Download Chapter 3.5: Derivatives of Trigonometric Functions

Derivatives of Trigonometric
Functions
Chapter 3.5
Proving that
sin 𝑥
lim
𝑥→0 𝑥
=1
• In section 2.1 you used a table of values approaching 0 from the left
sin 𝑥
and right that lim
= 1; but that was not a proof
𝑥→0
𝑥
• Because you will need to know this limit (and a related one for
cosine), we will begin this section by proving this through geometry
• We will need the following
• The Squeeze Theorem
• The formula for the area of a sector of a circle
2
Proving that
sin 𝑥
lim
𝑥→0 𝑥
=1
• The Squeeze Theorem says that if 𝑓, 𝑔, and ℎ are functions so that, for
𝑥-values in an open interval around some number 𝑐
if 𝑔 𝑥 ≤ 𝑓 𝑥 ≤ ℎ 𝑥
and
lim 𝑔(𝑥) = lim ℎ(𝑥) = 𝐿
𝑥→𝑐
𝑥→𝑐
then
lim 𝑓(𝑥) = 𝐿
𝑥→𝑐
3
Proving that
sin 𝑥
lim
𝑥→0 𝑥
=1
• If a circle of radius 1 is subtended by an angle 𝜃 in radians (creating a
sector or “pie slice”), then the area of the sector is
𝜃
𝜃
2
𝜋𝑟 ⋅
=
2𝜋 2
4
Proving that
sin 𝑥
lim
𝑥→0 𝑥
=1
5
Proving that
sin 𝑥
lim
𝑥→0 𝑥
=1
sin 𝜃 𝜃 tan 𝜃
≤ ≤
2
2
2
• We can simplify by multiplying through by 2; next divide through by
sin 𝜃 (we can do this without changing inequality signs; why?)
𝜃
1
1≤
≤
sin 𝜃 cos 𝜃
1
1
• For positive numbers 𝑥 and 𝑦, if 𝑥 ≥ 𝑦 then ≤
𝑥
1
3
• For example, 3 ≥ 2, but ≤
1
;
2
𝑦
we use this to take the reciprocals
6
Proving that
sin 𝑥
lim
𝑥→0 𝑥
=1
𝜃
1
1≤
≤
sin 𝜃 cos 𝜃
sin 𝜃
1≥
≥ cos 𝜃
𝜃
sin 𝜃
cos 𝜃 ≤
≤1
𝜃
sin 𝜃
• This brings in
and also puts it between two functions within the
𝜃
first quadrant; we are now able to use the Squeeze Theorem
7
Proving that
Since
sin 𝑥
lim
𝑥→0 𝑥
=1
sin 𝜃
cos 𝜃 ≤
≤1
𝜃
and since
lim+ cos 𝜃 = lim+ 1 = 1
𝜃→0
𝜃→0
then by the Squeeze Theorem
sin 𝜃
lim+
=1
𝜃→0
𝜃
8
Proving that
sin 𝑥
lim
𝑥→0 𝑥
=1
Since sine is an odd function, then
sin −𝜃
− sin 𝜃 sin 𝜃
=
=
−𝜃
−𝜃
𝜃
We can conclude that
sin 𝜃
sin 𝜃
lim−
= lim+
=1
𝜃→0
𝜃→0
𝜃
𝜃
Hence,
sin 𝜃
lim
=1
𝜃→0 𝜃
QED
9
Proving that
1−cos 𝑥
lim
𝑥
𝑥→0
=0
We can use the previous result to prove another special limit,
specifically
1 − cos 𝑥
lim
=0
𝑥→0
𝑥
We have
1 − cos 𝑥 1 + cos 𝑥
1 − cos2 𝑥
lim
⋅
= lim
𝑥→0
𝑥
1 + cos 𝑥 𝑥→0 𝑥 1 + cos 𝑥
We want to be able to use the previous theorem. Note that the
numerator is just sin2 𝑥
10
Proving that
1−cos 𝑥
lim
𝑥
𝑥→0
=0
1 − cos 𝑥 1 + cos 𝑥
1 − cos2 𝑥
lim
⋅
= lim
𝑥→0
𝑥
1 + cos 𝑥 𝑥→0 𝑥 1 + cos 𝑥
sin2 𝑥
= lim
𝑥→0 𝑥 1 + cos 𝑥
Now use the limit properties to get
sin 𝑥
sin 𝑥
0
= lim
⋅ lim
=1⋅
=0
𝑥→0 𝑥
𝑥→0 1 + cos 𝑥
1+1
11
Derivative of the Sine and Cosine Functions
sin 𝑥
lim
𝑥→0 𝑥
1−cos 𝑥
lim
𝑥
𝑥→0
• Having proved that
= 1 and that
= 0, we can now
use the derivative definition to determine the derivative of the sine
