Lesson 12: Linear Approximation and Differentials (Section 21 slides)

Section 2.8
Linear Approximation and Differentials

V63.0121.021, Calculus I

New York University

October 14, 2010

Announcements

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

. . . . . .

Announcements

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

. . . . . .

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

Midterm FAQ
Question
What sections are covered on the midterm?

. . . . . .


Midterm FAQ
Question

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

. . . . . .


Midterm FAQ
Question

Answer

Question
Is Section 2.6 going to be on the midterm?

. . . . . .


Midterm FAQ
Question

Answer

Question

Answer

. . . . . .


Midterm FAQ
Question

Answer

Question

Answer

Question
Is Section 2.8 going to be on the midterm?...
. . . . . .


Midterm FAQ, continued

Question
What format will the exam take?

. . . . . .



Question

Answer
There will be both fixed-response (e.g., multiple choice) and
free-response questions.

. . . . . .



Question

Answer

Question
Will explanations be necessary?

. . . . . .



Question

Answer

Question
Will explanations be necessary?

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



Question
Is (topic X) going to be tested?

. . . . . .



Question
Is (topic X) going to be tested?

Answer
Everything 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 the
exam. (This is not the same as saying all exam problems are similar to
class examples or homework problems.)

. . . . . .



Question
Will there be a review session?

. . . . . .



Question
Will there be a review session?

Answer
We’re working on it.

. . . . . .



Question
Will calculators be allowed?

. . . . . .



Question
Will calculators be allowed?

Answer
No. The exam is designed for pencil and brain.

. . . . . .



Question
How should I study?

. . . . . .



Question
How should I study?

Answer

The exam has problems; study by doing problems. If you get one
right, think about how you got it right. If you got it wrong or didn’t
get it at all, reread the textbook and do easier problems to build up
your understanding.
Break up the material into chunks. (related) Don’t put it all off until
the night before.
Ask questions.

. . . . . .



Question
How many questions are there?

. . . . . .



Question
How many questions are there?

Answer
Does this question contribute to your understanding of the material?

. . . . . .



Question
Will there be a curve on the exam?

. . . . . .



Question
Will there be a curve on the exam?

Answer

. . . . . .



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

. . . . . .



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

Answer

. . . . . .


Objectives
Use tangent lines to make
linear approximations to a
function.
Given a function and a
point in the domain,
compute the
linearization of the
function at that point.
Use linearization to
approximate values of
functions
Given a function, compute
the differential of that
function
Use the differential
notation to estimate error
in linear approximations. . . . . . .


Outline

The linear approximation of a function near a point
Examples
Questions

Differentials
Using differentials to estimate error

Advanced Examples

. . . . . .


The Big Idea

Question
Let f be differentiable at a. What linear function best approximates f
near a?

. . . . . .


The Big Idea

Question
near a?

Answer
The tangent line, of course!

. . . . . .


The Big Idea

Question
near a?

Answer

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

. . . . . .


The Big Idea

Question
near a?

Answer

Question
What 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 tangent line is a linear approximation

y
.

L(x) = f(a) + f′ (a)(x − a)

is a decent approximation to f L
. (x) .
near a. f
.(x) .

f
.(a) .
.
x−a

. x
.
a
. x
.

. . . . . .


The tangent line is a linear approximation

y
.

L(x) = f(a) + f′ (a)(x − a)

is a decent approximation to f L
. (x) .
near a. f
.(x) .
How decent? The closer x is to
a, the better the approxmation f
.(a) .
.
x−a
L(x) is to f(x)

. x
.
a
. x
.

. . . . . .


Example
.
Example
Estimate sin(61◦ ) = sin(61π/180) by using a linear approximation
(i) about a = 0 (ii) about a = 60◦ = π/3.

.

. . . . . .


Example
.
Example

Solution (i)

If f(x) = sin x, then f(0) = 0
and f′ (0) = 1.
So the linear approximation
near 0 is L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

.

. . . . . .


Example
.
Example

Solution (i) Solution (ii)
(π)
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3
and f′ (0) = 1. f′ π = .
3
near 0 is L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

.

. . . . . .


Example
.
Example

(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = .
3
near 0 is L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

.

. . . . . .


Example
.
Example

(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2
near 0 is L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

.

. . . . . .


Example
.
Example

(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2
So L(x) =
near 0 is L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

.

. . . . . .


Example
.
Example

(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2 √
So the linear approximation 3 1( π)
So L(x) = + x−
near 0 is L(x) = 0 + 1 · x = x. 2 2 3
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180

.

. . . . . .


Example
.
Example

(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2 √
So L(x) = + x−
near 0 is L(x) = 0 + 1 · x = x. 2 2 3
Thus Thus
( ) ( )
61π 61π 61π
sin ≈ ≈ 1.06465 sin ≈
180 180 180

.

