Download pdf - Lesson 11: Implicit Differentiation (handout)

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Sec on 2.6Implicit Differen a on

V63.0121.001: Calculus IProfessor Ma hew Leingang

New York University

February 28, 2011

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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

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Objectives

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

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. Sec on 2.6: Implicit Differen a on. V63.0121.001: Calculus I . February 28, 2011

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Outline

The big idea, by example

ExamplesBasic ExamplesVer cal and Horizontal TangentsOrthogonal TrajectoriesChemistry

The power rule for ra onal powers

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Motivating Example

ProblemFind the slope of the linewhich is tangent to thecurve

x2 + y2 = 1

at the point (3/5,−4/5).

.. x.

y

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Motivating Example, SolutionSolu on (Explicit)

I Isolate:y2 = 1− x2 =⇒ y = −

√1− x2.

(Why the−?)I Differen ate:

dydx

= − −2x2√1− x2

=x√

1− x2I Evaluate:

dydx

∣∣∣∣x=3/5

=3/5√

1− (3/5)2=

3/54/5

=34.

.. x.

y

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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)

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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.

.. x.

y

..

looks like a func on

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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:

dydx

∣∣∣∣( 35 ,−

45)

=3/54/5

=34.

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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.

.. x.

y

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Outline




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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

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Yet Another ExampleExample

Find y′ if y5 + x2y3 = 1+ y sin(x2).

Solu on

Differen 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)

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

5y4 + 3x2y2 − sin x2

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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).

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SolutionSolu on

I Differen ate: 2ydydx

= 3x2 + 2x, sodydx

=3x2 + 2x

2y, and

dydx

∣∣∣∣(3,−6)

=3 · 32 + 2 · 3

2(−6)= −33

12= −11

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

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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).

..

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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.

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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]

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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.

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Solution, continued

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

y2 =(−23

)3

+

(−23

)2

=427

=⇒ y = ±√

427

= ± 23√3,

so there are two horizontal tangents.

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Tangents

...

(−2

3 ,2

3√3

)..(

−23 ,−

23√3

)..node

..(−1, 0)

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Example

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

Solu on

I Tangent lines are ver cal whendxdy

= 0.

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

dxdy

+ 2xdxdy

, sodxdy

=2y

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]

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Solution, continued

I Solving the second equa on gives

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.

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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.

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Orthogonal Families of Curves

xy = cx2 − y2 = k .. x.

y

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xy=1

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xy=2

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xy=3

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xy=−1

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xy=−2

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xy=−3

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x2−

y2=

1

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x2−

y2=

2

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x2−

y2=

3

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x2 − y2 = −1

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x2 − y2 = −2

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x2 − y2 = −3

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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.

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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)

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Image credit: Sco Beale / Laughing Squid

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http://www.laughingsquid.com/

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CompressibilityDefini onThe isothermic compressibility of a fluid is defined by

β = −dVdP

1V

Approximately we have

∆V∆P

≈ dVdP

= −βV =⇒ ∆VV

≈ −β∆P

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

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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

dPdP

· V+ PdVdP

= 0 =⇒ dVdP

= −VP

So β = −1V· dVdP

=1P. Compressibility and pressure are inversely

related.

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Nonideal gassesNot that there’s anything wrong with thatExample

The van der Waals equa onmakesfewer simplifica ons:(

P+ an2

V2

)(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.

.H

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.H

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Oxygen

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H.

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H

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Oxygen .

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H.

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H

... Hydrogen bonds

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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(

P+an2

V2

)dVdP

+ (V− bn)(1− 2an2

V3dVdP

)= 0,

soβ = −1

VdVdP

=V2(V− nb)

2abn3 − an2V+ PV3

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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

dβdb

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

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Nasty derivativesAnswer

I We get the old (ideal) compressibilityI We have

dβdb

= −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).

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Outline




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Using implicit differentiation tofind derivatives

Example

Finddydx

if y =√x.

Solu onIf y =

√x, then

y2 = x,

so2y

dydx

= 1 =⇒ dydx

=12y

=1

2√x.

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

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

p−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

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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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