Lesson 12: Linear Approximations and Differentials (slides)

..

Sec on 2.8Linear Approxima ons and

Differen als

V63.0121.011: Calculus IProfessor Ma hew Leingang

New York University

March 2, 2011

Announcements

I Quiz in recita on thisweek on 1.5, 1.6, 2.1, 2.2

I Midterm March 7 in classon 1.1–2.5

I No quiz in recita on nextweek

Midterm FAQQues on

What sec ons are covered on the midterm?

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

Ques on

Is Sec on 2.6 going to be on the midterm?

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

Midterm FAQ, continued

Ques on

What format will the exam take?

AnswerThere will be both fixed-response (e.g., mul ple choice) andfree-response ques ons.


Ques on

Will explana ons be necessary?

AnswerYes, on free-response problems we will expect you to explainyourself. This is why it was required on wri en homework.

Midterm FAQ, continuedQues on

Is (topic X) going to be tested?

Answer

I Everything covered in class or on homework is fair game for theexam.

I No topic that was not covered in class nor on homework will beon the exam.

I (This is not the same as saying all exam problems are similar toclass examples or homework problems.)


Ques on

Will there be a review session?

AnswerThe recita ons this week will review for the exam.


Ques on

Will calculators be allowed?

AnswerNo. The exam is designed for pencil and brain.

Midterm FAQ, continuedQues on

How should I study?

Answer

I The exam has problems; study by doing problems. If you getone right, think about how you got it right. If you got it wrongor didn’t get it at all, reread the textbook and do easierproblems to build up your understanding.

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

I Ask ques ons.


Ques on

How many ques ons are there?

AnswerDoes this ques on contribute to your understanding of thematerial?


Ques on

Will there be a curve on the exam?



Ques on

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


ObjectivesI Use tangent lines to make linearapproxima ons to a func on.

I Given a func on and a point inthe domain, compute thelineariza on of the func on atthat point.

I Use lineariza on to approximatevalues of func ons

I Given a func on, compute thedifferen al of that func on

I Use the differen al nota on toes mate error in linearapproxima ons.

Outline

The linear approxima on of a func on near a pointExamplesQues ons

Differen alsUsing differen als to es mate error

Advanced Examples

The Big IdeaQues on

What linear func on best approximates f near a?

AnswerThe tangent line, of course!

Ques on

What is the equa on for the line tangent to y = f(x) at (a, f(a))?

AnswerL(x) = f(a) + f′(a)(x− a)

tangent line = linear approximation

The func on

L(x) = f(a) + f′(a)(x− a)

is a decent approxima on to fnear a.

How decent? The closer x is toa, the be er theapproxima on L(x) is to f(x)

.. x.

y

....

f(a)

.

f(x)

.

L(x)

.a

.x

.

x− a

tangent line = linear approximation

The func on

L(x) = f(a) + f′(a)(x− a)

is a decent approxima on to fnear a.How decent? The closer x is toa, the be er theapproxima on L(x) is to f(x) .. x.

y

....

f(a)

.

f(x)

.

L(x)

.a

.x

.

x− a

ExampleExample

Es mate sin(61◦) = sin(61π/180) by using a linear approxima on(i) about a = 0 (ii) about a = 60◦ = π/3.

Solu on (i)

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

I So the linear approxima onnear 0 is L(x) = 0+ 1 · x = x.

I sin(61π180

)≈ 61π

180≈ 1.06465

Solu on (ii)I We have f

(π3

)=

√32

andf′(π3

)=

12

.

I So the linear approxima on is

L(x) =

√32

+12

(x− π

3

)

I sin(61π180

)≈

0.87475

Calculator check: sin(61◦) ≈

0.87462.

ExampleExample


Solu on (i)



I sin(61π180

)≈ 61π

180≈ 1.06465


(π3

)=

√32 and

f′(π3

)=

12

.


L(x) =

√32

+12

(x− π

3

)

I sin(61π180

)≈

0.87475


0.87462.

ExampleExample


Solu on (i)



I sin(61π180

)≈ 61π

180≈ 1.06465


(π3

)=

√32 and

f′(π3

)= 1

2 .


L(x) =

√32

+12

(x− π

3

)

I sin(61π180

)≈

0.87475


0.87462.

ExampleExample


Solu on (i)



I sin(61π180

)≈ 61π

180≈ 1.06465


(π3

)=

√32 and

f′(π3

)= 1

2 .


L(x) =

√32

+12

(x− π

3

)I sin

(61π180

)≈

0.87475


0.87462.

ExampleExample


Solu on (i)



I sin(61π180

)≈ 61π

180≈ 1.06465


(π3

)=

√32 and

f′(π3

)= 1

2 .


