P1 Calculus I - Information Engineering Main/Home Pagedwm/Courses/1DF/1DF-L1.pdfP1 2017 1 / 73 P1 Calculus I Calculus of a Single Variable Prof David Murray [email protected]

P1 2017 1 / 73

P1 Calculus I

Calculus of a Single Variable

Prof David Murray

[email protected]/∼dwm/Courses/1DF

4 lectures, MT 2017

P1 2017 2 / 73

Motivation

Engineering involves the study of physical (and biological) systems,with the aim of turning them to the use of mankind.

Systems have inputs and outputs

x(t) y(t)System

and are often modelled by differential equations, eg

d2ydt2 + a

dydt

+ by(t) = x(t) ,

which relate the input x to the output y and its derivatives.

The calculus is exactly the mathematical machinery you need todescribe how the change in one quantity affects another.

P1 2017 3 / 73

P1 calculus courses

Four courses in P1 will advance your understanding of physicalsystems and their description using calculus.

é Calculus-I: of a single independent variable.é Ordinary Differential Equations.é Calculus-II: of multiple independent variables,é Modelling — cements how to apply calculus to physical

situations.

In Calculus-I, some of the material will be revision — but there aresubstantial new chunks to chew on.

However, the emphasis now is not on being able to do this or thatintegral or whatever — being able is a given! — but on understanding.

P1 2017 4 / 73

Course Contents

Inconveniently, the material seems to split into three topics to becovered over the four lecture slots.

é Functions, Limits and Derivativesé Integrationé Integer series and convergence; Taylor and MacLaurin

expansions

The tutorial sheet directly associated with this course is

é 1P1A Calculus I: Differentiation and Series

But performing just the examples in 1P1A is not really enough!

P1 2017 5 / 73

Reading

Read widely (don’t just rely on the notes); do try out more examples.Riley K.F., Hobson M.P., and Bence S.J. (2006)Mathematical Methods for Physics and EngineeringCambridge University Press, 3rd ed.

Stephenson, G. (1973)Mathematical Methods for Science StudentsLongman, 2nd ed, ISBN 0-582-44416-0

James, G. (2001)Modern Engineering MathematicsPearson, 5th ed, ISBN 0-13-018319-9

Stroud, K.A. (2013)Engineering MathematicsPalgrave Macmillan, 7th ed, ISBN 9781137031204

Also look at HLT (Howatson, Lund and Todd)Engineering Tables and Data

P1 2017 6 / 73

Course WWW Pages

Pdf copies of the notes, copies of the lecture slides, the tutorialsheets, corrections, answers to FAQs etc, will be accessible from

www.robots.ox.ac.uk/∼dwm/Courses

Only the notes and the tute sheets get put on weblearn.Follow the links from www.eng.ox.ac.uk

P1 2017 7 / 73

Some of the Cast

Cauchy D’Alembert De l’Hopital Holder

Leibnitz MacLaurin Newton Taylor

P1 2017 8 / 73

Lecture 1: Functions, Limits, and Derivatives

P1 2017 9 / 73

Functions, Limits, and Derivatives

In this course we are concerned with calculus of a single real variable.

Whether dealing with a single variable or, later, multiple variables,calculus is based upon — apart from the concept of number —two outstandingly important concepts:

the concept of the functionthe concept of the limit

Our aims in this first lecture areto use these concepts to define the derivativeto use that definition to establish the derivatives of functions,products, quotients, and so on.

P1 2017 10 / 73

Lecture contentsThe basics ...1.1 Definitions and jargon associated with functions1.2 Revise limits and continuity1.3 Definition of the derivative.

With the derivative defined we can handle ...1.4 Product Rule and Leibnitz’s Rule1.5 Quotient Rule1.6 The Chain Rule – function of a function

Carrying on ...1.7 Derivatives of standard functions, including1.8 Hyperbolic functions1.9 Stationary points

1.10 Sketching and Plotting

P1 2017 11 / 73

1.1 Functions

P1 2017 12 / 73

Functions of a single variable

Given an input x , a function f defines a recipe y = f (x) for evaluatingthe corresponding output y .

