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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
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