lesson 11: implicit differentiation (slides)

Sec on 2.6Implicit Differen a on

V63.0121.011: Calculus IProfessor Ma hew Leingang

New York University

February 28, 2011

Music Selection

“The Curse of Curves”by Cute is What WeAim For

Announcements

I Quiz 2 in recita on thisweek. Covers §§1.5, 1.6,2.1, 2.2

I Midterm next week.Covers §§1.1–2.5

Objectives

I Use implicit differenta onto find the deriva ve of afunc on definedimplicitly.

Outline

The big idea, by example

ExamplesBasic ExamplesVer cal and Horizontal TangentsOrthogonal TrajectoriesChemistry

The power rule for ra onal powers

Motivating Example

ProblemFind the slope of the linewhich is tangent to thecurve

x2 + y2 = 1

at the point (3/5,−4/5).

Motivating Example

x2 + y2 = 1

Motivating Example

x2 + y2 = 1

Motivating Example, SolutionSolu on (Explicit)

I Isolate:y2 = 1− x2 =⇒ y = −

√1− x2.

(Why the−?)

I Differen ate:dydx

= − −2x2√1− x2

1− x2I Evaluate:

∣∣∣∣x=3/5

=3/5√

1− (3/5)2=

3/54/5

I Isolate:y2 = 1− x2 =⇒ y = −

√1− x2.

(Why the−?)I Differen ate:

= − −2x2√1− x2

1− x2

I Evaluate:dydx

∣∣∣∣x=3/5

=3/5√

1− (3/5)2=

3/54/5

I Isolate:y2 = 1− x2 =⇒ y = −

√1− x2.

= − −2x2√1− x2

1− x2I Evaluate:

∣∣∣∣x=3/5

=3/5√

1− (3/5)2=

3/54/5

I Isolate:y2 = 1− x2 =⇒ y = −

√1− x2.

= − −2x2√1− x2

1− x2I Evaluate:

∣∣∣∣x=3/5

=3/5√

1− (3/5)2=

3/54/5

I Isolate:y2 = 1− x2 =⇒ y = −

√1− x2.

= − −2x2√1− x2

1− x2I Evaluate:

∣∣∣∣x=3/5

=3/5√

1− (3/5)2=

3/54/5

Motivating Example, another wayWe know that x2 + y2 = 1 does not define y as a func on of x, butsuppose it did.

I Suppose we had y = f(x), so that

x2 + (f(x))2 = 1

I We could differen ate this equa on to get

2x+ 2f(x) · f′(x) = 0

I We could then solve to get

f′(x) = − xf(x)

x2 + (f(x))2 = 1

2x+ 2f(x) · f′(x) = 0

f′(x) = − xf(x)

x2 + (f(x))2 = 1

2x+ 2f(x) · f′(x) = 0

f′(x) = − xf(x)

Yes, we can!The beau ful fact (i.e., deep theorem) is that this works!

I “Near” most points on the curvex2 + y2 = 1, the curve resemblesthe graph of a func on.

I So f(x) is defined “locally”,almost everywhere and isdifferen able

I The chain rule then applies forthis local choice.

looks like a func on

..looks like a func on

does not look like afunc on, but that’sOK—there are onlytwo points like this

Motivating Example, again, with Leibniz notationProblemFind the slope of the line which is tangent to the curve x2 + y2 = 1 atthe point (3/5,−4/5).

Solu on

I Differen ate: 2x+ 2ydydx

= 0Remember y is assumed to be a func on of x!

I Isolate:dydx

= −xy. Then evaluate:

∣∣∣∣( 35 ,−

=3/54/5

Solu on

Remember y is assumed to be a func on of x!

I Isolate:dydx

∣∣∣∣( 35 ,−

=3/54/5

Solu on

I Isolate:dydx

∣∣∣∣( 35 ,−

=3/54/5

Solu on

I Isolate:dydx

= −xy.

Then evaluate:dydx

∣∣∣∣( 35 ,−

=3/54/5

Solu on

I Isolate:dydx

∣∣∣∣( 35 ,−

=3/54/5

SummaryIf a rela on is given between x and y which isn’t a func on:I “Most of the me”, i.e., “atmost places” y can beassumed to be a func on of x

I we may differen ate therela on as is

I Solving fordydx

does give theslope of the tangent line tothe curve at a point on thecurve.

Outline

Another ExampleExample

Find y′ along the curve y3 + 4xy = x2 + 3.

