lesson 11: implicit differentiation
DESCRIPTION
With implicit differentiation we can find the rate of change of a relation which isn't necessarily a functionTRANSCRIPT
. . . . . .
Section2.6ImplicitDifferentiation
V63.0121.006/016, CalculusI
February23, 2010
Announcements
I Quiz2isFebruary26, covering§§1.5–2.3I MidtermisMarch4, covering§§1.1–2.5I OnHW 5, Problem2.3.46shouldbe2.4.46
..Imagecredit: TelstarLogistics
. . . . . .
Announcementsonwhitebackground
Announcements
I Quiz2isFebruary26, covering§§1.5–2.3I MidtermisMarch4, covering§§1.1–2.5I OnHW 5, Problem2.3.46shouldbe2.4.46
. . . . . .
Outline
Thebigidea, byexample
ExamplesBasicExamplesVerticalandHorizontalTangentsOrthogonalTrajectoriesChemistry
Thepowerruleforrationalpowers
. . . . . .
MotivatingExample
ProblemFindtheslopeofthelinewhichistangenttothecurve
x2 + y2 = 1
atthepoint (3/5,−4/5).
. .x
.y
.Solution(Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Whythe −?)
I Differentiate:dydx
= − −2x
2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
. . . . . .
MotivatingExample
ProblemFindtheslopeofthelinewhichistangenttothecurve
x2 + y2 = 1
atthepoint (3/5,−4/5).
. .x
.y
.Solution(Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Whythe −?)
I Differentiate:dydx
= − −2x
2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
. . . . . .
MotivatingExample
ProblemFindtheslopeofthelinewhichistangenttothecurve
x2 + y2 = 1
atthepoint (3/5,−4/5).
. .x
.y
.
Solution(Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Whythe −?)
I Differentiate:dydx
= − −2x
2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
. . . . . .
MotivatingExample
ProblemFindtheslopeofthelinewhichistangenttothecurve
x2 + y2 = 1
atthepoint (3/5,−4/5).
. .x
.y
.Solution(Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Whythe −?)
I Differentiate:dydx
= − −2x
2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
. . . . . .
MotivatingExample
ProblemFindtheslopeofthelinewhichistangenttothecurve
x2 + y2 = 1
atthepoint (3/5,−4/5).
. .x
.y
.Solution(Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Whythe −?)
I Differentiate:dydx
= − −2x
2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
. . . . . .
MotivatingExample
ProblemFindtheslopeofthelinewhichistangenttothecurve
x2 + y2 = 1
atthepoint (3/5,−4/5).
. .x
.y
.Solution(Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Whythe −?)
I Differentiate:dydx
= − −2x
2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
. . . . . .
MotivatingExample
ProblemFindtheslopeofthelinewhichistangenttothecurve
x2 + y2 = 1
atthepoint (3/5,−4/5).
. .x
.y
.Solution(Explicit)
I Isolate: y2 = 1− x2 =⇒ y = −√1− x2. (Whythe −?)
I Differentiate:dydx
= − −2x
2√1− x2
=x√
1− x2
I Evaluate:dydx
∣∣∣∣x=3/5
=3/5√
1− (3/5)2=
3/54/5
=34.
. . . . . .
MotivatingExample, anotherway
Weknowthat x2 + y2 = 1 doesnotdefine y asafunctionof x,butsupposeitdid.
I Supposewehad y = f(x), sothat
x2 + (f(x))2 = 1
I Wecoulddifferentiatethisequationtoget
2x+ 2f(x) · f′(x) = 0
I Wecouldthensolvetoget
f′(x) = − xf(x)
. . . . . .
MotivatingExample, anotherway
Weknowthat x2 + y2 = 1 doesnotdefine y asafunctionof x,butsupposeitdid.
I Supposewehad y = f(x), sothat
x2 + (f(x))2 = 1
I Wecoulddifferentiatethisequationtoget
2x+ 2f(x) · f′(x) = 0
I Wecouldthensolvetoget
f′(x) = − xf(x)
. . . . . .
MotivatingExample, anotherway
Weknowthat x2 + y2 = 1 doesnotdefine y asafunctionof x,butsupposeitdid.
