1a 1b

Table of Trig Functions

sin 𝑥 cos 𝑥

tan 𝑥 =sin 𝑥

cos 𝑥 cot 𝑥 =

cos 𝑥

sin(𝑥)

sec 𝑥 =1

cos(𝑥) csc 𝑥 =

1

sin(𝑥)

Obtain co-functions in right column from left column

by replacing sin 𝑥 → cos(𝑥) and cos 𝑥 → sin(𝑥).

Two Famous Triangles

sketch 45°-45°-90° triangle, label angles and sides

Pythagorean theorem

1

2

2+

1

2

2= 1 or

1

2+

1

2= 1

sin 𝜋

4 = 𝑐𝑜𝑠

𝜋

4 =

1

2, tan

𝜋

4 = 1

sketch 30°-60°-90° triangle, label angles and sides

Pythagorean theorem

3

2

2

+ 1

2

2= 1 or

3

4+

1

2= 1

sin 𝜋

6 =

1

2, cos

𝜋

6 =

3

2, tan

𝜋

6 =

1

3

sin 𝜋

3 =

3

2, cos

𝜋

3 =

1

2, tan

𝜋

3 = 3

2a 2b

Derivatives of Trig Functions

𝑑

𝑑𝑥tan 𝑥 =

𝑑

𝑑𝑥

sin 𝑥

cos 𝑥

= sin 𝑥 ′ (cos 𝑥)−(sin 𝑥) cos 𝑥 ′

cos 2 𝑥

=cos 𝑥 ⋅cos 𝑥 −sin 𝑥 ⋅ − sin 𝑥

cos 2 𝑥

=cos 2 𝑥 +sin 2 𝑥

cos 2 𝑥

=1

cos 2 𝑥 = sec2(𝑥)

where sec 𝑥 = 1/cos(𝑥).

𝑑

𝑑𝑥sec 𝑥 =

𝑑

𝑑𝑥

1

cos 𝑥

= 1 ′ ⋅cos 𝑥−1⋅ cos 𝑥 ′

cos 2 𝑥

=−(− sin 𝑥)

cos 2 𝑥

=1

cos 𝑥⋅

sin 𝑥

cos 𝑥

= sec 𝑥 ⋅ tan 𝑥

similarly, obtain the following table (to memorize)

𝑑

𝑑𝑥sin 𝑥 = cos 𝑥

𝑑

𝑑𝑥cos 𝑥 = − sin 𝑥

𝑑

𝑑𝑥tan 𝑥 = sec2 𝑥

𝑑

𝑑𝑥cot 𝑥 = − csc2 𝑥

𝑑

𝑑𝑥sec 𝑥 = sec 𝑥 ⋅ tan 𝑥

𝑑

𝑑𝑥csc 𝑥 = − csc 𝑥 ⋅ cot(𝑥)

to obtain the right hand column from the left hand

column, replace functions by co-functions and

multiply by −1.

?? 𝑑

𝑑𝑥 𝑥 sec(𝑥)

?? 𝑑

𝑑𝜃 tan (𝜃)

𝜃2+1

3a 3b

Example. Find an equation of the line tangent to the

curve

𝑦 = 𝑥 tan(𝑥)

at the point 𝜋

4,𝜋

4 .

Solution.

Point-slope form of tangent line

𝑦 − 𝑦0 = 𝑚(𝑥 − 𝑥0)

Need slope

𝑑𝑦

𝑑𝑥= tan 𝑥 + 𝑥 sec2(𝑥)

𝑑𝑦

𝑑𝑥 𝜋

4

= tan 𝜋

4 +

𝜋

4sec2

𝜋

4

where sec 𝜋

2 =

1

cos 𝜋

4

= 2

𝑑𝑦

𝑑𝑥 𝜋

4

= 1 +𝜋

4⋅ 2 = 1 +

𝜋

2

equation for tangent line

𝑦 −𝜋

4= 1 +

𝜋

2 𝑥 −

𝜋

4 ■

§2.5 The Chain Rule

Composition

Example. Consider 𝐹 𝑥 = 𝑥2 + 1

Let 𝑓 𝑢 = 𝑢, 𝑔 𝑥 = 𝑥2 + 1

𝐹 is the composition of 𝑓 and 𝑔

Function machines:

𝑥 → 𝑔 → 𝑥2 + 1 = 𝑢 → 𝑓 → 𝑢 = 𝑥2 + 1

■

Notation for composition

𝐹 𝑥 = 𝑓 𝑔 𝑥

𝑓 is the outer function,

𝑔 is the inner function

4a 4b

Derivatives of Compositions

Let 𝑦 = 𝐹 𝑥 = 𝑓 𝑔 𝑥

or 𝑦 = 𝑓(𝑢), 𝑢 = 𝑔(𝑥).