One might also call the function recipe a mapping, or atransformation, or an operator.

We should take care to define the type of the inputs and outputs —real numbers, integers, complex numbers, etc.Here we are concerned with a scalar real variables.

We should also define the interval in which the function definition isvalid.

y = f (x), a ≤ x ≤ b, x , y ∈ R

Programmers take similar care with the types of variables andintervals of validity.

P1 2017 13 / 73

Descriptions applied to functions

(a) Single-valued and multi-valued functions.

For a single value of the independent variable x , a function mightdeliver one or more values of y ...

y

x

y

x

y

x

(a) (b) (c)

P1 2017 14 / 73

Descriptions applied to functions(b) Continuous/discontinous functions.

Intuitively, a continuous function is one where the pen does not haveto be lifted to plot the function ...

y

x

y

x

(a) (b)

More precisely: for a continuous function defined within an interval,any change in y = f (x) can be kept within any arbitrary small boundby making the change in x suitably small.

P1 2017 15 / 73

Descriptions applied to functions

(c) Smooth functions.A continuous function does not have to be smooth!

A smooth function f is one where all its derivatives

f ′(x), f ′′(x), f ′′′(x), . . . f (∞)

are continuous.

Some functions might appear acceptably “smooth to the eye”because a good number of higher derivatives are continuous.

A function f is said to be of class Ck smooth when derivatives up tof (k) are continuous.

P1 2017 16 / 73

Descriptions applied to functions

(d) Monotonic increasing/decreasing functions.

If increasing x causes y = f (x) always to increase/decrease, then thefunction is monotonic increasing/decreasing.

The 1st derivative or gradient always has the same sign.

A monotonic increasing function, y = log x ...y

x

P1 2017 17 / 73

Monotonic functions ... may be useful in 1P1A?Consider y = f (x) = xe−x — obviously not monotonic ...

0 2 4 6 8 100

0.1

0.2

0.3

0.4

x

y=

x*e

xp(−

x)

−8 −6 −4 −2 0 2 40

0.1

0.2

0.3

0.4

log(x)

y=

x*e

xp(−

x)

0 1 2 3 40

0.1

0.2

0.3

0.4

sqrt(x)

y=

x*e

xp(−

x)

y vs x y vs log x y vs√

x

0 20 40 60 80 1000

0.1

0.2

0.3

0.4

x2

y=

x*e

xp(−

x)

−1 −0.5 0 0.5 10

0.1

0.2

0.3

0.4

sin(x)

y=

x*e

xp(−

x)

y vs x2 y vs sin x

P1 2017 18 / 73

Descriptions applied to functions(e) Inverse functions

The function y = f (x) defines a mapping between x and y ...... and hence also one between y and x as x = φ(y).

Functions f and φ are inverses, written f = φ−1.

The inverse might be easy to write down explicity, eg

y = ax + b ⇒ x = y/a− b/a

andy = exp x ⇒ x = ln(y)

or may be only implicit, eg

y = ex + ln(x) ⇒ x =?

P1 2017 19 / 73

Descriptions applied to functions

(f) Parametric functions.

It is sometimes convenient to define variables parametrically.

Instead of considering y = f (x), we are given y = y(p) and x = x(p).

Eg, instead of writing y = ±√

1− x2 to define a circle, one might write

x = cos(p) y = sin(p)

P1 2017 20 / 73

Descriptions applied to functions(g) Odd and even functions.

If

f (−x) ={−f (x) the function is ODD+f (x) the function is EVEN

x

y

x

y

(a) (b)

A function can, of course, be neither even nor odd.

P1 2017 21 / 73

Descriptions applied to functions

(h) Periodic functions.

A function whose output is the same if the input is increase by aperiod, T.

Eg, f (x) = cosωx is periodic with period T = 2π/ω

Hence for integer n

f (x + nT ) = f(

x + n2πω

)= f (x) .

