lesson 12: linear approximation and differentials (section 21 slides)

Section 2.8Linear Approximation and Differentials

V63.0121.021, Calculus I

New York University

October 14, 2010

Announcements

I Quiz 2 in recitation this week on §§1.5, 1.6, 2.1, 2.2I Midterm on §§1.1–2.5

. . . . . .

Announcements

I Quiz 2 in recitation thisweek on §§1.5, 1.6, 2.1,2.2

I Midterm on §§1.1–2.5

V63.0121.021, Calculus I (NYU) Section 2.8 Linear Approximation October 14, 2010 2 / 36

. . . . . .

Midterm FAQ

QuestionWhat sections are covered on the midterm?

AnswerThe midterm will cover Sections 1.1–2.5 (The Chain Rule).

QuestionIs Section 2.6 going to be on the midterm?

QuestionIs Section 2.8 going to be on the midterm?...

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Midterm FAQ

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Midterm FAQ

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Midterm FAQ

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Midterm FAQ

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Midterm FAQ, continued

QuestionWhat format will the exam take?

AnswerThere will be both fixed-response (e.g., multiple choice) andfree-response questions.

QuestionWill explanations be necessary?

AnswerYes, on free-response problems we will expect you to explain yourself.This is why it was required on written homework.

. . . . . .

QuestionIs (topic X) going to be tested?

AnswerEverything covered in class or on homework is fair game for the exam.No topic that was not covered in class nor on homework will be on theexam. (This is not the same as saying all exam problems are similar toclass examples or homework problems.)

. . . . . .

QuestionIs (topic X) going to be tested?

AnswerEverything covered in class or on homework is fair game for the exam.No topic that was not covered in class nor on homework will be on theexam. (This is not the same as saying all exam problems are similar toclass examples or homework problems.)

. . . . . .

QuestionWill there be a review session?

AnswerWe’re working on it.

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QuestionWill there be a review session?

AnswerWe’re working on it.

. . . . . .

QuestionWill calculators be allowed?

AnswerNo. The exam is designed for pencil and brain.

. . . . . .

QuestionWill calculators be allowed?

AnswerNo. The exam is designed for pencil and brain.

. . . . . .

QuestionHow should I study?

Answer

I The exam has problems; study by doing problems. If you get oneright, think about how you got it right. If you got it wrong or didn’tget it at all, reread the textbook and do easier problems to build upyour understanding.

I Break up the material into chunks. (related) Don’t put it all off untilthe night before.

I Ask questions.

. . . . . .

QuestionHow should I study?

Answer

I The exam has problems; study by doing problems. If you get oneright, think about how you got it right. If you got it wrong or didn’tget it at all, reread the textbook and do easier problems to build upyour understanding.

I Break up the material into chunks. (related) Don’t put it all off untilthe night before.

I Ask questions.

. . . . . .

QuestionHow many questions are there?

AnswerDoes this question contribute to your understanding of the material?

. . . . . .

QuestionHow many questions are there?

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QuestionWill there be a curve on the exam?

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QuestionWill there be a curve on the exam?

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QuestionWhen will you grade my get-to-know-you and photo extra credit?

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QuestionWhen will you grade my get-to-know-you and photo extra credit?

. . . . . .

Objectives

I Use tangent lines to makelinear approximations to afunction.

I Given a function and apoint in the domain,compute thelinearization of thefunction at that point.

I Use linearization toapproximate values offunctions

I Given a function, computethe differential of thatfunction

I Use the differentialnotation to estimate errorin linear approximations.

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Outline

The linear approximation of a function near a pointExamplesQuestions

DifferentialsUsing differentials to estimate error

Advanced Examples

. . . . . .

The Big Idea

QuestionLet f be differentiable at a. What linear function best approximates fnear a?

AnswerThe tangent line, of course!

QuestionWhat is the equation for the line tangent to y = f(x) at (a, f(a))?

Answer

L(x) = f(a) + f′(a)(x− a)

. . . . . .

The Big Idea

Answer

L(x) = f(a) + f′(a)(x− a)

. . . . . .

The Big Idea

Answer

L(x) = f(a) + f′(a)(x− a)

. . . . . .

The Big Idea

Answer

L(x) = f(a) + f′(a)(x− a)

. . . . . .

The tangent line is a linear approximation

L(x) = f(a) + f′(a)(x− a)

is a decent approximation to fnear a.

How decent? The closer x is toa, the better the approxmationL(x) is to f(x)

.f(x).L(x)

.x − a

. . . . . .