The Chain Rule

If 𝑔 is differentiable at 𝑥 and 𝑓 is differentiable at 𝑢,

then

𝐹′ 𝑥 = 𝑓 ′ 𝑢 𝑔′(𝑥)

The derivative of a composition of two functions is

the product of their derivatives

Write another way

𝐹′ 𝑥 = 𝑓 ′ 𝑔 𝑥 𝑔′ (𝑥)

Leibniz notation

Let 𝑦 = 𝑓(𝑢) and 𝑢 = 𝑔(𝑥) and

𝑦 = 𝐹 𝑥 = 𝑓 𝑔 𝑥

then

𝐹′ 𝑥 = 𝑓 ′ 𝑢 𝑔′(𝑥)

𝑑𝑦

𝑑𝑥 =

𝑑𝑦

𝑑𝑢

𝑑𝑢

𝑑𝑥

Easy to remember: looks as though a term “du”

cancels on the right hand side!

Example. Let 𝐹 𝑥 = 𝑥2 + 1. Find 𝐹′ (𝑥).

Write as a composition

𝑢 = 𝑔 𝑥 = 𝑥2 + 1

𝑦 = 𝑓 𝑢 = 𝑢 where 𝑢 = 𝑢1

2

apply the chain rule

𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑢

𝑑𝑢

𝑑𝑥

=1

2𝑢

−1

2 ⋅ 2𝑥

= 𝑥2 + 1 −1

2 𝑥 =𝑥

𝑥2+1 ■

5a 5b

Example. Let 𝑦 = cos 3𝑥 . Find 𝑑𝑦

𝑑𝑥.

Write as a composition

𝑦 = cos(𝑢) where 𝑢 = 3𝑥

apply the chain rule

𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑢

𝑑𝑢

𝑑𝑥

= − sin 𝑢 ⋅ 3

= −3 sin(3𝑥) ■

?? Class Practice

Let 𝑦 = 𝑥2 + 1 10 . Find 𝑑𝑦

𝑑𝑥.

Let 𝑦 = sin 𝑥 . Find 𝑑𝑦

𝑑𝑥.

Let 𝑦 = tan cos 𝑥

6a 6b

Plausibility Argument for the Chain Rule

Recall the limit definition of derivative

𝑑𝑦

𝑑𝑥= 𝑓 ′ 𝑥 = lim𝑕→0

𝑓 𝑥+𝑕 −𝑓(𝑥)

𝑕

Change notation

Let Δ𝑥 = 𝑕

Δ𝑦 = 𝑓 𝑥 + 𝑕 − 𝑓(𝑥)

then

𝑑𝑦

𝑑𝑥= limΔ𝑥→0

Δ𝑦

Δ𝑥

Statement of Chain Rule

Let 𝑦 = 𝐹 𝑥 = 𝑓 𝑔 𝑥 or 𝑦 = 𝑓(𝑢) with

𝑢 = 𝑔(𝑥)

If 𝑔 is differentiable at 𝑥, and 𝑓 is differentiable at 𝑢,

then

𝐹′ 𝑥 = 𝑓 ′ 𝑢 𝑔′(𝑥) or 𝑑𝑦

𝑑𝑥 =

𝑑𝑦

𝑑𝑢 𝑑𝑢

𝑑𝑥

Plausibility Argument

Let Δ𝑢 be the change in 𝑢 corresponding to Δ𝑥.

Δ𝑢 = 𝑔 𝑥 + Δ𝑥 − 𝑔(𝑥)

The corresponding change in 𝑦 is

Δ𝑦 = 𝑓 𝑢 + Δ𝑢 − 𝑓(𝑢)