function and the cosine function
• We will also need to be reminded of the following trigonometric
identity:
sin 𝐴 + 𝐵 = sin 𝐴 cos 𝐵 + sin 𝐵 cos 𝐴
12
Derivative of the Sine Function
THEOREM:
If 𝑓 𝑥 = sin 𝑥, then 𝑓 ′ 𝑥 = cos 𝑥
13
Derivative of the Sine Function
THEOREM:
PROOF
Using the definition of the derivative:
sin 𝑥 + ℎ − sin 𝑥
𝑓 𝑥 = lim
ℎ→0
ℎ
Use the previous identity to expand the numerator:
′
sin 𝑥 cos ℎ + sin ℎ cos 𝑥 − sin 𝑥
= lim
ℎ→0
ℎ
14
Derivative of the Sine Function
THEOREM:
PROOF
sin 𝑥 cos ℎ + sin ℎ cos 𝑥 − sin 𝑥
= lim
ℎ→0
ℎ
We can factor sin 𝑥
sin 𝑥 cos ℎ − 1 + sin ℎ cos 𝑥
= lim
ℎ→0
ℎ
Now use the limit properties
15
Derivative of the Sine Function
THEOREM:
PROOF
sin 𝑥 cos ℎ − 1 + sin ℎ cos 𝑥
= lim
ℎ→0
ℎ
− 1 − cos ℎ
sin ℎ
= lim sin 𝑥 ⋅ lim
+ lim cos 𝑥 ⋅ lim
ℎ→0
ℎ→0
ℎ→0
ℎ→0 ℎ
ℎ
= sin 𝑥 ⋅ 0 + cos 𝑥 ⋅ 1 = cos 𝑥
QED
16
Derivative of the Cosine Function
THEOREM:
If 𝑓 𝑥 = cos 𝑥, then 𝑓 ′ 𝑥 = − sin 𝑥
17
Derivative of the Cosine Function
THEOREM:
PROOF
We will need the angle sum identity for cosine which is
cos(𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
Use the definition of the derivative:
𝑓′
cos 𝑥 + ℎ − cos 𝑥
𝑥 = lim
ℎ→0
ℎ
Expand the numerator:
18
Derivative of the Cosine Function
THEOREM:
PROOF
𝑓′
cos 𝑥 + ℎ − cos 𝑥
𝑥 = lim
ℎ→0
ℎ
cos 𝑥 cos ℎ − sin 𝑥 sin ℎ − cos 𝑥
= lim
ℎ→0
ℎ
Factor out cos 𝑥
cos 𝑥 cos ℎ − 1 − sin 𝑥 sin ℎ
= lim
ℎ→0
ℎ
19
Derivative of the Cosine Function
THEOREM:
PROOF
cos 𝑥 cos ℎ − 1 − sin 𝑥 sin ℎ
= lim
ℎ→0
ℎ
Use the limit properties
− cos ℎ − 1
sin ℎ
= lim cos 𝑥 ⋅ lim
− lim sin 𝑥 ⋅ lim
ℎ→0
ℎ→0
ℎ→0
ℎ→0 ℎ
ℎ
= cos 𝑥 ⋅ 0 − sin 𝑥 ⋅ 1 = − sin 𝑥
QED
20
Example 1: Using the Derivatives of Sine and
Cosine Functions
Find the derivatives of
a) 𝑦 = 𝑥 2 sin 𝑥
b) 𝑢 =
cos 𝑥
1−sin 𝑥
21
Example 1: Using the Derivatives of Sine and
Cosine Functions
Find the derivatives of
a) 𝑦 = 𝑥 2 sin 𝑥
Use the Product Rule:
𝑑𝑦
𝑑
𝑑 2
2
=𝑥 ⋅
sin 𝑥 + sin 𝑥 ⋅
𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥
= 𝑥 2 cos 𝑥 + 2𝑥 sin 𝑥
22
Example 1: Using the Derivatives of Sine and
Cosine Functions
Find the derivatives of
cos 𝑥
b) 𝑢 =
1−sin 𝑥
Use the Quotient Rule
𝑑𝑢
=
𝑑𝑥
1 − sin 𝑥 ⋅
𝑑
𝑑
cos 𝑥 − cos 𝑥 ⋅
1 − sin 𝑥
𝑑𝑥
𝑑𝑥
1 − sin 𝑥 2
− sin 𝑥 1 − sin 𝑥 − cos 𝑥 ⋅ − cos 𝑥 − sin 𝑥 + sin2 𝑥 + cos 2 𝑥
=
=
1 − sin 𝑥 2
1 − sin 𝑥 2
1 − sin 𝑥
1
=
=
2
1 − sin 𝑥
1 − sin 𝑥
23
Simple Harmonic Motion
24
Example 2: The Motion of a Weight on a
Spring
A weight hanging from a spring is stretched 5 units beyond its rest
position (𝑠 = 0) and released at time 𝑡 = 0 to bob up and down. Its
position at any later time 𝑡 is
𝑠 𝑡 = 5 cos 𝑡
What are its velocity and acceleration at time 𝑡? Describe its motion.