In a later course, you will learn about Fourier series which provide amethod of describing all well-behaved periodic functions using sumsof cosines and sines.

P1 2017 22 / 73

Descriptions applied to functions(i) Piecewise functions.A piecewise function is one made by joining together differentfunctions defined over neighbouring intervals.

y

x

Notice that the piecewise linear function is continuous, but is only C0

smooth.

A piecewise quadratic function can be made C1 smooth — function +gradients are continuous

Piecewise cubic is C2 smooth, etc

P1 2017 23 / 73

Classes of Function

While the previous descriptions can be applied quite generally, weturn now to classify the small set of basic functions.

Functions are divided into two, perhaps three, classesAlgebraic, Transcendental (elementary), & Transcendental (special)

Algebraic: (a) Rational functions. A polynomial functiony = a0 + a1x + a2x2 + . . . is a rational integral function. If one takesratios of such functions, one ends up with a general rational function

y =a0 + a1x + a2x2 + . . .

b0 + b1x + b2x2 + . . .

Algebraic: (b) Roots of functions. Simple examples include rootsof general rational functions, eg

√x2 + 1.

P1 2017 24 / 73

Classes of FunctionTranscendental: (c) Exponential and logarithmic functions.Functions ex , ln x are the most obvious, but this class includes otherexponentiated functions like ax .

ln ax = x ln a → ax = ex ln a .

Transcendental: (d) Hyperbolic functions. These are made fromexponentials, and we deal with them in more detail later.

Transcendental: (e) Trignometric functions. These functions arethose involving sin and cos tan, cot, sec and cosec. They are alsocalled circular functions.

Transcendental: (f) Specials. Error functions, Bessel functions,Gamma functions, etc, are examples of smooth specials you maycome across.There are two discontinuous functions that are important for you, theHeaviside unit step u(x) and the Dirac delta function δ(x).

P1 2017 25 / 73

Heaviside and delta functions

x

u(x)

1

00

(x)δ

x00

To

infinity

Heaviside u(x) delta δ(x)

δ(x) is infinitely high, but infinitely narrow. It has the property that∫ 0+

0−δ(x)dx = 1

Can you think what the relationship between u(x) and δ(x) is?

P1 2017 26 / 73

1.2 Limits and Continuity

P1 2017 27 / 73

Limits and ContinuityConsider the function

f (x) =(

sin xx

).

Easy to evaluate, except when x = 0 — 0/0 might be anything).Indeed, at x = 0, f (x) is undefined.But if you put in decreasing values, sayx = 10−1,10−2,10−3..., your calculator will give(

1− sin xx

)= 1.7×10−3, 1.7×10−5, 1.7×10−7, ...

That is, our function f (x) gets ever closer to 1.The same happens when x increases towards zero from the negativeside, and empirically we can write

limx→0

(sin x

x

)= 1,

even though sin 0/0 remains undefined.

P1 2017 28 / 73

Limits and ContinuityWriting limx→a f (x) = ` means that as x gets closer and closer to a,the function value gets closer and closer to `.

For any arbitrarily small number ε > 0, it is possible to find an ηsuch that

|f (x)− `| < ε for all 0 < |x − a| < η . (1)

Note that when you consider x → a, x never reaches the value of a,so that is why 0 < |x − a|.

Indeed, as we saw with sin(x)/x , the function does not even have tobe defined at x = a

lε

a

η

ε/106

Keep zooming in

f(x)

x

P1 2017 29 / 73

Limits and ContinuityBecause the function does nothave to be defined at positiona, it is possible for differentlimits to exist either side of adiscontinuity,

If the discontinuity is at x = a,the limits from below andabove would be written

limx→a−

= p limx→a+

= q .

where a± indicates values an infinitesimally smaller or bigger than a.

