] given by P = 0, 4, 10, 12 × 1, 2, 4, 6.

x

y

E

0 4 10 12

12

4

6

(2,3)th subrectangle

A(3,2)=2(2)=4

Definition 7.6 Given function f : [a1, b1] × [a2, b2] → R and given partition P =x0, ..., xm × y0, ..., yn of [a1, b1]× [a2, b2], a Riemann sum associated to P for fis any expression of the form

m∑i=1

n∑j=1

f(ci,j)Ai,j

where for all i and j, ci,j belongs to the (i, j)th subrectangle of P . The points ci,j arecalled test points for the Riemann sum.

A Riemann sum approximates the volume under a function by adding up thevolumes of rectangular blocks:

371


As you choose partitions of smaller and smaller norm, you get skinnier andskinnier rectangular blocks (in both the x- and y-directions) whose total volumemore closely approximates the volume under the function. With that in mind, wedefine:

Definition 7.7 (Limit definition of the double integral) LetE = [a1, b1]×[a2, b2].Given function f : E → R, the double integral over E of f is

∫∫E

f(x, y) dA = lim||P||→0

m∑i=1

n∑j=1f(ci,j)Ai,j

if this limit exists (in Math 320, it always will). If the limit exists, we say f is inte-grable on E.)

Non-rectangular domains

Having defined the double integral of a function over a rectangular domain, wenow define double integrals over arbitrary bounded domains (recall from Chapter2 that a subset of Rn is bounded if you can draw a box around it). We do this with a“trick” of sorts:

Definition 7.8 Let E be a bounded subset of R2 and let f : E → R. Define f ext :R2 → R by setting

f ext(x, y) =f(x, y) if (x, y) ∈ E

0 if (x, y) /∈ E

f ext is called the extension of f to R2.

EXAMPLE 3Let f(x, y) = 1 + 1

6 sin 3x + 18 cos 2y and let E = (x, y) : x2 + y2 ≤ 1. Then f and

f ext have the graphs shown below:

372


Definition 7.9 Let E be a bounded subset of R2 and let f : E → R. Define thedouble integral of f over E to be∫∫

E

f(x, y) dA =∫∫

[a1,b1]×[a2,b2]

f ext(x, y) dA

where a1, a2, b1 and b2 are numbers such that E ⊆ [a1, b1]× [a2, b2].

Note: The integral on the right-hand side of this definition is the integral overa rectangle, so it has already been defined in Definition 7.7.

Properties of double integrals

Double integrals have lots of properties that are inherited from similar propertiesof Riemann integrals. I encourage you to read through these on your own time, toensure they make sense.

Theorem 7.10 (Continuity implies integrability) .

Riemann integrals: Let a < b. If f : [a, b] → R is continuous, then f is integrableon [a, b].

Double integrals: Let E ⊆ R2. If f : E → R is continuous, then f is integrable onE.

Definition 7.11 (Integral over no range is zero) .

Riemann integrals: Let f : [a, b]→ R. We define∫ aa f(x) dx = 0.

Double integrals: Let E be any subset of R2 which has zero area. Then for anyf : R2 → R,

∫∫Ef(x, y) dA = 0.

Theorem 7.12 (Linearity of integration) .

Riemann integrals: Let f and g be integrable on [a, b], and let k be any constant.Then:

•∫ ba [f(x) + g(x)] dx =

∫ ba f(x) dx+

∫ ba g(x) dx

•∫ ba [f(x)− g(x)] dx =

∫ ba f(x) dx−

∫ ba g(x) dx

•∫ ba k f(x) dx = k

∫ ba f(x) dx

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Double integrals: Let E ⊆ R2 and suppose f and g are integrable on E. Also, let kbe any constant. Then:

•∫∫E

[f(x, y) + g(x, y)] dA =∫∫Ef(x, y) dA+

∫∫Eg(x, y) dA

•∫∫E

[f(x, y)− g(x, y)] dx =∫∫Ef(x, y) dA−

∫∫Eg(x, y) dA

•∫∫Ek f(x, y) dA = k

∫∫Ef(x, y) dA

Theorem 7.13 (Additivity property) Riemann integrals: Suppose f is integrable.Then for any numbers a, b and c,

∫ b

af(x) dx+

∫ c

bf(x) dx =

∫ c

af(x) dx

Double integrals: Let E1 and E2 be disjoint subsets of R2. Then, for any integrablefunction f , ∫∫

E1∪E2

f(x, y) dA =∫∫E1

f(x, y) dA+∫∫E2

f(x, y) dA.

Here is a picture that illustrates additivity. Notice that if we set E = E1 ∪ E2,then E is the entire ellipse consisting of the blue and green regions. So the volumeunder f above E equals (the volume under f above E1) + (the volume under fabove E2).

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Theorem 7.14 (Positivity law) .

Riemann integrals: Suppose f is integrable on [a, b]. If f(x) ≥ 0 on [a, b], then∫ ba f(x) dx ≥ 0.

Double integrals: Let E ⊆ R2. If f is integrable on E and f(x, y) ≥ 0 on E, then∫∫Ef(x, y) dA ≥ 0.

Remember that with Riemann integrals, area above the x-axis contributes apositive amount to the integral, and area below the x-axis contributes a negativeamount to the integral. The same is true for double integrals: volume above thexy-plane contributes a positive amount to the integral, and volume below the xy-plane contributes a negative amount to the integral.

The next law says that integrals preserve inequalities: if f ≤ g, then the integralof f is less than or equal to the integral of g over the same set.

Theorem 7.15 (Monotonicity law) .

Riemann integrals: Suppose f and g are integrable on [a, b]. If f(x) ≤ g(x) on[a, b], then

∫ ba f(x) dx ≤

∫ ba g(x) dx.

Double integrals: Let E ⊆ R2. If f and g are integrable on E and f(x, y) ≤ g(x, y)on E, then

∫∫Ef(x, y) dA ≤

∫∫Eg(x, y) dA.

The last law says that if f always takes values between m and M , then we canbound the value of the integral using those values:

Theorem 7.16 (Max-Min inequality) .

Riemann integrals: Suppose f is integrable on [a, b]. If m ≤ f(x) ≤ M on [a, b],then m(b− a) ≤

∫ ba f(x) dx ≤M(b− a).

Double integrals: Let E ⊆ R2 and suppose f is integrable on E. Suppose m ≤f(x, y) ≤M on E. Then:

m(area of E) ≤∫∫E

f(x) dA ≤M(area of E).

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The easiest functions to integrate are constant functions:

Theorem 7.17 (Integral of a constant function) .

Riemann integrals: Let k be a constant. Then∫ ba k dx = k(b− a).

Double integrals: Let E ⊆ R2. Then, for any constant k,∫∫E

k dA = k(area of E).

To explain this formula, if f(x, y) = k, then∫∫Ef(x, y) dA gives the volume

shown in the picture below. This solid has height k, so its volume is k times thearea of its base (i.e. k times the area of E).

Summary so far

At this point, we have defined this theoretical object called a “double integral”,denoted ∫∫

E

f(x, y) dA

which gives the (signed) volume between the xy-plane and the graph of f : R2 → Rover region E (the region E sits in the xy-plane). What we don’t know is how tocompute such an object (other than doing limits of Riemann sums, which seemshorrible given what we know about the same procedure in Calculus 1).

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7.3. Fubini’s theorem

7.3 Fubini’s theoremRecall: we need an effective method to compute a double integral like∫∫

E

f(x, y) dA.

Big picture strategy

1. Divide E into pieces which are called “elementary regions”.

2. There are two types of elementary regions. For each type of elementaryregion, there will be a trick (called Fubini’s theorem) that turns the dou-ble integral over that elementary region into an “iterated integral” (thismeans one Calculus 1 integral inside another Calculus 1 integral).

3. Evaluate the iterated integrals you get from Step 2 using single-variablecalculus techniques (this means the FTC / antiderivatives / etc.).

4. Add up the integrals from the various pieces of E to get the double inte-gral over all of E.

Elementary regions in R2

Definition 7.18 Let E ⊆ R2. E is called vertically simple, a.k.a. x-simple, a.k.a.type 1, if it is of the form

E = (x, y) : l ≤ x ≤ r, b(x) ≤ y ≤ t(x)

for constants l and r and functions b, t : R→ R.

E

b(x)

t(x)

l rx

y

Equivalently, a region is vertically simple if all of its cross-sections parallel to they-axis (i.e. all of its vertical cross-sections) are intervals.

377


Definition 7.19 Let E ⊆ R2. E is called horizontally simple, a.k.a. y-simple,a.k.a. type 2, if it is of the form

E = (x, y) : b ≤ y ≤ t, l(y) ≤ x ≤ r(y)

for constants b and t and functions l, r : R→ R.

E

l(y)

r(y)

xb

t

y

Equivalently, a region is horizontally simple if all of its cross-sections parallel tothe x-axis (i.e. all of its horizontal cross-sections) are intervals.

Definition 7.20 Let E ⊆ R2. E is called an elementary region if E is horizontallysimple or vertically simple (or both).

EXAMPLE 4Classify each given pictured region as vertically simple, horizontally simple, both,or neither (check the box next to the best answer):

a)-2 -1 1 2 3 4

-2

-1

1

2

3

4

Horizontally simple Vertically simple Both horizontally and vertically simple Not an elementary region

b)-2 -1 1 2 3 4

-2

-1

1

2

3

4


378


c)-3 -2 -1 1 2 3

-3

-2

-1

1

2

3


d)-2 -1 1 2 3 4

-2

-1

1

2

3

4


e)

1 2

1

2 Horizontally simple Vertically simple Both horizontally and vertically simple Not an elementary region

Iterated integrals

The l, r, b and t in the definitions of elementary regions suggest how one mightthink of a double integral over an elementary region. More precisely:

Definition 7.21 An iterated integral is one Riemann integral inside another, i.e.an expression of the form

∫ r

l

∫ t(x)

b(x)f(x, y) dy dx or

∫ t

b

∫ r(y)

l(y)f(x, y) dx dy.

379


Note: Iterated integrals have invisible parenthesis in them. To evaluate them,evaluate the integral in the parenthesis, then evaluate the outside integral:

∫ r

l

∫ t(x)

b(x)f(x, y) dy dx or

∫ t

b

∫ r(y)

l(y)f(x, y) dx dy

Note: In an iterated integral, the “inside” limits can have the outside variable inthem, but the “outside” limits cannot have either the inside or the outside variablein them.

EXAMPLE 5Classify each of the following expressions as “legal” or “illegal”:

a)∫ 1

0

∫ 1

0f(x, y) dx dy Legal

Illegal

b)∫ 1

0

∫ w


Illegal

c)∫ y

0

∫ y


Illegal

d)∫ w

w3

∫ y4

−3f(x, y) dx dy Legal

Illegal

e)∫ x

0

∫ x

0f(x, y) dy dx Legal

Illegal

f)∫ w

0

∫ y


Illegal

g)∫ 1

0

∫ x


Illegal

h)∫ y

0

∫ 2x

−3xf(x, y) dy dx Legal

Illegal

i)∫ 1

0

∫ x+x2 sinx

−x3f(x, y) dy dx Legal

Illegal

j)∫ w

0

∫ x


Illegal

k)∫ y+3

y

∫ cosx


Illegal

380


EXAMPLE 6Evaluate each iterated integral:

a)∫ 2

0

∫ 4

2

(3xy2 + 2x− 4y

)dy dx

b)∫ 3

0

∫ y

112xy2 dx dy

381


c)∫ w

0

∫ y

y−1e−x dx dy

Solution: ∫ w

0

∫ y

y−1e−x dx dy =

∫ w

0

(∫ y

y−1e−x dx

)dy

=∫ w

0

[−e−x

]yy−1

dy

=∫ w

0

[−e−y − (−e−(y−1))

]dy

=∫ w

0

[−e−y + e1−y

]dy

=[e−y − e1−y

]w0

= e−w − e1−w − (1− e).

Remark: Integrals like the one in part (c), i.e. those having a third variablelike the “w” occurring as a limit in one or both of the double integrals are seenfrequently in MATH 414 and on the Society of Actuaries’ Exam P.

Fubini’s Theorem

Now put everything together that we have been doing with elementary regionsand iterated integrals. The next pair of important theorems say that you canrewrite a double integral over an elementary region as an iterated integral:

Theorem 7.22 (Fubini’s Theorem (vertically simple regions)) Let f : R2 → Rbe integrable, and let E be vertically simple. Then, writing

E = (x, y) : l ≤ x ≤ r, b(x) ≤ y ≤ t(x),

we have

∫∫E

f(x, y) dA =∫ r

l

∫ t(x)

b(x)f(x, y) dy dx.

b(x)

t(x)

l rx

y

382


Theorem 7.23 (Fubini’s Theorem (horizontally simple regions)) Let f : R2 →R be integrable, and let E be horizontally simple. Then, writing

E = (x, y) : b ≤ y ≤ t, l(y) ≤ x ≤ r(y),

we have

∫∫E

f(x, y) dA =∫ t

b

∫ r(y)

l(y)f(x, y) dx dy.

l(y) r(y)

xb

t

y

IMPORTANT:

A double integral over a region which is both vertically simple and hori-zontally simple can be rewritten using either version of Fubini’s theorem;you get the same answer either way.

However, one way can sometime produces an iterated integral which ismuch easier to evaluate than the other way.

Why Fubini’s Theorem works

Fubini’s Theorem has a very hard proof, which will not be discussed in these notes.To see the proof, go to graduate school in math or do a Google search. But the ideabehind Fubini’s Theorem is straight-forward. If you want to compute the volumeunder f above vertically simple region E, which is∫∫

E

f(x, y) dA,

then you can first integrate, for each x,

∫ t(x)

b(x)f(x, y) dy.

383


Each of these integrals gives you, in terms of x, the area of a “slice” through thesolid at x whose volume you want, parallel to the yz-plane:

So integrating these areas from l to r “sweeps out” the solid, so should (anddoes) give you the volume you want. See the Mathematica file on my website

m320fubinigraphic.nb

which has a graphic allowing you to see this in motion.

384


Putting all this togetherEXAMPLE 7

Let E ⊆ R2 be the region bounded by y = x and x = y2− 6. Find the volume of thesolid bounded below by E and bounded above by the graph of z = x+ 2y + 8.

385


EXAMPLE 8Let E ⊆ R2 be the triangular region with vertices (0, 0), (0, 1) and (1, 1). Considerthe double integral ∫∫

E

(x2 + xy) dA.

1. Compute the double integral above by thinking of E as a vertically simpleregion.

2. Compute the double integral above by thinking of E as a horizontally simpleregion.

386


EXAMPLE 9 (TAKEN FROM AN OLD ACTUARIAL EXAM)An insurance company insures a large number of drivers. Let X be the randomvariable representing the company’s losses under collision insurance, and let Yrepresent the company’s losses under liability insurance. X and Y have joint den-sity function

f(x, y) =

14(2x+ 2− y) for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 2

0 otherwise.

What is the probability that the total loss is at least 1?

What you will learn in Math 414: This problem is asking you to compute∫∫E

f(x, y) dA

where E is the region shown below:

0≤x≤1, 0≤y≤2

E

x+y = 1-1

2

1

21

3

22

-1

2

1

2

1

3

2

2

387

7.4. Triple integrals

7.4 Triple integralsDefinition of triple integral

Let f : R3 → R. To define the triple integral of f over some set E ⊆ R3, which hasthe notation ∫∫∫

E

f(x, y, z) dV,

take the machinery used to define a double integral and “bump everything up adimension”. Writing these definitions formally is a task I leave for you in the HW.

Fubini’s theorem for triple integrals

Fubini’s theorem works for triple integrals pretty much the way it works for dou-ble integrals, except that instead of writing a double integral as two iterated inte-grals, you write a triple integral as three iterated integrals.

You do have to be careful about what constitutes an elementary region, how-ever.

In this setting, there are 6 different types of elementary regions, depending onwhether or not you have to evaluate the triple integral in the order dx dy dz ordx dz dy or dy dx dz or dy dz dx or dz dx dy or dz dy dx.

Here are the details for one of these types of regions:

Definition 7.24 Let E ⊆ R3. E is called an elementary region of type 1 if E canbe described as

E = (x, y, z) : w ≤ x ≤ e, s(x) ≤ y ≤ n(x), b(x, y) ≤ z ≤ t(x, y)

for constants w and e (the “west” and “east” edges of E), functions s, n : R→ R (the“south” and “north” edges ofE) and functions b, t : R2 → R (the “bottom” and “top”of E).

388


Theorem 7.25 (Fubini’s Theorem for triple integrals (type 1 regions)) Supposethat f : R3 → R be integrable. Then for any elementary region E ⊆ R3 of type 1,where

E = (x, y) : w ≤ x ≤ e, s(x) ≤ y ≤ n(x), b(x, y) ≤ z ≤ t(x, y),

we have ∫∫∫E

f(x, y, z) dV =∫ e

w

∫ n(x)

s(x)

∫ t(x,y)

b(x,y)f(x, y) dz dy dx.

Triple integrals over other types of elementary regions can be rewritten in analo-gous ways.

EXAMPLE 10Compute ∫∫∫

E

4x dV

where E ⊆ R3 is the tetrahedron (i.e. triangular pyramid) formed by the coordi-nate planes and the plane 2x+ 3y + 4z = 12.

∫∫∫E

4x dv =∫ 6

0

∫ 4−2x/3

0

∫ 3−x/2−3y/4

04x dz dy dx

=∫ 6

0

∫ 4−2x/3

0

(12x− 2x2 − 3xy

)dy dx

=∫ 6

0

(23x

3 − 8x2 + 24x)dx

=[13x

4 − 83x

3 + 12x2]6

0= 72 .

389


What do triple integrals actually compute?

1. Probabilistic and statistical quantities

Suppose you have a situation like Example 9, except that the insurance companyhas losses under 3 types of insurance, rather than 2. The analogous problem (ofcomputing a probability associated to the sum of the three losses) requires a tripleintegral.

2. Averaging problems

One of the most important applications of integrals that you don’t learn muchabout in MATH 220 or 230 is that they compute averages.

Let f : R → R. The average value of f over the interval [a, b] should be somenumber h so that if you make a rectangle over [a, b] with height h, its area is thesame as the area under f from a to b. What should h be?

Definition 7.26 Let f : R → R be integrable, and let a < b. The average value off over [a, b] is ∫ b

a f(x) dxb− a

=∫ ba f(x) dx

length([a, b]) .

Similarly:

Definition 7.27 Let f : R2 → R be integrable, and let E ⊆ R2 be bounded. Theaverage value of f over E is ∫∫

Ef(x, y) dA

area(E) .

Definition 7.28 Let f : R3 → R be integrable, and let E ⊆ R3 be bounded. Theaverage value of f over E is ∫∫∫

Ef(x, y, z) dV

volume(E) .

390


EXAMPLE 11A solid block occupies E = [0, 2] × [0, 1] × [0, 2] in R3. If the temperature at point(x, y, z) in the block is T (x, y, z) = x + 4y + 2z C, find the average temperatureacross the block.

Solution: The average value is∫∫∫ET (x, y, z) dV

volume(E) =∫ 2

0∫ 1

0∫ 2

0 (x+ 4y + 2z) dz dy dx(2− 0)(1− 0)(2− 0)

= 204 (work omitted)

= 5 C.

WARNING:

In the previous example, it was easy to find the volume of E because Ewas a rectangular block. In general, you may need an integral (or a dou-ble or triple integral) to find the area or volume of E that constitutes thedenominator in an average value computation.

3. “4-dimensional volume”

A single integral computes the area of a planar (i.e. 2-D) shape; a double integralcomputes the volume of a 3-D solid. So a triple integral computes the “volume”of a 4-D region, which can be thought of as the region in R4 which is “above” xyz-space and below the “graph” of f(x, y, z). These types of quantities are useful insome settings in advanced physics.

4. Traditional 3-D volume

Triple integrals can be used to compute volumes of some 3-D solids, when thecorresponding double integral that gives the volume is too hard. We’ll see howlater in this chapter.

5. Centers of mass

You may remember from MATH 230 that you learned how to use integrals tocompute centers of mass of planar regions. We will see in the homework thatdouble and triple integrals are useful for computing the center of mass of three-dimensional solids.

391

7.5. Change of variables

7.5 Change of variablesRediscovering u-substitutions

EXAMPLE FROM MATH 220 (AND/OR 230)∫12x2 cosx3 dx

EXAMPLE FROM MATH 220 (AND/OR 230)

∫ π/4

0cos 2x dx

QUESTION

What is really going on in a u-substitution (besides a computational trick thatworks)?

392


Recall that formally, definite integrals are limits of Riemann sums:

∫ π/4

0cos 2x dx = lim

||P||→0

n∑j=1

cos(2cj) ∆xj

Each term being added in the Riemann sum corresponds to the area of a rectanglelike the one below, at left:

c j

f (c j)Δx j

π

4

x

1

f (x) = cos 2x

2c j

f (2c j)Δu j

π

2

u = 2x

1

f (u) = cos u

2c j

c j

Δu j

Δx j

π

4x

π

2

u = 2x

Notice that when you perform the u-substitution, the heights of the rectanglesin the Riemann sum for f stay the same, but the widths are distorted: they aremultiplied by du

dx= 2. This makes your new integral be, in some sense, “du

dx” times

too large, so to compensate, you have to divide by dudx

inside the integral.

This is why, for example, in an integral like∫cos 2x dx,

after performing the u-substitution u = 2x, your integral needs a 12 in it (since

dudx

= 2).

393


There’s another catch. Suppose the integral was∫ π/4

0cos(−2x) dx

Now your u-substitution is u = −2x, so dudx

= −2 so after the substitution, you get∫ −π/2

0cosu

(−12

)du

To get rid of the − sign on the 12 in the integral, reverse the limits of integration to

get ∫ 0

−π/2

12 cosu du

and proceed from there.

In general, if you want to end up with an integral whose lower limit is less thanits upper limit, you may have to use a − sign from the du

dxto reverse the limits. In

general, this means:

Theorem 7.29 (u-substitution in Riemann integrals) Let f be integrable. Then

∫ b

af(x) dx =

∫ d

cf(x) 1∣∣∣du

dx

∣∣∣ duwhere a ≤ x ≤ b corresponds to c ≤ u ≤ d.

Remark: When actually implementing this, you rewrite the integral on the right interms of u, so that it doesn’t have x in it any more.

Change of variables in multiple integrals

Goal: generalize Theorem 7.29 to double and triple integrals.

First question: What constitutes a substitution in this context?∫∫E

f(x, y) dA (Fubini)=∫ r

l

∫ t(x)

b(x)f(x, y) dy dx

With this in mind, we call substitutions in multiple integrals changes of variable(rather than “substitutions”).

394


Second question: What happens to Riemann sums when you change vari-ables?

To figure this out, let’s consider pictures similar to those we made earlier. Sup-pose we are performing the substitution φ(x, y) = (2x+y, x+3y), i.e. u = 2x+y, v =x+ 3y.