Then

𝑑𝑦

𝑑𝑥 = limΔ𝑥→0

Δ𝑦

Δ𝑥

= limΔ𝑥→0Δ𝑦

Δ𝑢

Δ𝑢

Δ𝑥 as long as Δ𝑢 ≠ 0

= limΔ𝑥→0Δ𝑦

Δ𝑢⋅ limΔ𝑥→0

Δ𝑢

Δ𝑥 product rule for limits

but Δ𝑢 → 0 as Δ𝑥 → 0, so

𝑑𝑦

𝑑𝑥= limΔ𝑢→0

Δ𝑦

Δ𝑢⋅ limΔ𝑥→0

Δ𝑢

Δ𝑥

=𝑑𝑦

𝑑𝑢⋅

𝑑𝑢

𝑑𝑥

The only problem is that we may have Δ𝑢 = 0. ■

7a 7b

Generalized Power Rule

Combine the power rule and the chain rule

Consider

𝐹 𝑥 = 𝑔 𝑥 𝑛

write as a composition

𝑢 = 𝑔(𝑥)

𝑦 = 𝑓 𝑢 = 𝑢𝑛

apply the chain rule

𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑢

𝑑𝑢

𝑑𝑥

= 𝑛 𝑢𝑛−1 𝑑𝑢

𝑑𝑥

this is frequently written

𝑑

𝑑𝑥𝑔 𝑥 𝑛 = 𝑛 𝑔 𝑥 𝑛−1𝑔′ (𝑥)

Example. Consider

𝐹 𝑥 = 𝑥3 + 4𝑥 10

this has the form given above, with

𝑔 𝑥 = 𝑥3 + 4𝑥

𝑛 = 10

then

𝐹′ 𝑥 = 10 𝑥3 + 10𝑥 9 𝑑

𝑑𝑥(𝑥3 + 4𝑥)

= 10 𝑥3 + 10𝑥 9 (3𝑥2 + 4) ■

Example.

𝐹 𝑥 =1

sin (𝑥)= sin 𝑥

−1

2

𝐹′ 𝑥 =−1

2 sin 𝑥

−3

2 cos(𝑥) ■

WARNING It is a common mistake to forget the

factor 𝑔′ (𝑥)!

8a 8b

Compositions of Three Functions

Consider

𝐹 𝑡 = 𝑓 𝑔 𝑕 𝑡

write as a composition

𝑦 = 𝑓 𝑢 , 𝑢 = 𝑔(𝑥), 𝑥 = 𝑕(𝑡)

chain rule

𝑑𝑦

𝑑𝑡=

𝑑𝑦

𝑑𝑢

𝑑𝑢

𝑑𝑥

𝑑𝑥

𝑑𝑡

it looks as though ‘𝑑𝑢’ and ‘𝑑𝑥’ factors cancel

Example

𝐹 𝑡 = sin2(4𝑡)

write as a composition

𝑦 = 𝑢2, 𝑢 = sin 𝑥 , 𝑥 = 4𝑡

chain rule

𝑑𝑦

𝑑𝑡=

𝑑𝑦

𝑑𝑢

𝑑𝑢

𝑑𝑥

𝑑𝑥

𝑑𝑡

= 2𝑢 ⋅ cos 𝑥 ⋅ 4

= 8 𝑢 cos(𝑥)

= 8 sin 𝑥 cos(𝑥)

= 8 sin 4𝑡 cos(4𝑡) ■

9a 9b

Example

Find the line tangent to the graph of 𝑦 = sin2 4𝑡 at

𝑡 =𝜋

16 .

Solution. Point slope form of a tangent line

𝑦 − 𝑦0 = 𝑚 𝑡 − 𝑡0

𝑡0 = 𝜋/16 , 𝑦0 = sin2 𝜋

4 =

1

2

2=

1

2

slope? 𝑚 = 𝑑𝑦

𝑑𝑥 𝑡=

𝜋

16

= 8 sin 4𝑡 cos 4𝑡 𝑡=𝜋

16

= 8 sin 𝜋

4 cos

𝜋

4

= 8 1

2

1

2

= 4

gives the tangent line

𝑦 −1

2= 4 𝑡 −

𝜋

16

■

?? Class practice differentiating!

Find 𝑑𝑦

𝑑𝑥

1. 𝑦 = 𝑥4 − 3𝑥2 + 6 3

2. 𝑦 = 2𝑥 𝑥2 + 1

10a 10b

3. 𝑦 = 𝑥 +1

𝑥2 7

4. 𝑦 =𝑡

1−𝑡2

5. 𝑦 = sin(sec(𝑥))

6. 𝑦 = cos tan 1 + 𝑡2

11a 11b

§2.6 Implicit Differentiation

circle of radius 1

… 0 … 𝑥-, … 0 … 𝑦-, circle

𝑥2 + 𝑦2 = 1

defines 𝑦 implicitly as a function of 𝑥

solve for 𝑦: 𝑦2 = 1 − 𝑥2

two possibilities

𝑦 = 1 − 𝑥2 1

2 = 1 − 𝑥2 [1a]

𝑦 = − 1 − 𝑥2 1

2 = − 1 − 𝑥2 [1b]

derivatives

𝑑𝑦

𝑑𝑥=

1

2 1 − 𝑥2

−1

2 −2𝑥 =−𝑥

1−𝑥2 [2a] [2a]

𝑑𝑦

𝑑𝑥= −

1

2 1 − 𝑥2

−1

2 −2𝑥 = 𝑥

1−𝑥2 [2b]

Using Implicit Differentiation is easier in many cases!