Note that if the limits are the same, p = q, we simply revert to theprevious case of one limit,

limx→a

= p = q

P1 2017 30 / 73

Limits and Continuity

To clarify the point further, consider the three functions

f (x) =sin(x)

x; g(x) =

{sin(x)/x x 6= 01 x = 0 ; h(x) =

{sin(x)/x x 6= 042 x = 0

The limits of all three functions as x → 0 are unity.

But f (0) is undefined, while g(0)=1 and h(0)=42.

Only g(x) is a continuous function;f (x) has a gap in it; whileh(x) is discontinuous.

P1 2017 31 / 73

A refined definition of continuity

Using the limit concept, we can revisit the definition of continuity.

A single-valued function g(x) is continuous at x=a iffThe function is defined at x=a, AND

The limit limx→a g(x) exists, AND

limx→a g(x) = g(a)

(You now see how the previous g(x) satisfies these conditions, whilefunctions f (x) and h(x) both violate them.)

P1 2017 32 / 73

Useful theorems on limits and hence continuityTheorems on limits:

limx→a{f (x) + g(x)} = lim

x→af (x) + lim

x→ag(x)

limx→a

{f (x)g(x)

}=

limx→a f (x)limx→a g(x)

limx→a{f (x)g(x)} = lim

x→af (x) lim

x→ag(x)

Then the connection between limits and continuity allow us to assertTheorems on continuity:

The sum and difference of two continuous functions iscontinuous.The quotient of two continuous functions is continuous at everypoint where the denominator is non-zero.The product of two continuous functions is continuous.

P1 2017 33 / 73

1.3 The Derivative

P1 2017 34 / 73

The DerivativeWith functions and limits defined, we can define the derivative.

The Derivative defined

dfdx

= limδx→0

[δfδx

]= lim

δx→0

[f (x + δx)− f (x)

δx

].

x+

x+f(

x

f(x)

x δx

δx)

δ x

δ f

Note the different meanings ofdf and δf , dx and δxδf is a small but finitedifference, resulting from asmall but finite difference in x .In the limit as δx tends to 0, thechanges becomesinfinitesimally small (that is, nolonger finite) and becomeδx → dx and δf → df .

P1 2017 35 / 73

The Derivative

Reminder ...

dfdx

= limδx→0

[δfδx

]= lim

δx→0

[f (x + δx)− f (x)

δx

].

Note that:f (x) is differentiable at x if this limit exists at x , and existsindependently of how δx → 0.

A differentiable function is continuous, but

A continuous function is not necessarily differentiable.

P1 2017 36 / 73

♣ Example:

Q: Derive dy/dx when y = f (x) = x2.A:

dfdx

= limδx→0

[(x + δx)2 − x2

δx

]=

[x2 + 2xδx + (δx)2 − x2

δx

]The x2 terms cancel, and we can neglect terms (δx)2 by comparisonwith terms in (δx), especially in the limit as δx → 0.

Hencedfdx

= limδx→0

[2xδxδx

]= 2x .

P1 2017 37 / 73

♣ Example:Q: Derive dy/dx when y = f (x) = xn.A: For f (x) = xn we need the binomial expansion for {x + δx}n

dfdx

= limδx→0

[{x + δx}n − xn

δx

]

=

{

xn + nxn−1δx + n(n−1)2! xn−2(δx)2 + . . .

}− xn

δx

=

[nxn−1δx

δx

]= nxn−1 .

This result allows use to handle any polynomial.

Also, if we had already proven the Quotient Rule, we would be able todifferentiate any general rational function.

P1 2017 38 / 73

1.4 The Product Rule and Leibnitz’s Rule

P1 2017 39 / 73

The Product Rule and Leibnitz’s RuleThe derivative of the product of two functions f (x)g(x) is given by

The Product Rule

ddx

(fg) = f ′g + fg′.

Proof:

ddx

(fg) = limδx→0f (x + δx)g(x + δx)− f (x)g(x)

δx

=[limδx→0f (x + δx)] [limδx→0g(x + δx)]− f (x)g(x)

limδx→0δx

But we have already shown in the definition of the derivative that

limδx→0f (x + δx) = f (x) + f ′(x)limδx→0δx

so that

ddx

(fg) = limδx→0fg + f ′gδx + fg′δx + f ′g′(δx)2 − fg

δx= f ′g + fg′.

