Section 2.6 Implicit Differentiation V63.0121.021, Calculus I New York University October 11, 2010 Announcements I Quiz 2 in recitation this week. Covers §§1.5, 1.6, 2.1, 2.2 I Midterm next week. Covers §§1.1–2.5 Announcements I Quiz 2 in recitation this week. Covers §§1.5, 1.6, 2.1, 2.2 I Midterm next week. Covers §§1.1–2.5 V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 2 / 34 Objectives I Use implicit differentation to find the derivative of a function defined implicitly. V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 3 / 34 Notes Notes Notes 1 Section 2.6 : Implicit Differentiation V63.0121.021, Calculus I October 11, 2010
Using implicit differentiation we can treat relations which are not quite functions like they were functions. In particular, we can find the slopes of lines tangent to curves which are not graphs of functions.
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Section 2.6Implicit Differentiation
V63.0121.021, Calculus I
New York University
October 11, 2010
Announcements
I Quiz 2 in recitation this week. Covers §§1.5, 1.6, 2.1, 2.2
I Midterm next week. Covers §§1.1–2.5
Announcements
I Quiz 2 in recitation thisweek. Covers §§1.5, 1.6, 2.1,2.2
I Midterm next week. Covers§§1.1–2.5
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 2 / 34
Objectives
I Use implicit differentation tofind the derivative of afunction defined implicitly.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 3 / 34
Notes
Notes
Notes
1
Section 2.6 : Implicit DifferentiationV63.0121.021, Calculus I October 11, 2010
Outline
The big idea, by example
ExamplesBasic ExamplesVertical and Horizontal TangentsOrthogonal TrajectoriesChemistry
The power rule for rational powers
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 4 / 34
Motivating Example
Problem
Find the slope of the linewhich is tangent to the curve
x2 + y 2 = 1
at the point (3/5,−4/5).
x
y
Solution (Explicit)
I Isolate: y 2 = 1− x2 =⇒ y = −√
1− x2. (Why the −?)
I Differentiate:dy
dx= − −2x
2√
1− x2=
x√1− x2
I Evaluate:dy
dx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/5
4/5=
3
4.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 5 / 34
Motivating Example, another way
We know that x2 + y 2 = 1 does not define y as a function of x , butsuppose it did.
I Suppose we had y = f (x), so that
x2 + (f (x))2 = 1
I We could differentiate this equation to get
2x + 2f (x) · f ′(x) = 0
I We could then solve to get
f ′(x) = − x
f (x)
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 6 / 34
Notes
Notes
Notes
2
Section 2.6 : Implicit DifferentiationV63.0121.021, Calculus I October 11, 2010
Yes, we can!
The beautiful fact (i.e., deep theorem) is that this works!
I “Near” most points on thecurve x2 + y 2 = 1, the curveresembles the graph of afunction.
I So f (x) is defined “locally”,almost everywhere and isdifferentiable
I The chain rule then appliesfor this local choice.
x
y
looks like a function
looks like a function
does not look like afunction, but that’sOK—there are onlytwo points like this
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 7 / 34
Motivating Example, again, with Leibniz notation
Problem
Find the slope of the line which is tangent to the curve x2 + y 2 = 1 at thepoint (3/5,−4/5).
Solution
I Differentiate: 2x + 2ydy
dx= 0
Remember y is assumed to be a function of x!
I Isolate:dy
dx= −x
y.
I Evaluate:dy
dx
∣∣∣∣( 3
5,− 4
5 )=
3/5
4/5=
3
4.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 8 / 34
Summary
If a relation is given between x and y which isn’t a function:
I “Most of the time”, i.e., “atmost places” y can be assumedto be a function of x
I we may differentiate the relationas is
I Solving fordy
dxdoes give the
slope of the tangent line to thecurve at a point on the curve.
x
y
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 9 / 34
Notes
Notes
Notes
3
Section 2.6 : Implicit DifferentiationV63.0121.021, Calculus I October 11, 2010
Outline
The big idea, by example
ExamplesBasic ExamplesVertical and Horizontal TangentsOrthogonal TrajectoriesChemistry
The power rule for rational powers
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 10 / 34
Another Example
Example
Find y ′ along the curve y 3 + 4xy = x2 + 3.
Solution
Implicitly differentiating, we have
3y 2y ′ + 4(1 · y + x · y ′) = 2x
Solving for y ′ gives
3y 2y ′ + 4xy ′ = 2x − 4y
(3y 2 + 4x)y ′ = 2x − 4y
=⇒ y ′ =2x − 4y
3y 2 + 4x
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 11 / 34
Yet Another Example
Example
Find y ′ if y 5 + x2y 3 = 1 + y sin(x2).
Solution
Differentiating implicitly:
5y 4y ′ + (2x)y 3 + x2(3y 2y ′) = y ′ sin(x2) + y cos(x2)(2x)
Collect all terms with y ′ on one side and all terms without y ′ on the other:
Find the isothermic compressibility of an ideal gas.
Solution
If PV = k (n is constant for our purposes, T is constant because of theword isothermic, and R really is constant), then
dP
dP· V + P
dV
dP= 0 =⇒ dV
dP= −V
P
So
β = − 1
V· dV
dP=
1
P
Compressibility and pressure are inversely related.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 27 / 34
Nonideal gassesNot that there’s anything wrong with that
Example
The van der Waals equationmakes fewer simplifications:(
P + an2
V 2
)(V − nb) = nRT ,
where P is the pressure, V thevolume, T the temperature, nthe number of moles of the gas,R a constant, a is a measure ofattraction between particles ofthe gas, and b a measure ofparticle size.
OxygenH
H
Oxygen
H
H
Oxygen H
H
Hydrogen bonds
Image credit: Wikimedia Commons
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 28 / 34
Compressibility of a van der Waals gas
Differentiating the van der Waals equation by treating V as a function ofP gives (
P +an2
V 2
)dV
dP+ (V − bn)
(1− 2an2
V 3
dV
dP
)= 0,
so
β = − 1
V
dV
dP=
V 2(V − nb)
2abn3 − an2V + PV 3
Question
I What if a = b = 0?
I Without taking the derivative, what is the sign ofdβ
db?
I Without taking the derivative, what is the sign ofdβ
da?
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 29 / 34
Notes
Notes
Notes
9
Section 2.6 : Implicit DifferentiationV63.0121.021, Calculus I October 11, 2010