1 2 3 4 5x

1

2

3

4

5

y

(u,v)=φ(x,y)−→

1 2 3 4 5x

1

2

3

4

5

y

Definition 7.30 Let φ : Rn → Rn. The Jacobian of φ, denoted J(φ), is the determi-nant of the total derivative of φ:

J(φ) = detDφ.

If n = 2 and φ(x, y) = (u, v), this is

J(φ) = det(ux uyvx vy

).

Theorem 7.31 (Change of variables in double integrals) Suppose that f : R2 →R is integrable on some set E ⊆ R2. Suppose φ : R2 → R2 is a function which mapsE to φ(E) in a one-to-one fashion. Then∫∫

E

f(x, y) dA =∫∫φ(E)

f(u, v) 1|J(φ)| dA.

The content of this theorem is often shorthanded as:

(u, v) = φ(x, y) ⇒ dA = dx dy = 1|J(φ)| du dv.

395


EXAMPLE 12Compute the integral ∫∫

E

cos(x− y) sin(x+ 3y) dA

where E is the triangle whose vertices are (0, 0), (8, 0) and (2, 2).

396



E

(x2 − y2)exy dA

where E is the region pictured below:

E

y =1x

y =6x

y = x

y = x+3

0 1 2 3 4 5

1

2

3

4

5

At this point, we can compute the iterated integral:∫ 6

1

∫ 3

0−veu dv du =

∫ 6

1

[−1

2v2eu]3

0du

=∫ 6

1

[−9

2eu]du

=[−9

2eu]6

1= −9

2e6 + 9

2e .

397



E

x dA

where E = (x, y) : x2 + y2 ≤ 1, 0 ≤ y ≤ x.

J(φ) = det(rx ryθx θy

)= det

1·2x

2√x2+y2

1·2y2√x2+y2

1·(−yx2 )

1+( yx)21· 1x

1+( yx)2

= det(

xr

yr−y

x2+y2x

x2+y2

)

= det(

xr

yr−y

r2xr2

)

= x2

r3 + y2

r3 = r2

r3 = 1r.

Applying the change of variable theorem, we have∫∫E

x dA =∫ π/4

0

∫ 1

0x

1|J(φ)| dr dθ =

∫ π/4

0

∫ 1

0r cos θ 1

1r

dr dθ

=∫ π/4

0

∫ 1

0r cos θ r dr dθ

=∫ π/4

0

∫ 1

0r2 cos θ dr dθ.

Finally, compute the iterated integral:∫ π/4

0

∫ 1

0r2 cos θ dr dθ =

∫ π/4

0

[r3

3 cos θ]1

0dθ

=∫ π/4

0

13 cos θ dθ

= 13 sin θ

∣∣∣∣π/4

0= 1

3

(√2

2 − 0)

=√

26 .

398


IMPORTANT:

You should remember the Jacobian that corresponds to polar coordinatesyou used in Example 14. (It’s given in the theorem below.)Furthermore, you should change variables in a double integral to polarcoordinates if your region E has some nice rotational description aroundthe origin (like Example 14), or if the rule for your function f can easily beconverted to polar coordinates (especially if it has x2 + y2 in it).

Theorem 7.32 (Double integrals in polar coordinates:) SupposeE ⊆ R2 is a re-gion which is described in polar coordinates by

E = (r, θ) : α ≤ θ ≤ β and f(θ) ≤ r ≤ g(θ).

Then: ∫∫EdA =

∫ β

α

∫ g(θ)

f(θ)f(r, θ) r dr dθ.

This theorem is written in shorthand as:

dA = dx dy = r dr dθ.

In case you forgot how to convert between polar and Cartesian coordinates:

r =√x2 + y2

tan θ = yx


EXAMPLE 15Compute the integral ∫∫∫

E

z dV

where E is the top half of the sphere of radius 1 centered at (0, 0, 0).

399


EXAMPLE 15, RESTATED

Compute the integral ∫∫∫E

z dV

where E is the top half of the sphere of radius 1 centered at (0, 0, 0).

When using spherical coordinates, J(φ) = 1ρ2 sinϕ (HW).

Work this integral out (you need the u-sub u = sinϕ for the ϕ integral) to get

∫ 1

0

∫ π/2

02πρ3 cosϕ sinϕdϕdρ =

∫ 1

0

∫ 1

02πρ3u du dρ

=∫ 1

0

[πρ3u2

]10dρ

=∫ 1

0πρ3 dρ =

[π

4ρ4]1

0= π

4 .

IMPORTANT:

You should remember the Jacobian that corresponds to spherical coordi-nates you used in Example 15. (It’s given in Theorem 7.33 on the nextpage.)

Furthermore, you should change variables in a triple integral to spheri-cal coordinates if your region E looks like part of a sphere (like Example15), or if the rule for your function f can easily be converted to sphericalcoordinates (especially if it has x2 + y2 + z2 in it).

400


Theorem 7.33 (Triple integrals in spherical coordinates:) Suppose E ⊆ R3 is aregion which is described in spherical coordinates by

E = (ρ, ϕ, θ) : α ≤ θ ≤ β, f(θ) ≤ ρ ≤ g(θ), F (ρ, θ) ≤ ϕ ≤ G(ρ, θ).

Then: ∫∫∫E

dV =∫ β

α

∫ g(θ)

f(θ)

∫ G(ρ,θ)

F (ρ,θ)f(ρ, ϕ, θ) ρ2 sinϕdϕdρ dθ.


dV = dx dy dz = ρ2 sinϕdϕdρ dθ.

In case you forgot how to convert between spherical and Cartesian coordinates:

ρ =√x2 + y2 + z2

tanϕ =√x2+y2

z

tan θ = yx

x = ρ sinϕ cos θy = ρ sinϕ sin θz = ρ cosϕ

Here’s the corresponding information for changes to cylindrical coordinates(which are just like polar coordinates, except z is along for the ride). Use cylindricalcoordinates when your function has x2 + y2 in it (but not x2 + y2 + z2), or when theregion over which you are integrating looks like a piece of a cylinder or cone.

Theorem 7.34 (Triple integrals in cylindrical coordinates:) Suppose E ⊆ R3 isa region which is described in cylindrical coordinates by

E = (r, ϕ, z) : α ≤ θ ≤ β, f(θ) ≤ r ≤ g(θ), F (r, θ) ≤ z ≤ G(r, θ).

Then: ∫∫∫E

dV =∫ β

α

∫ g(θ)

f(θ)

∫ G(r,θ)

F (r,θ)f(r, θ, z) r dz dr dθ.


dV = dx dy dz = r dz dr dθ.

In case you forgot:

r =√x2 + y2

tan θ = yx

z = z


401


EXAMPLE 16Compute the integral ∫∫∫

E

z√x2 + y2 dV

where E = (x, y, z) : x2 + y2 ≤ 1, 0 ≤ z ≤ 1− x2 − y2.

Solution:

Finally, evaluate the integral:∫∫∫E

z√x2 + y2 dV =

∫ 2π

0

∫ 1

0

∫ 1−r2

0zr2 dz dr dθ

= 2π∫ 1

0

[12z

2r2]1−r2

0dr

= 2π∫ 1

0

[12r

2(1− r2)2]dr

= π∫ 1

0(r2 − 2r4 + r6) dr

= π

[r3

3 −2r5

5 + r7

7

]1

0

= π[13 −

25 + 1

7

]= 8

105π.

402

7.6. New ways of computing area and volume

7.6 New ways of computing area and volumeMOTIVATING EXAMPLE

Find the area of the region E, defined to be the set of points (x, y) such that 0 ≤y ≤ 4x− x2.

E

y = 4x-x2

0 1 2 3 4

1

2

3

4

Solution # 1: (from Calculus 1)

∫ 4

0(4x− x2) dx =

[2x2 − 1

3x3]4

0=[32− 64

3

]= 32

3 .

Solution # 2: (new) think of the area of E as a volume computation:

∫∫E

1 dA =∫ 4

0

∫ 4x−x2

01 dy dx =

∫ 4

0[y]4x−x

2

0 dx =∫ 4

0

(4x− x2

)dx = 32

3 .

The method of Solution # 2 generalizes:

Theorem 7.35 Let E ⊆ R2. Then the area of E is∫∫EdA =

∫∫E

1 dA.

So what? This seems to be an unnecessarily hard way of doing a simple problem.

403


EXAMPLE 17Find the area of a circle of radius R. (Pretend you don’t know it’s πR2.)

E

x2+y2=R2

R

Calculus 1 Solution:2∫ 4

−4

√R2 − x2 dx =?

Calculus 3 Solution:

404


Translating this new perspective to volumeEXAMPLE 18

Find the volume of the solid consisting of points in R3 lying above the trianglewith vertices (0, 2), (2, 2) and (2, 0) and lying below the plane z = 3x+ 4y + 8.

Solution # 1: (double integral)

Let B ⊆ R2 be the base of E, which is a triangle in R2 with vertices (0, 2),(2, 2) and (2, 0).

Then the volume is

V =∫∫B

(3x+ 4y + 8) dA =∫ 2

0

∫ 2

2−x(3x+ 4y + 8) dy dx

=∫ 2

0

[3xy + 2y2 + 8y

]22−x

dx

=∫ 2

0

[6x+ 24− 3x(2− x)− 2(2− x)2 − 8(2− x)

]dx

=∫ 2

0

[x2 + 16x

]dx

=[x3

3 + 8x2]2

0

= 32 + 83

= 1043 .

405


Solution # 2: (triple integral)

Let E ⊆ R3 be the solid described in the problem. Then E can be describedusing inequalities as

Therefore

V =∫∫∫

EdV =

∫ 2

0

∫ 2

2−x

∫ 3x+4y+8

01 dz dy dx

=∫ 2

0

∫ 2

2−x[z]3x+4y+8

0 dz dy dx

=∫ 2

0

∫ 2

2−x(3x+ 4y + 8) dy dx

= 1043 (same integral as Solution # 1)

As with area, this method generalizes:

Theorem 7.36 Let E ⊆ R3. Then the volume of E is∫∫∫EdV =

∫∫∫E

1 dV .

So what?

EXAMPLE 19Find the volume of a sphere of radius R.

Solution # 1 (Double integral): LetB be the “base” of a (hemi)sphere of radiusR; thus B is a circle of radius R in R2. So the volume of the sphere is

V = 2∫∫B

√R2 − x2 − y2 dA = 2

∫ R

−R

∫ √R2−x2

−√R2−x2

√R2 − x2 − y2 dy dx = ?

406


Solution # 2 (Triple integral): LetE be the interior of the sphere whose volumeyou want to find. Then the volume is

V =∫∫∫E

1 dV = 2∫ R

−R

∫ √R2−x2

−√R2−x2

∫ √R2−x2−y2

−√R2−x2−y2

1 dz dy dx

The point is that when you are finding the area or volume of some region,it can be helpful to do an integral in one higher dimension, so that a coordinatechange (into polar, cylindrical or spherical coordinates) becomes available to youthat wasn’t before.

407


EXAMPLE 20Find the volume of the solid E ⊆ R3 consisting of the points that lie above thexy-plane, inside the cylinder x2 + y2 = 1, and below the hyperbolic paraboloidz = y2 − 1

2x2 + 2.

Solution: First, describe E with some inequalities:

Having done this, we compute

volume(E) =∫∫∫E

dV =∫ 2π

0

∫ 1

0

∫ 2+r2 sin2 θ− 12 r

2 cos2 θ

0r dz dr dθ

=∫ 2π

0

∫ 1

0[rz]2+r2 sin2 θ− 1

2 r2 cos2 θ

0 dr dθ

=∫ 2π

0

∫ 1

0

[2r + r3 sin2 θ − 1

2r3 cos2 θ

]dr dθ

=∫ 2π

0

∫ 1

0

[2r + r3

(1− cos 2θ

2

)− 1

2r3(

1 + cos 2θ2

)]dr dθ

=∫ 2π

0

∫ 1

0

[2r + 1

4r3 − 3

4r3 cos 2θ

]dr dθ

=∫ 2π

0

[r2 + 1

16r4 − 3

16r4 cos 2θ

]1

0dθ

=∫ 2π

0

[1 + 1

16 −316 cos 2θ

]dθ

=∫ 2π

0

[1716 −

316 cos 2θ

]dθ

=[1716θ −

332 sin 2θ

]2π

0

=[17

8 π − 0]− [0− 0]

= 178 π.

408



1. Consider the partition P = 0, 1, 5, 8, 10 × 0, 2, 6, 7 of [0, 10]× [0, 7].

a) How many subrectangles does P have?

b) What is the (1, 2)th subrectangle of P?

c) What are the coordinates of the point in the upper-right corner of the(2, 3)th subrectangle of P?

d) What is A(3,2)?

e) What is the norm of P?

2. (R) Consider the partition of [0, 30]× [0, 12] into 30 subrectangles, each havingequal size, that are arranged in a 5× 6 grid.

a) What is the length and width of each subrectangle in this partition?

b) What is the norm of this partition?

c) What are the coordinates of the upper-left corner of (4, 3)th subrectangleof this partition?

3. Consider the partition P of [0, 4]× [0, 4] into four equal-sized squares.

a) Compute the Riemann sum for f(x, y) = x2 − y associated to P , wherethe test points are chosen to be the lower-left corners of each subrectan-gle.

b) Compute the Riemann sum for g(x, y) = 4x+ 3y associated to P , wherethe test points are chosen to be in the exact center of each subrectangle.

Let B = [0, 4]× [0, 6]; C = [4, 6]× [0, 6]; and D = [0, 6]× [0, 4]. Suppose f : R2 → Rand g : R2 → R are integrable functions such that∫∫

B

f(x, y) dA = 5;∫∫C

f(x, y) dA = 9;∫∫D

f(x, y) dA = −6;

∫∫B

g(x, y) dA = −2;∫∫C

g(x, y) dA = 8;∫∫D

g(x, y) dA = 1.

In each part of Problems 4-6, you are given a double integral. Determine if thegiven information (together with basic properties of double integrals) is sufficientto evaluate that double integral; if it is, evaluate the integral.

409


4. a)∫∫C

4f(x, y) dA

b)∫∫D

[2f(x, y) + 10g(x, y)] dA

c)∫∫B∪C

f(x, y) dA

5. a)∫∫B∪D

f(x, y) dA

b)∫∫B∩C

f(x, y) dA

c)∫∫

[0,6]×[4,6]g(x, y) dA

6. a)∫∫

[0,4]×[4,6]g(x, y) dA

b)∫∫B∪C

[f(x, y)− g(x, y)] dA

c)∫∫C

[3 + 2g(x, y)] dA..

In each of Problems 7-12, you are given a set E ⊆ R2 and a double integral. Foreach problem, you are to

a) Sketch a picture of the solid in R3 whose volume is being computed by theintegral;

b) Give the geometric name of the solid you drew in part (a); andc) Find the value of the integral by using volume formulas from high-school

geometry.

7. E = [2, 5]× [4, 8]; ∫∫E

5 dA.

8. E = (x, y) : 0 ≤ x ≤ 4, 0 ≤ y ≤ 2x;∫∫E

6 dA.

9. E = (x, y) : x2 + y2 ≤ 9; ∫∫E

√9− x2 − y2 dA.

10. E = (x, y) : x2 + y2 ≤ 25; ∫∫E

4 dA.

11. E = (x, y) : x ≥ 0, 2x+ 5y ≤ 10;∫∫E

(2− 25x− y) dA.

410


12. E = (x, y) : 0 ≤ x ≤ 1, x ≤ y ≤ 1;∫∫E

2y dA.

13. (F) Consider the function f(x, y) = cos(xy) + 6.

a) What is the absolute maximum value of f (on all of R2)?

b) Based on your answer to part (a), what is the largest possible value of∫∫Ef(x, y), where E = (x, y) : x2 + y2 ≤ 16?

c) What is the absolute minimum value of f (on all of R2)?

d) Based on your answer to part (c), what is the smallest possible value of∫∫Ef(x, y), where E = (x, y) : x2 + y2 ≤ 16?


In part of Problems 14-18, determine if the given iterated integral is a legal expres-sion. If it is, evaluate it. If it isn’t, state that the expression is illegal.

14. a)∫ 2

1∫ 3

0 (xy2 + y4) dx dyb)

∫ 10∫ 1

0 xe−xy dy dx

15. a)∫ π/2

0∫ x

0 (2x4 − 3xy2) dx dyb)

∫ 2−3∫ 2y

0 (2x2 + y2) dx dy

16. a)∫ ln 5

1∫ x

0 ex+3y dy dx

b)∫ 4

1∫ x−1

0 y dy dx

17. a)∫ x

1∫ y

0 x sin y dx dyb)

∫ w0∫ y−w

0 2x2 dy dx

18. a)∫ 2

1∫ y

1yxdx dy

b)∫ w

0∫ 2xx 8y dy dx

.

.

In Problems 19-26, you are given a region E ⊆ R2. Describe how you would writethe double integral

∫∫Ef(x, y) dA as an iterated integral (not the sum of two or more

iterated integrals).

Example: if you were given E = [0, 1] × [0, 1], two different correct answerswould be

∫ 10∫ 1

0 f(x, y) dy dx and∫ 1

0∫ 1

0 f(x, y) dx dy.

19. E is the triangle with vertices (0, 0), (0, 6) and (3, 6).

20. E is the set of points in R2 lying above the graph of y = x and below thegraph of y = x4.

21. (R) E is the interior of the ellipse x2

4 + y2 = 1.

411


22. E is the set of points (x, y) satisfying x ≥ 0, y ≥ 0, and x+ y ≤ 9.

23. E = (x, y) : x ≥ 0, y ≥ 0, y ≤√x, 2y − x ≥ −8

24. E is the set of points in Quadrant I which have distance at most 5 from theorigin.

25. E is the set of points (x, y) satisfying x ≥ 0, x+ y ≥ 4, 2x+ y ≤ 8.

26. E is the rectangle with vertices (0, 0), (0, w), (4, 0) and (4, w).

27. Find the volume of the solid lying above the triangle in the xy-plane withvertices (0, 0), (2, 0) and (2, 2), and below the paraboloid z = x2 + y2.

28. Find the volume of the solid lying above the region in the xy-plane boundedby y = x2 and y =

√x, and below the plane 3x+ 2y + z = 6.

29. Find the volume of E = (x, y, z) : z ≥ 0, y ≥ x2, y + z ≤ 6.

In Problems 30-34, compute∫∫Ef(x, y) dA for the given function f and the given

region E ⊆ R2:

30. f(x, y) = y; E is the set of points pictured below:

y = 0

y = 9-x2

E

-3 3

9

31. f(x, y) = x2 − y2; E is the set of points lying above the graph of y = x2 andbelow the graph of y = x.

32. f(x, y) = 24x− y3; E is the triangle with vertices (0, 0), (0, 1) and (1, 1).

33. f(x, y) = 6y; E is the set of points lying above the x-axis, to the right of they-axis, to the left of x = 1, and below the graph of y = ex.

34. (R) f(x, y) = xy; E is the triangle with vertices (0, 0), (10, 0) and (4, 2).

35. (F) Let E be the triangle with vertices (0, 0), (2, 0) and (2, 2). Find the valueof c which makes

∫∫Ecx2y dA = 1.

Remark: Finding a constant like this c is an important step in solving certainkinds of problems from MATH 414 and on the P Actuarial Exam.

412


36. Compute∫∫E

(x2 + 2y2) dA, where E is the set of points bounded by the graphs

of y = x, y = 4x and xy = 64.

Hint: Partition E into two vertically simple regions; compute the integralover each region and add.

37. (R) Find the volume of the solid which lies above the polygonal region E ⊆R2 pictured below, and below the graph of z = 10x4 + 12y2.

E

1 2

1

2

38. (F) Recall that the graph of a function f : R2 → R is a surface in R3. Timepermitting, we will show in Chapter 9 that the surface area of the graph of f ,above a region E ⊆ R2, is given by∫∫

E

√1 + [fx(x, y)]2 + [fy(x, y)]2 dA.

Use this formula to find the area of the triangle whose vertices are (a, 0, 0),(0, b, 0) and (0, 0, c) (where a, b and c are positive constants).

In Problems 39-42, use Mathematica to compute a decimal approximation to theindicated double integral. To do this, you will need to write the double integralas an iterated integral and then use the NIntegrate[ ] command on the outsideintegral, but the Integrate[ ] command on any inside integral. For example, tocompute

∫ 10∫ 3x−2(x+ y) dy dx, you might execute:

NIntegrate[ Integrate[x+y, y, -2, 3x], x,0,1 ]

39.∫∫Ee−x

2−3xy−y2dA, where E = [−1, 1]× [0, 1]

40.∫∫Eex−y dA, where E = (x, y) : 0 ≤ x ≤ π

2 , 0 ≤ y ≤ sin x

41.∫∫E

arcsin x dA, where E = (x, y) : 0 ≤ x, tan x ≤ y ≤ 1

42.∫∫E

sin 2x dA, where E is the triangle with vertices (0, 0), (2, 4) and (4, 4)

43. Let E be the triangle in R2 with vertices (0, 0), (4, 0) and (4, 2).

413


a) Compute∫∫E

12xy dA, by thinking of E as a vertically simple region and

applying Fubini’s theorem.

b) Compute∫∫E

12xy dA, by thinking of E as a horizontally simple region

and applying Fubini’s theorem. Verify that you get the same answeryou got in part (a).

44. Consider the iterated integral∫ 4

0

∫ 2√x

√y3 + 1 dy dx.

a) Explain why you cannot compute this iterated integral directly.

b) This iterated integral represents a double integral∫∫E

√y3 + 1 dA for some

region E in the xy-plane. Sketch E, labelling any curves comprising theboundary of E with their equations.

c) In the context of the given iterated integral in this problem, are youthinking of E as being horizontally simple, or vertically simple?

d) Write an iterated integral which gives the same double integral by think-ing of E as being horizontally simple.

e) Evaluate the double integral you wrote in part (e).

In Problems 45-48, you are given an iterated integral of some unknown function f .Reverse the order of integration in the integral (i.e. write an integral with dx anddy in the opposite order, that gives the same value as the given iterated integral,akin to what you did in part (d) of Problem 44).

45. a)∫ 4

0∫ x2

0 f(x, y) dy dxb)

∫ 80∫ 8−y

0 f(x, y) dx dy

46. a)∫ 4

0∫√16−x2

−√

16−x2 f(x, y) dy dx

b)∫ 2

0∫ y

0 f(x, y) dx dy

47. a)∫ 1

0∫ 1x f(x, y) dy dx

b)∫ w

0∫ 3√x

0 f(x, y) dy dx

48. a)∫ 3

0∫ 3

3−x f(x, y) dy dxb)

∫ 10∫ yy2 f(x, y) dx dy

In Problems 49 and 50, evaluate the given iterated integral by changing order ofintegration, similar to what you did in Problem 44.