P1 2017 40 / 73

Higher derivatives of a product: Leibnitz

What happens if we differentiate again? Well, f ′(x) and g′(x) are stillfunctions of x , so that

d2

dx2 (fg) = f ′′g + f ′g′ + f ′g′ + fg′′ = f ′′g + 2f ′g′ + fg′′ ,

d3

dx3 (fg) = GRIND AWAY = f ′′′g + 3f ′′g′ + 3f ′g′′ + fg′′′ .

You see the binomial coefficients emerging, and can guess whathappens next. So did Leibnitz ...

Leibnitz’s Rule

dn

dxn fg =n∑

i=0

(n

n − i

)f (n−i)g(i)

P1 2017 41 / 73

Leibnitz Rule

Often∑

signs makes things harder to “see”.Unpacking this and the binomial coefficients gives

Perhaps the most immediately useful statement ofLeibnitz’s Rule

dn

dxn fg = f (n)g + nf (n−1)g(1) +n(n − 1)

2!f (n−2)g(2) + . . .

Note that we have started out at a particular end of the expression.

This has practical repercussions ...

It makes sense to choose g as the function that “disappears soonest”when differentiating ...

P1 2017 42 / 73

♣ Example: Leibnitz RuleQ: Find the nth derivative of x3e2x .A: Decide which function to choose as fand which as g.

Obviously x3 will disappear soonest — ase2x never does!

f (x) = e2x g(x) = x3

f ′(x) = 2e2x g′(x) = 3x2

f ′′(x) = 4e2x g′′(x) = 6xf ′′′(x) = 8e2x g′′′(x) = 6f (n)(x) = 2ne2x g(≥4)(x) = 0

dn

dxn (fg) = f (n)g + nf (n−1)g(1) +n(n − 1)

2!f (n−2)g(2) + . . .

= e2x{

2nx3 + n2n−13x2 +n(n − 1)

22n−26x +

n(n − 1)(n − 2)6

2n−36}

= e2x 2n{

x3 +3n2

x2 +3n(n − 1)

4x +

n(n − 1)(n − 2)8

}.

Suppose we wanted the n = 8th derivative. We would find it as

256e2x {x3 + 12x2 + 42x + 42}.

You might want to try this using just the product rule ...

P1 2017 43 / 73

Where were we yesterday?

Leibnitz’s Rule

dn

dxn fg =n∑

i=0

(n

n − i

)f (n−i)g(i) =

n∑i=0

n!(n − i)!i!

f (n−i)g(i)

The most immediately useful statement of ...

Leibnitz’s Rule

dn

dxn fg = f (n)g + nf (n−1)g(1) +n(n − 1)

2!f (n−2)g(2) + . . .

Choose g(x) to disappear quickly

P1 2017 44 / 73

♣ Example: Leibnitz Rule

Q: Find the nth derivative of (x3 + 2x)dydx

A: Choosef =

dydx

g = (x3 + 2x)

and notice that f (n) = y (n+1). Then

dn

dxn fg = y (n+1)(x3 + 2x) + ny (n)(3x2 + 2) +n(n − 1)

2y (n−1)(6x)

+n(n − 1)(n − 2)

6y (n−2)(6)

This example points to the possibility of taking the nth derivative of anordinary differential equation.

P1 2017 45 / 73

♣ Example: Leibnitz Rule (a bit awkard ...)Q: Find the nth derivative ofx cos(x).A: f , g and work out derivs:

f (0) = cos x g(0) = xf (1) = − sin x g(1) = 1f (2) = − cos x g(2) = 0f (3) = sin x ..

f (4) = cos x ..