I Supposewehad y = f(x), sothat
x2 + (f(x))2 = 1
I Wecoulddifferentiatethisequationtoget
2x+ 2f(x) · f′(x) = 0
I Wecouldthensolvetoget
f′(x) = − xf(x)
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
.lookslikeafunction
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
.
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
.
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
.
.lookslikeafunction
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
.
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
.
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
.
.does not look like afunction, but that’sOK—there are onlytwo points like this
.
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
.lookslikeafunction
. . . . . .
Yes, wecan!
Thebeautifulfact(i.e., deeptheorem)isthatthisworks!
I “Near”mostpointsonthecurve x2 + y2 = 1,thecurveresemblesthegraphofafunction.
I So f(x) is defined“locally”, almosteverywhere andisdifferentiable
I Thechainrulethenappliesforthislocalchoice.
. .x
.y
.
.lookslikeafunction
. . . . . .
MotivatingExample, again, withLeibniznotation
ProblemFindtheslopeofthelinewhichistangenttothecurvex2 + y2 = 1 atthepoint (3/5,−4/5).
Solution
I Differentiate: 2x+ 2ydydx
= 0
Remember y isassumedtobeafunctionof x!
I Isolate:dydx
= − xy.
I Evaluate:dydx
∣∣∣∣( 35 ,−
45)
=3/54/5
=34.
. . . . . .
MotivatingExample, again, withLeibniznotation
ProblemFindtheslopeofthelinewhichistangenttothecurvex2 + y2 = 1 atthepoint (3/5,−4/5).
Solution
I Differentiate: 2x+ 2ydydx
= 0
Remember y isassumedtobeafunctionof x!
I Isolate:dydx
= − xy.
I Evaluate:dydx
∣∣∣∣( 35 ,−
45)
=3/54/5
=34.
. . . . . .
MotivatingExample, again, withLeibniznotation
ProblemFindtheslopeofthelinewhichistangenttothecurvex2 + y2 = 1 atthepoint (3/5,−4/5).
Solution
I Differentiate: 2x+ 2ydydx
= 0
Remember y isassumedtobeafunctionof x!
I Isolate:dydx
= − xy.
I Evaluate:dydx
∣∣∣∣( 35 ,−
45)
=3/54/5
=34.
. . . . . .
MotivatingExample, again, withLeibniznotation
ProblemFindtheslopeofthelinewhichistangenttothecurvex2 + y2 = 1 atthepoint (3/5,−4/5).
Solution
I Differentiate: 2x+ 2ydydx
= 0
Remember y isassumedtobeafunctionof x!
I Isolate:dydx
= − xy.
I Evaluate:dydx
∣∣∣∣( 35 ,−
45)
=3/54/5
=34.
. . . . . .
MotivatingExample, again, withLeibniznotation
ProblemFindtheslopeofthelinewhichistangenttothecurvex2 + y2 = 1 atthepoint (3/5,−4/5).
Solution
I Differentiate: 2x+ 2ydydx
= 0
Remember y isassumedtobeafunctionof x!
I Isolate:dydx
= − xy.
I Evaluate:dydx
∣∣∣∣( 35 ,−
45)
=3/54/5
=34.
. . . . . .
Summary
Ifarelationisgivenbetween x and y whichisn’tafunction:
I “Mostofthetime”, i.e., “atmostplaces” y canbeassumedtobeafunctionofx
I wemaydifferentiatetherelationasis
I Solvingfordydx
doesgivethe
slopeofthetangentlinetothecurveatapointonthecurve.
. .x
.y
.
. . . . . .
Outline
Thebigidea, byexample
ExamplesBasicExamplesVerticalandHorizontalTangentsOrthogonalTrajectoriesChemistry
Thepowerruleforrationalpowers
. . . . . .
ExampleFind y′ alongthecurve y3 + 4xy = x2 + 3.
SolutionImplicitlydifferentiating, wehave
3y2y′ + 4(1 · y+ x · y′) = 2x
Solvingfor y′ gives
3y2y′ + 4xy′ = 2x− 4y
(3y2 + 4x)y′ = 2x− 4y
=⇒ y′ =2x− 4y3y2 + 4x
. . . . . .
ExampleFind y′ alongthecurve y3 + 4xy = x2 + 3.
