Lesson 11: Implicit Differentiation (Section 21 handout)

V63.0121.021, Calculus I Section 2.6 : Implicit Differentiation October 11, 2010

Notes
Section 2.6
Implicit Differentiation

V63.0121.021, Calculus I

New York University

October 11, 2010

Announcements

Quiz 2 in recitation this week. Covers §§1.5, 1.6, 2.1, 2.2
Midterm next week. Covers §§1.1–2.5

Announcements
Notes

Quiz 2 in recitation this
week. Covers §§1.5, 1.6, 2.1,
2.2
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
Notes

Use implicit differentation to
find the derivative of a
function defined implicitly.


1


Outline
Notes

The big idea, by example

Examples
Basic Examples
Vertical and Horizontal Tangents
Orthogonal Trajectories
Chemistry

The power rule for rational powers


Motivating Example
y Notes
Problem
Find the slope of the line
which is tangent to the curve
x
x2 + y2 = 1

at the point (3/5, −4/5).

Solution (Explicit)

Isolate: y 2 = 1 − x 2 =⇒ y = − 1 − x 2 . (Why the −?)
dy −2x x
Differentiate: =− √ =√
dx 2 1 − x2 1 − x2
dy 3/5 3/5 3
Evaluate: = = = .
dx x=3/5 1 − (3/5)2 4/5 4


Motivating Example, another way
Notes
We know that x 2 + y 2 = 1 does not define y as a function of x, but
suppose it did.
Suppose we had y = f (x), so that

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

We could differentiate this equation to get

2x + 2f (x) · f (x) = 0

We could then solve to get
x
f (x) = −
f (x)


2


Yes, we can!
Notes

The beautiful fact (i.e., deep theorem) is that this works!
y
“Near” most points on the
curve x 2 + y 2 = 1, the curve
resembles the graph of a
function. looks like a function
So f (x) is defined “locally”,
x
almost everywhere and is
differentiable does not look like a
The chain rule then applies function, but that’s
for this local choice. OK—there are only
two points like this looks like a function


Motivating Example, again, with Leibniz notation
Notes

Problem
Find the slope of the line which is tangent to the curve x 2 + y 2 = 1 at the
point (3/5, −4/5).

Solution
dy
Differentiate: 2x + 2y =0
dx
Remember y is assumed to be a function of x!
dy x
Isolate: =− .
dx y
dy 3/5 3
Evaluate: = = .
dx ( 3 ,− 4 ) 4/5 4
5 5


Summary
Notes

If a relation is given between x and y which isn’t a function:
“Most of the time”, i.e., “at y
most places” y can be assumed
to be a function of x
we may differentiate the relation x
as is
dy
Solving for does give the
dx
slope of the tangent line to the
curve at a point on the curve.


3


Outline
Notes


Examples
Basic Examples
Chemistry



Another Example
Notes
Example
Find y along the curve y 3 + 4xy = x 2 + 3.

Solution
Implicitly diﬀerentiating, we have

3y 2 y + 4(1 · y + x · y ) = 2x

Solving for y gives

3y 2 y + 4xy = 2x − 4y
(3y 2 + 4x)y = 2x − 4y
2x − 4y
=⇒ y = 2
3y + 4x


Yet Another Example
Notes
Example
Find y if y 5 + x 2 y 3 = 1 + y sin(x 2 ).

Solution


4


Finding tangent lines with implicit diﬀerentitiation
Notes
Example
Find the equation of the line tangent
to the curve

y 2 = x 2 (x + 1) = x 3 + x 2

at the point (3, −6).

Solution


Recall: Line equation forms
Notes

slope-intercept form
y = mx + b
where the slope is m and (0, b) is on the line.
point-slope form
y − y0 = m(x − x0 )
where the slope is m and (x0 , y0 ) is on the line.


Horizontal Tangent Lines
Notes
Example
Find the horizontal tangent lines to the same curve: y 2 = x 3 + x 2

Solution
We have to solve these two equations:

3x 2 + 2x
= 0
y2 = x3 + x2 2y
1 [(x, y ) is on the curve] 2 [tangent line
is horizontal]


5


Solution, continued
Notes
Solving the second equation gives

3x 2 + 2x
= 0 =⇒ 3x 2 + 2x = 0 =⇒ x(3x + 2) = 0
2y

(as long as y = 0). So x = 0 or 3x + 2 = 0.
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.
Substituting x = −2/3 into the ﬁrst equation gives
3 2
2 2 4 2
y2 = − + − = =⇒ y = ± √ ,
3 3 27 3 3
so there are two horizontal tangents.