Solu onImplicitly differen a ng, we have

3y2y′ + 4(1 · y+ x · y′) = 2x

Solving for y′ gives

3y2y′ + 4xy′ = 2x− 4y =⇒ (3y2 + 4x)y′ = 2x− 4y =⇒ y′ =2x− 4y3y2 + 4x

3y2y′ + 4(1 · y+ x · y′) = 2x

Yet Another ExampleExample

Find y′ if y5 + x2y3 = 1+ y sin(x2).

Solu on

Solu onDifferen a ng implicitly:

5y4y′ + (2x)y3 + x2(3y2y′) = y′ sin(x2) + y cos(x2)(2x)

Solu onDifferen a ng implicitly:

5y4y′ + (2x)y3 + x2(3y2y′) = y′ sin(x2) + y cos(x2)(2x)

Collect all terms with y′ on one side and all terms without y′ on theother:

5y4y′ + 3x2y2y′ − sin(x2)y′ = −2xy3 + 2xy cos(x2)

Solu onCollect all terms with y′ on one side and all terms without y′ on theother:

5y4y′ + 3x2y2y′ − sin(x2)y′ = −2xy3 + 2xy cos(x2)

Now factor and divide: y′ =2xy(cos x2 − y2)

5y4 + 3x2y2 − sin x2

Finding tangents with implicit differentitiation

Example

Find the equa on of the linetangent to the curve

y2 = x2(x+ 1) = x3 + x2

at the point (3,−6).

SolutionSolu on

I Differen ate: 2ydydx

= 3x2 + 2x, sodydx

=3x2 + 2x

2y, and

∣∣∣∣(3,−6)

=3 · 32 + 2 · 3

2(−6)= −33

12= −11

I Thus the equa on of the tangent line is y+ 6 = −114(x− 3).

SolutionSolu on

I Differen ate: 2ydydx

= 3x2 + 2x, sodydx

=3x2 + 2x

2y, and

∣∣∣∣(3,−6)

=3 · 32 + 2 · 3

2(−6)= −33

12= −11

I Thus the equa on of the tangent line is y+ 6 = −114(x− 3).

Finding tangents with implicit differentitiation

Example

Find the equa on of the linetangent to the curve

y2 = x2(x+ 1) = x3 + x2

at the point (3,−6).

Recall: Line equation formsI slope-intercept form

y = mx+ b

where the slope ism and (0, b) is on the line.I point-slope form

y− y0 = m(x− x0)

where the slope ism and (x0, y0) is on the line.

Horizontal Tangent LinesExample

Find the horizontal tangent lines to the same curve: y2 = x3 + x2

Solu onWe have to solve these two equa ons:

I y2 = x3 + x2 [(x, y) is on the curve]

I3x2 + 2x

2y= 0 [tangent line is horizontal]

Horizontal Tangent LinesExample

Find the horizontal tangent lines to the same curve: y2 = x3 + x2

Solu onWe have to solve these two equa ons:

I y2 = x3 + x2 [(x, y) is on the curve]

I3x2 + 2x

2y= 0 [tangent line is horizontal]

Solution, continuedI Solving the second equa on gives

3x2 + 2x2y

= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x+ 2) = 0

(as long as y ̸= 0). So x = 0 or 3x+ 2 = 0.

I Subs tu ng x = 0 into the first equa on gives

y2 = 03 + 02 = 0 =⇒ y = 0

which we’ve disallowed. So no horizontal tangents down thatroad.

3x2 + 2x2y

= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x+ 2) = 0

(as long as y ̸= 0). So x = 0 or 3x+ 2 = 0.

I Subs tu ng x = 0 into the first equa on gives

y2 = 03 + 02 = 0 =⇒ y = 0

3x2 + 2x2y

= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x+ 2) = 0

(as long as y ̸= 0). So x = 0 or 3x+ 2 = 0.I Subs tu ng x = 0 into the first equa on gives

y2 = 03 + 02 = 0 =⇒ y = 0

Solution, continued

I Subs tu ng x = −2/3 into the first equa on gives

y2 =(−23

(−23

=⇒ y = ±√

= ± 23√3,

so there are two horizontal tangents.

Horizontal Tangents

−23 ,−

23√3

..(−1, 0)

Horizontal Tangents

−23 ,−

23√3

)..node

..(−1, 0)

Example

Find the ver cal tangent lines to the same curve: y2 = x3 + x2

Solu on

I Tangent lines are ver cal whendxdy

I Differen a ng x implicitly as a func on of y gives2y = 3x2

+ 2xdxdy

, sodxdy

3x2 + 2x(no ce this is the

reciprocal of dy/dx).I We must solve y2 = x3 + x2 [(x, y) is on the curve] and

2y3x2 + 2x

= 0 [tangent line is ver cal]

Example

Solu on

+ 2xdxdy

, sodxdy

2y3x2 + 2x

Example

Solu on

+ 2xdxdy

, sodxdy

reciprocal of dy/dx).