𝑥2 + 𝑦2 = 1 [3] [3]

regard 𝑦 as a function of 𝑥

𝑦 = 𝑓(𝑥) or 𝑦2 = 𝑓 𝑥 2

recall the generalized power rule

𝑑

𝑑𝑥 𝑓 𝑥 2 = 2 𝑓 𝑥 𝑓 ′ (𝑥)

write this as

𝑑

𝑑𝑥𝑦2 = 2 𝑦 𝑦′

must remember 𝑦′ =𝑑𝑦

𝑑𝑥 !

12a 12b

differentiate equation [3] implicitly with respect to 𝑥

2𝑥 + 2𝑦𝑑𝑦

𝑑𝑥= 0

solve for 𝑑𝑦/𝑑𝑥

𝑦𝑑𝑦

𝑑𝑥= −𝑥

𝑑𝑦

𝑑𝑥= −

𝑥

𝑦

from [1], two possibilities

𝑑𝑦

𝑑𝑥=

−𝑥

1−𝑥2

𝑑𝑦

𝑑𝑥=

𝑥

1−𝑥2

these match [2]!

Example. Find 𝑑𝑦

𝑑𝑥 if

𝑥−1 + 𝑦−1 = 1. [1]

Solution. Regard 𝑦 as a function of 𝑥

𝑦 = 𝑓(𝑥)

Use the generalized power rule

𝑑

𝑑𝑥 𝑓 𝑥 −1 = −𝑓 𝑥 −2𝑓 ′ (𝑥)

𝑑

𝑑𝑥𝑦−1 = −𝑦−2 𝑑𝑦

𝑑𝑥

differentiate [1] implicitly with respect to 𝑥

−𝑥−2 − 𝑦−2 𝑑𝑦

𝑑𝑥= 0

solve for 𝑑𝑦

𝑑𝑥

−𝑦−2 𝑑𝑦

𝑑𝑥=

1

𝑥2

𝑑𝑦

𝑑𝑥= −

𝑦2

𝑥2 ■

13a 13b

Example. Find the equation of the line tangent to the

curve

2 cos 𝑥 sin 𝑦 = 1 [1]

at 𝑥0, 𝑦0 = 𝜋

4,𝜋

4 .

Solution. Recall sin 𝜋

4 = cos

𝜋

4 =

1

2.

think of 𝑦 as a function of 𝑥

𝑑

𝑑𝑥sin 𝑦 = cos 𝑦

𝑑𝑦

𝑑𝑥

differentiate [1] implicitly

−2 sin 𝑥 sin 𝑦 + 2 cos 𝑥 cos 𝑦 𝑑𝑦

𝑑𝑥= 0

solve for 𝑑𝑦

𝑑𝑥

2 cos 𝑥 cos 𝑦 𝑑𝑦

𝑑𝑥= 2 sin 𝑥 sin(𝑦)

𝑑𝑦

𝑑𝑥= tan 𝑥 tan(𝑦)

slope of curve at 𝑥 = 𝑦 = 𝜋/4

𝑑𝑦

𝑑𝑥 𝑥=𝑦=

𝜋

4

= tan 𝜋

4 tan

𝜋

4 = 1 ⋅ 1 = 1

point-slope form of tangent line

𝑦 − 𝑦0 = 𝑚(𝑥 − 𝑥0)

𝑦 −𝜋

4= 𝑥 −

𝜋

4

𝑦 = 𝑥 ■

14a 14b

?? Class practice with implicit differentiation

1. Let 𝑥3 + 𝑥2𝑦 + 4 𝑦2 = 𝑐, where 𝑐 is a constant.

Find 𝑑𝑦

𝑑𝑥.

Solution 𝑑𝑦

𝑑𝑥=

−(3𝑥2+2𝑥𝑦 )

𝑥2+8𝑦

2. Let sin 𝑥 + cos 𝑦 = sin 𝑥 cos(𝑦). Find 𝑑𝑦

𝑑𝑥.