Tangents
Notes

− 2 , 3√3
3
2

(−1, 0)

− 2 , − 3√3
3
2 node


Example
Find the vertical tangent lines to the same curve: y 2 = x 3 + x 2 Notes

Solution


6


Solution, continued
Notes


Examples
Notes
Example
Show that the families of curves

xy = c x2 − y2 = k

are orthogonal, that is, they intersect at right angles.

Solution
In the ﬁrst curve,
y
y + xy = 0 =⇒ y = −
x
In the second curve,
x
2x − 2yy = 0 = =⇒ y =
y
The product is −1, so the tangent lines are perpendicular wherever they
intersect.

Orthogonal Families of Curves
Notes
y
xy
xy
=
3
=
xy

2
=
x2 − y2 = 3
x −y =2
x2 − y2 = 1

1

xy = c
2

x2 − y2 = k x
1
−
2

− 2
=

x 2 − y 2 = −1
−
3
xy

=

x 2 − y 2 = −2
xy
=

x 2 − y 2 = −3
xy


7


Examples
Notes
Example
Show that the families of curves

xy = c x2 − y2 = k

are orthogonal, that is, they intersect at right angles.

Solution
In the first curve,
y
y + xy = 0 =⇒ y = −
x
In the second curve,
x
2x − 2yy = 0 = =⇒ y =
y
The product is −1, so the tangent lines are perpendicular wherever they
intersect.

Ideal gases
Notes

The ideal gas law relates
temperature, pressure, and
volume of a gas:

PV = nRT

(R is a constant, n is the amount
of gas in moles)

Image credit: Scott Beale / Laughing Squid

Compressibility
Notes

Definition
The isothermic compressibility of a fluid is defined by
dV 1
β=−
dP V
with temperature held constant.

Approximately we have
∆V dV ∆V
≈ = −βV =⇒ ≈ −β∆P
∆P dP V
The smaller the β, the “harder” the fluid.


8


Compressibility of an ideal gas
Notes

Example
Find the isothermic compressibility of an ideal gas.

Solution
If PV = k (n is constant for our purposes, T is constant because of the
word isothermic, and R really is constant), then

dP dV dV V
·V +P = 0 =⇒ =−
dP dP dP P
So
1 dV 1
·
β=− =
V dP P
Compressibility and pressure are inversely related.


Nonideal gasses
Not that there’s anything wrong with that Notes

Example
The van der Waals equation H
makes fewer simpliﬁcations:
Oxygen H
n2
P +a 2 (V − nb) = nRT ,
V H
Oxygen Hydrogen bonds
where P is the pressure, V the H
volume, T the temperature, n
the number of moles of the gas, Oxygen H
R a constant, a is a measure of
attraction between particles of H
the gas, and b a measure of
particle size.


Compressibility of a van der Waals gas
Notes
Diﬀerentiating the van der Waals equation by treating V as a function of
P gives

an2 dV 2an2 dV
P+ + (V − bn) 1 − = 0,
V2 dP V 3 dP
so
1 dV V 2 (V − nb)
β=− =
V dP 2abn3 − an2 V + PV 3

Question

What if a = b = 0?
dβ
Without taking the derivative, what is the sign of ?
db
dβ
Without taking the derivative, what is the sign of ?
da


9


Nasty derivatives
Notes

dβ (2abn3 − an2 V + PV 3 )(nV 2 ) − (nbV 2 − V 3 )(2an3 )
=−
db (2abn3 − an2 V + PV 3 )2
nV 3 an2 + PV 2
=− <0
(PV 3 + an2 (2bn − V ))2

dβ n2 (bn − V )(2bn − V )V 2
= >0
da (PV 3 + an2 (2bn − V ))2
(as long as V > 2nb, and it’s probably true that V 2nb).


Outline
Notes


Examples
Basic Examples
Chemistry



Using implicit diﬀerentiation to ﬁnd derivatives
Notes

Example
dy √
Find if y = x.
dx

Solution
√
If y = x, then
y 2 = x,
so
dy dy 1 1
2y = 1 =⇒ = = √ .
dx dx 2y 2 x


10


Notes
Theorem
p
If y = x p/q , where p and q are integers, then y = x p/q−1 .
q

Proof.


Summary
Notes

Implicit Diﬀerentiation allows us to pretend that a relation describes a
function, since it does, locally, “almost everywhere.”
The Power Rule was established for powers which are rational
numbers.


Notes

11

Lesson 11: Implicit Differentiation (Section 21 handout)

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