I We must solve y2 = x3 + x2 [(x, y) is on the curve] and2y

3x2 + 2x= 0 [tangent line is ver cal]

Example

Solu on

+ 2xdxdy

, sodxdy

2y3x2 + 2x

= 0 =⇒ 2y = 0 =⇒ y = 0

(as long as 3x2 + 2x ̸= 0).

I Subs tu ng y = 0 into the first equa on gives

0 = x3 + x2 = x2(x+ 1)

So x = 0 or x = −1.I x = 0 is not allowed by the first equa on, but x = −1 is.

2y3x2 + 2x

= 0 =⇒ 2y = 0 =⇒ y = 0

(as long as 3x2 + 2x ̸= 0).I Subs tu ng y = 0 into the first equa on gives

0 = x3 + x2 = x2(x+ 1)

So x = 0 or x = −1.

I x = 0 is not allowed by the first equa on, but x = −1 is.

2y3x2 + 2x

= 0 =⇒ 2y = 0 =⇒ y = 0

(as long as 3x2 + 2x ̸= 0).I Subs tu ng y = 0 into the first equa on gives

0 = x3 + x2 = x2(x+ 1)

So x = 0 or x = −1.I x = 0 is not allowed by the first equa on, but x = −1 is.

Tangents

−23 ,−

23√3

)..node

..(−1, 0)

ExamplesExample

Show that the families of curves xy = c, x2 − y2 = k are orthogonal,that is, they intersect at right angles.

Solu on

I In the first curve, y+ xy′ = 0 =⇒ y′ = −yx

I In the second curve, 2x− 2yy′ = 0 =⇒ y′ =xy

I The product is−1, so the tangent lines are perpendicularwherever they intersect.

Orthogonal Families of Curves

xy = cx2 − y2 = k .. x.

xy=−1

xy=−2

xy=−3

x2 − y2 = −1

x2 − y2 = −2

x2 − y2 = −3

xy = cx2 − y2 = k .. x.

xy=−1

xy=−2

xy=−3

x2 − y2 = −1

x2 − y2 = −2

x2 − y2 = −3

xy = cx2 − y2 = k .. x.

xy=−1

xy=−2

xy=−3

x2 − y2 = −1

x2 − y2 = −2

x2 − y2 = −3

xy = cx2 − y2 = k .. x.

xy=−1

xy=−2

xy=−3

x2 − y2 = −1

x2 − y2 = −2

x2 − y2 = −3

xy = cx2 − y2 = k .. x.

xy=−1

xy=−2

xy=−3

x2 − y2 = −1

x2 − y2 = −2

x2 − y2 = −3

xy = cx2 − y2 = k .. x.

xy=−1

xy=−2

xy=−3

x2 − y2 = −1

x2 − y2 = −2

x2 − y2 = −3

xy = cx2 − y2 = k .. x.

xy=−1

xy=−2

xy=−3

x2 − y2 = −1

x2 − y2 = −2

x2 − y2 = −3

xy = cx2 − y2 = k .. x.

xy=−1

xy=−2

xy=−3

x2 − y2 = −1

x2 − y2 = −2

x2 − y2 = −3

xy = cx2 − y2 = k .. x.

xy=−1

xy=−2

xy=−3

x2 − y2 = −1

x2 − y2 = −2

x2 − y2 = −3

xy = cx2 − y2 = k .. x.

xy=−1

xy=−2

xy=−3

x2 − y2 = −1

x2 − y2 = −2

x2 − y2 = −3

xy = cx2 − y2 = k .. x.

xy=−1

xy=−2

xy=−3

x2 − y2 = −1

x2 − y2 = −2

x2 − y2 = −3

xy = cx2 − y2 = k .. x.

xy=−1

xy=−2

xy=−3

x2 − y2 = −1

x2 − y2 = −2

x2 − y2 = −3

ExamplesExample

Solu on

ExamplesExample

Solu on

ExamplesExample

Solu on

Ideal gases

The ideal gas law relatestemperature, pressure, andvolume of a gas:

PV = nRT

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

..Image credit: Sco Beale / Laughing Squid

CompressibilityDefini onThe isothermic compressibility of a fluid is defined by

β = −dVdP

Approximately we have

∆V∆P

≈ dVdP

= −βV =⇒ ∆VV

≈ −β∆P

The smaller the β, the “harder” the fluid.