The tangent line is a linear approximation

L(x) = f(a) + f′(a)(x− a)

is a decent approximation to fnear a.How decent? The closer x is toa, the better the approxmationL(x) is to f(x)

.f(x).L(x)

.x − a

. . . . . .

Example.

Example

Estimate sin(61◦) = sin(61π/180) by using a linear approximation(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 So the linear approximationnear 0 is L(x) = 0+ 1 · x = x.

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

andf′(π3)=

I So L(x) =

(x− π

I Thus

sin(61π180

0.87475

Calculator check: sin(61◦) ≈

0.87462.

. . . . . .

Example.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

andf′(π3)=

I So L(x) =

(x− π

I Thus

sin(61π180

0.87475

0.87462.

. . . . . .

Example.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

andf′(π3)=

I So L(x) =

(x− π

I Thus

sin(61π180

0.87475

0.87462.

. . . . . .

Example.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32 and

f′(π3)=

I So L(x) =

(x− π

I Thus

sin(61π180

0.87475

0.87462.

. . . . . .

Example.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32 and

f′(π3)= 1

I So L(x) =

(x− π

I Thus

sin(61π180

0.87475

0.87462.

. . . . . .

Example.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32 and

f′(π3)= 1

I So L(x) =

(x− π

)I Thus

sin(61π180

0.87475

0.87462.

. . . . . .

Example.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32 and

f′(π3)= 1

I So L(x) =√32

(x− π

I Thus

sin(61π180

0.87475

0.87462.

. . . . . .

Example.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32 and

f′(π3)= 1

I So L(x) =√32

(x− π

)I Thus

sin(61π180

0.87475

0.87462.

. . . . . .

Example.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32 and

f′(π3)= 1

I So L(x) =√32

(x− π

)I Thus

sin(61π180

)≈ 0.87475

0.87462.

. . . . . .

Example.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32 and

f′(π3)= 1

I So L(x) =√32

(x− π

)I Thus

sin(61π180

)≈ 0.87475

0.87462.

. . . . . .

Example.

Example

Solution (i)

I Thus

sin(61π180

)≈ 61π

180≈ 1.06465

Solution (ii)

I We have f(π3)=

√32 and

f′(π3)= 1

I So L(x) =√32

(x− π

)I Thus

sin(61π180

)≈ 0.87475

Calculator check: sin(61◦) ≈ 0.87462.

. . . . . .

Illustration

.y = sin x

.61◦

.y = L1(x) = x

.big difference!

.y = L2(x) =√32 + 1

2(x− π

..π/3

.very little difference!

. . . . . .

Illustration

.y = sin x

.61◦

.y = L1(x) = x

.big difference!.y = L2(x) =

√32 + 1

2(x− π

..π/3

. . . . . .

Illustration

.y = sin x

.61◦

.y = L1(x) = x

.big difference!

.y = L2(x) =√32 + 1

2(x− π

..π/3

. . . . . .

Illustration

.y = sin x

.61◦

.y = L1(x) = x

.big difference!

.y = L2(x) =√32 + 1

2(x− π

..π/3

. . . . . .

Illustration

.y = sin x

.61◦

.y = L1(x) = x

.big difference!

.y = L2(x) =√32 + 1

2(x− π

..π/3

. .very little difference!

. . . . . .

Another Example

Example

Estimate√10 using the fact that 10 = 9+ 1.

SolutionThe key step is to use a linear approximation to f(x) =

√x near a = 9

to estimate f(10) =√10.

f(10) ≈ L(10) = f(9) + f′(9)(10− 9)

2 · 3(1) =

≈ 3.167

Check:(196

. . . . . .

Another Example

Example

√x near a = 9

f(10) ≈ L(10) = f(9) + f′(9)(10− 9)

2 · 3(1) =

≈ 3.167

Check:(196

. . . . . .

Another Example

Example

√x near a = 9

f(10) ≈ L(10) = f(9) + f′(9)(10− 9)

2 · 3(1) =

≈ 3.167

Check:(196

. . . . . .

Another Example

Example

√x near a = 9

f(10) ≈ L(10) = f(9) + f′(9)(10− 9)

2 · 3(1) =

≈ 3.167

Check:(196

. . . . . .

Another Example

Example

√x near a = 9

f(10) ≈ L(10) = f(9) + f′(9)(10− 9)

2 · 3(1) =

≈ 3.167

Check:(196

. . . . . .

Dividing without dividing?