49. a)∫ π/4

0∫ π/4x x2 cos(1 + y4) dy dx

b)∫ 2

0∫ 4y2

√5 + x3/2 dx dy

50. a)∫ π/2

0∫ π/2π/2−x

sin yydy dx

b)∫ 1

0∫ 4

4y e−y/x dx dy

414



51. (R) Let E = [a1, b1]× [a2, b2]× [a3, b3] ⊆ R3.

a) Define precisely what is meant by a partition P of E.Hint: As a prototype, use Definition 7.5, which precisely describes whatis meant by a partition of a rectangle in R2.

b) With the terminology you established in part (a), describe what is meantby the (i, j, k)th subrectangle of a partition P .

c) For the terminology you established in parts (a) and (b), let Vi,j,k be thevolume of the (i, j, k)th subrectangle. Give a formula for Vi,j,k, in termsof the language you established in part (a).

d) Define what is meant by the norm of the partition you wrote down inpart (a).

e) Let f : E → R, and suppose P is a partition of E as described in part(a). Define precisely what is meant by a Riemann sum associated to Pfor f .Hint: Definition 7.6 is an appropriate prototype here.

f) Define what is meant by the triple integral∫∫∫Ef(x, y, z) dV of a function

f : E → R.Hint: Definition 7.7 is the prototype.

52. (R) Let f(x, y, z) = x2 + 2y + 3z. Compute the Riemann sum for f associatedto the partition P = 0, 1, 2×1, 3, 5×0, 1, 3, 4 of E = [0, 2]× [1, 5]× [0, 4],where the test points are chosen from each subinterval so that x, y and z areall as small as possible.

In each part of Problems 53-55, determine if the given iterated integral is a legalexpression. If it is, evaluate it. If it isn’t, state that the expression is illegal.

53. a)∫ 4

1∫ x

1∫ xy

1 12xy dz dy dxb)

∫ 10∫ y+z

0∫ 1

0 2xz dx dz dy

54. a)∫ 2

0∫ y

1∫ x

1 4y dy dz dxb)

∫ w0∫ 1

0∫ z

0 ey dy dx dz

55. a)∫ 2

0∫ 1

0∫ 2

0 ey dz dx dz

b)∫ w

0∫ z

0∫ x

0 4x dz dx dy..

In Problems 56-61, compute∫∫∫Ef(x, y, z) dV for the given function f and the given

region E ⊆ R3:

56. f(x, y, z) = 10x4y + 12y2 + 6xz2; E = [0, 1]× [0, 4]× [0, 2].

57. f(x, y, z) = sin x cos y sin 2z; E =[0, π4

]×[π6 ,

π4

]×[0, π3

].

415


58. f(x, y, z) = yz; E is the pyramid with vertices (0, 0, 0), (2, 0, 0), (0, 2, 0) and(0, 0, 6).

59. f(x, y, z) = 18y; E = (x, y, z) : 0 ≤ x ≤ 4, 0 ≤ y ≤ x2, 0 ≤ z ≤ x+ y2.

60. f(x, y, z) = 4x; E is the set of points lying above the triangle in the xy-planewith vertices (0, 0), (6, 0) and (6, 3), and below the graph of z = x2 + 2y2.

61. f(x, y, z) = 2xy; E = (x, y, z) : 0 ≤ y ≤ 2, y3 ≤ x ≤ 8, 0 ≤ z, 2z + x ≤ 8.

62. Let E ⊂ R3 be the set of points in the first octant lying beneath the plane2x+ 4y + 3z = 24. Let f : R3 → R.

a) Write the triple integral∫∫∫Ef(x, y, z) dV as an iterated integral of the

form dx dy dz. (Your main task to determine the correct limits on theintegrals.)

b) Write the triple integral∫∫∫Ef(x, y, z) dV as an iterated integral of the

form dy dx dz.

c) Write the triple integral∫∫∫Ef(x, y, z) dV as an iterated integral of the

form dz dx dy.

d) (R) In how many different ways could the triple integral∫∫∫Ef(x, y, z) dV

be written as an iterated integral?

63. Use Mathematica to compute a decimal approximation to∫∫∫Ezexy

2dV , where

E is the set (x, y, z) : 0 ≤ x ≤ 1, 0 ≤ z ≤ 2x, 0 ≤ y ≤ x+ z.

64. Use Mathematica to compute a decimal approximation to∫∫∫Ez3 dV , where E

is the top half of the sphere x2 + y2 + z2 = 1.

65. Let f(x) = x2 + x3. Compute the average value of f over the interval [−2, 2].

66. Compute the average value of f(x, y) = x2− y2, over the set E = (x, y) : 0 ≤x, x2 ≤ y ≤ 4.

67. (R) Compute the average value of f(x, y, z) = 4x2 + 2y2 + z2, over the set[0, 3]× [0, 2]× [0, 1].


68. Compute the Jacobian J(φ) of each given function.

a) φ(x, y) = (u, v), where u = x+ y and v = x− 2y

416


b) φ(x, y) = (ex + ey, ex − ey)

c) (x, y) φ7−→ (u, v), where u = x+ y and v = xy

d) φ(x, y, z) = (x, x+ 3y − 4z, 3x− 2z)

69. Compute the Jacobian J(φ) of each given function, writing your answer interms of u and/or v (not x and/or y):

a) φ(x, y) = (x+ y, x− y)b) φ(x, y) = (u, v), where u = x

x+y and v = x+ y

c) (x, y) φ7−→ (u, v), where u = x+ y and v = xy

d) φ(x, y) = (x3y, y)

70. Compute∫∫E

(x + 5y) dA, where E is the parallelogram in R2 with vertices

(0, 6), (2, 0), (8, 10) and (10, 4).

71. (R) Compute∫∫E

sin y sin(x − y) dA, where E is the parallelogram in R2 with

vertices (0, 0),(π2 , 0

),(π2 ,

π2

)and

(π, π2

).

72. Compute∫∫Ey2 dA, where E is the region of points in R2 lying above the

graphs of y = ex and y = e−x, and below the graphs of y = 4ex and y = 4e−x.

73. Compute∫∫E

y2

x+y dA, where E is the region pictured below:

y=x

y=4x

y=5-x

y=1-x

E

-1 1 2 3 4 5-1

1

2

3

4

5

74. (R) LetE ⊆ R2 be the set of points in the first quadrant lying above the graphsof xy = 1 and y = x2, below the graphs of xy = 27 and y = 2x2. Find thevolume of the solid in R3 consisting of the points lying above E and belowthe graph of z = xy2.

75. Compute∫ 1

0∫√1−x2

0 (x2 + y2)5 dy dx.

417


76. Compute∫ 2−2∫√4−x2

−√

4−x2 cos(x2 + y2) dy dx.

77. (R) Compute∫∫E

2x dA, whereE is the region of points in R2 lying above y = 0,

below y = x and inside the circle x2 + y2 = 3.

78. Compute∫∫Eyx dA, where E is the region of points in R2 lying above y = 0,

below y =√

3x and inside the circle x2 + y2 = 1.

79. Find the volume of the solid consisting of the set of points above the xy-plane,inside the cylinder x2 + y2 = 4, and below the plane z = 2x+ y + 8.

80. (F) Let φ : R3 → R3 be the function that converts Cartesian coordinates(x, y, z) to spherical coordinates (ρ, ϕ, θ). Work out the Jacobian of φ and ver-ify that it is indeed equal to 1

ρ2 sinϕ , justifying the formula in Theorem 7.33.

81. Compute∫∫∫E

exp[(x2 + y2)3/2

]dV , where E = (x, y, z) : x2 + y2 ≤ 1, 0 ≤ z ≤

√x2 + y2.

82. Compute∫∫∫E

2z dA, whereE = (x, y, z) : x ≥ 0, y ≥ 0, z ≥ 0, x2 +y2 +z2 ≤ 1.

83. Compute∫ 1

0∫√1−y2

0∫√1−x2−y2√

x2+y2(x2 + y2 + z2)5/2 dz dx dy.

84. Compute the integral∫∫∫E

18x

2y dV , where E is the set of points in R3 between

the planes z = 0 and z = 2, between the planes y = 0 and y = 4, and betweenthe planes z = 1

4x and z = 14x− 4.

85. a) (F) Compute the integral∫∞−∞

∫∞−∞ e

−x2−y2dy dx by converting the inte-

gral to polar coordinates.

b) (F) Based on your answer to part (a), what is∫∞−∞ e

−x2dx?

Hint:∫∞−∞ e

−x2dx and

∫∞−∞ e

−y2dy must have the same value (since the

only difference between the two integrals is the name of their variableof integration). Multiply these two integrals together and combine theproduct into one double integral. Compare that integral with what youcomputed in part (a).

86. A gas chamber occupies the set E = (x, y, z) : x ≥ 0, y ≥ 0, 0 ≤ z ≤ 4− x2 −y2 in three-dimensional space. If the temperature at the point (x, y, z) in thechamber is 2xz C, what is the average temperature across the chamber?

87. (F) Use the surface area formula presented in Problem 38 to compute thesurface area of the graph of f(x, y) = 13 + x2 − y2 that lies above the circle ofradius 2 centered at the origin.

418



In each part of Problems 88-89, you are given a double integral which computesthe area of some region E ⊆ R2. Sketch a picture of E, and use basic area formulasfrom high-school geometry to evaluate the integral.

88. a)∫ 5

0∫ 3

1 dy dx

b)∫ 3π/4

0∫ 4

0 r dr dθ

89. a)∫ 8

0∫ 2y

0 dx dy

b)∫ 3−3∫ 5+

√9−x2

0 dy dx

In each part of Problems 90-94, you are given a triple integral which computes thevolume of some region E ⊆ R3. Sketch a picture of E, and use basic area formulasfrom high-school geometry to evaluate the integral.

90. a)∫ 5

0∫ 3

0∫ 2

0 dx dy dz

b)∫ 5

0∫ 3

0∫ 2

0 dz dx dy

91. a)∫ 4

0∫ 2

0∫ π

0 r dθ dr dz

b)∫ 4

0∫ 2

0∫ π

0 r dθ dz dr

92. a)∫ 2

0∫ 2

0∫ 2−y

0 dz dy dx

b)∫ 2

0∫ 2−x

0∫ 2−x−y

0 dz dy dx

93. a)∫ 3

0∫ 3r

∫ 2π0 r dθ dz dr

b)∫ 4−4∫√16−x2

−√

16−x2

∫√16−x2−y2

−√

16−x2−y2dz dy dx

94. a)∫ 2π

0∫ π

0∫ 9

0 ρ2 sinϕdρ dϕ dθ

b)∫ π/2

0∫ π/2

0∫ 3

0 ρ2 sinϕdρ dϕ dθ

95. Use a double integral, with a change of variables, to compute the area of theplanar region E, where

E =

(x, y) : 12√x ≤ y ≤ 4

√x, 4 ≤ xy ≤ 108

.

96. a) Suppose r = f(θ) is a polar function (these were described in Chapter2). Use an appropriate double integral to derive a formula for the areaenclosed by the polar graph of r = f(θ) from θ = α to θ = β. Youranswer should be an integral in terms of θ, f, α and β, but should notcontain an r.

b) Find the area of one petal of the rose whose polar equation is r = 2 sin 3θ.Hint: You will need the trig identity sin2 θ = 1

2(1− cos 2θ) to evaluate anappropriate integral.

c) Find the area enclosed by the polar function r = 1 + cos θ.Hint: You will need the trig identity cos2 θ = 1

2(1 + cos 2θ) to evaluate anappropriate integral.

97. Use an appropriate double or triple integral to derive a formula for volumeof a cone with radius R and height H .

419


Hint: Let E ⊆ R3 be the cone you want. The hard part of this is to figureout how to describe E using nice inequalities (choosing the right coordinatesystem helps a lot).

98. Use Mathematica to compute the volume of the region in R3 lying above theplane z = 2 and inside the sphere of radius 3 centered at the origin (you haveto set up the multiple integral yourself).

99. (R) Use an appropriate triple integral to derive a formula for the volume ofthe ellipsoid x2

a2 + y2

b2 + z2

c2 = 1.

Hint: Let E be the interior of the ellipsoid. Start with the variable changeu = x

a, v = y

b, w = z

c. This makes φ(E) a sphere, so you can do a second

variable change into spherical coordinates to evaluate the integral.

100. Compute the volume of the solid in R3 consisting of points inside the conez2 = x2 + y2 and inside the sphere x2 + y2 + z2 = 9.

101. Compute the volume of the region in R3 consisting of points inside the spherex2 + y2 + z2 = 1 and outside the cylinder x2 + y2 = 1

4 .

102. Compute the volume of the solid in R3 consisting of points lying above thexy-plane, having positive y-coordinate, lying outside the cylinder x2+

(y − 1

2

)2=

14 but inside the sphere x2 + y2 + z2 = 1.

Hint: To evaluate∫

cos3 t dt, write this integral as∫

(cos2 t) cos t dt =∫

(1 −sin2 t) cos t dt and use an appropriate u-substitution.

103. (F) Compute the volume of the region in R3 pictured below, which consistsof the points below the blue paraboloid z = 64 − x2 − 3y2 and above theorange paraboloid z = 3x2 + y2. (You will need one or more trig identitiesmentioned in hints earlier in this homework.)

104. (F) Use an appropriate double integral to derive a formula for the surfacearea of a sphere of radius R.

420


Hint: Compute the surface area of the top half of the sphere by using theformula presented in Problem 38, and changing the integral to polar coordi-nates.

105. (F) Suppose E ⊆ R2 is a planar region, and that some sort of material (likea sheet of metal) occupies set E, so that the density of the material at point(x, y) ∈ E is given by ρ(x, y). In this setting, we define the center of mass ofE to be the point the point (x, y) where

(x, y) =(Mx

M,My

M

).

In this formula,M =

∫∫E

ρ(x, y) dA

is the mass of the region, and Mx and My are the moments of inertia givenby

Mx =∫∫E

xρ(x, y) dA and My =∫∫E

yρ(x, y) dA.

Given these formulas, compute the center of mass of the region E describedin Problem 95, assuming that the density of the region is everywhere 1 kg/squnit.

106. (F) Suppose E ⊆ R3, and that some sort of material (like a block or a gascloud) occupies E, so that the density of the material at point (x, y, z) ∈ E isρ(x, y, z). In this setting, we define the center of mass of E to be the point(x, y, z) where

(x, y, z) =(Mx

M,My

M,Mz

M

).

In this formula,M =

∫∫∫E

1 dV

is the mass of the region, and Mx, My and Mz are the moments of inertiagiven by

Mx =∫∫∫E

x dV, My =∫∫∫E

y dV and Mz =∫∫∫E

z dV.

Given these formulas, compute the center of mass of the region in R3 consist-ing of points lying inside the sphere x2+y2+z2 = 12 and above the paraboloidz = z2 + y2, assuming that the density of the region at point (x, y, z) is zkg/cubic unit.

Hint: Two of the three coordinates of the center of mass are obvious, basedon symmetry.

421


Selected answers

1. c) (5, 7)d) 12

2. b) 6c) (18, 6)

3. b) 224

4. c) 14

5. b) insufficient info

6. b) 8c) 52

8. a)

6

4

8

b) triangular prism

c) 96

9. b) hemisphere

10. b) circular cylinder

c) 100π

11. a)2

5

2

12. c) 23

13. b) 112π

14. a) 29110

16. a) 112 (−e4 + 4e+ 605)

b) 92

17. a) illegal

19.∫ 3

0∫ 2x

0 f(x, y) dy dx or∫ 6

0∫ y/2

0 f(x, y) dx dy

21.∫ 2−2∫√1−x2/4

−√

1−x2/4f(x, y) dy dx or

∫ 1−1∫√4−4y2

−√

4−4y2f(x, y) dx dy

22.∫ 9

0∫ 9−x

0 f(x, y) dy dx or∫ 9

0∫ 9−y

0 f(x, y) dx dy

25.∫ 4

0∫ 8−2x

4−x f(x, y) dy dx

26.∫ 4

0∫ w

0 f(x, y) dy dx or∫ w

0∫ 4

0 f(x, y) dx dy

27. 163

30. 6485

31. 170

33. 32e

2 − 32

35. 516

36. 6912

422


38. abc2√a2+b2+c2

39. 1.51633

40. 1.9615

42. −.625741

44. a)E

y = x

y = 2

x = 0

-2 -1 1 2 3 4 5

1

2

e) 529

45. a)∫ 16

0∫ 4√

y f(x, y) dx dy

46. a)∫ 4−4∫√16−y2

0 f(x, y) dx dy

48. a)∫ 3

0∫ 3

3−y f(x, y) dx dy

b)∫ 1

0∫√xx f(x, y) dy dx

49. a) 112

[sin

(1 + π4

256

)− sin 1

]b) 4

9

(133/2 − 53/2

)51. b) The (i, j, k)th subrectangle of P is [xi−1, xi]× [yj−1, yj]× [zk−1, zk].

c) Vi,j,k = (xi − xi−1)(yj − yj−1)(zk − zk−1).

f) The triple integral is defined as

∫∫∫E

f(x, y, z) dV = lim||P||→0

m∑i=1

n∑j=1

p∑k=1

f(ci,j,k)Vi,j,k,

assuming this limit exists.

52. 264

53. a) 6752 (xy − 1)

b) illegal

54. a) illegalb) w(e− 2)

56. 576

57. 916

√2− 3

4

58. 125

59. 137216

60. 181445

61. 115211

62. a)∫ 8

0∫ 6−x/2

0∫ 12−2y−3z/2

0 f(x, y, z) dx dy dz

b)∫ 8

0∫ 12−3z/2

0∫ 6−x/2−3z/4

0 f(x, y, z) dy dx dz

63. 24.6647

64. .261799

65. 43

66. −21235

423


67. 15

68. a) −3b) −2ex+y

69. b) 1v

d) 3u2/3v1/3

70. 1680

71. 1

72. 989

73. 1243

(3310 + ln 4− 2 ln 5

)74. 2460( 3

√2− 1)

75. π24

76. 12π sin 4

77.√

6

78. 332

79. 32π

81. 23π(e− 1)

82. π4

32 −316π

2

83. 1524288π

9 sin2 12

84. 163843

85. a) π

b)√π

86. 512105π

C

87. π(

176

√17− 1

6

)

88. b) E

4

; 6π

89. a) E8

16

; 64

90. a)

5

2

3

; 30

91. a)

4

2

; 8π

92. a)

2

2

2 ; 4

93. a)

3

3; 9π

94. a)

9

9 ; 972π

95. 104 ln 4

96. a) A =∫ βα

12 [f(θ)]2 dθ

b) π3

c) 3π

424


98. 83π

99. πabc

100. 18− 9π√

2

101. π√

32

102. 49

103. 512π

104. 4πR2

105.(

108965 ln 4 ,

9013 ln 4

)106.

(0, 0, 12 + 8

5

√23 −

645√

3

)

425

Chapter 8

Line integrals

In MATH 220 you learn the Fundamental Theorem of Calculus (FTC), which tellsyou how to evaluate integrals. In particular, for any C1 function f : R → R (i.e.any function with a continuous derivative),

To understand how this might generalize, we need to bring in some of thetopology we learned back in Chapter 2. Recall that the boundary of a set E, de-noted ∂E, is the “edge” of E. As an example, if E = [a, b] ⊆ R, then

The important observation is that the left-hand side of the FTC depends onvalues of f ′ on all of E = [a, b], but the right-hand side depends only on values off at a and b, i.e. values of f on ∂E. So what the FTC says (in English with sometopology ideas incorporated) is this:

the integral of some derivative(in the FTC, “some derivative”

means f ′(x))

over some set E(in the FTC, E = [a, b])

=

the values of the original function(in the FTC, “somefunction” means f )

along the boundary of E(in the FTC, ∂E = a, b

and the values are f(b)− f(a))

426

So to get a version of the FTC that has to do with double integrals, our goalwould be to compute something like∫∫

E

(integrand) dA

where the integrand is some function of x and y. If the integrand can somehow,someway be thought of as a “derivative” of some sort, maybe (given how we haveseen the FTC works) this double integral can be rewritten as “something” over theboundary of E:∫∫

E

(some derivative

of something

)dA = something being

evaluated along ∂E

Potential problem: the boundary of a typical E ⊆ R2 is a curve (or a union ofcurves), so to get all the values of the “something” over ∂E, we can’t just add orsubtract two (or finitely many) values as in the good old FTC. We need to learnsomething else. As motivation, suppose g : [a, b] → R is some function. Howwould you “add” (or accumulate) all the values of g from a to b?

So to fix the potential problem, the FTC for multiple integrals probably shouldlook something like this:∫∫

E

(some derivative

of something

)dA =

∫∂E

(the something)

So to generalize the FTC, we probably need some mechanism to integrate a func-tion over a curve. The mechanism that does this is called a line integral, and isthe subject of this chapter (although we first need to detour and cover some setupmachinery).

427

8.1. Vector fields and flows

8.1 Vector fields and flowsDefinition 8.1 A vector field (on Rn) is a function f : Rn → Rn (same n in thedomain and range).

Long ago, we discussed what the graph of a vector field looks like (conceptu-ally). A good way to think about a vector field on R3 is to think of each input asbeing a point in the sky, and the output as being the direction and magnitude ofthe wind at that point.

Vector fields are also used to represent the force f(x) acting on a particle at po-sition x coming from a gravitational, electrical or magnetic field.

EXAMPLE 1Sketch a graph of the vector field f : R2 → R2 defined by f(x, y) = (x+ y, x− y).

Solution: If doing this by hand, choose some xs and ys, and compute f(x, y).Then, draw the vector f(x, y) starting at each (x, y):

(choose these) (solve for this)x y f(x, y) = (x+ y, x− y)0 0 (0 + 0, 0− 0) = (0, 0)0 1 (0 + 1, 0− 1) = (1,−1)1 0 (1 + 0, 1− 0) = (1, 1)3 −2 (1, 5)2 2 (2, 0)−3 1 (−2,−4)

-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

Mathematica will sketch this vector field (with rescaled arrows to emphasize theirdirection rather than their magnitude) with the following code:

VectorPlot[x + y, x - y, x, -4, 4, y, -6, 6,. VectorScale -> .05, Automatic, None,. VectorStyle -> Red, Arrowheads[.03]]

-4 -2 0 2 4

-6

-4

-2

0

2

4

6

For a vector field f : R3 → R3, use VectorPlot3D[ ] in Mathematica.

428


Conservative vector fields

The most common and most important examples of vector fields are gradients ofscalar functions:

Definition 8.2 Let f : Rn → Rn be a vector field. f is called conservative, or agradient field, if there is a function f : Rn → R such that∇f = f .

In this setting, f is called a potential (or potential function) for f , and levelcurves of the potential f are called equipotential sets of f .

Idea: Think of f(x) as describing a force that acts on an object at position x. If fis a gradient field, f is (±) the “potential energy” for the force. Since f = ∇f ,

• f points in the direction in which f most rapidly increases; and

• f is orthogonal to the equipotential sets.

EXAMPLE 2Let f : R2 → R be f(x, y) = y2 − x2y.

1. Compute the gradient field of f .

2. Write an equation for the equipotential set at z = 3.

3. The contour plot of f is shown below, with graph of the equipotential set to off at z = 3 indicated in bold. Sketch∇f at various points on this equipotentialset.