CompressibilityDefini onThe isothermic compressibility of a fluid is defined by

β = −dVdP

Approximately we have

∆V∆P

≈ dVdP

= −βV =⇒ ∆VV

≈ −β∆P

The smaller the β, the “harder” the fluid.

Compressibility of an ideal gasExample

Find the isothermic compressibility of an ideal gas.

Solu onIf PV = k (n is constant for our purposes, T is constant because of theword isothermic, and R really is constant), then

· V+ PdVdP

= 0 =⇒ dVdP

= −VP

So β = −1V· dVdP

=1P. Compressibility and pressure are inversely

related.

Compressibility of an ideal gasExample

Find the isothermic compressibility of an ideal gas.

Solu onIf PV = k (n is constant for our purposes, T is constant because of theword isothermic, and R really is constant), then

· V+ PdVdP

= 0 =⇒ dVdP

= −VP

So β = −1V· dVdP

=1P. Compressibility and pressure are inversely

related.

Nonideal gassesNot that there’s anything wrong with thatExample

The van der Waals equa onmakesfewer simplifica ons:(

P+ an2

)(V− nb) = nRT,

where a is a measure of a rac onbetween par cles of the gas, and b ameasure of par cle size.

...Oxygen.

Oxygen

Oxygen .

... Hydrogen bonds

Nonideal gassesNot that there’s anything wrong with that

Example

The van der Waals equa onmakesfewer simplifica ons:(

P+ an2

)(V− nb) = nRT,

where a is a measure of a rac onbetween par cles of the gas, and b ameasure of par cle size.

Wikimedia Commons

Compressibility of a van der Waals gas

Differen a ng the van der Waals equa on by trea ng V as afunc on of P gives(

+ (V− bn)(1− 2an2

V3dVdP

soβ = −1

=V2(V− nb)

2abn3 − an2V+ PV3

Compressibility of a van der Waals gas

Differen a ng the van der Waals equa on by trea ng V as afunc on of P gives(

+ (V− bn)(1− 2an2

V3dVdP

soβ = −1

=V2(V− nb)

2abn3 − an2V+ PV3

Nonideal compressibility,continued

β = −1VdVdP

=V2(V− nb)

2abn3 − an2V+ PV3

Ques on

I What if a = b = 0?I Without taking the deriva ve, what is the sign of

I Without taking the deriva ve, what is the sign ofdβda

β = −1VdVdP

=V2(V− nb)

2abn3 − an2V+ PV3

Ques on

I What if a = b = 0?

I Without taking the deriva ve, what is the sign ofdβdb

β = −1VdVdP

=V2(V− nb)

2abn3 − an2V+ PV3

Ques on

β = −1VdVdP

=V2(V− nb)

2abn3 − an2V+ PV3

Ques on

Nasty derivativesAnswer

I We get the old (ideal) compressibilityI We have

= −nV3

(an2 + PV2

)(PV3 + an2(2bn− V)

)2 < 0

I We havedβda

=n2(bn− V)(2bn− V)V2(PV3 + an2(2bn− V)

)2 > 0 (as long as

V > 2nb, and it’s probably true that V ≫ 2nb).

Outline

Using implicit differentiation tofind derivatives

Example

Finddydx

if y =√x.

Solu onIf y =

√x, then

y2 = x,

= 1 =⇒ dydx

2√x.

Using implicit differentiation tofind derivatives

Example

Finddydx

if y =√x.

Solu onIf y =

√x, then

y2 = x,

= 1 =⇒ dydx

2√x.

The power rule for rational powersTheoremIf y = xp/q, where p and q are integers, then y′ =

pqxp/q−1.

Proof.

pqxp/q−1.

Proof.First, raise both sides to the qth power:

y = xp/q =⇒ yq = xp

pqxp/q−1.

Proof.First, raise both sides to the qth power:

y = xp/q =⇒ yq = xp

Now, differen ate implicitly:

qyq−1dydx

= pxp−1 =⇒ dydx

=pq· x

yq−1

pqxp/q−1.

Proof.Now, differen ate implicitly:

qyq−1dydx

= pxp−1 =⇒ dydx

=pq· x

yq−1

pqxp/q−1.

Proof.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

Summary

I Using implicit differen a on we can treat rela ons which arenot quite func ons like they were func ons.

I In par cular, we can find the slopes of lines tangent to curveswhich are not graphs of func ons.

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