SolutionImplicitlydifferentiating, wehave
3y2y′ + 4(1 · y+ x · y′) = 2x
Solvingfor y′ gives
3y2y′ + 4xy′ = 2x− 4y
(3y2 + 4x)y′ = 2x− 4y
=⇒ y′ =2x− 4y3y2 + 4x
. . . . . .
ExampleFind y′ alongthecurve y3 + 4xy = x2 + 3.
SolutionImplicitlydifferentiating, wehave
3y2y′ + 4(1 · y+ x · y′) = 2x
Solvingfor y′ gives
3y2y′ + 4xy′ = 2x− 4y
(3y2 + 4x)y′ = 2x− 4y
=⇒ y′ =2x− 4y3y2 + 4x
. . . . . .
ExampleFind y′ if y5 + x2y3 = 1+ y sin(x2).
SolutionDifferentiatingimplicitly:
5y4y′ + (2x)y3 + x2(3y2y′) = y′ sin(x2) + y cos(x2)(2x)
Collectalltermswith y′ ononesideandalltermswithout y′ ontheother:
5y4y′ + 3x2y2y′ − sin(x2)y′ = −2xy3 + 2xy cos(x2)
Nowfactoranddivide:
y′ =2xy(cos x2 − y2)
5y4 + 3x2y2 − sin x2
. . . . . .
ExampleFind y′ if y5 + x2y3 = 1+ y sin(x2).
SolutionDifferentiatingimplicitly:
5y4y′ + (2x)y3 + x2(3y2y′) = y′ sin(x2) + y cos(x2)(2x)
Collectalltermswith y′ ononesideandalltermswithout y′ ontheother:
5y4y′ + 3x2y2y′ − sin(x2)y′ = −2xy3 + 2xy cos(x2)
Nowfactoranddivide:
y′ =2xy(cos x2 − y2)
5y4 + 3x2y2 − sin x2
. . . . . .
ExampleFindtheequationofthelinetangenttothecurve
y2 = x2(x+ 1) = x3 + x2
atthepoint (3,−6).
.
.
SolutionDifferentiatingtheexpressionimplicitlywithrespectto x gives
2ydydx
= 3x2 + 2x, sodydx
=3x2 + 2x
2y, and
dydx
∣∣∣∣(3,−6)
=3 · 32 + 2 · 3
2(−6)= −33
12= −11
4.
Thustheequationofthetangentlineis y+ 6 = −114(x− 3).
. . . . . .
ExampleFindtheequationofthelinetangenttothecurve
y2 = x2(x+ 1) = x3 + x2
atthepoint (3,−6).
.
.SolutionDifferentiatingtheexpressionimplicitlywithrespectto x gives
2ydydx
= 3x2 + 2x, sodydx
=3x2 + 2x
2y, and
dydx
∣∣∣∣(3,−6)
=3 · 32 + 2 · 3
2(−6)= −33
12= −11
4.
Thustheequationofthetangentlineis y+ 6 = −114(x− 3).
. . . . . .
ExampleFindtheequationofthelinetangenttothecurve
y2 = x2(x+ 1) = x3 + x2
atthepoint (3,−6).
.
.SolutionDifferentiatingtheexpressionimplicitlywithrespectto x gives
2ydydx
= 3x2 + 2x, sodydx
=3x2 + 2x
2y, and
dydx
∣∣∣∣(3,−6)
=3 · 32 + 2 · 3
2(−6)= −33
12= −11
4.
Thustheequationofthetangentlineis y+ 6 = −114(x− 3).
. . . . . .
Lineequationforms
I slope-interceptform
y = mx+ b
wheretheslopeis m and (0,b) isontheline.I point-slopeform
y− y0 = m(x− x0)
wheretheslopeis m and (x0, y0) isontheline.
. . . . . .
ExampleFindthehorizontaltangentlinestothesamecurve: y2 = x3 + x2
SolutionWehavetosolvethesetwoequations:
.
.y2 = x3 + x2
[(x, y) is onthe curve]
.1.3x2 + 2x
2y= 0
[tangent lineis horizontal]
.2
. . . . . .
ExampleFindthehorizontaltangentlinestothesamecurve: y2 = x3 + x2
SolutionWehavetosolvethesetwoequations:
.