-6 -6-2-2

00

1

1

3

3

6

6

12

12

3

3

-3 -2 -1 0 1 2 3-3

-2

-1

0

1

2

3-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

x

y

429


EXAMPLE 3Newton’s law of gravitation says the magnitude of the gravitational force exertedby an object at the origin on a particle at position x = (x, y, z) is

||f(x)|| = Gm1m2

||x||2

where m1 is the mass of the object at the origin, m2 is the mass of the other particle,and G ≈ 6.674× 10−11 Nm2/kg2 is a universal constant).

1. Find a formula for f(x).

2. Verify that f = ∇f , where f : R3 → R is

f(x) = Gm1m2

||x||.

Solution: Start by writing the potential f coordinate-wise:

f(x, y, z) = Gm1m2√x2 + y2 + z2 = Gm1m2

(x2 + y2 + z2

)−1/2.

Then compute the gradient:

∇f(x, y, z) = (fx, fy, fz)

= Gm1m2

(−12 (x2 + y2 + z2)−3/2(2x), ..., −1

2 (x2 + y2 + z2)−3/2(2z))

= −Gm1m2

(x(x2 + y2 + z2)−3/2, y(x2 + y2 + z2)−3/2, z(x2 + y2 + z2)−3/2

)= −Gm1m2

(x, y, z)||x||3

= −Gm1m2x||x||3 .

This equals f(x), so f is a potential for f .

430


Differential equations and flows

Question: Suppose a leaf (or other small object) is located at point x0 ∈ Rn at timet = 0. The wind (whose direction and magnitude at any point given by a vectorfield f ) blows the leaf around as time passes. What is the function x : R→ Rn thatgives the leaf’s position x(t) at time t?

EXAMPLE 4Suppose the wind at position (x, y) is given by f(x, y) = (x − 2y + 1, y − x2 + 2)(for simplicity, we ignore the affect of elevation on the wind). The vector field fis sketched below. If a particle blown by the wind is at position (−1, 0) at time 0,sketch the trajectory of the particle as it is blown around.

-4 -2 0 2 4

-4

-2

0

2

4

What you sketched in Example 4 is an example of something called a flow line.What is the relationship between the flow line and the vector field?

Definition 8.3 Let f : Rn → Rn be a vector field. A flow line, a.k.a. stream line,of f is a differentiable function x : R→ Rn satisfying

x′(t) = f(x(t))

for all t.

NOTE

Flow “lines” aren’t usually lines (they may not be straight).

431


So flow lines are exactly the solutions of the differential equation x′ = f(x), andeach flow line lies tangent to the vector field at every point on the flow line. InMATH 330, you learn a lot about differential equations of this type (in particular,these are called autonomous equations). Among other things, you learn:

Theorem 8.4 (Existence/Uniqueness Theorem for ODEs) Let f : Rn → Rn bea C1 vector field. Then, given any x0 ∈ Rn, there is one and only one C1 functionx : R→ Rn such that x′(t) = f(x(t)) for all t, and f(0) = x0.

This theorem means that given any C1 vector field and any point x0 ∈ Rn, thereis exactly one flow line to the vector field passing through x0.

EXAMPLE 5Verify that the function x(t) = (et cos t, et sin t) is a flow line for the vector fieldf(x, y) = (x− y, x+ y).

432


Given a vector field f , to find the flow line through a point you have to solve an“initial value problem” of the form

x′(t) = f(x(t))x(0) = x0

For instance, if your vector field is f(x, y) = (3x− 2y, x+ 5y) and x0 = (2, 1), you’dhave to solve

x′(t) = 3x(t)− 2y(t)y′(t) = x(t) + 5y(t)x(0) = 2y(0) = 1

If the vector field is not too bad (i.e. linear), then you learn how to explicitly solvefor flow lines in MATH 330 (Differential Equations). If the vector field is supersimple, you can solve it in this class. Here is the key fact you should know:

Theorem 8.5 (Exponential growth/decay) Let r and x0 be constants. The onlysolution x(t) of the initial value problem

x′(t) = rx(t)x(0) = x0

is the function x(t) = x0ert.

PROOF Clearly, x(t) = x0ert works in both equations of the system, so this is a

solution. To show that this is the only solution, suppose y(t) is some other solution.Then let g(t) = y(t)

x(t) .

g′(t) = d

dt

(y(t)x(t)

)= y′(t)x(t)− y(t)x′(t)

(x(t))2 = [ry(t)]x(t)− y(t)[rx(t)](x(t))2 = 0

so g is constant, i.e. y(t) = const x(t). But since x(0) = y(0) = x0, the constant mustbe 1. Therefore y isn’t a solution different from x, so this x is the only solution.

EXAMPLE 6Compute the flow line to f(x, y, z) = (3x,−2y, 7) satisfying x(0) = (1, 4,−2).

433


Definition 8.6 Let f : Rn → Rn be a vector field. The collection of all the flow linesto the vector field is called the flow associated to the vector field.

The definition above is vague. Formally, a flow is a function F : Rn × R → Rn

represented by F(x0, t) = x(t), where x : R → Rn is the unique flow line to fsatisfying f(0) = x0. Come ask me if you want to know more about this formaldefinition.

EXAMPLE 7Describe the flow associated to the vector field f(x, y) = (−y, x) (pictured below),and verify that your description is correct.

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

434

8.2. Operators on vector fields

8.2 Operators on vector fieldsDivergence

Definition 8.7 Let f = (f1, f2, ..., fn) : Rn → Rn be a differentiable vector field. Thedivergence of f , denoted div f or∇ · f , is the function div f : Rn → R defined by

div f(x) =n∑j=1

∂fj∂xj

= ∂f1

∂x1+ ...+ ∂fn

∂xn

where x = (x1, ..., xn).

Note: The “·” in ∇ · f is important. Without the dot,∇means gradient.

EXAMPLE 8Compute the divergence of f(x, y, z) = (x4y3, xy3z, 4xz2).

Solution: div f =

EXAMPLE 9Let f(x, y) = cos(x+ y)i + sin(x− y)j. Compute∇ · f

(π6 ,

π3

).

435


What does divergence measure?

f(x, y) = (x, y) f(x, y) = (−x,−y) f(x, y) = (−y, x)

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

∇ · f(x, y) = 1 + 1 = 2 ∇ · f(x) = −1− 1 = −2 ∇ · f(x) = 0 + 0 = 0

∇ · f > 0 ∇ · f < 0 ∇ · f = 0

Notice: the arrowsending at each x

are longer than thearrow starting at

the same x

Notice: the arrowending at each x

is the same lengthas the arrow starting

at the same x

Conclusion: div f measures the “net flow” in and out of each point.

• div f(x) > 0 ⇒ more “volume of stuff” flowing out from x than into x(relatively speaking);

• div f(x) < 0 ⇒ more “volume of stuff” flowing into x than out of x(relatively speaking);

• div f(x) = 0⇒ flow in at x = flow out at x

If you are familiar with homogeneous differential equations of the form x′ = f(x),the connection is:

• div f > 0 at sources (a.k.a. unstable equilibria);• div f < 0 at sinks (a.k.a. stable equilibria);• div f = 0 at centers (semistable equilibria).

Definition 8.8 Let f : Rn → Rn be a differentiable vector field. f is called incom-pressible, a.k.a. solenoidal, if div f = 0 everywhere.

436


Curl

Definition 8.9 Let f = (f1, f2, f3) : Rn → Rn be a differentiable vector field. Thecurl of f , denoted curl f or ∇× f , is the vector field curl f : R3 → R3 defined by

curl f(x) =(∂f3

∂y− ∂f2

∂z,∂f1

∂z− ∂f3

∂x,∂f2

∂x− ∂f1

∂y

)

= det

i j k∂∂x

∂∂y

∂∂z

f1 f2 f3

where x = (x1, x2, x3).

EXAMPLE 10Compute the curl of f(x, y, z) = (4x2y,−8z, x− y + 2z2).

EXAMPLE 11Let f(x, y, z) = (y3 − 2x, 3y2x+ z2, 2yz). Compute∇× f .

Solution:

curl f = det

i j k∂∂x

∂∂y

∂∂z

f1 f2 f3

= det

i j k∂∂x

∂∂y

∂∂z

y3 − 2x 3y2x+ z2 2yz

= i(2z − 2z) + j(0− 0) + k(3y2 − 3y2)= (0, 0, 0) = 0.

437


What does curl measure?

f(x) = (x, y, 0) f(x) = (−y, x, 0) f(x) = (y,−x, 0)

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

curl f

= det

i j k∂∂x

∂∂y

∂∂z

x y 0

curl f

= det

i j k∂∂x

∂∂y

∂∂z

−y x 0

= k− (−k)= (0, 0, 2).

curl f

= det

i j k∂∂x

∂∂y

∂∂z

y −x 0

= −k− (k)= (0, 0,−2).

Notice: as time passes,we see clockwise rotation

Conclusion: curl f measures the tendency of an object to rotate when movedby the vector field f . Take your right hand, make a “thumbs up” symbol. Twistyour hand so that your thumb points in the direction of the curl; the object willrotate in the direction your fingers close. The larger the magnitude of curl f ,the faster the object rotates.

Definition 8.10 Let f : Rn → Rn be a differentiable vector field. We say f is irrota-tional if curl f = 0 everywhere.

A preview: We will see that being irrotational is a very important property of avector field.

438


Two theorems relating these operators

Theorem 8.11 (Conservative vector fields are irrotational) Let f : R3 → R beC2. Then

curl (∇f) = 0.

PROOF Recall that∇f = (fx, fy, fz). Then

curl (∇f) = det

i j k∂∂x

∂∂y

∂∂z

fx fy fz

= (fzy − fyz, fxz − fzx, fyx − fxy)

By Clairaut’s Theorem, this simplifies to 0.

Theorem 8.12 (Curls are incompressible) Let f : R3 → R3 be a C2-vector field.Then

div (curl f) = 0.

PROOF Write f = (f1, f2, f3). Then

curl f(x) =(∂f3

∂y− ∂f2

∂z,∂f1

∂z− ∂f3

∂x,∂f2

∂x− ∂f1

∂y

)= ((f3)y − (f2)z, (f1)z − (f3)x, (f2)x − (f1)y)

⇒ div (curl f) = [(f3)y − (f2)z]x + [(f1)z − (f3)x]y + [(f2)x − (f1)y]z

= (f3)yx − (f2)zx + (f1)zy − (f3)xy + (f2)xz − (f1)yz

439

8.3. Paths and parametrizations

8.3 Paths and parametrizationsBack in Chapter 5, I told you that a path γ is the image of a function x : R→ Rn.

I also told you that I was lying to you when I wrote that. In fact, I lied to you intwo ways:

1. First, paths do not generally go to ±∞ (although there are exceptions). Theyshould be associated to the image of a closed, bounded interval [a, b] undersome function x.

2. Second, a path isn’t the image of a function, it is the function itself (meaningthat x is the path, not γ):

Definition 8.13 Let a < b. A path is a function x : [a, b]→ Rn. In this setting, x iscalled a parametrization of the path.

EXAMPLE 12Sketch the image of the path x : R→ R2 given by x(t) = (cos t, sin t) for 0 ≤ t ≤ π.

Definition 8.14 Let x : [a, b]→ Rn be a path.

• The initial point of x is x(a) (i.e. where x starts).

• The terminal point of x is x(b) (i.e. where x ends).

• x is called closed if its x(a) = x(b) (i.e. x ends where it begins).

• x is called simple if x does not intersect or cross itself, except possibly at itsinitial and terminal points).

Closed and simple Not simple Closed, but not simple

440


As we learned in Chapter 5, paths have lots of different parametrizations.

Definition 8.15 Let x : [a, b] → Rn and y : [c, d] → Rn be paths. We say y is areparametrization of x if there is an invertible function φ : [c, d] → [a, b] such thaty = x φ.

A reparametrization is called orientation-preserving if φ(c) = a and φ(d) = b;it is called orientation-reversing if φ(c) = b and φ(d) = a.

Recall from §5.3 that reparametrizations change the speed at which the pathis traversed (by a factor of |φ′(t)|; the direction the path is traversed is reversed ifφ′(t) < 0).

EXAMPLE 13Show that y(t) = (cos 2t, sin 2t) for 0 ≤ t ≤ π

2 is a reparametrization of x(t) =(cos t, sin t) for 0 ≤ t ≤ π. Is the reparametrization orientation-preserving ororientation-reversing?

EXAMPLE 14Show that y(t) = (−t3,−t3) for −1 ≤ t ≤ 1 is a reparametrization of x(t) = (t, t)for −1 ≤ t ≤ 1. Is the reparametrization orientation-preserving or orientation-reversing?

441


Path classes

Often, we want to:

• think of a path that we have not yet specified a parametrization of, or

• perform a computation where the answer does not depend on the choice ofparametrization of the path.

With this in mind, we make the following definitions:

Definition 8.16 Let x and y be paths in Rn. We say x and y are equivalent if y isan orientation-preserving reparametrization of x.

The set of paths which are equivalent to x is called the path class of x.

You should think of a path class as a “path where the parametrization eitherhasn’t been chosen yet, or where the specific parametrization doesn’t matter”.

Path classes are denoted by Greek letters like γ. (Back in Chapter 5, when I toldyou γ was a path, my lie was that the γ being studied back then were actually pathclasses, not paths.)

Arithmetic on path classes

Every path class γ has a natural “orientation-reversed” path class −γ:

Definition 8.17 Let γ be a path class parametrized by x : [a, b]→ Rn. The oppositepath class of γ, denoted −γ, is the path class of the path xopp : [a, b]→ Rn defined by

xopp(t) = x(a+ b− t).

t ↦ a+b-t

a bt for x

a

b

t for xopp

442


We can also sometimes “add” path classes:

Definition 8.18 Let γ1, γ2 be two path classes with parametrizations x1 : [a, b]→ Rn

and x2 : [c, d]→ Rn, such that x1(b) = x2(c). The path class γ1 + γ2 is the path classof the concatenation of x1 and x2, i.e. the path class of x : [0, b + d − c − a] → Rn

where

x(t) =

x1(t+ a) if t ≤ b− ax2(t− b+ a+ c) if t > b− a

A path class γ is called piecewise C1 if it contains a parametrization which is theconcatenation of finitely many C1 paths (piecewise C2, piecewise C∞, etc. definedsimilarly).

WARNING

This addition on paths is not commutative:

γ1

γ2

1

1

1

1

1

1

When computing γ1 + γ2, γ1 comes first. (In fact, just because γ1 + γ2 existsmay not even guarantee that γ2 + γ1 even exists.)

443

8.4. Line integrals

8.4 Line integralsScalar line integrals

Definition 8.19 Let γ be a path class in Rn parametrized by x : [a, b] → Rn, and letf : Rn → R. The (scalar) line integral of f along γ is

∫γf ds = lim

||P||→0

n∑j=1

f(x(cj))∆sj.

where P = t0, ..., tn is a partition of [a, b] and cj ∈ [tj−1, tj] for all j.

Where the heck does this come from?

1. Line integrals are mass computations Imagine a piece of wire bent to form acurve in R3. The density of the wire at point x = (x, y, z) on the wire is given byf(x). What is the mass of the wire?

2. Line integrals in R2 are area computations Suppose f : R2 → R and let γ bea path class in R2. What is the area of the (2-D) region in R3 consisting of pointslying above γ but under the graph of f?:

444

8.4. Line integrals

How do you compute scalar line integrals?

Theorem 8.20 Let γ be a path class in Rn parametrized by x : R → Rn, and letf : Rn → R.

1. For any parametrization x : [a, b]→ Rn of γ, then

∫γf ds =

∫ b

af(x(t)) ||x′(t)|| dt.

2. The value of∫γ f ds does not depend on the choice of parametrization x.

Importance of this theorem

Statement (1) tells you how to compute line integrals: once you choose aparametrization of γ, you can rewrite the line integral as a Riemann integralusing the formula in the theorem.

Statement (2) is important because it ensures that line integrals depend onthe class γ, and not on a particular choice of parametrization.

PROOF First, for statement (1), choose a parametrization x : [a, b] → Rn of γ, andlet s = s(t) be the arc length parametrization

s(t) =∫ t

a||x′(u)|| du

as studied in Chapter 5. Recall that

ds

dt= ||x′(t)|| ⇒ ds = ||x′(t)|| dt.

Therefore, by substituting the parametrization into the integral, we see∫γf ds =

∫ b

af(x(t)) ds =

∫ b

af(x(t))||x′(t)|| dt.

Now, we prove statement (2). Suppose x : [a, b] → Rn parametrizes γ and y :[c, d] → Rn is a reparametrization of x. That means there is an invertible functionφ such that y φ = x. By the Chain Rule, this means

x′(t) = y′(φ(t))φ′(t).

From statement (1), we have∫γf ds =

∫ b

af(x(t))||x′(t)|| dt

445

8.4. Line integrals

so by substitution x = y φ, this becomes

∫γf ds =

∫ b

af(y(φ(t)) ||y′(φ(t))φ′(t)|| dt.

Now perform the u-substitution u = φ(t), du = φ′(t) dt. There are two cases:

• Case 1: φ is orientation-preserving, i.e. φ′(t) > 0. Here,

∫γf ds =

∫ d

cf(y(u))||y′(u)|| du

as wanted.

• Case 2: φ is orientation-reversing, i.e. φ′(t) < 0. Here,∫γf ds =

∫ c

df(y(u))||y′(u)|| (−1) du

=∫ d

cf(y(u))||y′(u)|| du

as wanted.

What is “ds”?

1. It comes from arc length parametrization of the path: as seen in the preceding proof,if s(t) is the arc length parametrization, then

ds

dt= ||x′(t)|| ⇒ ds = ||x′(t)|| dt .

2. It is a “small amount of arc length”: from the definition of line integral,

∫γf ds = lim

||P||→0

n∑j=1

f(x(cj))∆sj.

446

8.4. Line integrals

NOTE∫γ f dx is very different from

∫γ f ds, as seen in the following example:

EXAMPLE 15Let γ be the path class of the line segment starting at (0, 0) and ending at (0, 1), andlet f(x, y) = x+ y. Compute

∫γ f ds and

∫γ f dx.

447

8.4. Line integrals

EXAMPLE 16Compute

∫γ(x2 − y + 3z) ds where γ is the line segment connecting the origin with

(3, 5, 2).

Now we can compute the line integral:∫γf ds =

∫ 1

0f(3t, 5t, 2t)||x′(t)|| dt

=∫ 1

0

((3t)2 − 5t+ 3(2t)

)√38 dt

=∫ 1

0(9t2 + t)

√38 dt

=√

38[3t3 + 1

2t2]1

0

=√

38[3 + 1

2

]= 7

2√

38 .

448

8.4. Line integrals

Vector line integrals

Definition 8.21 Let γ be a path class in Rn and let f be a vector field on Rn. The(vector) line integral of f along γ is

∫γ

f · ds = lim||P||→0

n∑j=1

f(x(cj)) ·∆sj

where x : [a, b]→ Rn is a parametrization of γ, P = t0, ..., tn is a partition of [a, b],and for all j we have cj ∈ [tj−1, tj] and ∆sj = x(tj)− x(tj−1).

Where the heck does this come from?

Essentially we are copying the formula which defines a scalar line integral, butinstead of multiplying two scalars, we are taking the dot product of two vectorsinside the integral.

What the heck does this do?

1. Vector line integrals compute work Suppose f describes a force coming fromsomething like an electrical or gravitational field (so that f(x) is the force impartedon an object at position x). If you move an object along γ, what is the work done?

2. Vector line integrals compute circulation, which is the instantaneous rate offluid moved along γ, where the velocity vector field of the fluid is f .

3. Vector line integrals play a role in the “FTC for double integrals”, which, ifyou recall, was the point of this chapter. Thus vector line integrals can be used tocompute double integrals, and therefore areas and volumes, etc.

449

8.4. Line integrals

The four versions of a vector line integral

Theorem 8.22 (Equivalent forms of a vector line integral) Let γ be a path classin Rn, parametrized by C1 function x : [a, b] → Rn. Let f = (f1, ..., fn) be a vectorfield on Rn. Then, the following four quantities are all equal:

1. The vector line integral ∫γ

f · ds,

defined as a Riemann sum as on the previous page.

2. The scalar line integral ∫γ

f ·T ds.

3. The Riemann integral ∫ b

af(x(t)) · x′(t) dt.

4. The differential form of the line integral∫γ

(f1 dx1 + f2 dx2 + ...+ fn dxn) .

Importance of this theorem

This theorem tells you how to compute vector line integrals: you convertthem to one of the other three equivalent forms, and then evaluate that version.

PROOF To justify that the first and third quantities are equal, notice that

∆s = ∆(x1, ..., xn) = (∆x1, ...,∆xn).

By linear approximation, ∆xj ≈ x′j(t)∆t for all j, so

∆s ≈ (x′1(t), ..., x′n(t))∆t = x′(t)∆t.

Taking limits as ||P|| → 0, we get

ds = x′(t) dt,

showing that the first and third quantities are equal.

To show that the second quantity is the same as the third, multiply and divideby ||x′(t)|| inside the Riemann integral to get∫ b

af(x(t)) · x′(t)

||x′(t)|| ||x′(t)|| dt =

∫ b

af(x(t)) ·T(t) ds,

450

8.4. Line integrals

showing that the second and third quantities are equal.

To show the differential form is equal to the first quantity, notice from the defi-nition of vector line integral that

∆sj = f(x(tj))− f(x(tj−1)) = ∆(x1, ..., xn) = (∆x1, ...,∆xn).

So

f(x(cj)) ·∆sj = f(x(cj)) · (∆x1, ...,∆xn)= f1(x(cj))∆x1 + ...+ fn(x(cj))∆xj.

Add up to get

n∑j=1

f(x(cj)) ·∆sj =n∑j=1

f1(x(cj))∆x1 + ...+ fn(x(cj))∆xj.

Take limits as ||P|| → 0 on both sides of this equation to get

lim||P||→0

n∑j=1

f(x(cj)) ·∆sj = lim||P||→0

n∑j=1

f1(x(cj))∆x1 + ...+ fn(x(cj))∆xj

∫γ

f · ds =∫γ

(f1 dx1 + f2 dx2 + ...+ fn dxn)

as wanted.

Remarks on the notation associated to differential forms

In two dimensions, if f = (M,N), we write∫γ

f · ds =∫γ

(M dx+N dy) .

In three dimensions, if f = (M,N,P ), we write∫γ

f · ds =∫γ

(M dx+N dy + P dz) .

451

8.4. Line integrals

EXAMPLE 17Consider the line integral

∫γ f · ds where f(x, y) = (y,−x) and γ is parametrized by

x(t) = (t2 + 1, 1− t2) for 0 ≤ t ≤ 1.

1. Compute∫γ f · ds by rewriting the vector line integral as a Riemann integral.

2. Write the scalar line integral associated to∫γ f · ds.

3. Compute∫γ f · ds by computing the associated differential form.

452

8.4. Line integrals

EXAMPLE 18Compute ∫

γ[(2y + 3z) dx+ x dy + (y − x) dz]

where γ is the parametrized by x(t) = (t, 3t, 2t2) for 0 ≤ t ≤ 2.

