Lesson 12: Linear Approximation

. . . . . .

Section2.8LinearApproximationand

Differentials

V63.0121.006/016, CalculusI

February26, 2010

Announcements

I Quiz2isFebruary26, covering§§1.5–2.3I MidtermisMarch4, covering§§1.1–2.5

. . . . . .

Announcements

I Quiz2isFebruary26, covering§§1.5–2.3I MidtermisMarch4, covering§§1.1–2.5

. . . . . .

Outline

ThelinearapproximationofafunctionnearapointExamplesQuestions

MidtermReview

DifferentialsUsingdifferentialstoestimateerror

AdvancedExamples

. . . . . .

TheBigIdea

QuestionLet f bedifferentiableat a. Whatlinearfunctionbestapproximates f near a?

AnswerThetangentline, ofcourse!

QuestionWhatistheequationforthelinetangentto y = f(x) at (a, f(a))?

Answer

L(x) = f(a) + f′(a)(x− a)

. . . . . .

TheBigIdea




Answer

L(x) = f(a) + f′(a)(x− a)

. . . . . .

TheBigIdea




Answer

L(x) = f(a) + f′(a)(x− a)

. . . . . .

TheBigIdea




Answer

L(x) = f(a) + f′(a)(x− a)

. . . . . .

Thetangentlineisalinearapproximation

L(x) = f(a) + f′(a)(x− a)

isadecentapproximationtof near a.

Howdecent? Thecloser x isto a, thebettertheapproxmation L(x) isto f(x)

. .x

.y

.

.

.

.f(a)

.f(x).L(x)

.a .x

.x − a

. . . . . .

Thetangentlineisalinearapproximation

L(x) = f(a) + f′(a)(x− a)

isadecentapproximationtof near a.Howdecent? Thecloser x isto a, thebettertheapproxmation L(x) isto f(x)

. .x

.y

.

.

.

.f(a)

.f(x).L(x)

.a .x

.x − a

. . . . . .

Example

ExampleEstimate sin(61◦) = sin(61π/180) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.

Solution(i)

I If f(x) = sin x, then f(0) = 0and f′(0) = 1.

I Sothelinearapproximationnear 0 isL(x) = 0+ 1 · x = x.

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution(ii)

I Wehave f(π3

)=

√32

andf′(π3

)=

12

.

I So L(x) =

√32

+12

(x− π

3

)

I Thus

sin(61π180

)≈

0.87475

Calculatorcheck: sin(61◦) ≈

0.87462.

. . . . . .

Example


Solution(i)



I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution(ii)

I Wehave f(π3

)=

√32

andf′(π3

)=

12

.

I So L(x) =

√32

+12

(x− π

3

)

I Thus

sin(61π180

)≈

0.87475


0.87462.

. . . . . .

Example


Solution(i)



I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution(ii)

I Wehave f(π3

)=

√32

andf′(π3

)=

12

.

I So L(x) =

√32

+12

(x− π

3

)

I Thus

sin(61π180

)≈

0.87475


0.87462.

. . . . . .

Example


Solution(i)



I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution(ii)

I Wehave f(π3

)=

√32 and

f′(π3

)=

12

.

I So L(x) =

√32

+12

(x− π

3

)

I Thus

sin(61π180

)≈

0.87475


0.87462.

. . . . . .

Example


Solution(i)



I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution(ii)

I Wehave f(π3

)=

√32 and

f′(π3

)= 1

2 .

I So L(x) =

√32

+12

(x− π

3

)

I Thus

sin(61π180

)≈

0.87475


0.87462.

. . . . . .

Example


Solution(i)



I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution(ii)

I Wehave f(π3

)=

√32 and

f′(π3

)= 1

2 .

I So L(x) =

√32

+12

(x− π

3

)I Thus

sin(61π180

)≈

0.87475


0.87462.

. . . . . .

Example


Solution(i)



I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution(ii)

I Wehave f(π3

)=

√32 and

f′(π3

)= 1

2 .

I So L(x) =

√32

+12

(x− π

3

)

I Thus

sin(61π180

)≈

0.87475


0.87462.

. . . . . .

Example


Solution(i)



I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution(ii)

I Wehave f(π3

)=

√32 and

f′(π3

)= 1

2 .

I So L(x) =

√32

+12

(x− π

3

)I Thus

sin(61π180

)≈

0.87475


0.87462.

. . . . . .

Example


Solution(i)



I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution(ii)

I Wehave f(π3

)=

√32 and

f′(π3

)= 1

2 .

I So L(x) =

√32

+12

(x− π

3

)I Thus

sin(61π180

)≈ 0.87475


0.87462.

. . . . . .