. . . . . .


Example
.
Example

(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2 √
So L(x) = + x−
near 0 is L(x) = 0 + 1 · x = x. 2 2 3
Thus Thus
( ) ( )
61π 61π 61π
sin ≈ ≈ 1.06465 sin ≈ 0.87475
180 180 180

.

. . . . . .


Example
.
Example

(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2 √
So L(x) = + x−
near 0 is L(x) = 0 + 1 · x = x. 2 2 3
Thus Thus
( ) ( )
61π 61π 61π
sin ≈ ≈ 1.06465 sin ≈ 0.87475
180 180 180

Calculator check: sin(61◦ ) ≈
.

. . . . . .


Example
.
Example

(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2 √
So L(x) = + x−
near 0 is L(x) = 0 + 1 · x = x. 2 2 3
Thus Thus
( ) ( )
61π 61π 61π
sin ≈ ≈ 1.06465 sin ≈ 0.87475
180 180 180

Calculator check: sin(61◦ ) ≈ 0.87462.
.

. . . . . .


Illustration

y
.

y
. = sin x

. x
.
. 1◦
6
. . . . . .


Illustration

y
.
y
. = L1 (x) = x

y
. = sin x

. x
.
0
. . 1◦
6
. . . . . .


Illustration

y
.
y
. = L1 (x) = x

b
. ig difference! y
. = sin x

. x
.
0
. . 1◦
6
. . . . . .


Illustration

y
.
y
. = L1 (x) = x

√ ( )
y
. = L2 (x) = 2
3
+ 1
2 x− π
3
y
. = sin x
.

. . x
.
0
. .
π/3 . 1◦
6
. . . . . .


Illustration

y
.
y
. = L1 (x) = x

√ ( )
y
. = L2 (x) = 2
3
+ 1
2 x− π
3
y
. = sin x
. . ery little difference!
v

. . x
.
0
. .
π/3 . 1◦
6
. . . . . .


Another Example

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

. . . . . .


Another Example

Example
√

Solution
√
The key step is to use a linear approximation to f(x) =
√ x near a = 9
to estimate f(10) = 10.

. . . . . .


Another Example

Example
√

Solution
√
√ x near a = 9

f(10) ≈ L(10) = f(9) + f′ (9)(10 − 9)
1 19
=3+ (1) = ≈ 3.167
2·3 6

. . . . . .


Another Example

Example
√

Solution
√
√ x near a = 9

f(10) ≈ L(10) = f(9) + f′ (9)(10 − 9)
1 19
=3+ (1) = ≈ 3.167
2·3 6
( )2
19
Check: =
6

. . . . . .


Another Example

Example
√

Solution
√
√ x near a = 9

f(10) ≈ L(10) = f(9) + f′ (9)(10 − 9)
1 19
=3+ (1) = ≈ 3.167
2·3 6
( )2
19 361
Check: = .
6 36

. . . . . .


Dividing without dividing?
Example
A student has an irrational fear of long division and needs to estimate
577 ÷ 408. He writes
577 1 1 1
= 1 + 169 = 1 + 169 × × .
408 408 4 102
1
Help the student estimate .
102

. . . . . .


Dividing without dividing?
Example
A student has an irrational fear of long division and needs to estimate
577 ÷ 408. He writes
577 1 1 1
= 1 + 169 = 1 + 169 × × .
408 408 4 102
1
Help the student estimate .
102

Solution
1
Let f(x) = . We know f(100) and we want to estimate f(102).
x
1 1
f(102) ≈ f(100) + f′ (100)(2) = − (2) = 0.0098
100 1002
577
=⇒ ≈ 1.41405
408 . . . . . .


Questions

Example
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?

. . . . . .


Answers

Example

. . . . . .


Answers

Example

Answer

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

. . . . . .


Questions

Example

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

. . . . . .


Answers

Example
lots? 12 more lots?

Answer

$100
$150
$600 (?)

. . . . . .


Questions

Example

Example
lots? 12 more lots?

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

. . . . . .


Answers

Example

. . . . . .


Answers

Example

Answer
The slope of the line is
rise
m=
run
We are given a “run” of dx, so the corresponding “rise” is m dx.

. . . . . .


Outline

Examples
Questions

Differentials

Advanced Examples

. . . . . .