25
Example 2: The Motion of a Weight on a
Spring
A weight hanging from a spring is stretched 5 units beyond its rest
position (𝑠 = 0) and released at time 𝑡 = 0 to bob up and down. Its
position at any later time 𝑡 is
𝑠 𝑡 = 5 cos 𝑡
What are its velocity and acceleration at time 𝑡? Describe its motion.
Recall that velocity is the first derivative of position and acceleration is the second
derivative of position (or the first derivative of velocity).
𝑣 𝑡 = 𝑠 ′ 𝑡 = −5 sin 𝑡
𝑎 𝑡 = 𝑣 ′ 𝑡 = 𝑠 ′′ 𝑡 = −5 cos 𝑡
26
Example 2: The Motion of a Weight on a
Spring
Time
𝒔 𝒕 = 𝟓 cos 𝒕
𝒗 𝒕 = −𝟓 sin 𝒕
𝒂 𝒕 = −𝟓 cos 𝒕
0
𝜋
2
𝜋
3𝜋
2
2𝜋
5𝜋
2
3𝜋
7𝜋
2
4𝜋
5
0
−5
0
−5
0
−5
0
5
0
5
0
5
0
−5
0
−5
0
−5
0
5
0
5
0
5
0
−5
27
Jerk
• A sudden change in acceleration is called a jerk
• When a ride in a vehicle is jerky, it is caused by sudden changes in
acceleration
• Hence, jerk, sudden changes in acceleration, is the first derivative of
acceleration, or the third derivative of acceleration
28
Example 3: A Couple of Jerks
a) What is the jerk caused by the acceleration due to gravity?
b) What is the jerk of the simple harmonic motion from Example 2?
29
Example 3: A Couple of Jerks
a) What is the jerk caused by the acceleration due to gravity?
𝑑2 𝑠
𝑑𝑡 2
The acceleration due to gravity is constant (near the Earth’s surface):
= −32
feet per second per second. Since the derivative of a constant is zero, then
𝑑3𝑠
=0
3
𝑑𝑡
b) What is the jerk of the simple harmonic motion from Example 2?
𝑑𝑎
= 5 sin 𝑡
𝑑𝑡
Jerk is maximum when sin 𝑡 = ±1, which occurs when the weight is at the rest
position and acceleration changes sign
30
Derivatives of the Other Basic Trigonometric
Functions
• We can use the derivatives of sine and cosine as well as the Product
and/or Quotient Rules to determine the derivatives of the other basic
trigonometric functions
• Find
𝑑
[tan 𝑥]
𝑑𝑥
𝑑
𝑑 sin 𝑥
tan 𝑥 =
𝑑𝑥
𝑑𝑥 cos 𝑥
cos 𝑥 ⋅ cos 𝑥 − sin 𝑥 ⋅ − sin 𝑥
1
2
=
=
=
sec
𝑥
2
2
cos 𝑥
cos 𝑥
31
Derivatives of the Other Basic Trigonometric
Functions
• Find
•
𝑑
𝑑𝑥
cot 𝑥
•
𝑑
𝑑𝑥
sec 𝑥
•
𝑑
[csc 𝑥]
𝑑𝑥
32
Derivatives of the Other Basic Trigonometric
Functions
THEOREM:
•
𝑑
𝑑𝑥
tan 𝑥 = sec 2 𝑥
•
𝑑
𝑑𝑥
cot 𝑥 = − csc 2 𝑥
•
𝑑
𝑑𝑥
sec 𝑥 = tan 𝑥 sec 𝑥
•
𝑑
𝑑𝑥
csc 𝑥 = − cot 𝑥 csc 𝑥
33
Example 4: Finding Tangent and Normal
Lines
Find equations for the lines that are tangent and normal to the graph of
tan 𝑥
𝑓 𝑥 =
𝑥
at 𝑥 = 2.
34
Example 4: Finding Tangent and Normal
Lines
Find equations for the lines that are tangent and normal to the graph of
tan 𝑥
𝑓 𝑥 =
𝑥
at 𝑥 = 2.
2 𝑥 − tan 𝑥 ⋅ 1
2 𝑥 − tan 𝑥
𝑥
⋅
sec
𝑥
sec
𝑓′ 𝑥 =
=
2
𝑥
𝑥2
The slope of the tangent line at 𝑥 = 2 is 𝑓 ′ 2 ≈ 3.433 and of the normal line is
1
− ′ ≈ −0.291. The equations are 𝑦 ≈ 3.433 𝑥 − 2 − 1.093 and 𝑦 ≈
𝑓 2
− 0.291 𝑥 − 2 − 1.093
35
Example 5: A Trigonometric Second
Derivative
Find 𝑦′′ if 𝑦 = sec 𝑥.
36
Example 5: A Trigonometric Second
Derivative
Find 𝑦′′ if 𝑦 = sec 𝑥.
𝑦 ′ = tan 𝑥 sec 𝑥
𝑦 ′′ = tan 𝑥 ⋅ tan 𝑥 sec 𝑥 + sec 𝑥 ⋅ sec 2 𝑥
= tan2 𝑥 sec 𝑥 + sec 3 𝑥
37
Exercise 3.5
38