Example


Solution(i)



I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution(ii)

I Wehave f(π3

)=

√32 and

f′(π3

)= 1

2 .

I So L(x) =

√32

+12

(x− π

3

)I Thus

sin(61π180

)≈ 0.87475


0.87462.

. . . . . .

Example


Solution(i)



I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution(ii)

I Wehave f(π3

)=

√32 and

f′(π3

)= 1

2 .

I So L(x) =

√32

+12

(x− π

3

)I Thus

sin(61π180

)≈ 0.87475

Calculatorcheck: sin(61◦) ≈ 0.87462.

. . . . . .

Illustration

. .x

.y

.y = sin x

.61◦

.y = L1(x) = x

..0

.

.bigdifference!

.y = L2(x) =√32 + 1

2

(x− π

3

)

..π/3

.

.verylittledifference!

. . . . . .

Illustration

. .x

.y

.y = sin x

.61◦

.y = L1(x) = x

..0

.

.bigdifference!.y = L2(x) =

√32 + 1

2

(x− π

3

)

..π/3

.


. . . . . .

Illustration

. .x

.y

.y = sin x

.61◦

.y = L1(x) = x

..0

.

.bigdifference!

.y = L2(x) =√32 + 1

2

(x− π

3

)

..π/3

.


. . . . . .

Illustration

. .x

.y

.y = sin x

.61◦

.y = L1(x) = x

..0

.

.bigdifference!

.y = L2(x) =√32 + 1

2

(x− π

3

)

..π/3

.


. . . . . .

Illustration

. .x

.y

.y = sin x

.61◦

.y = L1(x) = x

..0

.

.bigdifference!

.y = L2(x) =√32 + 1

2

(x− π

3

)

..π/3

. .verylittledifference!

. . . . . .

AnotherExample

ExampleEstimate

√10 usingthefactthat 10 = 9+ 1.

SolutionThekeystepistousealinearapproximationto f(x) =

√x near

a = 9 toestimate f(10) =√10.

√10 ≈

√9+

ddx

√x∣∣∣∣x=9

(1)

= 3+1

2 · 3(1) =

196

≈ 3.167

Check:(196

)2

=36136

.

. . . . . .

AnotherExample

ExampleEstimate



√x near


√10 ≈

√9+

ddx

√x∣∣∣∣x=9

(1)

= 3+1

2 · 3(1) =

196

≈ 3.167

Check:(196

)2

=36136

.

. . . . . .

AnotherExample

ExampleEstimate



√x near


√10 ≈

√9+

ddx

√x∣∣∣∣x=9

(1)

= 3+1

2 · 3(1) =

196

≈ 3.167

Check:(196

)2

=36136

.

. . . . . .

AnotherExample

ExampleEstimate



√x near


√10 ≈

√9+

ddx

√x∣∣∣∣x=9

(1)

= 3+1

2 · 3(1) =

196

≈ 3.167

Check:(196

)2

=

36136

.

. . . . . .

AnotherExample

ExampleEstimate



√x near


√10 ≈

√9+

ddx

√x∣∣∣∣x=9

(1)

= 3+1

2 · 3(1) =

196

≈ 3.167

Check:(196

)2

=36136

.

. . . . . .

Dividingwithoutdividing?ExampleSupposeI haveanirrationalfearofdivisionandneedtoestimate577÷ 408. I write

577408

= 1+ 1691

408= 1+ 169× 1

4× 1

102.

ButstillI havetofind1

102.

SolutionLet f(x) =

1x. Weknow f(100) andwewanttoestimate f(102).

f(102) ≈ f(100) + f′(100)(2) =1

100− 1

1002(2) = 0.0098

=⇒ 577408

≈ 1.41405

Calculatorcheck:577408

≈ 1.41422.

. . . . . .

Dividingwithoutdividing?ExampleSupposeI haveanirrationalfearofdivisionandneedtoestimate577÷ 408. I write

577408

= 1+ 1691

408= 1+ 169× 1

4× 1

102.

ButstillI havetofind1

102.

SolutionLet f(x) =

1x. Weknow f(100) andwewanttoestimate f(102).

f(102) ≈ f(100) + f′(100)(2) =1

100− 1

1002(2) = 0.0098

=⇒ 577408

≈ 1.41405

Calculatorcheck:577408

≈ 1.41422.

. . . . . .

Questions

ExampleSupposewearetravelinginacarandatnoonourspeedis50mi/hr. Howfarwillwehavetraveledby2:00pm? by3:00pm?Bymidnight?

ExampleSupposeourfactorymakesMP3playersandthemarginalcostiscurrently$50/lot. Howmuchwillitcosttomake2morelots? 3morelots? 12morelots?