Solution 𝑑𝑦

𝑑𝑥=

cos (x)

sin x −1

cos y −1

sin (y)

15a 15b

Find an equation of the line tangent to the following

curve at the given point.

2 𝑥2 + 𝑦2 2 = 25(𝑥2 − 𝑦2) at the point (3,1)

Solution. Differentiate implicitly with respect to 𝑥

4 𝑥2 + 𝑦2 2𝑥 + 2𝑦𝑦′ = 25(2𝑥 − 2𝑦𝑦′ )

Solve for 𝑦′

4 𝑥2 + 𝑦2 2𝑦𝑦′ + 50𝑦𝑦′ = −8 𝑥2 + 𝑦2 𝑥 + 50𝑥

𝑦′ (8𝑦 𝑥2 + 𝑦2) + 50𝑦 = −8𝑥3 − 8𝑥𝑦2 + 50𝑥

𝑦′ =−8𝑥3 − 8𝑥𝑦2 + 50𝑥

8𝑦 𝑥2 + 𝑦2 + 50𝑦

=2𝑥(−4𝑥2 − 4𝑦2 + 25)

2𝑦(4𝑥2 + 4𝑦2 + 25)

evaluate at 𝑥 = 3 and 𝑦 = 1

𝑦′ =6(−36 − 4 + 25)

2(36 + 4 + 25)=

6(−15)

2(65)=

−45

65=

−9

13

point slope form of tangent line

𝑦 − 𝑦0 = 𝑚 𝑥 − 𝑥0

𝑦 − 1 =−9

13 𝑥 − 3

in slope intercept form

𝑦 =−9

13𝑥 +

27

13+ 1

=−9

13𝑥 +

40

13

show Maple image of curve and tangent line ■

16a 16b

§2.7 Related Rates

Example. A descending balloonist lets helium escape

at a rate of 10 cubic feet / minute. Assume the

balloon is spherical. How fast is the radius decreasing

when the radius is 10 feet?

sketch balloon, basket, escaping gas

Given

𝑉 = volume of balloon

𝑑𝑉

𝑑𝑡= −10

feet 3

minute

𝑟 = radius of balloon = 10 feet

Find 𝑑𝑟

𝑑𝑡 when 𝑟 = 10

Equation relating these quantities

𝑉 =4

3𝜋𝑟3 [1]

differentiate with respect to time using the chain rule

generalized power rule: 𝑑

𝑑𝑡 𝑔 𝑡 𝑛 = 𝑛 𝑔 𝑡 𝑛−1

consider radius 𝑟 a function of time 𝑡

𝑑

𝑑𝑡𝑟3 = 3𝑟2 𝑑𝑟

𝑑𝑡

Differentiate [1] with respect to time to get

𝑑𝑉

𝑑𝑡=

4

3𝜋 3 𝑟2

𝑑𝑟

𝑑𝑡 = 4𝜋𝑟2

𝑑𝑟

𝑑𝑡

Solve for 𝑑𝑟/𝑑𝑡

𝑑𝑟

𝑑𝑡=

1

4𝜋𝑟2

𝑑𝑉

𝑑𝑡

𝑑𝑟

𝑑𝑡 𝑟=10

=1

4 𝜋 100

1

feet 2 −10

feet 3

min

= −1

40 𝜋

feet

min≈ −0.008

feet

min≈ −0.1

inches

min

■

17a 17b

Problem Solving Strategy

1. Read problem carefully

2. Draw picture if possible

3. Assign notation to given information and

unknown rate

4. Write down an equation relating the given

information and the function whose rate is

unknown

5. Differentiate with respect to time

6. Solve for the unknown rate

Example. The length of a rectangle is increasing at a

rate of 7 cm/s and its width is increasing at a rate of 7

cm/s. When the length is 6cm and the width is 4 cm,

how fast is the area of the rectangle increasing?

2. picture

3. Given information

𝐿(𝑡) = 6cm, 𝑊(𝑡) = 4cm

𝐿′ 𝑡 = 7 cm/s, 𝑊 ′ 𝑡 = 7 cm/s

Unknown rate is 𝐴′ (𝑡), where 𝐴 = 𝐿𝑊.