Reparametrization of vector line integrals

Recall that for scalar line integrals, reparametrization has no effect on the value of∫γ f ds. What about vector line integrals?

Theorem 8.23 Let γ ⊆ Rn be a piecewise C1 path class and let f be a continuousvector field on Rn. If γ is parametrized by x : [a, b] → Rn and y : [c, d] → Rn is areparametrization of x, then:

1. if y is orientation-preserving, then the value of the line integral is preserved;

2. if y is orientation-reversing, then the value of the line integral is multiplied by−1. In particular: ∫

−γf · ds = −

∫γ

f · ds.

The second part of this theorem should remind you of something about Riemannintegrals you learn in Calculus 1:

453

8.4. Line integrals

PROOF As before, there is some function u so that x = yφ, i.e. x′(t) = y′(φ(t))φ′(t).From Theorem 8.22, we have∫

γf · ds =

∫ b

af(x(t)) · x′(t) dt

=∫ b

af(y(φ(t)) y′(φ(t))φ′(t) dt.

Now we make the substitution u = φ(t), du = φ′(t) dt.

Case 1: y is orientation-preserving. That means φ(a) = b and φ(c) = d, so theintegral becomes

∫ d

cf(y(u)) y′(u) du =

∫γ

f · ds

as desired.

Case 2: y is orientation-reversing. That means φ(a) = d and φ(b) = c, so theintegral becomes

∫ c

df(y(u)) y′(u) du = −

∫ d

cf(y(u)) y′(u) du = −

∫γ

f · ds

as desired.

Review

Recall that the goal of this chapter was to generalize the FTC to double integrals,deriving a formula that looks something like this:

∫∫E

(some derivative

of something

)dA =

∫∂E

(the something) ds

We know understand that what the notation on the right-hand side means: it is avector line integral. What we don’t know is what the “something” is; that is thetask of the next section.

454

8.5. Green’s Theorem

8.5 Green’s TheoremA motivating example

Let E = (x, y) : x2 ≤ y ≤ x. Then ∂E = γ1 + γ2 where

γ1 = (t, t2) : 0 ≤ t ≤ 1γ2 = (1− t, 1− t) : 0 ≤ t ≤ 1

γ1

γ2

E

1

1

Remember that our goal is to get∫∫E

(derivative of something) dA =∫∂E

(the something) · ds

=∫∂E

[something dx+ (something else) dy]

Punchline # 1: ∫∫EMy dA = −

∫∂EM dx.

i.e.−∫∫

EMy dA =

∫∂EM dx.

455


Now, do the same double integral as on the previous page, but in the oppositeorder, by thinking of E as horizontally simple rather than vertically simple:∫∫

ENx dA =

∫ 1

0

∫ √yy

Nx dx dy

=∫ 1

0[N(√y)−N(y)] dy

=∫ 1

0N(√y) dy −

∫ 1

0N(y) dy

(t = √y) (t = 1− y)

=∫ 1

0N dy −

∫ 1

0N(1− t) (−1) dt

(u = 1− t, du = −dt)

=∫γ1N dy −

∫ 0

1N(u) du

=∫γ1N dy +

∫ 1

0N(u) du

=∫γ1N dy +

∫γ2N dy

=∫∂EN dy.

Punchline # 2: ∫∫ENx dA =

∫∂EN dy.

Punchline # 1 (restated from the previous page)

−∫∫

EMy dA =

∫∂EM dx.

Adding the two punchlines together: we obtain∫∫E

(Nx −My) dA =∫∂E

(M dx+N dy)

This formula is called Green’s Theorem, and works not just for the nice region Ewe chose, but for any reasonable region E. Green’s Theorem gives the gener-alization of the FTC to double integrals we have been looking for.

456


Green’s Theorem

Theorem 8.24 (Green’s Theorem) Let E ⊆ R2 be a closed and bounded regionwhose boundary ∂E consists of finitely many simple, closed piecewise C1 curves. Ori-ent the curves comprising ∂E so that as you move along ∂E, E is on the left. Let

f(x, y) = (M,N) = M i +N j

be a C1-vector field on E. Then∮∂E

f · ds =∮∂E

(M dx+N dy) =∫∫

E(Nx −My) dA.

Extra notation: If γ is a closed path (or a union of closed paths), we often write∮γ

f · ds;

the circle on the integral sign is a reminder that the path(s) is (are) closed.

I am not going to prove Green’s Theorem here (go to grad school in math orread a proof online). Here is a sketch of how the logic goes:

1. Break E up into lots of pieces Ej where each piece is both Type 1 and Type 2.2. Compute the double integral of −My over Ej by writing it as an iterated

integral∫ ∫−My dy dx, and show that this works out to be

∮∂EM dx (like what

we did in the previous example).3. Compute the double integral of Nx over Ej by writing it as an iterated inte-

gral∫ ∫

Nx dx dy, and show that this works out to be∮∂E N dy (like what we

did in the previous example).4. Add the two equations from the previous two steps to show that Green’s

Theorem holds on Ej .5. Add up over all the Ej (some care needs to be taken here to make sure the

orientations of each ∂Ej work well together) to get Green’s Theorem for E.

Note

For Green’s Theorem, to work, the boundary ∂E has to be oriented cor-rectly:

457


EXAMPLE 19Verify that Green’s Theorem works in the situation where E is the triangle withvertices (0, 0), (2, 0) and (2, 2), and the vector field is f(x, y) = (x2y, xy + y3).

=∫ 1

00 dt+

∫ 1

0

[2(2t) + (2t)3

]2 dt+

∫ 1

0

[(2− 2t)3 + (2− 2t)2 + (2− 2t)3

](−2) dt

= 0 +∫ 1

0

[8t+ 16t3

]dt− 2

∫ 1

0

[2(2− 2t)3 + (2− 2t)2

]dt

= 0 +[4t2 + 4t4

]10− 2

[2(2− 2t)4

4(−2) + (2− 2t)3

3(−2)

]1

0

= 0 + 8 − 2[0− 2(2)4

4(−2) −(2)3

3(−2)

]1

0

= 8 − 2[0− 2(2)4

4(−2) −(2)3

3(−2)

]

= 8 − 2[4 + 4

3

]= −8

3 .

RHS =∫∫

E(Nx −My) dA =

∫∫E

(y − x2) dA =∫ 2

0

∫ x

0(y − x2) dy dx

=∫ 2

0

[y2

2 − x2y

]x0dx

=∫ 2

0

[12x

2 − x3]dx

=[16x

3 − 14x

4]2

0= 8

6 −164 = −8

3 .

458


Reformulations of Green’s Theorem

Thinking of f as a vector field on R3 by setting f = (M,N, 0), recall that

curl f = (something, something, Nx −My),

so the integrand in the double integral of Green’s Theorem is really

Theorem 8.25 (Green’s Theorem, vector version) Let E ⊆ R2 be a closed andbounded region whose boundary ∂E consists of finitely many simple, closed piecewiseC1 curves. Orient the curves comprising ∂E so that as you move along ∂E, E is onthe left. Let f be a C1-vector field on E. Then∮

∂Ef · ds =

∫∫E

curl f · (0, 0, 1) dA.

There is a higher-dimensional version of this theorem called Stokes’ Theorem whichconnects something called a “surface integral” (like a line integral but over a 2-dimensional patch rather than a curve) to a line integral. This theorem says∮

∂Ef · ds =

∫∫E

curl f · dS.

In R2, the thing called “dS” is like “(0, 0, 1)dA”, so you get Green’s Theorem.Stokes’ Theorem allows you to do similar computations for 2-dimensional subsetsof R3 that aren’t necessarily planes (e.g. curved surfaces, etc.).

Theorem 8.26 (Divergence Theorem (in the plane)) Let E ⊆ R2 be a closed andbounded region whose boundary ∂E consists of finitely many simple, closed piecewiseC1 curves. Orient the curves comprising ∂E so that as you move along ∂E, E is onthe left. Let f be a C1-vector field on E. Then∮

∂Ef · n ds =

∫∫E

div f dA

where n is the outward unit normal to E.

PROOF Write f = (M,N) = (M(x, y), N(x, y)) as usual. Suppose x(t) = (x(t), y(t))for a ≤ t ≤ b parametrizes ∂E. Then the unit vector n can be obtained by taking aunit vector in the direction of x′(t) and rotating it π

2 clockwise. Rotating any vector(a, b) clockwise by π

2 gives (b,−a), so

n = (y′(t),−x′(t))||x′(t)|| .

459


Now ∮∂E

f · n ds =∫ b

af(x(t)) · n(t) ||x′(t)|| dt

=∫ b

a(M,N) · (y′(t),−x′(t))

||x′(t)|| ||x′(t)|| dt

=∫ b

a[My′(t)−Nx′(t)] dt

=∮∂E

(−N dx+M dy)

=∫∫

E(Mx − (−N)y) dA (by Green’s Theorem)

=∫∫

E(Mx +Ny) dA

=∫∫

Ediv f dA.

There is also a higher-dimensional version of the Divergence Theorem, calledGauss’ Theorem, which surface integrals to triple integrals. This theorem says∫∫

∂Ef · dS =

∫∫∫E

div f dV.

All this stuff extends to higher-dimensions as well (this is the field of mathcalled differential geometry).

Computing area with Green’s TheoremEXAMPLE 20

Find the area of a circle of radius R (call this region E):

Method 1 (bad): Calculus 2-style Riemann integral:

area(E) =∫ R

−R2√R2 − x2 dx

Method 2 (bad): Ugly double integral:

area(E) =∫∫

EdA =

∫ R

−R

∫ √R2−x2

−√R2−x2

1 dy dx

Method 3: Change variables to polar coordinates:

area(E) =∫∫

EdA =

∫ R

0

∫ 2π

0r dθdr

460


Method 4 (new): Green’s Theorem!

Theorem 8.27 (Area formula via Green’s Theorem) Let E ⊆ R2 be a closed andbounded region whose boundary ∂E consists of finitely many simple, closed piecewiseC1 curves. Orient the curves comprising ∂E so that as you move along ∂E, E is onthe left. Then, for any C1-vector field f = (M,N) on R2 such that Nx −My = 1,

area(E) =∮∂E

f · ds =∮∂E

(M dx+N dy) .

Examples of f that work here include:

f = (0, x)⇒ area(E) =∮∂Ex dy

f = (−y, 0)⇒ area(E) = −∮∂Ey dx

f =(−y2 ,−

x

2

)⇒ area(E) = 1

2

∮∂E

(x dy − y dx)

The last choice of f above often makes the line integral easiest.

461


EXAMPLE 20The folium of Descartes is the planar curve with Cartesian equation x3 + y3 = xy,pictured below:

E

12

12

1. Find a parametrization of the folium of Descartes.

Hint: Let y = tx; solve for x in terms of t and then y in terms of t.

2. Find the area of the regionE enclosed by the “loop” of the folium of Descartes.

Ultimately we obtain

area(E) = 12

∫ ∞0

t2

(1 + t3)2 dt

(u = 1 + t3; du = 3t2 dt)

= 12

∫ ∞1

13 ·

1u2 du = 1

6 ·−1u

∣∣∣∣∞1

= 16 .

462

8.6. Conservativity and path independence

8.6 Conservativity and path independenceDefinition 8.28 Suppose f is a continuous vector field on Rn. We say f has path-independent line integrals if∫

γ1f · ds =

∫γ2

f · ds

for any two simple, piecewise C1 oriented curves γ1 and γ2 in the domain of f whichhave the same initial and terminal points (i.e. begin and end in the same places)

EXAMPLE 21Show that the vector field f(x, y) = (y,−x) does not have path-independent lineintegrals, by integrating f over a straight line from (0, 0) to (1, 1) and integrating fover the parabola y = x2 from (0, 0) to (1, 1).

QUESTION

How would you show that a vector field does have path-independent line inte-grals?

463


Theorem 8.29 (Fundamental Theorem of Line Integrals (FTLI)) Suppose f is avector field on Rn. Then:

1. f is conservative if and only if f has path-independent line integrals.

2. If f = ∇f is conservative, then for any piecewise C1 curve γ with initial pointa and terminal point b, ∫

γf · ds = f(b)− f(a).

This theorem is also called the Gradient Theorem.

PROOF We begin by proving statement (2). Assume f is conservative, i.e. f = ∇ffor some C1 function f : Rn → R. Parametrize γ by x(t) for a ≤ t ≤ b; then

∫γ

f · ds =∫γ∇f · ds =

∫ b

a∇f(x(t)) · x′(t) dt

=∫ b

a[∇f(x(t))]T x′(t) dt

=∫ b

aDf(x(t))Dx(t) dt

=∫ b

aD(f x)(t) dt (by the Chain Rule)

=∫ b

a(f x)′(t) dt

= f(x(b))− f(x(a)) (by the FTC)= f(b)− f(a).

Having proven statement (2), we observe that the (⇒) direction of (1) follows: sincethe value of the line integral depends only on the values of f at the endpoints, fmust have path-independent line integrals, as wanted.

That leaves the (⇐) direction of statement (1). To show this, assume f has path-independent line integrals. Fix a point x0 ∈ E and and define f : E → R by

f(x) =∫γ

f · ds

where γ is any path in E starting at x0 and ending at x. We will show∇f = f .

To do this, fix j ∈ 1, ..., d. For each h ∈ R, let γ(h) be the straight-line path inE starting at x and ending at x + hej ; this path is parametrized by t 7→ x + tej fort ∈ [0, h]. Next, use the limit definition of partial derivative to show that ∂f

∂xj, the

464


jth partial of f , is equal to fj , the jth component function of f :

∂f

∂xj(x) = lim

h→0

f(x + hej)− f(x)h

= limh→0

1h

[∫γ+γ(h)

f · ds−∫γ

f · ds]

= limh→0

1h

∫γ(h)

f · ds

= limh→0

1h

∫ h

0f(x + hej) · ej dt

= limh→0

∫ h0 fj(x + hej) dt

hL= limh→0

fj(x + hej)1

= fj(x).

So

∇f(x) =(∂f

∂x1(x), ..., ∂f

∂xn(x)

)= (f1(x), ..., fn(x)) = f(x)

as wanted.

Theorem 8.30 Suppose f is a vector field on Rn. Then f is conservative if and onlyif, for any piecewise C1, closed path γ,∮

γf · ds = 0.

PROOF (⇒) Suppose f is conservative. Let γ be a closed path. Write γ = γ1 + γ2where γ1 begins at a and ends at b, and γ2 begins at b and ends at a:

465


Then ∮γ

f · ds =∫γ1

f · ds +∫γ2

f · ds

=∫γ1

f · ds−∫−γ2

f · ds

=∫γ1

f · ds−∫γ1

f · ds (by path independence)

= 0.

(⇐) Let γ1 and γ2 be two paths, both starting at a and ending at b. Noticeγ1 + (−γ2) is a closed path, so by hypothesis,∮

γ1+(−γ2)f · ds = 0.

Splitting the integral by additivity gives∫γ1

f · ds +∫−γ2

f · ds = 0,

i.e. ∫γ1

f · ds = −∫−γ2

f · ds =∫γ2

f · ds.

In other words, the value of the line integral does not depend on the choice of path.Therefore f has path-independent line integrals.

Theorem 8.31 (Conservativity Test in R3) Let f be a C1-vector field on R2 or R3

whose domain is simply connected (“simply connected” means it has no holes). Thenf is conservative if and only if curl f = 0 everywhere.

PROOF (⇒) This was Theorem 8.11 (curl (∇f) = 0).

(⇐) Suppose curl (f) = 0 everywhere. Let γ be any piecewiseC1, simple, closedcurve in Rn. Let E be the region bounded by γ, so that γ = ±∂E. Then∮

γf · ds = ±

∫∫E

curl f · (0, 0, 1) dA =∫∫

E0 · (0, 0, 1) dA = 0.

Therefore the line integral of f around any simple closed curve is 0. By addingtogether simple closed curves if necessary, the line integral of f around any closed(not necessarily simple) curve must also be 0. By Theorem 8.30, this means f isconservative.

466


EXAMPLE 22For each given vector field f :

a) Determine whether or not f is conservative.

b) If f is conservative, find a potential function for f .

c) If f is conservative, compute∫γ f · ds where γ is parametrized by the curve

(et2−t cosπt, e2t sin πt) for 0 ≤ t ≤ 1.

1. f(x, y, z) = (x, y, xz)

2. f(x, y) = (2xy + cos 2y, x2 − 2x sin 2y)

467


3. f(x, y, z) = (ex sin y − yz, ex cos y − xz, z − xy)(Ignore part (c) of the directions, since this is a vector field on R3.)

Enrichment: The method of finding potential functions we use in these exam-ples is the same as the method used to solve exact equations in MATH 330.

This is because exact equations are exactly those arising from studying the rateof change of y with respect to x, as you move along an equipotential curve of apotential function.

468


8.7 Summary of Chapter 8Vector fields A vector field on Rn is a function f : Rn → Rn.

Divergence

div f(x) = ∇ · f = ∂f1

∂x1+ ...+ ∂fn

∂xn.

This measures the “net flow” of a vector field in and out of each point.

Curl (only exists for vector fields on R3 or R2)

curl f(x) = ∇× f =(∂f3∂y− ∂f2

∂z,∂f1∂z− ∂f3∂x

,∂f2∂x− ∂f1

∂y

)= det

i j k∂∂x

∂∂y

∂∂z

f1 f2 f3

.

The curl measures the tendency of an object to rotate when moved by f .

For any vector field f , curl (∇f) = 0 and div (curl f) = 0.

Scalar line integral ∫γf ds =

∫ b

af(x(t)||x′(t)|| dt.

These are unchanged under reparametrization of the path γ.

Vector line integral

∫γ

f · ds =∫γ

f ·T ds =∫ b

af(x(t)) · x′(t) dt =

∫γ

n∑j=1

fj dxj.

These are unchanged under an orientation-preserving reparametrization ofγ, but multiplied by −1 under an orientation-reversing reparametrization.

Green’s Theorem If ∂E is oriented so that E is on the left as you move along ∂E,then ∮

∂Ef · ds =

∫∫E

(Nx −My) dA.

Vector version: ∮∂E

f · ds =∫∫

Ecurl f · (0, 0, 1) dA.

Divergence Theorem If n is an outward unit normal to E and ∂E is oriented as itis in Green’s Theorem, then∮

∂Ef · n ds =

∫∫E

div f dA.

469


Area formulas coming from Green’s Theorem:

area(E) =∮∂Ex dy = −

∮∂Ey dx = 1

2

∮∂E

(x dy − y dx) .

Conservativity A vector field f is called conservative if any of the following equiv-alent properties hold:

• f is a gradient field, i.e. there is a potential function f : Rn → R suchthat f = ∇f ;

• f has path-independent line integrals;

• The Fundamental Theorem of Line Integrals holds, i.e. for any poten-tial function f of f , ∫

γf · ds = f(b)− f(a)

where γ starts at a and ends at b;

•∮γ f ds = 0 for any closed curve γ;

• (in R3) f is irrotational, i.e. curl f = 0 everywhere.


1. Let f(x, y, z) = x sin yz + z cosxy. Compute the gradient field of f .

2. Let f(x, y) = xy ln(x+ y). Compute ∇f .

3. (R) Show that the vector field f : R2 → R2 given by f(x, y) = (yexy, xexy) isconservative, by establishing that it is the gradient field of f(x, y) = exy.

4. Show that the vector field f : R3 → R3 given by f(x, y, z) = (yz, xz, xy) isconservative, by finding a function f : R3 → R such that f = ∇f .

5. Show that the vector field f : R2 → R2 given by f(x, y) = 1x

i− 1y

j is conserva-tive, by finding a function f : R2 → R such that f = ∇f .

6. a) Suppose f : Rn → Rn is a C1 vector field. Write the inputs and outputsof f as (x1, ..., xn) f7−→ (f1, ..., fn). Suppose there exist i, j ∈ 1, ..., n suchthat ∂fi

∂xj6= ∂fj

∂xi. Explain why it is impossible for f to be conservative.

470


b) Use the preceding result to explain why f : R2 → R2 given by f(x, y) =(6x2y2, 2x3y) is not conservative.

7. Let f : R2 → R2 be f(x, y) = (2x, 8y).

a) Show that f is conservative, by establishing that it is the gradient fieldof f(x, y) = x2 + 4y2.

b) Sketch the equipotential set to f at height 16.

c) What shape do all the equipotential curves have?

d) Find parametric equations for the flow line of f passing through (0, 3).

e) Find parametric equations for the flow line of f passing through (2, 1).

f) Sketch the flow lines you computed in parts (d) and (e) on the pictureyou drew in part (b).

8. The picture of some vector field f on R2 is shown below, at left. Sketch graphsof the flow lines to f passing through the points (−4, 0), (3,−3) and (0,−1).

-4 -2 0 2 4

-4

-2

0

2

4

-4 -2 0 2 4

-4

-2

0

2

4

9. The picture of some vector field g on R2 is shown above, at right. Sketch theequipotential sets to g passing through the points (2, 2), (3,−3) and (0,−2).

10. Consider the vector field f(x, y) = (3x, x − 2y) on R2. Determine whether ornot the parametric equations x(t) =

(2e3t, 3

5e−2t + 2

5e3t)

describe a flow lineof f .

11. (R) Consider the vector field f(x, y, z) = (−2x+2y−3z,−x+y−z, 2x−2y+3z)on R3. Determine whether or not the parametric equations

x(t) =(6− 4et + 5tet, 6− 5et, 4et + 10tet

)describe a flow line of f .

471


12. Let f(x, y) = (9x, y).

a) Find parametric equations for the flow line x of f passing through (5,−2).

b) Find Cartesian equations of the curve which is the image of x in R2.

13. Let f(x, y, z) = (2x, y, 4z). Find parametric equations for the flow line of fpassing through (−3, 6, 1).

14. In each part of this problem, you are given the picture of a vector field on R2.For each picture, describe the general shape of (most of) the flow lines to thatvector field.

a)

-4 -2 0 2 4

-4

-2

0

2

4

b)

-4 -2 0 2 4

-4

-2

0

2

4

c)

-4 -2 0 2 4

-4

-2

0

2

4

d)

-4 -2 0 2 4

-4

-2

0

2

4

e)

-4 -2 0 2 4

-4

-2

0

2

4

f)

-4 -2 0 2 4

-4

-2

0

2

4


15. Compute the divergence of each given vector field:

a) f(x, y) = (xex, yey)b) f(x, y) = y cos y i + x sin x j

c) (R) f(x, y, z) =(

yx+z ,

zy+x ,

xz+y

)472


d) f(x, y, z) = (x2y2 + y2z, xy2 + yz2, xy2 − z2x)

16. Let f(x, y, z) =(ex−z, ey

2−x, ez2−x2

). Compute div f(4, 2, 2).

17. Compute∇ · f , where f is the gradient field of f(x, y) = ln(x− y).

18. (R) Compute div f , where f : R2 → R2 is a vector field with ∂f1∂x

= 2xy cos(x2y),∂f1∂y

= x2 cos(x2y), ∂f2∂x

= −y2 sin(xy2) and ∂f2∂y

= −2xy sin(xy2).