ExampleSupposealinegoesthroughthepoint (x0, y0) andhasslope m. Ifthepointismovedhorizontallyby dx, whilestayingontheline,whatisthecorrespondingverticalmovement?

. . . . . .

Answers


Answer

I 100miI 150miI 600mi(?) (Isitreasonabletoassume12hoursatthesamespeed?)

. . . . . .

Answers


Answer

I 100miI 150miI 600mi(?) (Isitreasonabletoassume12hoursatthesamespeed?)

. . . . . .

Questions




. . . . . .

Answers


Answer

I $100I $150I $600(?)

. . . . . .

Questions




. . . . . .

Answers


AnswerTheslopeofthelineis

m =riserun

Wearegivena“run”of dx, sothecorresponding“rise”is mdx.

. . . . . .

Answers


AnswerTheslopeofthelineis

m =riserun

Wearegivena“run”of dx, sothecorresponding“rise”is mdx.

. . . . . .

Outline


MidtermReview


AdvancedExamples

. . . . . .

MidtermFacts

I Coverssections1.1–2.5(Limits, Derivatives,DifferentiationuptoChainRule)

I CalculatorfreeI 20multiple-choicequestionsand4free-responsequestions

I Tostudy:I outlineI doproblemsI metacognitionI askquestions!(maybeinrecitation?)

. . . . . .

Outline


MidtermReview


AdvancedExamples

. . . . . .

Differentialsareanotherwaytoexpressderivatives

f(x+∆x)− f(x)︸︷︷︸∆y

≈ f′(x)∆x︸︷︷︸dy

Rename ∆x = dx, sowecanwritethisas

∆y ≈ dy = f′(x)dx.

AndthislooksalotliketheLeibniz-Newtonidentity

dydx

= f′(x) . .x

.y

.

.

.x .x+∆x

.dx = ∆x

.∆y.dy

Linearapproximationmeans ∆y ≈ dy = f′(x0)dx near x0.

. . . . . .

Differentialsareanotherwaytoexpressderivatives

f(x+∆x)− f(x)︸︷︷︸∆y

≈ f′(x)∆x︸︷︷︸dy

Rename ∆x = dx, sowecanwritethisas

∆y ≈ dy = f′(x)dx.

AndthislooksalotliketheLeibniz-Newtonidentity

dydx

= f′(x) . .x

.y

.

.

.x .x+∆x

.dx = ∆x

.∆y.dy

Linearapproximationmeans ∆y ≈ dy = f′(x0)dx near x0.

. . . . . .

Usingdifferentialstoestimateerror

If y = f(x), x0 and ∆x isknown, andanestimateof∆y isdesired:

I Approximate: ∆y ≈ dyI Differentiate:

dy = f′(x)dxI Evaluateat x = x0 and

dx = ∆x.

. .x

.y

.

.

.x .x+∆x

.dx = ∆x

.∆y.dy

. . . . . .

ExampleA sheetofplywoodmeasures 8 ft× 4 ft. Supposeourplywood-cuttingmachinewillcutarectanglewhosewidthisexactlyhalfitslength, butthelengthispronetoerrors. Ifthelengthisoffby 1 in, howbadcantheareaofthesheetbeoffby?

SolutionWrite A(ℓ) =

12ℓ2. Wewanttoknow ∆A when ℓ = 8 ft and

∆ℓ = 1 in.

(I) A(ℓ+∆ℓ) = A(9712

)=

9409288

So

∆A =9409288

− 32 ≈ 0.6701.

(II)dAdℓ

= ℓ, so dA = ℓdℓ, whichshouldbeagoodestimatefor

∆ℓ. When ℓ = 8 and dℓ = 112 , wehave

dA = 812 = 2

3 ≈ 0.667. Sowegetestimatesclosetothehundredthofasquarefoot.

. . . . . .




∆ℓ = 1 in.

(I) A(ℓ+∆ℓ) = A(9712

)=

9409288

So

∆A =9409288

− 32 ≈ 0.6701.

(II)dAdℓ



dA = 812 = 2


. . . . . .




∆ℓ = 1 in.

(I) A(ℓ+∆ℓ) = A(9712

)=

9409288

So

∆A =9409288

− 32 ≈ 0.6701.

(II)dAdℓ



dA = 812 = 2


. . . . . .




∆ℓ = 1 in.

(I) A(ℓ+∆ℓ) = A(9712

)=

9409288

So

∆A =9409288

− 32 ≈ 0.6701.

(II)dAdℓ



dA = 812 = 2


. . . . . .

Why?

Whyuselinearapproximations dy whentheactualdifference ∆yisknown?

I Linearapproximationisquickandreliable. Finding ∆yexactlydependsonthefunction.

I Theseexamplesareoverlysimple. Seethe“AdvancedExamples”later.