.y2 = x3 + x2
[(x, y) is onthe curve]
.1.3x2 + 2x
2y= 0
[tangent lineis horizontal]
.2
. . . . . .
Solution, continuedI Solvingthesecondequationgives
3x2 + 2x2y
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x+ 2) = 0
(aslongas y ̸= 0). So x = 0 or 3x+ 2 = 0.
I Substituting x = 0 intothe first equationgives
y2 = 03 + 02 = 0 =⇒ y = 0
whichwe’vedisallowed. Sonohorizontaltangentsdownthatroad.
I Substituting x = −2/3 intothefirstequationgives
y2 =
(−23
)3
+
(−23
)2
=427
=⇒ y = ± 2
3√3,
sotherearetwohorizontaltangents.
. . . . . .
Solution, continuedI Solvingthesecondequationgives
3x2 + 2x2y
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x+ 2) = 0
(aslongas y ̸= 0). So x = 0 or 3x+ 2 = 0.I Substituting x = 0 intothe first equationgives
y2 = 03 + 02 = 0 =⇒ y = 0
whichwe’vedisallowed. Sonohorizontaltangentsdownthatroad.
I Substituting x = −2/3 intothefirstequationgives
y2 =
(−23
)3
+
(−23
)2
=427
=⇒ y = ± 2
3√3,
sotherearetwohorizontaltangents.
. . . . . .
Solution, continuedI Solvingthesecondequationgives
3x2 + 2x2y
= 0 =⇒ 3x2 + 2x = 0 =⇒ x(3x+ 2) = 0
(aslongas y ̸= 0). So x = 0 or 3x+ 2 = 0.I Substituting x = 0 intothe first equationgives
y2 = 03 + 02 = 0 =⇒ y = 0
whichwe’vedisallowed. Sonohorizontaltangentsdownthatroad.
I Substituting x = −2/3 intothefirstequationgives
y2 =
(−23
)3
+
(−23
)2
=427
=⇒ y = ± 2
3√3,
sotherearetwohorizontaltangents.
. . . . . .
HorizontalTangents
..
.(−2
3 ,2
3√3
).
.(−2
3 ,−2
3√3
)
.
.node
..(−1, 0)
. . . . . .
HorizontalTangents
..
.(−2
3 ,2
3√3
).
.(−2
3 ,−2
3√3
) .
.node
..(−1, 0)
. . . . . .
ExampleFindthe vertical tangentlinestothesamecurve: y2 = x3 + x2
Solution
I Tangentlinesareverticalwhendxdy
= 0.
I Differentiating x implicitlyasafunctionof y gives
2y = 3x2dxdy
+ 2xdxdy
, sodxdy
=2y
3x2 + 2x(noticethisisthe
reciprocalof dy/dx).I Wemustsolve
.
.y2 = x3 + x2
[(x, y) is onthe curve]
.1.
2y3x2 + 2x
= 0
[tangent lineis vertical]
.2
. . . . . .
ExampleFindthe vertical tangentlinestothesamecurve: y2 = x3 + x2
Solution
I Tangentlinesareverticalwhendxdy
= 0.
I Differentiating x implicitlyasafunctionof y gives
2y = 3x2dxdy
+ 2xdxdy
, sodxdy
=2y
3x2 + 2x(noticethisisthe
reciprocalof dy/dx).I Wemustsolve
.
.y2 = x3 + x2
[(x, y) is onthe curve]
.1.
2y3x2 + 2x
= 0
[tangent lineis vertical]
.2
. . . . . .
ExampleFindthe vertical tangentlinestothesamecurve: y2 = x3 + x2
Solution
I Tangentlinesareverticalwhendxdy
= 0.
I Differentiating x implicitlyasafunctionof y gives
2y = 3x2dxdy
+ 2xdxdy
, sodxdy
=2y
3x2 + 2x(noticethisisthe
reciprocalof dy/dx).
I Wemustsolve
.
.y2 = x3 + x2
[(x, y) is onthe curve]
.1.
2y3x2 + 2x
= 0
[tangent lineis vertical]
.2
. . . . . .
ExampleFindthe vertical tangentlinestothesamecurve: y2 = x3 + x2
Solution
I Tangentlinesareverticalwhendxdy
= 0.