19. a) Compute div f , where f : R3 → R3 is a vector field whose total deriva-tive is

Df =

4x3y2z3 2x4yz3 3x4y2z2

8xy2z 8x2yz 4x2y2

3yz3 3xz3 9xyz2

.b) For any C1 vector field f : Rn → Rn, what matrix operation, applied to

Df , gives div f?

20. A picture of some unknown vector field f : R2 → R2 is given below:

-6 -4 -2 0 2 4 6

-6

-4

-2

0

2

4

6

Use this picture to answer the following questions:

a) Is div f(−5,−3) positive, negative or zero?

b) Is div f(−1, 5) positive, negative or zero?

c) Is div f(5, 0) positive, negative or zero?

d) Is div f(−4, 5) positive, negative or zero?

e) Find a point (x, y) ∈ [−6, 6]× [−6, 6] where div f(x, y) = 0.

473


In Problems 21-22, use the following contour plots, labelled (A)-(H):

A)

-7

-6

-5

-4-3

-2-1 01 2

34

5

6

7

-10 -5 0 5 10-10

-5

0

5

10

x

y

B)

-7

-6

-5

-4-3

-2 -10 123

4

5

6

7

-10 -5 0 5 10-10

-5

0

5

10

x

y

C)

-6

-6

-5

-5

-4

-4

-3

-3

-2

-2-1

-1

-10 -5 0 5 10-10

-5

0

5

10

x

y

D)1

12

2

3

3

4

4

5

5

6

6

-10 -5 0 5 10-10

-5

0

5

10

x

y

E)

-5

-4

-3

-2

-1

0

1

2

3

4

5

-10 -5 0 5 10-10

-5

0

5

10

x

y

F)

-5

-4

-3

-2

-1

0

1

2

3

4

5

-10 -5 0 5 10-10

-5

0

5

10

x

y

G)

1

1

2

2

3

3

4

4

5

5

-10 -5 0 5 10-10

-5

0

5

10

x

y

H)

-5

-5

-4

-4

-3

-3

-2

-2

-1

-1-10 -5 0 5 10

-10

-5

0

5

10

x

y

474


21. In each part of this problem, you are given the picture of a C1 vector field fon R2. Choose the picture (A)-(H) above which gives a contour plot of div f .

a)

-10 -5 0 5 10

-10

-5

0

5

10

x

y

b)

-10 -5 0 5 10

-10

-5

0

5

10

x

y

c)

-10 -5 0 5 10

-10

-5

0

5

10

x

y

d)

-10 -5 0 5 10

-10

-5

0

5

10

x

y

22. Same directions as the previous problem:

a)

-10 -5 0 5 10

-10

-5

0

5

10

x

y

b)

-10 -5 0 5 10

-10

-5

0

5

10

x

y

475


c)

-10 -5 0 5 10

-10

-5

0

5

10

x

yd)

-10 -5 0 5 10

-10

-5

0

5

10

x

y

23. Determine whether or not each given vector field is incompressible:

a) f : R3 → R3 given by f(x) = x||x||

b) The vector field pictured below:

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

x

y

c) f : R2 → R2, where f(x, y) = (f(y), g(x)) for C1 functions f, g : R→ Rd) The vector field pictured below:

-6 -4 -2 0 2 4 6-6

-4

-2

0

2

4

6

x

y

476


24. Compute the curl of each given vector field on R3:

a) f(x, y, z) = (y + z − x, y2z,−2yx)b) f(x) = y2z i + exyz j + x2y kc) f(x) = x× (1, 2, 3)

25. Let f(x, y, z) = (x2yz, 2xyz2, 3y2z) and g(x, y, z) = (2y2z, xz3, 2x2y3). Com-pute∇× (f × g).

26. Mathematica computes the curl of a vector field with the Curl[ ] command (thesyntax is Curl[f[x,y,z], x,y,z], assuming you first save the vector field as f).Use Mathematica to compute and simplify the curl of each of these vectorfields:

a) f(x, y, z) = (sin(x− y), sin(y − z), sin(z − x))b) f(x, y, z) =

(arctan x

y, ln√x2 + y2, 1

)27. Suppose f = (f1, f2, f3) is a C1 vector field on R3 such that

Df(x, y, z) =

z sin y xz cos y x sin yy sin z x sin z xy cos zyz cosx z sin x y sin x

.Compute∇× f

(π2 ,

π2 ,

π2

).

28. Compute curl (curl f), where f(x, y, z) = (xyz, xyz, xyz).

29. In each part of this problem, you are given a picture of a vector field f onR2. Thinking of f as a vector field on R3 by setting its z-coordinate equal tozero, determine (by studying the picture) which one of the following threestatements is true about the curl of f at the origin:

• curl f(0) = 0;• curl f(0) is a vector pointing upward, out of the xy-plane;• curl f(0) is a vector pointing downward, below the xy-plane.

477


a) (0,0)

-4 -2 0 2 4

-4

-2

0

2

4

b) (0,0)

-4 -2 0 2 4

-4

-2

0

2

4

c) (0,0)

-4 -2 0 2 4

-4

-2

0

2

4

d) (0,0)

-4 -2 0 2 4

-4

-2

0

2

4

30. Determine whether or not each vector field is irrotational:

a) The vector field f : R3 → R3 given by f(x) = x||x||

b) The vector field in Problem 24, part (a).

c) The vector field pictured in Problem 29, part (d).

d) f(x, y, z) = (x+ 2y − z, 3x− y + 4z, x+ y + 3z)

31. Determine whether or not each vector field is irrotational:

a) The vector field on R3 given by f(x) = x× (1, 2, 3)b) ∇f , where f : R3 → R is f(x, y, z) = cos(x sin y) sin(y cos z)√

1+x2+z4y2.

c) The vector field pictured below:

-4 -2 0 2 4

-4

-2

0

2

4

x

y

478


d) The vector field pictured below:

-4 -2 0 2 4

-4

-2

0

2

4

x

y

Throughout Problems 32-36, assume f and g are C∞ vector fields on R3, and that fis a C∞ function from R3 to R. For each given expression, determine if the expres-sion is a well-defined function (in which case you should specify its domain andcodomain), or whether the expression is nonsense:

32. a) ∇fb) div f

c) curl f

d) ∇fe) div ff) curl f

33. a) div (f + g)b) div f + gc) div (f · g)

d) (div f) · ge) (div f)gf) div (f × g)

34. a) f× curl gb) f×div gc) curl f× curl gd) (curl f)ge) (curl f)× gf) curl (f · g)

35. a) f ·div f

b) f ·div f

c) f · curl f

d) f · curl f

36. a) curl (div f)

b) div (curl f)

c) div (div f)

d) curl (curl f)

37. (F) Let f ,g be C2 vector fields on R3. Prove the following identity:

div (f × g) = (curl f) · g− f · (curl g).

38. (F) If f and g are both conservative vector fields on R3, what is div (f × g)?Explain.


In each part of Problems 39-44, sketch each given path. Be sure to indicate thedirection of motion, and the initial and terminal points of the path.

479


39. a) x(t) = (3 cos t, 3 sin t),for 0 ≤ t ≤ 2π

b) x(t) = (2− 3t, 1 + t),for 0 ≤ t ≤ 2

40. a) x(t) = 2 cos t i + 3 sin t j,for 0 ≤ t ≤ 2π

b) x(t) = (0, 2t, t), for 0 ≤ t ≤ 1

41. a) (R) x(t) = (3 cos t, 3 sin t),for π

4 ≤ t ≤ 3π4

b) (R) x(t) = (t, et),for −4 ≤ t ≤ 4

42. a) (R) x(t) = (sin t, t),for 0 ≤ t ≤ 2π

b) (R) x(t) = (cos t, 0, sin t),for 0 ≤ t ≤ π

43. a) x(t) = i + t j + t2 k,for 0 ≤ t ≤ 2

b) x(t) = (4 cos t, sin t),for π ≤ t ≤ 2π

44. a) x(t) = (cos t, sin t, t),for 0 ≤ t ≤ 2π

b) x(t) = (cos t, sin t, sin t),for 0 ≤ t ≤ 2π

45. Show that y(t) = (ln t, ln t2) for 1 ≤ t ≤ e is a reparametrization of x(t) =(t, 2t) for 0 ≤ t ≤ 1, by finding a function φ such that y = x φ. Is thisreparametrization orientation-preserving, or orientation-reversing?

46. (R) Show that y(t) = (cos t, cos2 t) for 0 ≤ t ≤ π2 is a reparametrization of

x(t) = (et−1, e2t−2) for 0 ≤ t ≤ 2, by finding a function φ such that y = x φ.Is this reparametrization orientation-preserving, or orientation-reversing?

In each part of Problems 47-51, you are given a path class γ. Find a parametrizationof γ, i.e. a function x : [a, b]→ Rn whose image is in the class γ. Be sure to specifynot just a formula for x but the values of a and b as well.

47. a) γ is the line segment starting at (0, 0, 0) and ending at (3,−2, 4).

b) γ is the portion of the graph of y = ex between (0, 1) and (5, e5), orientedfrom left to right.

48. a) γ is the line segment with initial point (4,−1) and terminal point (−2, 3).

b) γ is the portion of the graph y2 = x between (1,−1) and (1, 1), orientedfrom bottom to top.

49. a) γ is the circle of radius 4 centered at the origin (in R2), oriented counter-clockwise.

b) γ is the circle of radius 4 centered at the origin (in R2), oriented clock-wise.

50. a) (R) γ is the portion of the circle of radius 4 centered at the origin (in R2),starting at (0, 4) and ending at (−4, 0).

480


b) γ is the circle of radius 4 in the xy-plane centered at the origin (this isa path class in R3), oriented counterclockwise (in the xy-plane, whenviewed from the positive z-axis).

51. a) γ is the circle of radius 4 in the xz-plane centered at the origin (this isa path class in R3), oriented counterclockwise (in the xz-plane, whenviewed from the positive y-axis).

b) γ is the portion of the ellipse x2

4 + y2

49 = 1 in the first quadrant, orientedcounterclockwise.

52. In each part of this problem, you are given a path class γ. Find a parametriza-tion of the opposite path class −γ.

a) γ is as in Problem 47 (a).

b) γ is parametrized by x(t) = (cos 2t, sin t), for 0 ≤ t ≤ π.

c) γ is parametrized by x(t) = (et + e−t, et + 3e−t, 2et − e−t), for 0 ≤ t ≤ 1.


In each part of Problems 53-55, you are given a line integral of a function R2 → Rwhich represents the area of an easy-to-describe geometric shape. In each prob-lem, identify the shape whose area is given by the integral, and then compute theintegral by finding the area of the shape using high-school geometry area formulas.

53. a)∫γ 3 ds, where γ is the line segment from (0, 0) to (0, 8)

b)∫γ 4 ds, where γ is the line segment from (0, 0) to (2, 3)

54. a)∫γ(3x+ 2y) ds, where γ is the line segment from (1, 0) to (1, 5)

b)∫γ(x+ 4y) ds, where γ is the line segment from (0, 0) to (3, 1)

55. a)∫γ 2 ds, where γ is the circle x2 + y2 = 9

b)∫γ(2x+ y) ds, where γ is the line segment from (1, 2) to (3, 6)

In Problems 56-60, compute each given line integral by hand:

56.∫γ 4xy ds, where γ is the quarter of the circle x2 + y2 = 25 lying in the first

quadrant, oriented counterclockwise

57.∫γ(x3− 2xy2) ds, where γ is the line segment beginning at (1, 2) and ending at

(5, 3)

58.∫γ x ds, where γ is parametrized by x(t) = (3t3, t) for 0 ≤ t ≤ 4

59.∫γ(x + z) ds, where γ is parametrized by x(t) = 2 cos ti+, 2 sin t j + 4tk for

0 ≤ t ≤ 2π

481


60. (R)∫γ(xyz + z2y) ds, where γ is the line segment beginning at (0, 0, 0) and

ending at (4, 1, 3)

61. Let E ⊆ R3 be the set of points on the cylinder x2 + y2 = 1 which lie abovethe xy-plane but under the paraboloid z = (x − 1)2 + y2 + 2. Compute the(surface) area of E.

In Problems 62-64, find a decimal approximation to each scalar line integral byfirst converting the scalar line integral to a Riemann integral, and then using theNIntegrate[ ] command with Mathematica to estimate the Riemann integral:

62.∫γ y ds, where γ is parametrized by x(t) = (t− sin t, 1− cos t) for 0 ≤ t ≤ 2π

63.∫γ xe

y ds, where γ is the portion of y = x2 starting at the point (−1, 1) andending at the point (2, 4)

64.∫γ e

x+2y ds, where γ is the top half of the ellipse x2

25 + y2

16 = 1, oriented counter-clockwise

In Problems 65-66, write each given vector line integral as a Riemann integral (theonly variable allowed in your answers is t):

65.∫γ f · ds, where γ is the circle of radius 1 in R2 centered at the origin and

f(x) = (ex−y, ln(x2 + y2 + 1)).

66. (R)∫γ f · ds, where γ is the line segment beginning at (0, 0, 0) and ending at

(2,−1, 2) and f(x, y, z) = (cos xz, ey−z, arcsin x2y).

67. Consider the line integral∫γ f · ds, where f : R2 → R2 is given by f(x, y) =

(x2, y2) and γ is the circle of radius 4 in R2 centered at the origin, orientedcounterclockwise.

a) Compute this line integral by rewriting it as a Riemann integral.

b) Write the differential form of the line integral.

c) Compute the line integral by evaluating the differential form you wrotedown in part (b).

68. Consider the line integral∫γ f · ds, where f : R3 → R3 is given by f(x, y, z) =

(x + yz, x2y, xy + z2) and γ is the line segment in R3 starting at (0, 0, 0) andending at (3, 2, 5).

a) Compute this line integral by rewriting it as a Riemann integral.

b) Write the differential form of the line integral.

c) Compute the line integral by evaluating the differential form you wrotedown in part (b).

482


In Problems 69-73, compute each given line integral by hand:

69.∫γ∇f · ds, where f(x, y) = 4x2y− 3xy3 and γ is the line segment beginning at

(0, 0) and ending at (1, 2)

70.∫γ f · ds, where f(x, y, z) = (ex, xexy, xyexy) and γ is parametrized by x(t) =

(t, t2, t3) for 0 ≤ t ≤ 1

71.∫γ (y dx+ z dy + x dz), where γ is the line segment beginning at (0, 0, 0) and

ending at (1, 5, 2)

72.∫γ (y dx+ x dy), where γ is the portion of the graph of y = x2 beginning at

(0, 0) and ending at (1, 1)

73.∫γ f · ds, where f(x, y) = (xy, x) and γ is the portion of the circle x2 + y2 = 1

starting at (0, 0) and ending at(√

32 ,

12

)In Problems 74-77, find a decimal approximation to each given vector line integral,by first converting the integral to a Riemann integral and then using the NInte-grate[ ] command in Mathematica:

74.∫γ f · ds, where f(x, y) = sin(x − y) i + cos(x + y) j and γ is the line segment

beginning at (0, 0) and ending at (0, π)

75.∫γ f · ds, where f(x, y, z) = (ex−y, ey−2z, ez−z) and γ is parametrized by x(t) =

(cos 2t, 3 sin t, t) for 0 ≤ t ≤ 2π

76.∫γ (sin x dx+ cos y dy), where γ is the portion of the circle x2 + y2 = 16 lying

in the first quadrant, oriented counterclockwise

77.∫γ (xz2 dx+ 2yz dy + 4xyz3 dz), where γ is parametrized by x(t) =

(t+ 1

t, t− 1

t, t2)

for 1 ≤ t ≤ 2

78. In Theorem 8.23, we learned that the following fact about Riemann integrals:∫ a

bf(x) dx = −

∫ b

af(x) dx

generalizes to this fact about vector line integrals:∫−γ

f · ds = −∫γ

f · ds

Another fact you learn about Riemann integrals in Calculus 1 is that∫ c

af(x) dx =

∫ b

af(x) dx+

∫ c

bf(x) dx.

There is a corresponding fact about vector line integrals. Formulate this factas a “theorem” (you don’t have to prove it).

483


79. Evaluate∫γ (x2y3 dx− xy2 dy), where γ is the square with vertices (±1,±1),

oriented counterclockwise.

Hint: Write γ as the sum of four different curves; evaluate the line integralover each curve and use the result of Problem 78.

80. Evaluate∫γ x ds, where γ is the piecewise-smooth curve shown below at left.

y=2-x2

y=x

γ

1 2

1

2

y= x

y=x2

γ

1

1

81. Evaluate∫γ (x2 + y2,−2xy) ·ds, where γ is the piecewise-smooth closed curve

shown above at right.

82. (R) Evaluate∫γ (z dx+ x dy + y dz), where γ ⊆ R3 consists of the line seg-

ments from (0, 0, 0) to (2, 0, 0), from (2, 0, 0) to (2, 0, 3), and (2, 0, 3) to (2, 1, 3).

83. (F) Let f : R2 → R and suppose γ is a level set of f . Prove that∫γ∇f · ds = 0.

84. An object is moved along the image of x(t) = (cos t, sin t, 2t) from (1, 0, 0) to(1, 0, 4π). If the object is exposed to a force f(x, y, z) =

(−1

3x,−13y,

25

)as it

moves, what is the work done by the force field on the object?

85. Suppose γ is the circle of radius 3 centered at (0, 0), oriented counterclock-wise. For each vector field pictured in Problem 29, determine, based on thepicture, if

∫γ f · ds is positive, negative or zero. Explain your reasoning.


In each of Problems 86-90, verify that Green’s Theorem holds for the given regionE ⊆ R2 and the given vector field f on R2:

86. E = (x, y) : x2 + y2 ≤ 1; f(x, y) = (x2y,−xy2).

87. E is the square with vertices (0, 0), (1, 0), (1, 1) and (0, 1); f(x, y) = (y2, x3).

88. E = (x, y) : y ≤ x and y ≥ 14x

2; f(x, y) = (x− y, x+ 2y).

89. E = (x, y) : 0 ≤ y ≤ 4− x2 f(x, y) = 2xy i + (x+ y) j.

484


90. E is the annular (ring-shaped) region (x, y) : 1 ≤ x2 + y2 ≤ 9; f(x, y) =(x, 2x).

91. Use Green’s Theorem to find the area enclosed by the curve whose paramet-ric equations are x(t) = (2 cos3 t, 2 sin3 t) for 0 ≤ t ≤ 2π. (The graph of thiscurve is called an astroid, and is pictured below at left.)

-1 1x

-1

1

y

π 2π 3πx

12y

92. A cycloid, introduced in a previous lab, is the curve which is parametrizedby x(t) = (t− sin t, 1−cos t). The graph of this curve is shown above, at right.Find the area above the x-axis and beneath one “hump” of the cycloid.

93. Use Green’s Theorem to find the area of the region in R2 consisting of thepoints which are outside the ellipse x2

25 + y2

144 = 1 but inside the circle x2 +y2 =169.

94. (F) Verify that the area of a rectangle with vertices (0, 0), (a, 0), (a, b) and(0, b) is ab by computing an appropriate line integral.

95. (F) Find the area of the region of points in R2 lying inside the circle x2 + y2 =10 and inside the ellipse (x− 3)2 + 2y2 = 22. (This region is the shaded regionin the picture below.) While I want an exact answer, you can use Mathematicato compute any integrals.

x2+y2=9

(x-3)2+2y2=22

E

-4 -3 -2 -1 1 2 3 4 5 6 7 8

-4

-3

-2

-1

1

2

3

4

96. Suppose E ⊆ R2 is a bounded region so that∮∂E (3x2y dx+ (x3 + 3x) dy) =

30. What is the area of E?

485


97. The graph of the polar function r = 2 cos3 θ is called a simple folium.

a) Use the PolarPlot[ ] command in Mathematica to graph this simple folium(for 0 ≤ t ≤ π; the graph starts repeating itself after this). Sketch thegraph you obtain.

b) Find a Cartesian equation of this simple folium.

c) Find a parametrization of this simple folium, by setting y = tx in youranswer to part (b) and solving for x in terms of t, similar to how wefound a parametrization of the folium of Descartes in the notes.

d) Use Green’s Theorem to find the area of the region enclosed by this sim-ple folium.

e) In (the homework problems of) Chapter 7, we found this formula whichgives the area enclosed by an arbitrary polar function r = f(θ):

A =∫ β

α

12[f(θ)]2 dθ

Use this formula, together with your answer to part (d), to find the valueof∫ π

0 cos6 θ dθ.

98. A lemniscate is a curve in R2 which looks like a figure 8,and is parametrizedby

x(t) =(a√

2 sin tcos2 t+ 1 ,

−a√

2 cos t sin tcos2 t+ 1

)for 0 ≤ t ≤ 2π, where a > 0 is a constant.

a) Use Mathematica to sketch the lemniscate when a = 2.

b) Find the total area (in terms of a) enclosed by a lemniscate.CAUTION: You need to be careful about the orientation of the lemnis-cate.

99. (F) Let γ ⊆ R2 be the circle of radius 2 centered at the origin. Evaluate∫γex cos 2y dx− 2ex sin 2y dy.

Hint: Use Green’s Theorem to rewrite this line integral as a double integral.

486


100. Let E be the region in R2 pictured below:

E

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

Evaluate the line integral ∫∂E

5x4ey dx+ ey dy,

where ∂E is oriented so that as you move along ∂E, E is on the left.

101. Compute∫∫E x

2 dA, where E ⊆ R2 is the region bounded by the ellipse x2

25 +y2

16 = 1.


In each of Problems 102-105, you are given a vector field f on Rn and two pointsa,b ∈ Rn. For each problem:

• determine whether or not f is conservative;• if f is conservative, find a potential function for f ; and• if f is conservative, compute

∫γ f · ds, where γ is a path class in Rn with initial

point a and terminal point b.

102. f(x, y, z) = (4xyz−3z, 2x2z+8yz2, 2x2y+8y2z−3x); a = (0, 0, 0); b = (1, 2,−1)

103. f(x, y, z) = (y2 + 3x, z3 − 2y, x2 + z); a = (0, 0, 0); b = (3,−1, 0)

104. f(x, y) = − sin x sin y i + cosx cos y j; a = (0, π); b = (π, 0)

105. (R) f(x, y) = (ex−y,−ex−y); a = (0, ln 2); b = (ln 5, 0)

106. Compute∫γ f · ds, where γ is parametrized by x(t) =

(sin πt, et2 , cos(πt2)

)for

0 ≤ t ≤ 1 and f(x, y, z) = (2xy + z, x2 − 2yz, x− y2).

107. Let f(x, y) = (x+ y, 2y − x).

a) Compute∫γ f · ds, where γ is the circle of radius 1.

487


b) Based solely on the computation you did in part (a), which statement ismost accurate?A. f is definitely a gradient vector field.B. f may or may not be a gradient vector field.C. f is definitely not a gradient vector field.