Differentials are another way to express derivatives
The fact that the the tangent
line is an approximation means
that
y
.
f(x + ∆x) − f(x) ≈ f′ (x) ∆x
∆y dy

Rename ∆x = dx, so we can
write this as .
.
dy
.
∆y
′
∆y ≈ dy = f (x)dx. .
.
dx = ∆x

Note this looks a lot like the
Leibniz-Newton identity
. x
.
dy x x
. . + ∆x
= f′ (x)
dx
. . . . . .



y
.
Estimating error with
differentials
If y = f(x), x0 and ∆x is known,
and an estimate of ∆y is
desired: .
.
Approximate: ∆y ≈ dy
dy
.
∆y

Differentiate: dy = f′ (x) dx .
.
dx = ∆x
Evaluate at x = x0 and
dx = ∆x.
. x
.
x x
. . + ∆x

. . . . . .


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

. . . . . .


Example

Solution
1 2
Write A(ℓ) = ℓ . We want to know ∆A when ℓ = 8 ft and ∆ℓ = 1 in.
2

. . . . . .


Example

Solution
1 2
2 ( )
97 9409 9409
(I) A(ℓ + ∆ℓ) = A = So ∆A = − 32 ≈ 0.6701.
12 288 288

. . . . . .


Example

Solution
1 2
2 ( )
97 9409 9409
(I) A(ℓ + ∆ℓ) = A = So ∆A = − 32 ≈ 0.6701.
12 288 288
dA
(II) = ℓ, so dA = ℓ dℓ, which should be a good estimate for ∆ℓ.
dℓ
When ℓ = 8 and dℓ = 12 , we have dA = 12 = 2 ≈ 0.667. So we
1 8
3
get estimates close to the hundredth of a square foot.

. . . . . .


Why?

Why use linear approximations dy when the actual difference ∆y is
known?
Linear approximation is quick and reliable. Finding ∆y exactly
depends on the function.
These examples are overly simple. See the “Advanced Examples”
later.
In real life, sometimes only f(a) and f′ (a) are known, and not the
general f(x).

. . . . . .


Outline

Examples
Questions

Differentials

Advanced Examples

. . . . . .


Gravitation
Pencils down!

Example

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

. . . . . .


Gravitation
Pencils down!

Example

Drop a 1 kg ball off the roof of the Silver Center (50m high). We
usually say that a falling object feels a force F = −mg from gravity.
In fact, the force felt is
GMm
F(r) = − ,
r2
where M is the mass of the earth and r is the distance from the
center of the earth to the object. G is a constant.
GMm
At r = re the force really is F(re ) = = −mg.
r2
e
What is the maximum error in replacing the actual force felt at the
top of the building F(re + ∆r) by the force felt at ground level
F(re )? The relative error? The percentage error? . . . . . .


Gravitation Solution
Solution
We wonder if ∆F = F(re + ∆r) − F(re ) is small.
Using a linear approximation,

dF GMm
∆F ≈ dF = dr = 2 3 dr
dr re re
( )
GMm dr ∆r
= 2
= 2mg
re re re

∆F ∆r
The relative error is ≈ −2
F re
re = 6378.1 km. If ∆r = 50 m,
∆F ∆r 50
≈ −2 = −2 = −1.56 × 10−5 = −0.00156%
F re 6378100
. . . . . .


Systematic linear approximation

√ √
2 is irrational, but 9/4 is rational and 9/4 is close to 2.

. . . . . .



√ √
2 is irrational, but 9/4 is rational and 9/4 is close to 2. So
√ √ √ 1 17
2 = 9/4 − 1/4 ≈ 9/4 + (−1/4) =
2(3/2) 12

. . . . . .



√ √
√ √ √ 1 17
2 = 9/4 − 1/4 ≈ 9/4 + (−1/4) =
2(3/2) 12

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

. . . . . .



√ √
√ √ √ 1 17
2 = 9/4 − 1/4 ≈ 9/4 + (−1/4) =
2(3/2) 12

This is a better approximation since (17/12)2 = 289/144
Do it again!
√ √ √ 1
2 = 289/144 − 1/144 ≈ 289/144 + (−1/144) = 577/408
2(17/12)
( )2
577 332, 929 1
Now = which is away from 2.
408 166, 464 166, 464

. . . . . .


Illustration of the previous example

.

. . . . . .



.
2
.

. . . . . .



.

.
2
.

. . . . . .



. 2, 17 )
( 12
. .

.
2
.

. . . . . .



.
. 2, 17/12)
(
. . 4, 3)
(9 2

. . . . . .



.
. 2, 17/12)
(
.. ( . 9, 3)
(
)4 2
289 17
. 144 , 12

. . . . . .



.
. 2, 17/12)
(
.. ( . 9, 3)
(
( 577 ) )4 2
. 2, 408 289 17
. 144 , 12

. . . . . .


Summary

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

Lf,a (x) = f(a) + f′ (a)(x − a)

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

dy dy
∆y ≈ dy = · dx = · ∆x
dx dx
Don’t buy plywood from me.

. . . . . .


Lesson 12: Linear Approximation and Differentials (Section 21 slides)

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