I Differentiating x implicitlyasafunctionof y gives
2y = 3x2dxdy
+ 2xdxdy
, sodxdy
=2y
3x2 + 2x(noticethisisthe
reciprocalof dy/dx).I Wemustsolve
.
.y2 = x3 + x2
[(x, y) is onthe curve]
.1.
2y3x2 + 2x
= 0
[tangent lineis vertical]
.2
. . . . . .
Solution, continued
I Solvingthesecondequationgives
2y3x2 + 2x
= 0 =⇒ 2y = 0 =⇒ y = 0
(aslongas 3x2 + 2x ̸= 0).
I Substituting y = 0 intothe first equationgives
0 = x3 + x2 = x2(x+ 1)
So x = 0 or x = −1.I x = 0 isnotallowedbythefirstequation, but
dxdy
∣∣∣∣(−1,0)
= 0,
sohereisaverticaltangent.
. . . . . .
Solution, continued
I Solvingthesecondequationgives
2y3x2 + 2x
= 0 =⇒ 2y = 0 =⇒ y = 0
(aslongas 3x2 + 2x ̸= 0).I Substituting y = 0 intothe first equationgives
0 = x3 + x2 = x2(x+ 1)
So x = 0 or x = −1.
I x = 0 isnotallowedbythefirstequation, but
dxdy
∣∣∣∣(−1,0)
= 0,
sohereisaverticaltangent.
. . . . . .
Solution, continued
I Solvingthesecondequationgives
2y3x2 + 2x
= 0 =⇒ 2y = 0 =⇒ y = 0
(aslongas 3x2 + 2x ̸= 0).I Substituting y = 0 intothe first equationgives
0 = x3 + x2 = x2(x+ 1)
So x = 0 or x = −1.I x = 0 isnotallowedbythefirstequation, but
dxdy
∣∣∣∣(−1,0)
= 0,
sohereisaverticaltangent.
. . . . . .
Tangents
..
.(−2
3 ,2
3√3
).
.(−2
3 ,−2
3√3
) .
.node
..(−1, 0)
. . . . . .
ExamplesExampleShowthatthefamiliesofcurves
xy = c x2 − y2 = k
areorthogonal, thatis, theyintersectatrightangles.
SolutionInthefirstcurve,
y+ xy′ = 0 =⇒ y′ = −yx
Inthesecondcurve,
2x− 2yy′ = 0 = =⇒ y′ =xy
Theproductis −1, sothetangentlinesareperpendicularwherevertheyintersect.
. . . . . .
OrthogonalFamiliesofCurves
xy = cx2 − y2 = 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 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
. . . . . .
OrthogonalFamiliesofCurves
xy = cx2 − y2 = 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 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
. . . . . .
OrthogonalFamiliesofCurves
xy = cx2 − y2 = 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 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
. . . . . .
OrthogonalFamiliesofCurves
xy = cx2 − y2 = 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 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
. . . . . .
OrthogonalFamiliesofCurves
xy = cx2 − y2 = 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 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
. . . . . .
OrthogonalFamiliesofCurves
xy = cx2 − y2 = 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 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
. . . . . .
OrthogonalFamiliesofCurves
xy = cx2 − y2 = 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 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
. . . . . .
OrthogonalFamiliesofCurves
xy = cx2 − y2 = 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 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
. . . . . .
OrthogonalFamiliesofCurves
xy = cx2 − y2 = 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 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
. . . . . .
OrthogonalFamiliesofCurves
xy = cx2 − y2 = 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 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
. . . . . .
OrthogonalFamiliesofCurves
xy = cx2 − y2 = 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 − y2 = −1
.x2 − y2 = −2
.x2 − y2 = −3
. . . . . .
OrthogonalFamiliesofCurves
xy = cx2 − y2 = 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 − y2 = −1.x2 − y2 = −2
.x2 − y2 = −3
. . . . . .
OrthogonalFamiliesofCurves
xy = cx2 − y2 = 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 − y2 = −1.x2 − y2 = −2.x2 − y2 = −3
. . . . . .
ExamplesExampleShowthatthefamiliesofcurves
xy = c x2 − y2 = k
areorthogonal, thatis, theyintersectatrightangles.