108. Let g(x, y) = (2x+ y)i + (x− 2y)j.

a) Compute∫γ g · ds, where γ is the circle of radius 1.

b) Based solely on the computation you did in part (a), which statement ismost accurate?A. g is definitely a gradient vector field.B. g may or may not be a gradient vector field.C. g is definitely not a gradient vector field.

109. Describe all functions N : R3 → R so that the vector field f(x, y, z) = (x3y −3x2z,N(x, y, z), 2yz − x3) is conservative. If there are no such functions, ex-plain why no such function exists.

110. Describe all functions N : R3 → R so that the vector field f(x, y, z) = (2xy +2x2z2, N(x, y, z), 2zy − 2x2z2) is conservative. If there are no such functions,explain why no such function exists.

111. Let f(x, y) = M(x, y)i + N(x, y)j be a C1 vector field on R2. Prove that if∂M∂y

= ∂N∂x

, then f is conservative.

488


Selected answers

1. (−yz sin xy + sin yz, xz cos yz − xz sin xy, cosxy + xy cos yz)

4. f(x, y, z) = xyz

5. f(x, y) = ln x− ln y

7. b,f)-5 -4 -3 -2 -1 1 2 3 4 5

-3

-2

-1

1

2

3

4

5

c) They are ellipses.

d) x(t) = (0, 3e8t)e) x(t) = (2e2t, e8t)

8.

-4 -2 0 2 4

-4

-2

0

2

4

9.

-4 -2 0 2 4

-4

-2

0

2

4

10. They describe a flow line of f .

11. They do not describe a flow line off .

12. a) x(t) = (5e9t,−2et)b) x

5 = −y9

29

13. x(t) = (−3e2t, 6et, e4t)

14. a) circles (centered at (0, 0), ori-ented clockwise)

b) hyperbolas

c) straight lines (of slope ≈ −12 )

15. a) xex + ex + yey + ey

d) 2xy + 2xy2 − 2xz + z2

16. 4 + 4e−12 + e2

18. 2xy cos(x2y)− 2xy sin(xy2)

19. a) 4x3y2z3 + 8x2yz + 9xyz2

20. a) negative

b) positive

e) (2, 0) or (−2, 0)

21. a) F

b) C

c) B

d) H

23. a) not incompressible

b) not incompressible

24. a) (−2x− y2, 1 + 2y,−1)c) (−2,−4,−6)

489


25. (2x4y4− 12y4z− 12xy2z3 +x3z4, 8x3y4z− 12xy2z3 + 4y3z3− 3x2yz4,−8x3y4z−16x3y3z2 + 6xyz4)

26. a) f(x, y, z) = (cos(y − z), cos(x− z), cos(x− y))b)

(0, 0, 3x

x2+y2

)27.

(π2 ,

π2 ,

π2

)28. (y + z, x+ z, x+ y)

29. a) curl f(0) is a vector pointing downward, below the xy-plane

c) curl f(0) = 0

30. a) irrotational

31. a) not irrotationalb) irrotational

32. a) function R3 → R3

d) nonsensee) function R3 → Rf) function R3 → R3

33. c) nonsensee) function R3 → R3

34. a) function R3 → R3

f) nonsense

35. c) nonsensed) function R→ R

36. a) nonsenseb) function R3 → R

38. 0

39. a) (3,0)-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

b) (2,1)

(-4,3)

-4 -2 2

-1

1

2

3

4

41. a)(3 22,3 22

)(-3 22,3 22

)

-3 -2 -1 1 2 3

0.5

1.0

1.5

2.0

2.5

3.0

3.5

42. a)

(0,2π)

(0,0)-2 -1 1 2

2

4

6

b)

490


43. a)

b)(-4,0) (4,0)

-4 -2 2 4

-2.0-1.5-1.0-0.5

0.51.0

45. φ(t) = ln t; orientation-preserving

47. a) x(t) = (3t,−2t, 4t), for 0 ≤ t ≤ 1b) x(t) = (t, et), for 0 ≤ t ≤ 5

48. a) x(t) = (4− 6t,−1 + 4t), for 0 ≤ t ≤ 1

50. a) x(t) = (4 cos t, 4 sin t), for π2 ≤ t ≤ π

b) x(t) = (4 cos t, 4 sin t, 0), for 0 ≤ t ≤ 2π

51. a) x(t) = (4 cos t, 0, 4 sin t), for 0 ≤ t ≤ 2πb) x(t) = (2 cos t, 7 sin t), for 0 ≤ t ≤ π

2

52. a) y(t) = (3− 3t,−2 + 2t, 4− 4t) for 0 ≤ t ≤ 1c) y(t) = (e1−t + e1+t, e1−t + 3e1+t, 2e1−t − e1+t), for 0 ≤ t ≤ 1.

53. a) rectangle; 24

54. a) trapezoid; 32

55. a) circular cylinder; 12π

56. 250

57. −73

√17

60. 21√

132√

2

61. 8π

62.∫ 2π

0√

2(1− cos t)3/2 dt ≈ 10.6667

64.∫ π

0 e5 cos t+8 sin t

√16 + 9 sin2 t dt ≈ 48497.2

65.∫ 2π

0 [−ecos t−sin t sin t+ (ln 2) cos t] dt

66.∫ 1

0

[23 cos 4t2 − 1

3e−3t + 2

3 arcsin(−4t3)]dt

67. a) 0b)

∫γ x

2 dx+ y2 dy

491


69. 16

70. 13(5e− 2)

71. 172

74. 0

75. 563.81

78. If γ1 and γ2 are path classes in Rn so that the terminal point of γ1 equals theinitial point of γ2, then∫

γ1+γ2f · ds =

∫γ1f · ds +

∫γ2f · ds.

77. −83

78. 112

(−1− 6

√2 + 5

√5)

79. −35

84. 85π

85. b) zero

86. Both sides of Green’s Theoremevaluate to −π

2 .

87. Both sides of Green’s Theoremevaluate to 0.

90. Both sides of Green’s Theoremevaluate to 16π.

91. 3π2

93. 109π

95. 2710 + 9 arctan 3− 15

√1113 + 11

√2( 6

13 + π4)

97. a)1 2

-1

1

b) (x2 + y2)2 = 2x3

98. a)-2 2 2 2

100. 64e2 − 64e−2 + 2e− 2e−1

101. 125π

102. f = ∇f where f(x, y, z) = 2x2yz − 3xz + 4y2z2;∫γ f · ds = −37

30 .

104. f = ∇f where f(x, y) = cos x sin y;∫γ f · ds = 0.

106. e2 + 1

107. a) −2πb) C

108. a) 0b) B

109. N(x, y, z) = 14x

4 + z2 + g(y), where g is any differentiable function R→ R.

492



Statements of important theorems: Precisely write statements of Fubini’s Theo-rem, Green’s Theorem, the Divergence Theorem, and the Fundamental The-orem of Line Integrals

Operators on vector fields: Compute divergence and curl; determine whether ornot a vector field is conservative (and if so, find a potential for the vectorfield); compute flow lines to easy vector fields

Computations of integrals: Compute double and triple integrals (possibly with achange of variables); compute scalar and vector line integrals; compute areasand volumes


1. Green’s Theorem is an important fact that asserts the equality of a doubleintegral and a line integral.

a) What are all the hypotheses that have to be true for Green’s Theorem towork?

b) Write the equation that is the conclusion of Green’s Theorem.

c) Use Green’s Theorem to find the area of a circle of radius R.

2. Suppose f : R2 → R is some continuous function such that∫ 2

0

∫ 2

0f(x, y) dy dx = 8

∫ 2

0

∫ 4

2f(x, y) dy dx = 4

∫ 4

2

∫ 4

0f(x, y) dy dx = 10.

Compute each quantity:

a)∫ 2

0∫ 2

0 f(x, y) dx dyb)

∫ 20∫ 4

2 5f(x, y) dy dxc)∫ 4

2∫ 4

0 [f(x, y) + 3] dy dxd)

∫ 40∫ 4

0 f(x, y) dy dx

3. Compute the following integrals:

a)∫ π

0∫ 1

0 x sin y dx dyb)

∫∫E xy

3 dA, where E = (x, y) : x ≥ 0, y ≥ 0, x+ y ≤ 3.c)∫∫E e

x−y dA, where E is the square with vertices (1, 0), (0, 1), (−1, 0) and(0,−1).

d)∫∫E x dA, where E = (x, y) : x2 + y2 ≤ 1, x ≤ y ≤

√3x.

493


e)∫ 5

0∫ 4−2∫ 2

1 xy2z3 dx dy dz

f)∫∫∫

E(x2+2y) dV , whereE is the triangular pyramid with vertices (0, 0, 0),(4, 0, 0), (0, 2, 0) and (0, 0, 6).

g)∫∫∫

E(x2 + y2 + z2)2 dV , where E is the sphere of radius 2 centered at theorigin.

h)∫γ(x3 + y) ds, where γ is the curve parametrized by x(t) = (2t, t3) for

0 ≤ t ≤ 1.

i)∫γ f ·ds, where f(x, y, z) = (8x3y, x2z, 4y2z) and γ is the curve parametrized

by x(t) = (t, t2, t3) for 0 ≤ t ≤ 1.

j)∫γ [(x+ 2y) dx+ (x− y) dy], where γ is the line segment from (1, 1) to

(3, 5).

k)∫γ f ·ds, where f(x, y) = (ex+2y−ey−x, 2ex+2y+ey−x) and γ is parametrized

by x(t) = (sin t, cos 2t) for 0 ≤ t ≤ π2 .

4. Find the volume of each of these subsets of R3:

a) E is the set of points lying above the rectangle with vertices (0, 0, 0),(1, 0, 0), (0, 2, 0) and (1, 2, 0), but below the surface z = 4− x2 − y.

b) E = (x, y, z) : 0 ≤ z ≤ 2, z2 ≤ y ≤ 2z, 0 ≤ x ≤ y + z.c) E = (x, y, z) : y ≥ 0, x2 + y2 ≤ z ≤

√x2 + y2.

d) E is a sphere of radius R.

5. Let f(x, y, z) = x2 i− 2xy j + yz2 k.

a) Compute div f(3, 1, 4).

b) What does your answer to part (a) mean?

c) Compute the curl of f .

d) Is f conservative? Why or why not?

e) Does f have path-independent line integrals? Why or why not?

6. You may assume that each vector field given below is conservative. Find apotential function for each vector field.

a) f(x, y) = (8xy + 2y3 + 2x+ 1, 4x2 + 6xy2 − 2)b) f(x, y, z) = (y cosxy, x cosxy − 2z sin 2yz,−2y sin 2yz)

7. Compute the area of the ellipse x2

a2 + y2

b2 = 1.

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My answers

1. a) There are three important hypotheses: (1) E ⊆ R2 is a closed, boundedregion with a piecewise C1 boundary; (2) ∂E is oriented so that as youtravel along ∂E, E is on the left; (3) f(x, y) = (M,N) is a C1 vector fieldon E.

b) ∮∂E

f · ds =∫∫

E

(∂N

∂x− ∂M

∂y

)dA.

c) Parametrize the boundary ∂E of the circle by x(t) = (R cos t, R sin t) for0 ≤ t ≤ 2π. Then by Green’s Theorem,

area(E) = 12

∮∂Ex dy − y dx = 1

2

∫ 2π

0(R2 cos2 t+R2 sin2 t) dt

= R2

2

∫ 2π

0dt

= R2

2 (2π)

= πR2.

2. a) 8b) 20

c) 28d) 22

3. a) 1b) 243

40

c)√

2(e− e−1)d) 1

6(√

3−√

2)e) 5625f) 104

5

g) 5127 π

h) 16

(13√

13− 8)

i) 16760

j) 12

k) e−2 + e−1 − e− e2

Hint: Show f is conservative anduse the Fundamental Theoremof Line Integrals.

4. a) 163

b) 5215

c) π12

d) 43πR

3

5. a) 12b) Since div f(3, 1, 4) > 0, the instantaneous flow rate out of point (3, 1, 4)

exceeds the rate in.

c) curl f = (z2, 0,−2y).

d) No, because curl f 6= 0.

495


e) No, because f is not conservative.

6. a) f(x, y) = 4x2y + 2xy3 + x2 + x− 2y.

b) f(x, y, z) = sin xy + cos 2yz.

7. πab

496

Appendix A

Mathematica reference

A.1 What is Mathematica?Mathematica is an extremely useful and powerful software package / program-

ming language invented by a mathematician named Stephen Wolfram. Early ver-sions of Mathematica came out in the late 1980s and early 1990s; as of Spring 2020,the most recent version available to you is Mathematica 12.

Mathematica does symbolic manipulation of mathematical expressions; it solvesall kinds of equations; it has a library of important functions from mathemat-ics which it recognizes while doing computations; it does 2− and 3−dimensionalgraphics; it has a built-in word processor tool; it works well with Java and C++;etc. One thing it doesn’t do is prove theorems, so it is less useful for a theoreticalmathematician than it is for an engineer or college student.

How to obtain Mathematica: Go to my webpage http://mcclendonmath.com,choose your course from the left-hand menu, then follow the link that says

“Students: to obtain your copy of Mathematica, click here.”

A bit about how Mathematica works: When you use the Mathematica program,you are actually running two programs. The “front end” of Mathematica is the partthat you type on and the part you see. The “kernel” is the part of Mathematicathat actually does the calculations. If you type in 2 + 2 and hit [ENTER] (actually[SHIFT]+[ENTER]; see below), the front end “sends” that information to the kernelwhich actually does the computation. The kernel then “sends” the result back tothe front end, which displays 4 on the screen.

About Mathematica notebooks and cells: The actual files that Mathematicaproduces that you can edit and save are called notebooks and carry the file designa-

497

A.1. What is Mathematica?

tion *.nb; they take up little space and can easily be saved to Google docs or on aflash drive, or emailed to yourself if you want them somewhere you can retrievethem. Suggestion: when saving any file, include the date in the file name (so it iseasier to remember which file you are supposed to be open).

A Mathematica notebook is broken into cells. A cell can contain text, input, oroutput. A cell is indicated by a dark blue, right bracket (a “]”) on the right-handside of the notebook. To select a cell, click that bracket. This highlights the “]”in blue. Once selected, you can cut/copy/paste/delete cells as you would high-lighted blocks of text in a Word document.

To change the formatting of a cell, select the cell, then click “Format / Style” andselect the style you want. You may want to play around with this to see what thevarious styles look like. There are three particularly important styles:

• input: this is the default style for new cells you type• output: this is the default style for cells the kernel produces from your com-

mands• text: changing a cell to text style allows you to make comments in between

the calculations

To execute an input cell, put the cursor anywhere in the cell and hit [SHIFT]+[ENTER](or the [ENTER] on the numeric keypad at the far-right edge of the keyboard). The[ENTER] next to the apostrophe key (a.k.a. [RETURN]) gives you only a carriagereturn.

498

A.2. Important general concepts re: Mathematica syntax

A.2 Important general concepts re: Mathematica syntaxExecuting mathematical commands: To execute an input cell, put the cursor any-

where in the cell and hit [SHIFT]+[ENTER] (or the [ENTER] on the numerickeypad at the far-right edge of the keyboard). The [ENTER] next to the apos-trophe key (a.k.a. [RETURN]) gives you only a carriage return.

Multiplication: use a star or a space: 2 * 3 or 2 3 will multiply numbers; a x meansa times x; ax means the variable ax (in Mathematica, variables do not have tobe named after one letter; they can be named by words or other strings ofcharacters as well).

Parentheses: used for grouping only. Parentheses mean “times” in Mathematica.

Brackets: used to enclose all functions and Mathematica commands. For example,to evaluate a function f(x), you would type f[x]; for sin x you type Sin[x]; etc..Brackets mean “of” in Mathematica and cannot be used for multiplication.

Capitalization: All Mathematica commands and built-in functions begin with cap-ital letters. For example, to find the sine of π, typing sin(pi) does you no good(this would be the variable “sin” times the variable “pi”). The correct syntaxis Sin[Pi].

Spaces: Mathematica commands do not have spaces in them; for example, theinverse function of sine is ArcSin, not Arc Sin or Arcsin.

Pallettes: Lots of useful commands are available on the Basic Math AssistantPallette, which can be brought up by clicking “Pallettes / Basic Math Assistant”on the toolbar. If you click on a button in the pallette, what you see appearsin the cell.

Commands Mathematica knows: Sqrt, Sin, Cos, Tan, Csc, Cot, Sec, ArcSin, ArcCos,ArcTan, ArcCsc, ArcSec, ArcCot, ! (for factorial). It knows what Pi and E are(but not pi or e).

Logarithms: Log[ ] means natural logarithm (base e); Log10[ ] means commonlogarithm (base 10).

% refers to the last output (like ANS on a TI-calculator).

Exact answers versus decimal approximations: Mathematica gives exact answersfor everything if possible. If you need a decimal approximation, click “numericalvalue" or use the command N[ ]. For example, N[Pi] spits out 3.14159...

To solve an equation: make sure there are two equals signs (“==”) in your equa-tion.

499

A.2. Important general concepts re: Mathematica syntax

Getting help from the program: To get help on a command, type ? followed bythe command you don’t understand (with no space between the ? and thecommand).

To export graphics: Once Mathematica produces a graphic, you can right-click thegraphic, and select “Copy Graphic”. Then you can go in a Word document ora PowerPoint, and paste the graphic. You can subsequently resize it and/ormove it around as you see fit.

Troubleshooting: For a command to run correctly, you usually want everythingin your command to be black. If anything is purple or red, that suggestswhere the problem is. Variables that don’t have values should be blue. Next,check that everything is capitalized appropriately. Next, check that you aren’tmissing a space if you are trying to multiply two variables. Next, if you areusing variables in your code, try clearing the variables by executing some-thing like Clear[x] (if your variable is x). Then re-run the command that isgiving you trouble.

If Mathematica freezes up in the middle of a calculation and you see “Run-ning..." at the top of your screen, click “Evaluation / Abort Evaluation” onthe toolbar. If this doesn’t help, kill the program and restart it.

To get help: Email me, and attach your Mathematica file to your email. I can trou-bleshoot things pretty quickly if the file is attached. If the file isn’t attached,it is hard for me to figure out what you are doing wrong. Alternatively, seekassistance from another math major who has experience with Mathematica.

500

A.3. Entering vectors and matrices in Mathematica

A.3 Entering vectors and matrices in MathematicaTyping in vectors

To type in a vector, use a set of squiggly braces to surround the components,and separate the components with commas: for example, to save the vector v =(2,−3, 5, 0, 1), execute

v = 2,-3,5,0,1

Typing in matrices

To type in a matrix, use one of two methods:

1. Use squiggly braces and commas to separate the entries. Each row should besurrounded by a squiggly brace, and the entire matrix should be surroundedby a set of squiggly braces, and everything should be separated by commas.For example, to define A as (

1 23 4

)one could execute

A = 1, 2, 3, 4

Note that if you have a column matrix like

123

, this matrix can be defined

by thinking of B as a vector and typing something like B = 1, 2, 3, insteadof having to type B = 1, 2,3.

2. On the Basic Math Assistant palette, under Basic Commands, click the matrix.Then type A =, then click the matrix in the palette. To add rows and columns,click AddRow or AddColumn until the matrix is the appropriate size. Then gointo the matrix and type in each entry, moving between the locations usingthe [TAB] key or clicking on the location you want.

If the entries of the matrix are functions, then define the matrix as a function byexecuting A[t_] = ... or A[x_, y_] = ... instead of A = ...

501

A.4. Mathematica quick reference guides

A.4 Mathematica quick reference guidesGeneral tasks

TASK MATHEMATICA SYNTAX

To call the preceding output %To get a decimal approximation to the N[%]

preceding output (or click numerical value)

Algebraic manipulations

TASK MATHEMATICA SYNTAX

To factor an expression Factor[ ]To multiply out an expression Expand[ ]

(i.e. FOIL an expression)Partial fraction decomposition Apart[ ]To combine rational terms Together[ ]

(i.e. “undo” a partial fraction decomp)To simplify an answer Simplify[ ] (or FullSimplify[ ])

Solving equations

GOAL MATHEMATICA SYNTAX

Find exact solution(s) to equation Solve[lhs == rhs, x]of form lhs = rhs (two equals signs)

(assuming the variable is x) (works only with polynomials or otherrelatively “easy” equations)

Find decimal approx. to solutions NSolve[lhs == rhs, x]of equation lhs = rhs (two equals signs)

(works only with “easy” equations)Find decimal approx. to solutions FindRoot[lhs == rhs, x, guess]of equation lhs = rhs (two equals signs)Solve two (or more) equations Solve[lhs1==rhs1, lhs2==rhs2, x,y]together, like lhs1 = rhs1

lhs2 = rhs2

(assuming variables are x and y)

502


Precalculus operations

EXPRESSION MATHEMATICA SYNTAXSP

EC

IAL

SYM

BO

LS

e E (not e) (or use Basic Math Assistant pallette)π Pi (or use Basic Math Assistant)∞ Infinity (or use Basic Math Assistant)

(or type [Esc] inf [Esc])i =√−1 I (not i) (or use Basic Math Assistant)

AR

ITH

ME

TIC

3 + 4x 3 + 4x5− 27 5 - 27

12x 12x or 12 x or 12 * xxy x y (don’t forget the space)xy

x/y (or use Basic Math Assistant pallette)(or type [CTRL]+/ to get )√

32 Sqrt[32](or use Basic Math Assistant)(or type [CTRL]+2 for the√ sign)

4√

40 40ˆ(1/4) (or use Basic Math Assistant)|x− 3| Abs[x-3]

30! (factorial) 30!

TR

IG

sin π Sin[Pi]cos(x(y + 1)) Cos[x(y+1)]

cos 60 Cos[60 Degree](or use Basic Math Assistant)

cot(

2π3 + 3π

4

)Cot[2 Pi/3 + 3 Pi/4]

sin2 x Sin[x]ˆ2 (not Sinˆ2[x])arctan 1 ArcTan[1]

EX

PS

/L

OG

S

ln 3 Log[3]log6 63 Log[6,63]log 18 Log10[18] or Log[10, 18]

27y 2ˆ(7y) (or use Basic Math Assistant)(or type [CTRL]+6 to get )

ex−5+x2 E^(x-5+xˆ2) or Exp[x-5+xˆ2](or use Basic Math Assistant)

503


Defining functions

CLASS OF FUNCTION SYNTAX TO DEFINE FUNCTION

Calculus 1 function f : R→ Rx

f7−→ y f[x_] = formula

(one equals sign, underscore after the x)Ex: f(x) = 3 cos(x2−x) f[x_] = 3 Cos[xˆ(2-x)]

θf7−→ r f[t_] = formula

Ex: f(θ) = 2 + 3 sin θ f[t_] = 2 + 3 Sin[t]Parametrized curve f : R→ Rm

tf7−→ (x, y) f[t_] = x formula, y formula

Ex: f(t) = (cos t, t− 4) f[t_] = Cos[t], t-4t

f7−→ (x, y, z) f[t_] = x formula, y formula, z form.Ex: x(t) = (2t+ 1, t

t−1 , 7t5) x[t_] = 2t+1, t/(t-1), 7tˆ5

Surface/hypersurface f : Rn → R(x, y) f7−→ z f[x_, y_] = formula

(or f[x_, y_] = formula)(use the second version above if you are

composing f with another function)Ex: f(x, y) = 3x2 − 2xy f[x_, y_] = 3xˆ2 - 2 x y

(x, y, z) f7−→ w f[x_, y_, z_] = formula

Generic Calc 3 function f : Rn → Rm

(x, y, z) f7−→ (u, v) f[x_, y_, z_] = u formula, v formula(or f[x_, y_, z_] = u formula,

v formula)(use the second version above if you are

composing f with another function)Ex: f(x, y) = (3xy, ex−2y, x sin y) f[x_, y_] = 3x y, Eˆ(x-2y), x Sin[y]

504


Algebraic operations on functions

All these commands assume you have previously defined the function(s) as out-lined on the previous page.