So ...

f (m) =

{(−1)m/2 cos x m even

(−1)(m+1)/2 sin x m odd

NB! There are different overall results forn-odd and n-even.NB! if n is even, the second term, f (n−1),requires you to set m = (n − 1) in the ODDderivative expression, and so on.

Applying Leibnitz’s rule for n even

dn

dxn x cos x = (−1)n/2 cos x(x) + n(−1)(n−1+1)/2 sin x (1)

= (−1)n/2 (x cos x + n sin x) ,

but for n odddn

dxn x cos x = (−1)(n+1)/2 sin x (x) + n(−1)(n−1)/2 cos x (1)

= (−1)(n+1)/2 (x sin x − n cos x) .

P1 2017 46 / 73

Check that last result by hand ... for n = 6f = x cos x

f ′ = cos x − x sin xf ′′ = − sin x − sin x − x cos x = −(2 sin x + x cos x)

f ′′′ = −(2 cos x + cos x − x sin x) = −(3 cos x − x sin x)f (4) = −(−3 sin x − sin x − x cos x) = (4 sin x + x cos x)

f (5) = 4 cos x + cos x − x sin x) = (5 cos x − x sin x)f (6) = −5 sin x − sin x − x cos x = −(6 sin x + x cos x)

P1 2017 47 / 73

Check that last result by hand ... for n = 6f = x cos x

f ′ = cos x − x sin xf ′′ = − sin x − sin x − x cos x = −(2 sin x + x cos x)

f ′′′ = −(2 cos x + cos x − x sin x) = −(3 cos x − x sin x)f (4) = −(−3 sin x − sin x − x cos x) = (4 sin x + x cos x)

f (5) = 4 cos x + cos x − x sin x) = (5 cos x − x sin x)f (6) = −5 sin x − sin x − x cos x = −(6 sin x + x cos x)

Now use the formula derived. n = 6 is even, so

dn

dxn x cos x = (−1)n/2 (x cos x + n sin x)

= (−1)6/2 (x cos x + 6 sin x)= − (x cos x + 6 sin x)

Agreement!

P1 2017 48 / 73

1.5 The Quotient Rule

P1 2017 49 / 73

The Quotient RuleUse the basic definition ...

ddx

(1v

)= limδx→0

1δx

{1

v(x + δx)− 1

v(x)

}= limδx→0

1δx

{v(x)− v(x + δx)

v(x + δx)v(x)

}= limδx→0

1δx

{−v ′(x)δx

(v(x) + v ′δx)v(x)

}=

{−v ′(x)

v2

}

Then use the product rule

ddx

(uv

)=

u′

v+ u

{−v ′

v2

}=

u′v − uv ′

v2

P1 2017 50 / 73

1.6 The Chain Rule

P1 2017 51 / 73

Function of a function: the chain ruleSuppose f = f (g) and g = g(x).

If you are told x , you can work out g, and knowing g you can find f .This must mean that f is also a function of x ,⇒df/dx exists.

Chain ruleIf f = f (g) and g = g(x), then

dfdx

=dfdg

dgdx

When δx → 0 and becomes dx , the infinitesimal dg in the numeratoris exactly the same as that in the denominator giving rise to df/dg.

We’re allowed to treat d(whatever) as a divisible quantity.

But this won’t apply to partials ∂(whatever) when we come to partialdifferentiation.

P1 2017 52 / 73

Function of a function: the chain rule

♣ Example:Q: Differentiate y = exp(x2) with respect to x .

A:y = exp(g) g = x2

dydg

= exp(g)dgdx

= 2x ⇒dydx

= 2x exp(x2).

P1 2017 53 / 73

Function of a function: the chain rule

♣ Example:Q: Differentiate y = exp(cos(x2)) with respect to x .

A: This is a triple decker ...

y = exp(g) g = cos(h) h = x2

dydg

= exp(g)dgdh

= − sin(h)dhdx

= 2x

⇒dydx

=dydg

dgdh

dhdx

= exp(g)(− sin(h))2x= exp(cos(x2))(− sin(x2))2x .