SolutionInthefirstcurve,
y+ xy′ = 0 =⇒ y′ = −yx
Inthesecondcurve,
2x− 2yy′ = 0 = =⇒ y′ =xy
Theproductis −1, sothetangentlinesareperpendicularwherevertheyintersect.
. . . . . .
ExamplesExampleShowthatthefamiliesofcurves
xy = c x2 − y2 = k
areorthogonal, thatis, theyintersectatrightangles.
SolutionInthefirstcurve,
y+ xy′ = 0 =⇒ y′ = −yx
Inthesecondcurve,
2x− 2yy′ = 0 = =⇒ y′ =xy
Theproductis −1, sothetangentlinesareperpendicularwherevertheyintersect.
. . . . . .
MusicSelection
“TheCurseofCurves”byCuteisWhatWeAimFor
. . . . . .
Idealgases
The idealgaslaw relatestemperature, pressure, andvolumeofagas:
PV = nRT
(R isaconstant, n istheamountofgasinmoles)
.
.Imagecredit: ScottBeale/LaughingSquid
. . . . . .
Compressibility
DefinitionThe isothermiccompressibility ofafluidisdefinedby
β = −dVdP
1V
withtemperatureheldconstant.
Approximatelywehave
∆V∆P
≈ dVdP
= −βV =⇒ ∆VV
≈ −β∆P
Thesmallerthe β, the“harder”thefluid.
. . . . . .
Compressibility
DefinitionThe isothermiccompressibility ofafluidisdefinedby
β = −dVdP
1V
withtemperatureheldconstant.
Approximatelywehave
∆V∆P
≈ dVdP
= −βV =⇒ ∆VV
≈ −β∆P
Thesmallerthe β, the“harder”thefluid.
. . . . . .
ExampleFindtheisothermiccompressibilityofanidealgas.
SolutionIf PV = k (n isconstantforourpurposes, T isconstantbecauseoftheword isothermic, and R reallyisconstant), then
dPdP
· V+ PdVdP
= 0 =⇒ dVdP
= −VP
So
β = −1V· dVdP
=1P
Compressibilityandpressureareinverselyrelated.
. . . . . .
ExampleFindtheisothermiccompressibilityofanidealgas.
SolutionIf PV = k (n isconstantforourpurposes, T isconstantbecauseoftheword isothermic, and R reallyisconstant), then
dPdP
· V+ PdVdP
= 0 =⇒ dVdP
= −VP
So
β = −1V· dVdP
=1P
Compressibilityandpressureareinverselyrelated.
. . . . . .
NonidealgassesNotthatthere’sanythingwrongwiththat
ExampleThe vanderWaalsequationmakesfewersimplifications:(P+ a
n2
V2
)(V− nb) = nRT,
where P isthepressure, V thevolume, T thetemperature, nthenumberofmolesofthegas, R aconstant, a isameasureofattractionbetweenparticlesofthegas,and b ameasureofparticlesize.
...Oxygen
..H
..H
..Oxygen
..H
..H
..Oxygen ..H
..H
.
.
.Hydrogenbonds
.
.Imagecredit: WikimediaCommons
. . . . . .
NonidealgassesNotthatthere’sanythingwrongwiththat
ExampleThe vanderWaalsequationmakesfewersimplifications:(P+ a
n2
V2
)(V− nb) = nRT,
where P isthepressure, V thevolume, T thetemperature, nthenumberofmolesofthegas, R aconstant, a isameasureofattractionbetweenparticlesofthegas,and b ameasureofparticlesize. .
.Imagecredit: WikimediaCommons
. . . . . .
Let’sfindthecompressibilityofavanderWaalsgas.DifferentiatingthevanderWaalsequationbytreating V asafunctionof P gives(
P+an2
V2
)dVdP
+ (V− bn)(1− 2an2
V3dVdP
)= 0,
so
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V+ PV3
I Whatif a = b = 0?
I Withouttakingthederivative, whatisthesignofdβdb
?
I Withouttakingthederivative, whatisthesignofdβda
?
. . . . . .
Let’sfindthecompressibilityofavanderWaalsgas.DifferentiatingthevanderWaalsequationbytreating V asafunctionof P gives(
P+an2
V2
)dVdP
+ (V− bn)(1− 2an2
V3dVdP
)= 0,
so
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V+ PV3
I Whatif a = b = 0?