L(x) =√32

+12

(x− π

3

)

I sin(61π180

)≈

0.87475


0.87462.

ExampleExample


Solu on (i)



I sin(61π180

)≈ 61π

180≈ 1.06465


(π3

)=

√32 and

f′(π3

)= 1

2 .


L(x) =√32

+12

(x− π

3

)I sin

(61π180

)≈

0.87475


0.87462.

ExampleExample


Solu on (i)



I sin(61π180

)≈ 61π

180≈ 1.06465


(π3

)=

√32 and

f′(π3

)= 1

2 .


L(x) =√32

+12

(x− π

3

)I sin

(61π180

)≈ 0.87475


0.87462.

ExampleExample


Solu on (i)



I sin(61π180

)≈ 61π

180≈ 1.06465


(π3

)=

√32 and

f′(π3

)= 1

2 .


L(x) =√32

+12

(x− π

3

)I sin

(61π180

)≈ 0.87475

Calculator check: sin(61◦) ≈ 0.87462.

Illustration

.. x.

y

.

y = sin x

.61◦

.

y = L1(x) = x

..0

.

.

big difference!

.

y = L2(x) =√32 + 1

2

(x− π

3

)

..π/3

.

.

very li le difference!

Illustration

.. x.

y

.

y = sin x

.61◦

.

y = L1(x) = x

..0

..

big difference!

.

y = L2(x) =√32 + 1

2

(x− π

3

)

..π/3

.

.


Illustration

.. x.

y

.

y = sin x

.61◦

.

y = L1(x) = x

..0

.

.

big difference!

.

y = L2(x) =√32 + 1

2

(x− π

3

)

..π/3

.

.


Illustration

.. x.

y

.

y = sin x

.61◦

.

y = L1(x) = x

..0

.

.

big difference!

.

y = L2(x) =√32 + 1

2

(x− π

3

)

..π/3

..


Another ExampleExample

Es mate√10 using the fact that 10 = 9+ 1.

Solu onThe key step is to use a linear approxima on to f(x) =

√x near a = 9 to es mate

f(10) =√10.

f(10) ≈ L(10) = f(9) + f′(9)(10− 9)

= 3+1

2 · 3(1) =

196

≈ 3.167

Check:(196

)2

=36136

.





f(10) =√10.

f(10) ≈ L(10) = f(9) + f′(9)(10− 9)

= 3+1

2 · 3(1) =

196

≈ 3.167

Check:(196

)2

=

36136

.





f(10) =√10.

f(10) ≈ L(10) = f(9) + f′(9)(10− 9)

= 3+1

2 · 3(1) =

196

≈ 3.167

Check:(196

)2

=36136

.

Dividing without dividing?

Example

A student has an irra onal fear of long division and needs toes mate 577÷ 408. He writes

577408

= 1+ 1691

408= 1+ 169× 1

4× 1

102.

Help the student es mate1

102.

Dividing without dividing?Solu on

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

f(102) ≈ f(100) + f′(100)(2) =1

100− 1

1002(2) = 0.0098

=⇒ 577408

≈ 1.41405

Calculator check:577408

≈ 1.41422.

QuestionsExample

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

Answer

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

QuestionsExample

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

Answer

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

QuestionsExample

Suppose a line goes through the point (x0, y0) and has slopem. Ifthe point is moved horizontally by dx, while staying on the line, whatis the corresponding ver cal movement?

AnswerThe slope of the line is

m =riserun

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

Outline



Advanced Examples

Differentials are derivativesThe fact that the the tangent line is anapproxima on means that

f(x+∆x)− f(x)︸︷︷︸∆y

≈ f′(x)∆x︸︷︷︸dy

Rename∆x = dx, so we can write this as

∆y ≈ dy = f′(x)dx.

Note this looks a lot like the Leibniz-Newtoniden ty

dydx

= f′(x) .. x.

y

...x

.x+∆x

.

dx = ∆x

.

∆y

.

dy

Using differentials to estimate error

Es ma ng error withdifferen alsIf y = f(x), x0 and∆x isknown, and an es mate of∆yis desired:

I Approximate: ∆y ≈ dyI Differen ate:dy = f′(x) dx

I Evaluate at x = x0 anddx = ∆x.

.. x.

y

...x

.x+∆x

.

dx = ∆x

.

∆y

.

dy

Using differentials to estimate error

Example

A regular sheet of plywoodmeasures 8 ft× 4 ft. Supposea defec ve plywood-cu ngmachine will cut a rectanglewhose width is exactly half itslength, but the length is proneto errors. If the length is off by1 in, how bad can the area ofthe sheet be off by?