P1 2017 54 / 73

1.7. Derivatives of the elementary functions

P1 2017 55 / 73

Derivatives of the elementary functions

For completeness, we should apply the basic definition ofdifferentiation to the set of simple functions.

(a) General rational functions.We have already proved that the derivative of xn is nxn−1,and hence we can deal with any polynomial.

Using the quotient rule we can handle any general rational function.

P1 2017 56 / 73

Derivatives ...

(b) nth roots.Helpful knowledge: the n-th root of “1 plus a small thing” is approximated as:

(1 + h)1/n ≈(

1 +hn

)

Then write

dydx

= limδx→0(x + δx)1/n − x1/n

δx= limδx→0

1δx

[x1/n

(1 +

δxx

)1/n

−x1/n]

Using the approximation this becomes

limδx→01δx

[x1/n

(1 +

δxnx

)−x1/n

]= limδx→0

1δx

[x1/nδx

nx

]=

[1n

x (1/n)−1]

P1 2017 57 / 73

Derivatives ...(c) Exponentials.Helpful knowledge: when h is small eh ≈ 1 + h.Then

ddx

ex = limδx→01δx

[e(x+δx) − ex

]= limδx→0

1δx

[ex (eδx − 1

) ]= limδx→0

1δx

[ex (1 + δx − 1)

]= ex .

Looks great — but this is a circular argument. As we see in Lecture 3,one uses the derivative to show that eh ≈ 1 + h — so we can’t use itto find the derivative!

Instead, e is defined as the number for which

ddx

ex = ex

The value of the function ex at every point is the same as its gradi-ent.

P1 2017 58 / 73

♣ Example: Chain Rule and ExponentialQ: Determine the derivative of ax .A: Method 1

Let u = ax

ln(u) = x ln ad

dxln(u) = ln a

dudx

ddu

ln(u) = ln a

dudx

1u

= ln a

dudx

= u ln a

ddx

ax = ax ln a

A: Method 2

ln(ax) = x ln aax = e(x ln a)

ddx

ax =d

dx(x ln a) e(x ln a)

= (ln a) e(x ln a)

= (ln a) ax

P1 2017 59 / 73

Derivatives ...

(d) Trig functions.A DIY exercise.

P1 2017 60 / 73

1.8 Hyperbolic functions

P1 2017 61 / 73

Hyperbolic FunctionsThe hyperbolic functions are defined in terms of exponentials, buthave properties similar to, but subtly different from, trig functions.Why hyperbolic? They can be generated in a manner similiar totrignometric functions — but instead of being based on a circlex2 + y2 = 1, they are based on a hyperbola x2 − y2 = 1.In parametric form

x = cos(t), y = sin(t), −π < t ≤ πx = cosh(t), y = sinh(t), −∞ ≤ t ≤ ∞ .

cos(t)

sin(t)

y

xcosh(t)

sinh(t)

x

y

P1 2017 62 / 73

Hyperbolic FunctionsThe explicit definitions of the hyperbolic cosine and sine are

cosh x =12(ex + e−x) sinh x =

12(ex − e−x)

These are defined for −∞ ≤ x ≤ ∞

−4 −2 0 2 40

5

10

15

20

25

30

x

y=

co

sh

(x)

−4 −2 0 2 4−30

−20

−10

0

10

20

30

x

y=

sin

h(x

)

(cosh) (sinh)

P1 2017 63 / 73

Hyperbolic FunctionsThe other hyperbolic functions like sech, cosech, etc, are defined asexpected in terms of sinh and cosh.

tanh x =sinh xcosh x

=ex − e−x

ex + e−x

−4 −2 0 2 4−1

−0.5

0

0.5

1

x

y=

tan

h(x

)

(tanh)

P1 2017 64 / 73

Hyperbolic FunctionsPropertiesThe parametric definition of sinh and cosh indicate that

cosh2 x − sinh2 x = 1 .