Example

A student has an irrational fear of long division and needs to estimate577÷ 408. He writes

577408

= 1+ 1691

408= 1+ 169× 1

Help the student estimate1

Solution

Let f(x) =1x. We know f(100) and we want to estimate f(102).

f(102) ≈ f(100) + f′(100)(2) =1

100− 1

1002(2) = 0.0098

=⇒ 577408

≈ 1.41405

Calculator check:577408

≈ 1.41422.

. . . . . .

Dividing without dividing?

Example

A student has an irrational fear of long division and needs to estimate577÷ 408. He writes

577408

= 1+ 1691

408= 1+ 169× 1

Help the student estimate1

Solution

Let f(x) =1x. We know f(100) and we want to estimate f(102).

f(102) ≈ f(100) + f′(100)(2) =1

100− 1

1002(2) = 0.0098

=⇒ 577408

≈ 1.41405

Calculator check:577408

≈ 1.41422.

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Questions

Example

Suppose we are traveling in a car and at noon our speed is 50mi/hr.How far will we have traveled by 2:00pm? by 3:00pm? By midnight?

Example

Suppose our factory makes MP3 players and the marginal cost iscurrently $50/lot. How much will it cost to make 2 more lots? 3 morelots? 12 more lots?

Example

Suppose a line goes through the point (x0, y0) and has slope m. If thepoint is moved horizontally by dx, while staying on the line, what is thecorresponding vertical movement?

. . . . . .

Answers

Example

Answer

I 100miI 150miI 600mi (?) (Is it reasonable to assume 12 hours at the same

speed?)

. . . . . .

Answers

Example

Answer

I 100miI 150miI 600mi (?) (Is it reasonable to assume 12 hours at the same

speed?)

. . . . . .

Questions

Example

. . . . . .

Answers

Example

Answer

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

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Questions

Example

. . . . . .

Answers

Example

AnswerThe slope of the line is

m =riserun

We are given a “run” of dx, so the corresponding “rise” is mdx.

. . . . . .

Answers

Example

AnswerThe slope of the line is

m =riserun

We are given a “run” of dx, so the corresponding “rise” is mdx.

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Outline

Advanced Examples

. . . . . .

Differentials are another way to express derivatives

The fact that the the tangentline is an approximation meansthat

f(x+∆x)− f(x)︸︷︷︸∆y

≈ f′(x)∆x︸︷︷︸dy

Rename ∆x = dx, so we canwrite this as

∆y ≈ dy = f′(x)dx.

Note this looks a lot like theLeibniz-Newton identity

= f′(x)

.x .x+∆x

.dx = ∆x

.∆y.dy

. . . . . .

Using differentials to estimate error

Estimating error withdifferentialsIf y = f(x), x0 and ∆x is known,and an estimate of ∆y isdesired:

I Approximate: ∆y ≈ dyI Differentiate: dy = f′(x)dxI Evaluate at x = x0 and

dx = ∆x.. .x

.x .x+∆x

.dx = ∆x

.∆y.dy

. . . . . .

Example

A sheet of plywood measures 8 ft× 4 ft. Suppose our plywood-cuttingmachine will cut a rectangle whose width is exactly half its length, butthe length is prone to errors. If the length is off by 1 in, how bad can thearea of the sheet be off by?

Solution

Write A(ℓ) =12ℓ2. We want to know ∆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, which should be a good estimate for ∆ℓ.

When ℓ = 8 and dℓ = 112 , we have dA = 8

12 = 23 ≈ 0.667. So we

get estimates close to the hundredth of a square foot.

. . . . . .

Example

Solution

(I) A(ℓ+∆ℓ) = A(9712

9409288

So ∆A =9409288

− 32 ≈ 0.6701.

(II) dAdℓ

12 = 23 ≈ 0.667. So we

. . . . . .

Example

Solution

(I) A(ℓ+∆ℓ) = A(9712

9409288

So ∆A =9409288

− 32 ≈ 0.6701.

(II) dAdℓ

12 = 23 ≈ 0.667. So we

. . . . . .

Example

Solution

(I) A(ℓ+∆ℓ) = A(9712

9409288

So ∆A =9409288

− 32 ≈ 0.6701.

(II) dAdℓ

12 = 23 ≈ 0.667. So we

. . . . . .

Why use linear approximations dy when the actual difference ∆y isknown?

I Linear approximation is quick and reliable. Finding ∆y exactlydepends on the function.

I These examples are overly simple. See the “Advanced Examples”later.