I Inreallife, sometimesonly f(a) and f′(a) areknown, andnotthegeneral f(x).

. . . . . .

Outline


MidtermReview


AdvancedExamples

. . . . . .

GravitationPencilsdown!

Example

I Dropa1 kgballofftheroofoftheSilverCenter(50mhigh).Weusuallysaythatafallingobjectfeelsaforce F = −mgfromgravity.

I Infact, theforcefeltis

F(r) = −GMmr2

,

where M isthemassoftheearthand r isthedistancefromthecenteroftheearthtotheobject. G isaconstant.

I At r = re theforcereallyis F(re) =GMmr2e

= −mg.

I Whatisthemaximumerrorinreplacingtheactualforcefeltatthetopofthebuilding F(re +∆r) bytheforcefeltatgroundlevel F(re)? Therelativeerror? Thepercentageerror?

. . . . . .

GravitationPencilsdown!

Example

I Dropa1 kgballofftheroofoftheSilverCenter(50mhigh).Weusuallysaythatafallingobjectfeelsaforce F = −mgfromgravity.

I Infact, theforcefeltis

F(r) = −GMmr2

,

where M isthemassoftheearthand r isthedistancefromthecenteroftheearthtotheobject. G isaconstant.

I At r = re theforcereallyis F(re) =GMmr2e

= −mg.

I Whatisthemaximumerrorinreplacingtheactualforcefeltatthetopofthebuilding F(re +∆r) bytheforcefeltatgroundlevel F(re)? Therelativeerror? Thepercentageerror?

. . . . . .

SolutionWewonderif ∆F = F(re +∆r)− F(re) issmall.

I Usingalinearapproximation,

∆F ≈ dF =dFdr

∣∣∣∣re

dr = 2GMmr3e

dr

=

(GMmr2e

)drre

= 2mg∆rre

I Therelativeerroris∆FF

≈ −2∆rre

I re = 6378.1 km. If ∆r = 50m,

∆FF

≈ −2∆rre

= −2 506378100

= −1.56×10−5 = −0.00156%

. . . . . .

Systematiclinearapproximation

I√2 isirrational, but

√9/4 isrationaland 9/4 iscloseto 2.

So

√2 =

√9/4− 1/4 ≈

√9/4+

12(3/2)

(−1/4) =1712

I Thisisabetterapproximationsince (17/12)2 = 289/144

I Doitagain!

√2 =

√289/144− 1/144 ≈

√289/144+

12(17/12)

(−1/144) = 577/408

Now(577408

)2

=332, 929166, 464

whichis1

166, 464awayfrom 2.

. . . . . .



√9/4 isrationaland 9/4 iscloseto 2. So

√2 =

√9/4− 1/4 ≈

√9/4+

12(3/2)

(−1/4) =1712


I Doitagain!

√2 =

√289/144− 1/144 ≈

√289/144+

12(17/12)

(−1/144) = 577/408

Now(577408

)2

=332, 929166, 464

whichis1

166, 464awayfrom 2.

. . . . . .




√2 =

√9/4− 1/4 ≈

√9/4+

12(3/2)

(−1/4) =1712


I Doitagain!

√2 =

√289/144− 1/144 ≈

√289/144+

12(17/12)

(−1/144) = 577/408

Now(577408

)2

=332, 929166, 464

whichis1

166, 464awayfrom 2.

. . . . . .




√2 =

√9/4− 1/4 ≈

√9/4+

12(3/2)

(−1/4) =1712


I Doitagain!

√2 =

√289/144− 1/144 ≈

√289/144+

12(17/12)

(−1/144) = 577/408

Now(577408

)2

=332, 929166, 464

whichis1

166, 464awayfrom 2.

. . . . . .

Illustrationofthepreviousexample

.

.2

.

(94 ,32)

..(2, 1712)

. . . . . .


.

.2

.

(94 ,32)

..(2, 1712)

. . . . . .


..2

.

(94 ,32)

..(2, 1712)

. . . . . .


..2

.

(94 ,32)

..(2, 1712)

. . . . . .


..2

.

(94 ,32)

..(2, 1712)

. . . . . .


..2

.

(94 ,32)

..(2, 1712)

. . . . . .


..2

.

(94 ,32)

..(2, 1712)

. . . . . .


..2

..(94 ,

32).

.(2, 17/12)

..(289144 ,

1712

)..(2, 577408

)

. . . . . .


..2

..(94 ,

32).

.(2, 17/12)..(289144 ,

1712

)

..(2, 577408

)

. . . . . .


..2

..(94 ,

32).

.(2, 17/12)..(289144 ,

1712

)..(2, 577408

)

Education

Lesson 12: Linear Approximation