Their derivatives, found by differentiating the ex ’s,

ddx

cosh x = sinh xd

dxsinh x = cosh x

ddx

tanh x = sech2x

Other properties are

cosh(−a) = cosh(a)sinh(−a) = − sinh(a)

cosh(a + b) = sinh(a) sinh(b) + cosh(a) cosh(b)sinh(a + b) = cosh(a) sinh(b) + sinh(a) cosh(b)

cosh(a) + sinh(a) = ea

cosh(a)− sinh(a) = e−a

P1 2017 65 / 73

Hyperbolic FunctionsBut why is there such interest in these functions?

There are two related reasons.

First, if you give cosh and sinh an imaginary argument, sin and cospop out. For example,

sinh(iy) = i sin(y) .

Second, cos t & sin t and cosh t & sinh t are solutions to thedifferential equation

ad2ydt2 + b

dydt

+ cy = 0 .

So while cos t and sin t handle oscillations in a system ...

cosh t and sinh t can be use to describing time-decaying solutions.

P1 2017 66 / 73

1.9 Stationary and inflexion points

P1 2017 67 / 73

Stationary and inflexion pointsThe following table summarizes properties of stationary and inflexionpoints.

Minimum f ′(x) = 0 f ′′(x) > 0Maximum f ′(x) = 0 f ′′(x) < 0Saddle f ′(x) = 0 f ′′(x) = 0Inflexion f ′(x) 6= 0 f ′′(x) = 0

Here a some examples:

−1 −0.5 0 0.5 1−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

x

y=

x2 +

x3

Max

Min

Inflexion

−1 −0.5 0 0.5 1−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

x

y=

x3

Saddle

f (x) = x2 + x3 f (x) = x3.

P1 2017 68 / 73

1.10 Sketching and Plotting

P1 2017 69 / 73

Sketching functions

It is very important to become skilled at sketching functions.♣ Example:

Sketch the function f (ω) =1√

(1 + α2ω2).

Spot that f (ω) ≈ 1 when αω � 1, and f (ω) ≈ 1/(αω) when αω � 1.

Now sketch on a log-log plot ...

On one side log f = 0, and on the other log f = − logα− logω or“y = − logα− x”, a graph of slope −1 on a log-log plot.

The crossover point is when ω = 1/α. The function has no stationary points.

P1 2017 70 / 73

Sketching functionsReminder:

f (ω) =1√

(1 + α2ω2)

Reminder: f (ω) ≈ 1 when αω � 1, and f (ω) ≈ 1/(αω) when αω � 1.

On one side log f = 0, and on the other log f = − logα− logω or“y = − logα− x”, a graph of slope −1 on a log-log plot.

P1 2017 71 / 73

Sketching functions♣ Example: Sketch the function f (t) = t cos

(1t

).

We know that cos(ωt) oscillates with angular frequency ω.

In our case the (instantaneous) angular frequency ω is such that ωt = 1/t ,⇒ω = 1/t2 – it oscillates with infinite frequency at t = 0 then slows down.

As t → ±∞, cos(1/t)→ 1. We multiply all that by t , a straight line throughthe origin. There are stationary points, and you may wish to locate these.

P1 2017 72 / 73

Plotting functions

You will learn Matlab this year as part of your P5 laboratory work. Youare encouraged to use it to help with your work outside the lab too.

Matlab, which is available to load onto your own machines, is aninteractive computing language and it contains some powerfulgraphing tools.

For example, some base-level code to plot y = x exp(−x) is just

% The % lines are just comments% Use 200 equally spaced points between -0.5 and 4x= linspace(-0.5,4.0,200);y= x.* exp(-x);% plot y against xplot(x,y);

P1 2017 73 / 73

Summary

In this lecture we have

Reviewed aspects of functions of a single variableReviewed the limit conceptDefined the derivativeDerived the Product Rule and Leibnitz’s RuleDerived the Quotient ruleDerived the Chain ruleApplied our definitions to find from first principles the derivativesof basic functionsSpent some time introducing hyperbolic functions

Next lecture — integration.