4. 𝐴 𝑡 = 𝐿 𝑡 𝑊(𝑡)

5. 𝐴′ 𝑡 = 𝐿′ 𝑡 𝑊 𝑡 + 𝐿 𝑡 𝑊 ′ (𝑡)

6. = 7cm

s⋅ 4 cm + 6 cm ⋅ 7

cm

s

= 28cm2

s+ 42

cm2

s= 70

cm2

𝑠 ■

18a 18b

Example. One end of a 13 foot ladder is on the

ground and the other end rests on a vertical wall. If

the bottom end is drawn from the wall at 3

feet/second, how fast is the top of the ladder sliding

down the wall when the bottom is 5 feet from the

wall?

… 𝑥 … 𝑥-, … 𝑦 … 𝑦-, ladder betw. 𝑥 & 𝑦, 𝑙 = 13 ft

given information

𝑙 = 13 ft, 𝑥 = 5ft, 𝑑𝑥

𝑑𝑡= 3 ft./sec.

unknown rate

𝑑𝑦

𝑑𝑡= ?

Pythagorean theorem

𝑥2 + 𝑦2 = 𝑙2

differentiate with respect to time

2 𝑥𝑑𝑥

𝑑𝑡+ 2 𝑦

𝑑𝑦

𝑑𝑡= 0

solve for 𝑦

𝑑𝑦

𝑑𝑡= −

𝑥

𝑦

𝑑𝑥

𝑑𝑡

to evaluate 𝑑𝑦/𝑑𝑡 we need 𝑦

𝑦2 = 𝑙2 − 𝑥2 = 169 − 25 = 144

𝑦 = 144 = 12

then

𝑑𝑦

𝑑𝑡 𝑥=5 ft

= −5

12⋅ 3

feet

sec= −

5

4

feet

sec

■

19a 19b

Example. Pat walks 5 feet/second towards a street

light whose lamp is 20 feet above the ground. If Pat

is 6 feet tall, find how rapidly Pat’s shadow changes

in length.

ground, lamp, Pat, 𝑥, 𝑠, 𝑦, 𝑕

given information

𝑦 = 20 feet, 𝑕 = 6 feet, 𝑑𝑥

𝑑𝑡= −5

feet

sec

unknown rate

𝑑𝑠

𝑑𝑡= ?

by similar triangles

𝑦

𝑥 + 𝑠=

𝑕

𝑠

𝑦 𝑠 = 𝑕 (𝑥 + 𝑠)

𝑦 − 𝑕 𝑠 = 𝑕𝑥

𝑠 = 𝑕

𝑦 − 𝑕 𝑥

differentiate wrt time

𝑑𝑠

𝑑𝑡=

𝑕

𝑦 − 𝑕

𝑑𝑥

𝑑𝑡

= 6

20 − 6 −5

feet

sec

= − 15

7

feet

sec ■

20a 20b

Example. The beacon of a lighthouse 1 mile from the

shore makes 5 rotations per minute. Assuming that

the shoreline is straight, calculate the speed at which

the spotlight sweeps along the shoreline as it lights

up sand 2 miles from the lighthouse.

0 … 𝑥 … 𝑥-, 0 … 𝑦 … 𝑦-, lighthouse at (0, 𝑦), beam to

(𝑥, 0), length 𝐿, angle 𝜃

given information

𝑦 = 1 mile 𝐿 = 2 miles

𝑑𝜃

𝑑𝑡= 5

revolutions

minute = 5

revolutions

minute 2𝜋

radians

revolution

= 10𝜋 radians

minute

unknown rate 𝑑𝑥

𝑑𝑡=?

from the definition of tangent

tan 𝜃 =𝑥

𝑦

differentiate wrt time

sec2 𝜃 𝑑𝜃

𝑑𝑡=

1

𝑦

𝑑𝑥

𝑑𝑡 using that 𝑦 is a constant

𝑑𝑥

𝑑𝑡= 𝑦 sec2 𝜃

𝑑𝜃

𝑑𝑡

we need sec2 𝜃

cos 𝜃 =𝑦

𝐿 sec 𝜃 =

𝐿

𝑦= 2

then

𝑑𝑥

𝑑𝑡= 1 mile ⋅ 4 ⋅ 10𝜋

radians

minute

= 40𝜋miles

minute

= 40𝜋miles

minute 60

minutes

hour

= 40 60 𝜋 miles

hour

≈ 7539 miles

hour ■ STOP

21a 21b

§2.8 Linear Approximations and Differentials

Linear or Tangent Line Approximation

0 … 𝑎 … 𝑥-, 0 … 𝑦-, 𝑦 = 𝑓(𝑥), 𝑃 𝑎, 𝑓 𝑎

Point-slope form of the tangent line

𝑦 = 𝑦0 = 𝑚(𝑥 − 𝑥0)

