lesson 11: implicit differentiation (section 21 handout)
DESCRIPTION
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.TRANSCRIPT
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