I Withouttakingthederivative, whatisthesignofdβdb
?
I Withouttakingthederivative, whatisthesignofdβda
?
. . . . . .
Let’sfindthecompressibilityofavanderWaalsgas.DifferentiatingthevanderWaalsequationbytreating V asafunctionof P gives(
P+an2
V2
)dVdP
+ (V− bn)(1− 2an2
V3dVdP
)= 0,
so
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V+ PV3
I Whatif a = b = 0?
I Withouttakingthederivative, whatisthesignofdβdb
?
I Withouttakingthederivative, whatisthesignofdβda
?
. . . . . .
Let’sfindthecompressibilityofavanderWaalsgas.DifferentiatingthevanderWaalsequationbytreating V asafunctionof P gives(
P+an2
V2
)dVdP
+ (V− bn)(1− 2an2
V3dVdP
)= 0,
so
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V+ PV3
I Whatif a = b = 0?
I Withouttakingthederivative, whatisthesignofdβdb
?
I Withouttakingthederivative, whatisthesignofdβda
?
. . . . . .
Let’sfindthecompressibilityofavanderWaalsgas.DifferentiatingthevanderWaalsequationbytreating V asafunctionof P gives(
P+an2
V2
)dVdP
+ (V− bn)(1− 2an2
V3dVdP
)= 0,
so
β = −1VdVdP
=V2(V− nb)
2abn3 − an2V+ PV3
I Whatif a = b = 0?
I Withouttakingthederivative, whatisthesignofdβdb
?
I Withouttakingthederivative, whatisthesignofdβda
?
. . . . . .
Nastyderivatives
I
dβdb
= −(2abn3 − an2V+ PV3)(nV2)− (nbV2 − V3)(2an3)(2abn3 − an2V+ PV3)2
= −nV3 (an2 + PV2)(
PV3 + an2(2bn− V))2 < 0
Idβda
=n2(bn− V)(2bn− V)V2(PV3 + an2(2bn− V)
)2 > 0
(aslongas V > 2nb, andit’sprobablytruethat V ≫ 2nb).
. . . . . .
Outline
Thebigidea, byexample
ExamplesBasicExamplesVerticalandHorizontalTangentsOrthogonalTrajectoriesChemistry
Thepowerruleforrationalpowers
. . . . . .
Usingimplicitdifferentiationtofindderivatives
Example
Finddydx
if y =√x.
SolutionIf y =
√x, then
y2 = x,
so
2ydydx
= 1 =⇒ dydx
=12y
=1
2√x.
. . . . . .
Usingimplicitdifferentiationtofindderivatives
Example
Finddydx
if y =√x.
SolutionIf y =
√x, then
y2 = x,
so
2ydydx
= 1 =⇒ dydx
=12y
=1
2√x.
. . . . . .
ThepowerruleforrationalpowersTheoremIf y = xp/q, where p and q areintegers, then y′ =
pqxp/q−1.
Proof.First, raisebothsidestothe qthpower:
y = xp/q =⇒ yq = xp
Now, differentiateimplicitly:
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
. . . . . .
ThepowerruleforrationalpowersTheoremIf y = xp/q, where p and q areintegers, then y′ =
pqxp/q−1.
Proof.First, raisebothsidestothe qthpower:
y = xp/q =⇒ yq = xp
Now, differentiateimplicitly:
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
. . . . . .
ThepowerruleforrationalpowersTheoremIf y = xp/q, where p and q areintegers, then y′ =
pqxp/q−1.
Proof.First, raisebothsidestothe qthpower:
y = xp/q =⇒ yq = xp
Now, differentiateimplicitly:
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
. . . . . .
ThepowerruleforrationalpowersTheoremIf y = xp/q, where p and q areintegers, then y′ =
pqxp/q−1.
Proof.First, raisebothsidestothe qthpower:
y = xp/q =⇒ yq = xp
Now, differentiateimplicitly:
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
. . . . . .
Whathavewelearnedtoday?
I ImplicitDifferentiationallowsustopretendthatarelationdescribesafunction, sinceitdoes, locally, “almosteverywhere.”
I ThePowerRulewasestablishedforpowerswhicharerationalnumbers.