Point 𝑥0, 𝑦0 = 𝑎, 𝑓 𝑎

Slope 𝑚 = 𝑓 ′ 𝑎

Then

𝑦 − 𝑓 𝑎 = 𝑓 ′ 𝑎 𝑥 − 𝑎 or

𝑦 = 𝐿 𝑥 = 𝑓 𝑎 + 𝑓 ′ 𝑎 (𝑥 − 𝑎)

𝐿(𝑥) approximates 𝑓(𝑥) near 𝑥 = 𝑎

Example. Find the linear approximation to

𝑓 𝑥 = 1 + 𝑥 near 𝑥 = 0.

𝑓 𝑥 = 1 + 𝑥 1

2

𝑓 ′ 𝑥 =1

2 1 + 𝑥 −

1

2

𝑓 ′ 0 =1

2

𝐿 𝑥 = 𝑓 0 + 𝑓 ′ 0 𝑥 − 0

= 1 +1

2𝑥

this straight line approximates 1 + 𝑥 near 𝑥 = 0.

−1 … 0 … 1 … 𝑥-, 0 … 1 … 𝑦-, 𝑓(𝑥), 𝐿(𝑥)

■

22a 22b

Example (continued). Use linear approximation to

estimate 1.1.

1.1 = 1 + 0.1 = 𝑓(0.1)

𝐿 0.1 = 1 +1

2 0.1 = 1.05

Exact value 1.1 = 10.488 …■

Example (continued). For what values of 𝑥 is the

linear approximation

1 + 𝑥 ≈ 1 +1

2𝑥

accurate to within 0.05? This means

𝐿 𝑥 − 𝑓 𝑥 < 0.05

or

𝐿 𝑥 − 0.05 < 𝑓 𝑥 < 𝐿 𝑥 + 0.05

show Maple output

From the plot, the linear approximation is accurate to

within 0.05 for

−0.54 < 𝑥 < 0.74■

Realistic Example. Estimate the amount of paint

required to paint a spherical tank of radius 20 feet

with a coat 0.01 inches thick.

Exact calculation: volume of sphere

𝑉 𝑟 =4

3 𝜋𝑟3

volume between two concentric spheres of radii 𝑟

and 𝑎

Δ𝑉 = 𝑉 𝑟 − 𝑉 𝑎 =4

3 𝜋 𝑟3 −

4

3 𝜋 𝑎3

𝑎 = 20feet = 240 inches

𝑟 = 240.01 inches

Δ𝑉 ≈ 7238.53 cubic inches ≈ 31 gallons

1 gallon = 231 cubic inches

23a 23b

Approximate calculation:

Linear approximation, usual notation

𝑓 𝑥 ≈ 𝑓 𝑎 + 𝑓 ′ (𝑎)(𝑥 − 𝑎)

Use 𝑉(𝑟) instead of 𝑓(𝑥)

𝑉 𝑟 ≈ 𝑉 𝑎 + 𝑉 ′ 𝑎 (𝑥 − 𝑎)

Δ𝑉 = 𝑉 𝑟 − 𝑉 𝑎 ≈ 𝑉 ′ 𝑎 (𝑟 − 𝑎)

𝑉 ′ 𝑟 =𝑑

𝑑𝑟

4

3 𝜋 𝑟3 =

4

3 𝜋 3 𝑟2 = 4 𝜋 𝑟2

and

𝑉 ′ 𝑎 = 4𝜋𝑎2

then

Δ𝑉 = 4 𝜋𝑎2 𝑟 − 𝑎

= surface area of sphere × thickness of paint

= 4 𝜋 240 inches 2 (0.01 inches)

≈ 7237.23 cubic inches

the error is only 0.3 cubic inches! ■

STOP Differentials

0 … 𝑥-, 0 … 𝑦-, 𝑓, 𝑃, 𝐿, 𝑥, 𝑥 + Δ𝑥

tangent line approximation

𝐿 𝑥 + Δ𝑥 = 𝑓 𝑥 + 𝑓 ′ 𝑥 Δ𝑥

write

𝑑𝑥 = Δ𝑥

𝑑𝑦 = 𝐿 𝑥 + Δ𝑥 − 𝑓(𝑥)

= 𝑓 ′ 𝑥 𝑑𝑥

add 𝑑𝑥 = Δ𝑥, Δ𝑦, 𝑑𝑦 = 𝑓 ′ 𝑥 𝑑𝑥

Δ𝑦 = 𝑓 𝑥 + Δ𝑥 − 𝑓(𝑥)

is approximated by

𝑑𝑦= 𝑓 ′ 𝑥 𝑑𝑥

Note that 𝑟𝑖𝑠𝑒

𝑟𝑢𝑛=

𝑑𝑦

𝑑𝑥= 𝑓 ′ (𝑥). This is clever notation.