Generate table of values for f Table[x, f[x], x, xmin, xmax, step](put //TableForm after this command to

format the output in a table)f(x+ 3) f[x+3]xf(2x)− x2f(x) x f[2x] - xˆ2 f[x]

(spaces important)Composition (f g)(x) . f[g[x]]Addition (f + g)(x) f[x] + g[x]Multiplication (fg)(x) f[x] g[x]Powers fn(x) (f[x])ˆn (or just f[x]ˆn)

Ex: sin2 x Sin[x]ˆ2

Single-variable calculus


limx→4

f(x) Limit[f[x], x -> 4]f ′(3) f’[3]h′(x) D[h[x], x]

ddx

(cosx) D[Cos[x], x]g′′′(x) g’ ’ ’[x] or D[g[x], x,3]∫x2 dx Integrate[xˆ2, x] (or use Basic Math Assistant pallette)

Note: answer will be missing the “+C”∫ 52 cosx dx For an exact answer:

Integrate[Cos[x], x, 2, 5](or use Basic Math Assistant)

For a decimal approximation:NIntegrate[Cos[x], x, 2, 5]

12∑k=1

f(k) Sum[f[k], k, 1, 12]

(or use Basic Math Assistant)∞∑n=3

blah Sum[blah, n, 3, Infinity]

(or use Basic Math Assistant)

505


Vector operations


Vector addition / subtraction:(2, 5,−1) + (5, 0, 2) 2,5,-1 + 5,0,2v−w v - w

Scalar multiplication:3(1, 2,−3,−4) 31,2,-3,-45v− 3w 5v - 3w

Dot product v ·w v.w(3,−4) · (2, 10) 3,-4.2,10

Norm ||v|| Norm[v]Unit vector in same direction as v Normalize[v]

(i.e. v||v|| )

Projection of v onto w Projection[v,w](i.e. πwv)

Angle between two vectors VectorAngle[v,w] (answer is in radians)(to get degrees, click degree measure

in the suggestions bar)Cross product v×w Cross[v,w]To get the number of components of v Length[v]To get the ith component of v v[[i]]

506


Matrix operations

To make Mathematica display an answer as a matrix:

1. follow your command with // MatrixForm, or2. once you’ve executed the command, choose Display as... matrix from the sug-

gestions bar.


Matrix addition / subtractionA+B A + BA−B A - B

Scalar multiplication3A 3AnA n A (space important)−5A+ 1

2B -5A + (1/2)BMatrix product AB A.B (the period is important)

A2 A.A or MatrixPower[A,2] (not Aˆ2)A7 MatrixPower[A,7]

Trace tr(A) Tr[A]Determinant detA Det[A]Transpose AT Transpose[A]To get the entry of matrix A in the ith row A[[i,j]]

and jth columnTo call the n× n identity matrix I IdentityMatrix[n]Find derivative of a matrix of functions A’[t]

term-by-termMatrix exponential eA = exp(A) MatrixExp[A]Eigenvalues and eigenvectors of A Eigensystem[A]

Just the eigenvalues of A Eigenvalues[A]Just the eigenvectors of A Eigenvectors[A]Find det(A− xI) CharacteristicPolynomial[A,x]

Determine if A is diagonalizable DiagonalizableMatrixQ[A]Determine if A is positive definite PositiveDefiniteMatrixQ[A]Determine if A is negative definite NegativeDefiniteMatrixQ[A]

507


Graphs and other pictures

CLASS OF FUNCTION BASIC MATHEMATICA COMMAND(S)calculus 1/2 function Plot[f(x), x, xmin, xmax]f : R→ Ry = f(x)

polar function r = f(θ) PolarPlot[f(θ), θ, θmin, θmax]image (in a plane) of ParametricPlot[f(t), t, tmin, tmax]

f : R→ R2 ParametricPlot[x(t), y(t), t, tmin, tmax]i.e. (x, y) = f(t)

image (in 3D space) of ParametricPlot3D[f(t), t, tmin, tmax]f : R→ R3 ParametricPlot3D[x(t), y(t), z(t),i.e. (x, y, z) = f(t) t, tmin, tmax]

surfacef : R2 → Ri.e. z = f(x, y)

graph: Plot3D[f(x, y), x, xmin, xmax,y, ymin, ymax]

level curve at height z = c:ContourPlot[f(x, y) == c,x, xmin, xmax, y, ymin, ymax]

basic contour plot (shaded, without labels):ContourPlot[f(x, y),x, xmin, xmax, y, ymin, ymax]

contour plot with z-values from a to b (unshaded,with labels):

ContourPlot[f(x, y),x, xmin, xmax, y, ymin, ymax,Contours -> Range[a,b],ContourShading -> None,ContourLabels -> True]

f : R3 → R level surface for w = c:i.e. w = f(x, y, z) ContourPlot3D[f(x, y, z) == c, x, xmin, xmax,

y, ymin, ymax, z, zmin, zmax]planar vector field VectorPlot[f(x, y), x, xmin, xmax,

f : R2 → R2 y, ymin, ymax]i.e. (u, v) = f(x, y)

3D vector field VectorPlot3D[f(x, y, z), x, xmin, xmax,f : R3 → R3 y, ymin, ymax, z, zmin, zmax]i.e. (u, v, w) = f(x, y, z)

To superimpose more than one different type of these pictures, type the com-mands in the same cell, separated by commas, and surround the list of commandswith Show[ ].

508


“Add-ons” to graphics commands

The examples below describe how to adapt the Plot[ ] command, but the othercommands on the previous page can be adapted in similar ways.

GOAL HOW TO ADAPT THE Plot[ ] COMMAND

Plot multiple graphs at once Plot[formula, formula, ..., formula,x, xmin, xmax]

Plot the graph of f(x) = formula Plot[formula, x, xmin, xmax,with range of y−values specified PlotRange -> ymin, ymax]Plot the graph of f(x) = formula Plot[formula, x,xmin,xmax,with x- and y-axes on same scale PlotRange -> ymin,ymax,

AspectRatio -> Automatic]Plot the graph of f(x) = formula Plot[formula, x,xmin,xmax,with a red, dashed curve PlotStyle -> Red, Dashed]

Differentiation of functions of several variables

The next page gives commands to evaluate the various types of derivatives westudy. To substitute numerical values for x, y and z in these answers, do one oftwo things:

1. Define the derivative as a function of x, y and z, then ask Mathematica to plugin the values of x, y and z to your newly-defined function:

Example: Suppose you wanted to compute fxy(3, 2,−5). You could executethese commands, one at a time:h[x_,y_,z_] = D[f[x,y,z], x, y]h[3,2,-5]

2. Follow any of the commands above with some syntax that causes Mathemat-ica to substitute in numbers for the variables:

Example: Suppose you wanted to compute fxy(3, 2,−5). You could executethis single command:D[f[x,y,z], x, y] /.x -> 3 /.y -> 2 /.z -> -5

In general, you follow the command with a series of /.var->number com-mands; this plugs in number to variable var in the preceding expression.

509



Partial derivatives of f : Rn → Rfx = ∂f

∂xD[f[x,y], x] or D[f[x,y,z], x], etc.

fy = ∂f∂y

D[f[x,y], y] or D[f[x,y,z], y], etc.

Partial derivatives of f : Rn → Rm

∂f1∂x

D[f[x,y][[1]],x] or D[f[x,y,z][[1]], x], etc.∂f2∂x

D[f[x,y][[2]],x] or D[f[x,y,z][[2]], x], etc.

(alternatively, compute total derivativeusing the command given below, andread off the answer)

Higher-order partial derivativesof f : Rn → Rfxx = ∂2f

∂x2 D[f[x,y], x, x] or D[f[x,y], x, 2]fyyyyy = ∂5f

∂y5 D[f[x,y,z], y, 5]fxy = ∂2f

∂x∂yD[f[x,y], x,y]

∂10f∂x3∂y2∂z5 D[f[x,y,z], x,3, y,2, z,5]

Total derivative Df of D[f[x,y], x,y] or D[f[x,y,z], x,y,z], etc.f : Rn → Rm (to get the answer as matrix, click

MatrixForm)

Total derivative Df of f’[x] or f’[t], etc.f : R→ Rm (f”[x] works as well)

Gradient∇f Grad[f[x,y], x,y] or Grad[f[x,y,z], x,y,z], etc.(or use total derivative command)

Directional derivative Duf(x) Grad[f[x,y], x,y].Normalize[u]or Grad[f[x,y,z], x,y,z].Normalize[u]

Hessian Hf D[f[x,y], x,y,2] orD[f[x,y,z], x,y,z,2], etc.

Jacobian J(f) = detDf Det[D[f[x,y], x,y] orDet[D[f[x,y,z], x,y,z], etc.

510


Integration

Mathematica only computes Riemann and iterated integrals, so if you are given adouble integral or line integral, you must first convert it to an iterated integral orRiemann integral, then use commands similar to those described below:

EXAMPLE MATHEMATICA SYNTAX∫ 83 cosx dx Integrate[Cos[x], x, 3,8]

(or use the∫

sign from the Basic Math Assistant pallette)∫ 10∫ x

0 y dy dx Integrate[Integrate[y, y, 0, x], x, 0, 1](or use one

∫sign inside another)

Vector fieldsGOAL MATHEMATICA SYNTAX

Divergencevector field on R2 Div[f[x,y], x,y]vector field on R3 Div[f[x,y,z], x,y,z]

Curl Curl[f[x,y,z], x,y,z]Picture of vector field on R2 VectorPlot[f[x,y], x, xmin, xmax,

y, ymin, ymax]Picture of vector field on R2 VectorPlot[f[x,y], x, xmin, xmax,with flow line through (a, b) y, ymin, ymax,graphed StreamPoints -> a,b,

StreamScale -> Full]Picture of vector field on R2 VectorPlot[f[x,y], x, xmin, xmax,with several flow lines y, ymin, ymax,graphed StreamPoints -> 20,

StreamScale -> Full](this will sketch 20 flow lines)

Picture of vector field on R3 VectorPlot3D[f[x,y,z], x, xmin, xmax,y, ymin, ymax, z, zmin, zmax]

511

Index

Ai,j , 371Bε(x), 95C1−smooth, 220C2−smooth, 223C∞−smooth, 223Cr−smooth, 223Df , 211Duf(x), 242Dom(f), 130Hf , 335Mmn(R), 550, 390m×n, 58∆xk, 368Rn, 37T(t), 292ej , 43∅, 32∂f∂x

, 219γ1 + γ2, 443i, 44∈, 32∫γ f · ds, 449∫γ f ds, 444

j, 44k, 44N(t), 292∇ · f , 435∇× f , 437∇f , 243

∂E, 96⊆, 32xopp, 442aN(t), 295aT (t), 295aij , 55ds, 446f(A), 130f−1(B), 130f ext, 372fx, 221s, 304

absolute extrema, 342absolute maximum/minimum, 342absolute value (of a vector), 47absolutely convergent (series), 25acceleration, 279acceleration, components of, 295acceleration, Pythagorean Theorem for,

295addition (of matrices), 57addition (of vectors), 39additivity property of integrals, 374angle (between vectors), 52antiderivative, 20, 284Antiderivative Theorem, 284arc length, 282arc length parametrization, 304

512

Index

arc length parametrization, formula for,306

arc length parametrization, propertiesof, 306

area (of a subrectangle), 371area formula via Green’s Theorem, 461astroid, 485average value, 390

ball, open, 95binormal vector, 320boundary (of a set), 96boundary point, 96bounded set, 100bracket rule, 68

Cartesian ↔ cylindrical conversions,90

Cartesian↔ polar conversions, 85Cartesian↔ spherical conversions, 93Cartesian coordinates, 84Cartesian product (of sets), 35Cauchy-Schwarz Inequality, 53center of mass, 421Chain Rule, higher-dimensional, 236change of variables in double integrals,

395circle, 97circle, osculating, 323Clairaut’s Theorem, 223classification of critical points, 332clopen set, 98closed interval, 95closed path, 440closed set, 98codomain, 130column vector, 57compact set, 100complement (of a set), 34component functions, 131computing cross product, 69concatenation (of paths), 443concavity, 17

conditionally convergent (series), 25cone, 169conic section, 155conic section, degenerate, 164conic section, nondegenerate, 163conic sections, summary of, 163connected set, 100conservative (vector field), 429Conservativity Test in R3, 466Constant Multiple Rule, 213constrained optimization problem, 346constraint, 346continuous (function R→ R), 12continuous (function Rn → Rm), 182contour plot, 145convergent (series), 25converges absolutely, 25converges conditionally, 25coordinate planes, 83coordinate projections, 131coordinate-wise integration, 284coordinates, Cartesian, 84coordinates, cylindrical, 89coordinates, polar, 84coordinates, rectangular, 84coordinates, spherical, 92critical point, 330Critical Point Theorem, 330critical points, classification of, 332cross product, 67Cross Product Rule, 214cross product, computing, 69cross product, properties, 67curl, 437curvature, 309curvature formulas, 313curve, 299cycloid, 485cylinder, 145cylindrical ↔ Cartesian conversions,

90cylindrical coordinates, 89

513

Index

definite integral, 20, 369definite integral (of function R→ Rn),

284definiteness test for symmetric 2 × 2

matrices, 337definiteness test using minors, 338degenerate conic, 164derivative (of function R→ R), 16derivative (of function Rn → Rm), 211derivative mapping, 211derivative, directional, 242derivative, partial, 219derivative, total, 211determinant, 64determinants, properties of, 65Difference Rule, 213differentiable (function R→ R), 16differentiable (function Rn → Rm), 211differential geometry, 460differentiation rules, 18, 213differentiation, implicit, 240direction of maximum rate of change,

248direction vector (of a line), 72direction vectors (of a plane), 76directional derivative, 242directional derivative, formula for, 244directional derivative, maximum and

minimum values, 248disconnected set, 100disjoint (sets), 34displacement, 282distance (between vectors), 47div f , 435divergence, 435Divergence Theorem (in the plane), 459divergent (series), 25domain, 130dot product, 49Dot Product Rule, 213dot product, interpretation of, 53double integral, 372

double integrals in polar coordinates,399

double integrals, properties of, 373

eigenvalue, 337element (of a set), 32elementary u-substitution, 23elementary region, 378, 388ellipse, 158ellipse, parametric equations of, 159ellipsoid, 166elliptic paraboloid, 170empty set, 32entrema, local (relative), 330equality (of matrices), 55equality (of sets), 32equality of mixed partials, 223equation of sphere, 97equipotential set, 429equivalent (paths), 442Euclidean space, 37Existence/Uniqueness Theorem for ODEs,

432exponential growth and decay, 433extension (of a function), 372extrema, absolute (global), 342extrinsic property, 304

first derivative, 17flow (of a vector field), 434flow line, 431folium of Descartes, 462folium, simple, 486Frenet-Serret formulas, 298Fubini’s Theorem (horizontally simple

regions), 383Fubini’s Theorem (triple integrals), 389Fubini’s Theorem (vertically simple re-

gions), 382function of several variables, 131function, definition of, 130functions, component, 131

514

Index

Fundamental Theorem of Calculus, 21,284, 367

Fundamental Theorem of Line Integrals,464

Gauss’ Theorem, 460global maximum/minimum, 342gradient, 243gradient field, 429Gradient Theorem, 464gradient, applications of, 244gradients as vector fields, 250graph (of a function Rn → Rm), 136graphing planes in R3, 82graphing polar functions, 86Green’s Theorem, 457Green’s Theorem area formula, 461Green’s Theorem, vector version, 459

Hessian, 335higher-dimensional Chain Rule, 236higher-order partial derivatives, 222horizontally simple region, 378hyperbola, 160hyperbola, parametric equations of, 160hyperbolic paraboloid, 170hyperboloid, 167hyperplane, 77

identity matrix, 56image, 130implicit differentiation, 240incompressible, 436indefinite integral, 20, 284infinite series, 25initial point, 440integrable, 284, 369, 372integral of a constant function, 376integral, additivity property of, 374integral, definite, 20, 369integral, double, 372integral, indefinite, 20, 284integral, iterated, 379

integral, triple, 388integration by parts, 24integration rules, 21interpretation of dot product, 53intersection, 34interval, 95intrinsic property, 304inverse image, 130irrotational, 438iterated integral, 379

Jacobian, 395

L’Hôpital’s Rule, 14Lagrange multiplier, 351Lagrange’s method, 351lemniscate, 486length (of a curve), 282length (of a vector), 47level curve, 145level set, 149level set, normal vector to, 245level set, tangent hyperplane to, 248level surface, 149level surface, tangent plane to, 247limit (of function R→ R), 12limit (of function Rn → R), 173limit (of function Rn → Rm), 182limit (one-sided), 14line integrals, path independent, 463linear approximation, 229Linear Replacement Principle, 24linearity of differentiation, 213linearity of integration, 373linearization, 229local extrema, 330local maximum, 330local maximum/minimum, 330

magnitude (of a vector), 47matrix, 55matrix multiplication, 58matrix multiplication, properties of, 61

515

Index

matrix operations, 57Max-Min Existence Theorem, 343Max-Min inequality (for integrals), 375maximum, local (relative), 330meaning of ds, 446minimum, absolute (global), 342minimum, local (relative), 330mixed partials, equality of, 223moments of inertia, 421Monotonicity law (for integrals), 375motion, projectile, 286multiplication of matrices, 58

n−dimensional Euclidean space, 37nth derivative, 17negative definite (matrix), 336neighborhood, 95non-uniqueness of parametrizations,

303nondegenerate conic, 163norm (of a partition), 368, 371norm (of a vector), 47normal component of acceleration, 295,

310normal equation (of a hyperplane), 77normal equation (of a line), 79normal equation (of a plane), 79normal line (to a function R2 → R),

properties of, 233normal line (to a graph), 232normal vector (to a hyperplane), 77normalized version (of a vector), 50

one-sided limit, 14open ball, 95open interval, 95open set, 98operation on sets, 34operations of matrices, 57opposite path class, 442optimization problem, 327optimization problem, constrained, 346orientation-preserving, 441

orientation-reversing, 441orthogonal, 51osculating circle, 323osculating plane, 320

parabola, 162parabola, parametric equations of, 162paraboloid, 170parallelepiped, 64parametric equations (of ellipses), 159parametric equations (of hyperbolas),

160parametric equations (of lines in Rn),

73parametric equations (of parabolas), 162parametric equations (of planes in Rn),

76parametrization, 299, 440parametrization, arc length, 304parametrizations, non-uniqueness of,

303parametrized curves, 137partial derivative, 219partial derivatives, geometric interpre-

tation of, 224partial derivatives, higher-order, 222partition, 368, 371path, 299, 440path class, 442path class, opposite, 442path independent line integrals, 463perpendicular, 51piecewise C1 path, 443piecewise C2 path, 443piecewise C∞ path, 443plane, 76plane, coordinate, 83plane, graph of, 82point-slope equation, 71polar↔ Cartesian conversions, 85polar coordinates, 84polar coordinates trick, 179polar functions, 86

516

Index

polynomial, Taylor, 26position, 279positive definite (matrix), 336Positivity law (for integrals), 375potential (function), 429power series, uniqueness of, 26principal minors (of a matrix), 337principal unit normal vector, 292product (of matrices), 58product rules, 213product, Cartesian, 35projectile motion, 286projections, coordinate, 131properties of cross product, 67properties of determinants, 65properties of double integrals, 373properties of matrix multiplication, 61properties of Riemann integrals, 22Pythagorean Theorem, 48Pythagorean Theorem for acceleration,

295

quadric surface, 165quadric surfaces, summary of, 172Quotient Rule, 214

range, 130Rearrangement Theorem, 25rectangular coordinates, 84region, elementary, 378, 388relative maximum/minimum, 330reparametrization, 441Riemann integral, 20Riemann integrals, properties of, 22Riemann sum, 368, 371Rule of Sarrus, 66

saddle point, 330scalar, 39scalar line integral, 444scalar line integral, computation of, 445scalar line integral, reparametrization

of, 445

scalar multiplication (by a matrix), 57scalar multiplication (by a vector), 39Scalar Product Rule, 213scalar triple product, 68second derivative, 17Second Derivative Test, 338second-order partial derivatives, 222series, infinite, 25series, Taylor, 26set, 32set operations, 34set-builder notation, 33simple folium, 486simple path, 440simply connected, 466size (of a matrix), 55slope-intercept equation, 71smooth (function), 220, 223solenoidal, 436speed, 279sphere, 97sphere, equation of, 97spherical↔ Cartesian conversions, 93spherical coordinates, 92spherical coordinates trick, 181square (matrix), 55Squeeze Theorem, 175standard basis of Rn, 43standard equation (of a hyperplane),

77standard equation (of a line), 71Stokes’ Theorem, 459stream line, 431subinterval, 368, 371subrectangle, 371subset, 32sum (of an infinite series), 25Sum Rule, 213surface, 141symmetric equations, 72

tangent hyperplane to a level set, 248tangent line, 231

517

Index

tangent plane, 232tangent plane (to function R2 → R),

equations of, 233tangent plane to a level surface, 247tangential component of acceleration,

295, 307Taylor polynomial, 26Taylor series, 26terminal point, 440test point, 368, 371tone, 17topology, 95total derivative, 211trace (of a function), 142trace (of a matrix), 55transpose, 55Triangle Inequality, 54triple integral, 388triple integrals in cylindrical coordi-

nates, 401triple integrals in spherical coordinates,

401triple product, 68type 1 region, 377type 2 region, 378

u-substitution (in integrals), 23, 394union, 34uniqueness of power series, 26unit tangent vector, 292unit vector, 50utility, 327, 346

vector, 39vector field, 150, 428vector line integral, 449vector line integral, computing, 450vector line integral, equivalent forms,

450vector line integrals, reparametrization

of, 453vector space, 39vector triple product, 68

velocity, 279Vertical Line Test, 144vertically simple region, 377

width (of a subinterval), 368, 371

x-simple region, 377xy-plane, 83xz-plane, 83

y-simple region, 378yz-plane, 83

zero matrix, 58zero vector, 39zeroth derivative, 17

518

Vector Calculus Lecture Notes - David M. McClendon

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