SolutionSolu on

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 = ℓ dℓ, which should be a good es mate for∆ℓ.When ℓ = 8 and dℓ = 1

12 , we have dA = 812 = 2

3 ≈ 0.667.So we get es mates close to the hundredth of a square foot.

SolutionSolu on


∆ℓ = 1 in.

(I) A(ℓ+∆ℓ) = A(9712

)=

9409288

So∆A =9409288

− 32 ≈ 0.6701.

(II)dAdℓ

= ℓ, so dA = ℓ dℓ, which should be a good es mate for∆ℓ.

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

12 = 23 ≈ 0.667.

So we get es mates close to the hundredth of a square foot.

SolutionSolu on


∆ℓ = 1 in.

(I) A(ℓ+∆ℓ) = A(9712

)=

9409288

So∆A =9409288

− 32 ≈ 0.6701.

(II)dAdℓ


12 , we have dA = 812 = 2

3 ≈ 0.667.

So we get es mates close to the hundredth of a square foot.

SolutionSolu on


∆ℓ = 1 in.

(I) A(ℓ+∆ℓ) = A(9712

)=

9409288

So∆A =9409288

− 32 ≈ 0.6701.

(II)dAdℓ


12 , we have dA = 812 = 2

3 ≈ 0.667.So we get es mates close to the hundredth of a square foot.

Why should we care?

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

I Linear approxima on is quick and reliable. Finding∆y exactlydepends on the func on.

I With more complicated func ons, linear approxima on muchsimpler. See the “Advanced Examples” later.

I In real life, some mes only f(a) and f′(a) are known, and notthe general f(x).

Outline



Advanced Examples

GravitationExample

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

I In fact, the force felt is F(r) = −GMmr2

,whereM is the mass of the earth andr is the distance from the center of the earth to the object. G is a constant.

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

, and g is defined to beGMr2e

I What is the maximum error in replacing the actual force felt at the top ofthe building F(re +∆r) by the force felt at ground level F(re)? The rela veerror? The percentage error?

Gravitation SolutionSolu onWe wonder if∆F = F(re +∆r)− F(re) is small.

I Using a linear approxima on,

∆F ≈ dF =dFdr

∣∣∣∣redr = 2

GMmr3e

dr

=

(GMmr2e

)drre

= 2mg∆rre

I The rela ve error is∆FF

≈ −2∆rre

Solution continued

re = 6378.1 km. If∆r = 50m,

∆FF

≈ −2∆rre

= −250

6378100= −1.56× 10−5 = −0.00156%

Systematic linear approximationI√2 is irra onal, but

√9/4 is ra onal and 9/4 is close to 2.

So√2 =

√9/4 − 1/4 ≈

√9/4 +

12(3/2)

(−1/4) =1712

I This is a be er approxima on since (17/12)2 = 289/144I Do it again!

√2 =

√289/144 − 1/144 ≈

√289/144 +

12(17/12)

(−1/144) = 577/408

Now(577408

)2

=332, 929166, 464

which is1

166, 464away from 2.


√9/4 is ra onal and 9/4 is close to 2. So

√2 =

√9/4 − 1/4 ≈

√9/4 +

12(3/2)

(−1/4) =1712


√2 =

√289/144 − 1/144 ≈

√289/144 +

12(17/12)

(−1/144) = 577/408

Now(577408

)2

=332, 929166, 464

which is1




√2 =

√9/4 − 1/4 ≈

√9/4 +

12(3/2)

(−1/4) =1712

I This is a be er approxima on 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

)2

=332, 929166, 464

which is1




√2 =

√9/4 − 1/4 ≈

√9/4 +

12(3/2)

(−1/4) =1712


√2 =

√289/144 − 1/144 ≈

√289/144 +

12(17/12)

(−1/144) = 577/408

Now(577408

)2

=332, 929166, 464

which is1


Illustration of the previousexample

.

.2

.

(94 ,32)

..

(2, 1712)


..2

.

(94 ,32)

..

(2, 1712)


..2

..

(94 ,32)

..

(2, 17/12)

..

(289144 ,

1712

)

..

(2, 577408

)

SummaryI Linear approxima on: If f is differen able at a, the best linearapproxima on to f near a is given by

Lf,a(x) = f(a) + f′(a)(x− a)

I Differen als: If f is differen able at x, a good approxima on to∆y = f(x+∆x)− f(x) is

∆y ≈ dy =dydx

· dx = dydx

·∆x

I Don’t buy plywood from me.

Lesson 12: Linear Approximations and Differentials (slides)

Economy & Finance

Transcript of Lesson 12: Linear Approximations and Differentials (slides)