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:
5y 4y ′ + 3x2y 2y ′ − sin(x2)y ′ = −2xy 3 + 2xy cos(x2)
Now factor and divide:
y ′ =2xy(cos x2 − y 2)
5y 4 + 3x2y 2 − sin x2
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 12 / 34
Notes
Notes
Notes
4
Section 2.6 : Implicit DifferentiationV63.0121.021, Calculus I October 11, 2010
Finding tangent lines with implicit differentitiation
Example
Find the equation of the line tangentto the curve
y 2 = x2(x + 1) = x3 + x2
at the point (3,−6).
Solution
Differentiate: 2ydy
dx= 3x2 + 2x , so
dy
dx=
3x2 + 2x
2y, and
dy
dx
∣∣∣∣(3,−6)
=3 · 32 + 2 · 3
2(−6)= −33
12= −11
4.
Thus the equation of the tangent line is y + 6 = −11
4(x − 3).
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 13 / 34
Recall: Line equation forms
I slope-intercept formy = mx + b
where the slope is m and (0, b) is on the line.
I point-slope formy − y0 = m(x − x0)
where the slope is m and (x0, y0) is on the line.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 14 / 34
Horizontal Tangent Lines
Example
Find the horizontal tangent lines to the same curve: y 2 = x3 + x2
Solution
We have to solve these two equations:
y 2 = x3 + x2
[(x , y) is on the curve]13x2 + 2x
2y= 0
[tangent lineis horizontal]
2
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 15 / 34
Notes
Notes
Notes
5
Section 2.6 : Implicit DifferentiationV63.0121.021, Calculus I October 11, 2010
Solution, continued
I Solving the second equation gives
3x2 + 2x
2y= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x + 2) = 0
(as long as y 6= 0). So x = 0 or 3x + 2 = 0.
I Substituting x = 0 into the first equation gives
y 2 = 03 + 02 = 0 =⇒ y = 0
which we’ve disallowed. So no horizontal tangents down that road.
I Substituting x = −2/3 into the first equation gives
y 2 =
(−2
3
)3
+
(−2
3
)2
=4
27=⇒ y = ± 2
3√
3,
so there are two horizontal tangents.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 16 / 34
Tangents
(−2
3 ,2
3√
3
)
(−2
3 ,−2
3√
3
)node
(−1, 0)
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 17 / 34
Example
Find the vertical tangent lines to the same curve: y 2 = x3 + x2
Solution
I Tangent lines are vertical whendx
dy= 0.
I Differentiating x implicitly as a function of y gives
2y = 3x2 dx
dy+ 2x
dx
dy, so
dx
dy=
2y
3x2 + 2x(notice this is the reciprocal
of dy/dx).
I We must solve
y 2 = x3 + x2
[(x , y) is onthe curve]
12y
3x2 + 2x= 0
[tangent lineis vertical]
2
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 18 / 34
Notes
Notes
Notes
6
Section 2.6 : Implicit DifferentiationV63.0121.021, Calculus I October 11, 2010
Solution, continued
I Solving the second equation gives
2y
3x2 + 2x= 0 =⇒ 2y = 0 =⇒ y = 0
(as long as 3x2 + 2x 6= 0).
I Substituting y = 0 into the first equation gives
0 = x3 + x2 = x2(x + 1)
So x = 0 or x = −1.
I x = 0 is not allowed by the first equation, but
dx
dy
∣∣∣∣(−1,0)
= 0,
so here is a vertical tangent.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 19 / 34
Examples
Example
Show that the families of curves
xy = c x2 − y 2 = k
are orthogonal, that is, they intersect at right angles.
Solution
In the first curve,
y + xy ′ = 0 =⇒ y ′ = −y
x
In the second curve,
2x − 2yy ′ = 0 = =⇒ y ′ =x
y
The product is −1, so the tangent lines are perpendicular wherever theyintersect.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 21 / 34
Orthogonal Families of Curves
xy = cx2 − y 2 = k x
y
xy=
1
xy=
2
xy=
3
xy=−1
xy=−2
xy=−3
x2−
y2
=1
x2−
y2
=2
x2−
y2
=3
x2 − y 2 = −1x2 − y 2 = −2x2 − y 2 = −3
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 22 / 34
Notes
Notes
Notes
7
Section 2.6 : Implicit DifferentiationV63.0121.021, Calculus I October 11, 2010
Examples
Example
Show that the families of curves
xy = c x2 − y 2 = k
are orthogonal, that is, they intersect at right angles.
Solution
In the first curve,
y + xy ′ = 0 =⇒ y ′ = −y
x
In the second curve,
2x − 2yy ′ = 0 = =⇒ y ′ =x
y
The product is −1, so the tangent lines are perpendicular wherever theyintersect.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 23 / 34
Ideal gases
The ideal gas law relatestemperature, pressure, andvolume of a gas:
PV = nRT
(R is a constant, n is the amountof gas in moles)
Image credit: Scott Beale / Laughing SquidV63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 25 / 34
Compressibility
Definition
The isothermic compressibility of a fluid is defined by
β = −dV
dP
1
V
with temperature held constant.
Approximately we have
∆V
∆P≈ dV
dP= −βV =⇒ ∆V
V≈ −β∆P
The smaller the β, the “harder” the fluid.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 26 / 34
Notes
Notes
Notes
8
Section 2.6 : Implicit DifferentiationV63.0121.021, Calculus I October 11, 2010
Compressibility of an ideal gas
Example
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
Nasty derivatives
I
dβ
db= −(2abn3 − an2V + PV 3)(nV 2)− (nbV 2 − V 3)(2an3)
(2abn3 − an2V + PV 3)2
= −nV 3
(an2 + PV 2
)(PV 3 + an2(2bn − V ))2
< 0
I
dβ
da=
n2(bn − V )(2bn − V )V 2
(PV 3 + an2(2bn − V ))2> 0
(as long as V > 2nb, and it’s probably true that V � 2nb).
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 30 / 34
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 31 / 34
Using implicit differentiation to find derivatives
Example
Finddy
dxif y =
√x .
Solution
If y =√
x, theny 2 = x ,
so
2ydy
dx= 1 =⇒ dy
dx=
1
2y=
1
2√
x.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 32 / 34
Notes
Notes
Notes
10
Section 2.6 : Implicit DifferentiationV63.0121.021, Calculus I October 11, 2010
The power rule for rational powers
Theorem
If y = xp/q, where p and q are integers, then y ′ =p
qxp/q−1.
Proof.
First, raise both sides to the qth power:
y = xp/q =⇒ yq = xp
Now, differentiate implicitly:
qyq−1 dy
dx= pxp−1 =⇒ dy
dx=
p
q· xp−1
yq−1
Simplify: yq−1 = x (p/q)(q−1) = xp−p/q so
xp−1
yq−1=
xp−1
xp−p/q = xp−1−(p−p/q) = xp/q−1
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 33 / 34
Summary
I Implicit Differentiation allows us to pretend that a relation describes afunction, since it does, locally, “almost everywhere.”
I The Power Rule was established for powers which are rationalnumbers.
V63.0121.021, Calculus I (NYU) Section 2.6 Implicit Differentiation October 11, 2010 34 / 34
Notes
Notes
Notes
11
Section 2.6 : Implicit DifferentiationV63.0121.021, Calculus I October 11, 2010