24a 24b

Example. Estimate the amount of paint required to

paint a spherical tank of radius 20 feet with a coat of

paint 0.01 inches thick.

Volume of sphere

𝑉 𝑟 =4

3 𝜋 𝑟3

approximate amount of paint

𝑑𝑉 = 𝑉 ′ 𝑟 𝑑𝑟 = 4 𝜋 𝑟2 𝑑𝑟

if 𝑑𝑟 = 0.01 inches, then

𝑑𝑉 = 4 𝜋 240 inches 2 (0.01 inches)

the same result as above. This is a shorthand for

linear approximation. ■

Newton’s Method – An Application of Linear

Approximation

0 … 𝑟 … 𝑥-, 0 … 𝑦-, 𝑓

A root of 𝑓 is a number 𝑟 such that 𝑓 𝑟 = 0.

How to find roots geometrically?

guess 𝑥1add

let 𝐿1 be the line tangent to 𝑓 at 𝑥1add

follow 𝐿1 to the 𝑥 axis, get an intersection at 𝑥2add

repeat this process

if the initial guess was good, the sequence 𝑥1 , 𝑥2, …

rapidly converges to 𝑟

How to find roots algebraically?

Make an initial guess 𝑥1.

Find the line tangent to 𝑓 at 𝑥 = 𝑥1.

𝐿1 𝑥 = 𝑓 𝑥1 + 𝑓′(𝑥1)(𝑥 − 𝑥1) by the linear or

tangent line approximation!

𝐿1 intersects the 𝑥 axis at 𝑥2. Find 𝑥2.

𝐿1 𝑥2 = 0 = 𝑓 𝑥1 + 𝑓′(𝑥1)(𝑥2 − 𝑥1)

−𝑓 𝑥1 = 𝑓′(𝑥1)(𝑥2 − 𝑥1)

25a 25b

−𝑓(𝑥1)

𝑓′(𝑥1)= 𝑥2 − 𝑥1

𝑥2 = 𝑥1 −𝑓(𝑥1)

𝑓′(𝑥1)

Find the line tangent to 𝑓 at 𝑥 = 𝑥2.

𝐿2 𝑥 = 𝑓 𝑥2 + 𝑓′(𝑥2)(𝑥 − 𝑥2)

𝐿2 intersects the 𝑥 axis at 𝑥3. Find 𝑥3.

𝐿2 𝑥3 = 0 = 𝑓 𝑥2 + 𝑓′(𝑥2)(𝑥3 − 𝑥2)

−𝑓 𝑥2 = 𝑓′(𝑥2)(𝑥3 − 𝑥2)

−𝑓(𝑥2)

𝑓′(𝑥2)= 𝑥3 − 𝑥2

𝑥3 = 𝑥2 −𝑓(𝑥2)

𝑓′(𝑥2)

repeat this procedure to obtain a general formula for

𝑛 = 1, 2, …

𝑥𝑛+1

= 𝑥𝑛 −𝑓(𝑥𝑛)

𝑓′(𝑥𝑛)

If the initial guess 𝑥1 was good, the iterates 𝑥2, 𝑥3, …

will quickly approach 𝑟.

Example. 𝑓 𝑥 = 𝑥2 − 2

−2 … − 1 … 0 … 1 … 2 … 𝑥-, −2 … 0 … 2 … 𝑦-,

𝑓, 𝑟 = 2

𝑓 ′ 𝑥 = 2𝑥

from the boxed formula above

𝑥𝑛+1 = 𝑥𝑛 −𝑥𝑛

2 − 2

2 𝑥𝑛

guess 𝑥1 = 2

𝑥2 = 2 −4 − 2

4= 2 −

1

2=

3

2

𝑥3 =3

2−

9

4− 2

3=

3

2−

1

4

3=

3

2−

1

12=

18 − 1

12=

17

12= 1.4167 …

Exact value of positive root

26a 26b

𝑟 = 2 = 1.4142 …

Transparency – How Newton’s Method can go

wrong!