I In real life, sometimes only f(a) and f′(a) are known, and not thegeneral f(x).

. . . . . .

Outline

Advanced Examples

. . . . . .

GravitationPencils down!

Example

I Drop a 1 kg ball off the roof of the Silver Center (50m high). Weusually say that a falling object feels a force F = −mg from gravity.

I In fact, the force felt is

F(r) = −GMmr2

where M is the mass of the earth and r is the distance from thecenter of the earth to the object. G is a constant.

I At r = re the force really is F(re) =GMmr2e

= −mg.

I What is the maximum error in replacing the actual force felt at thetop of the building F(re +∆r) by the force felt at ground levelF(re)? The relative error? The percentage error?

. . . . . .

GravitationPencils down!

Example

I Drop a 1 kg ball off the roof of the Silver Center (50m high). Weusually say that a falling object feels a force F = −mg from gravity.

I In fact, the force felt is

F(r) = −GMmr2

where M is the mass of the earth and r is the distance from thecenter of the earth to the object. G is a constant.

I At r = re the force really is F(re) =GMmr2e

= −mg.

I What is the maximum error in replacing the actual force felt at thetop of the building F(re +∆r) by the force felt at ground levelF(re)? The relative error? The percentage error?

. . . . . .

Gravitation Solution

SolutionWe wonder if ∆F = F(re +∆r)− F(re) is small.

I Using a linear approximation,

∆F ≈ dF =dFdr

∣∣∣∣redr = 2

GMmr3e

(GMmr2e

= 2mg∆rre

I The relative error is∆FF

≈ −2∆rre

I re = 6378.1 km. If ∆r = 50m,

≈ −2∆rre

= −250

6378100= −1.56× 10−5 = −0.00156%

. . . . . .

Systematic linear approximation

I√2 is irrational, but

√9/4 is rational and 9/4 is close to 2.

√2 =

√9/4 − 1/4 ≈

√9/4 +

12(3/2)

(−1/4) =1712

I This is a better approximation since (17/12)2 = 289/144

I Do it again!

√2 =

√289/144 − 1/144 ≈

√289/144 +

12(17/12)

(−1/144) = 577/408

Now(577408

332,929166,464

which is1

166,464away from 2.

. . . . . .

√9/4 is rational and 9/4 is close to 2. So

√2 =

√9/4 − 1/4 ≈

√9/4 +

12(3/2)

(−1/4) =1712

I Do it again!

√2 =

√289/144 − 1/144 ≈

√289/144 +

12(17/12)

(−1/144) = 577/408

Now(577408

332,929166,464

which is1

166,464away from 2.

. . . . . .

√2 =

√9/4 − 1/4 ≈

√9/4 +

12(3/2)

(−1/4) =1712

I Do it again!

√2 =

√289/144 − 1/144 ≈

√289/144 +

12(17/12)

(−1/144) = 577/408

Now(577408

332,929166,464

which is1

166,464away from 2.

. . . . . .

√2 =

√9/4 − 1/4 ≈

√9/4 +

12(3/2)

(−1/4) =1712

I Do it again!

√2 =

√289/144 − 1/144 ≈

√289/144 +

12(17/12)

(−1/144) = 577/408

Now(577408

332,929166,464

which is1

166,464away from 2.

. . . . . .

Illustration of the previous example

(94 ,32)

..(2, 1712)

. . . . . .

(94 ,32)

..(2, 1712)

. . . . . .

(94 ,32)

..(2, 1712)

. . . . . .

(94 ,32)

..(2, 1712)

. . . . . .

(94 ,32)

..(2, 1712)

. . . . . .

(94 ,32)

..(2, 1712)

. . . . . .

(94 ,32)

..(2, 1712)

. . . . . .

..(94 ,

32)..(2, 17/12)

..(289144 ,

)..(2, 577408

. . . . . .

..(94 ,

32)..(2, 17/12) .

.(289144 ,

..(2, 577408

. . . . . .

..(94 ,

32)..(2, 17/12) .

.(289144 ,

)..(2, 577408

. . . . . .

Summary

I Linear approximation: If f is differentiable at a, the best linearapproximation to f near a is given by

Lf,a(x) = f(a) + f′(a)(x− a)

I Differentials: If f is differentiable at x, a good approximation to∆y = f(x+∆x)− f(x) is

∆y ≈ dy =dydx

· dx =dydx

·∆x

I Don’t buy plywood from me.

lesson 12: linear approximation and differentials (section 21 slides)

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