α
β
A
B
7
(a) ∠ BAC = 1
(b) Triangle △ ABC
2 ∠ C AD, ∠ ABC = 1
2 ∠ CBD
Figure 4.3.6
By symmetry, we see that ∠ BAC = 12 ∠ CAD and ∠ ABC = 12 ∠ CBD. So let α = ∠ BAC and β =
∠ ABC, as in Figure 4.3.6(b). By the Law of Cosines, we have
72 + 52 − 42
cos α =
= 0.8286
⇒
α = 0.594 rad
⇒
∠ C AD = 2(0.594) = 1.188 rad
2 (7) (5)
72 + 42 − 52
cos β =
= 0.7143
⇒
β = 0.775 rad
⇒
∠ CBD = 2(0.775) = 1.550 rad
2 (7) (4)
Thus, the area K is
K = (Area of segment CD in circle at A) + (Area of segment CD in circle at B)
= 1 (5)2 (1.188
(4)2 (1.550
2
− sin 1.188) + 12
− sin 1.550)
= 7.656 cm2 .
Exercises
For Exercises 1-3, find the area of the sector for the given angle θ and radius r.
1. θ = 2.1 rad, r = 1.2 cm
2. θ = 3 π rad, r
7
= 3.5 ft
3. θ = 78◦, r = 6 m
4. The centers of two belt pulleys, with radii of 3 cm and 6 cm, respectively, are 13 cm apart. Find
the total area K enclosed by the belt.
5. In Exercise 4 suppose that both belt pulleys have the same radius of 6 cm. Find the total area K
enclosed by the belt.
Area of a Sector • Section 4.3
99
6. Find the area enclosed by the figure eight in Exercise 8 from Section 4.2.
For Exercises 7-9, find the area of the sector for the given radius r and arc length s.
7. r = 5 cm, s = 2 cm
8. r = a, s = a
9. r = 1 cm, s = π cm
For Exercises 10-12, find the area of the segment formed by a chord of length a in a circle of radius r.
10. a = 4 cm, r = 4 cm
11. a = 1 cm, r = 5 cm
12. a = 2 cm, r = 5 cm
13. Find the area of the shaded region in Figure 4.3.7.
4
7
8
3
9
5
6
6
5
Figure 4.3.7 Exercise 13
Figure 4.3.8 Exercise 14
Figure 4.3.9 Exercise 15
14. Find the area of the shaded region in Figure 4.3.8. ( Hint: Draw two central angles. )
15. Find the area of the shaded region in Figure 4.3.9.
16. The centers of two circles are 4 cm apart, with one circle having a radius of 3 cm and the other a
radius of 2 cm. Find the area of their intersection.
17. Three circles with radii of 4 m, 2 m, and 1 m are externally tangent to each other. Find the area
of the curved region between the circles, as in Figure 4.3.10. ( Hint: Connect the centers of the
circles. )
1
r
4
2
r
r
Figure 4.3.10
Exercise 17
Figure 4.3.11
Exercise 18
18. Show that the total area enclosed by the loop around the three circles of radius r in Figure 4.3.11
is ( π + 6 + 3) r 2.
19. For a fixed central angle θ, how much does the area of its sector increase when the radius of the
circle is doubled? How much does the length of its intercepted arc increase?
100
Chapter 4 • Radian Measure
§4.4
4.4 Circular Motion: Linear and Angular Speed
Radian measure and arc length can be applied to the study
time t > 0
of circular motion. In physics the average speed of an object
is defined as:
distance s
θ
distance traveled
average speed =
r
time elapsed
time t = 0
So suppose that an object moves along a circle of radius r,
traveling a distance s over a period of time t, as in Figure
Figure 4.4.1
4.4.1. Then it makes sense to define the (average) linear
speed ν of the object as:
s
ν =
(4.8)
t
Let θ be the angle swept out by the object in that period of time. Then we define the
(average) angular speed ω of the object as:
θ
ω =
(4.9)
t
Angular speed gives the rate at which the central angle swept out by the object changes as
the object moves around the circle, and it is thus measured in radians per unit time. Linear
speed is measured in distance units per unit time (e.g. feet per second). The word linear is
used because straightening out the arc traveled by the object along the circle results in a line
of the same length, so that the usual definition of speed as distance over time can be used.
We will usually omit the word average when discussing linear and angular speed here.4
Since the length s of the arc cut off by a central angle θ in a circle of radius r is s = r θ, we see that
s
r θ
θ
ν =
=
=
· r ,
t
t
t
so that we get the following relation between linear and angular speed:
ν = ω r
(4.10)
4Many trigonometry texts assume uniform motion, i.e. constant speeds. We do not make that assumption. Also,
many texts use the word velocity instead of speed. Technically they are not the same; velocity has a direction
and a magnitude, whereas speed is just a magnitude.
Circular Motion: Linear and Angular Speed • Section 4.4
101
Example 4.14
An object sweeps out a central angle of π radians in 0.5 seconds as it moves along a circle of radius 3
3
m. Find its linear and angular speed over that time period.
Solution: Here we have t = 0.5 sec, r = 3 m, and θ = π rad. So the angular speed ω is
3
π rad
θ
3
2 π
ω =
=
⇒
ω =
rad/sec ,
t
0.5 sec
3
and thus the linear speed ν is
2 π
ν = ω r =
rad/sec (3 m)
⇒
ν = 2 π m/sec .
3
Note that the units for ω are rad/sec and the units of ν are m/sec. Recall that radians are actually
unitless, which is why in the formula ν = ω r the radian units disappear.
Example 4.15
An object travels a distance of 35 ft in 2.7 seconds as it moves along a circle of radius 2 ft. Find its
linear and angular speed over that time period.
Solution: Here we have t = 2.7 sec, r = 2 ft, and s = 35 ft. So the linear speed ν is
s
35 feet
ν =
=
⇒
ν = 12.96 ft/sec ,
t
2.7 sec
and thus the angular speed ω is given by
ν = ω r
⇒
12.96 ft/sec = ω(2 ft)
⇒
ω = 6.48 rad/sec .
Example 4.16
An object moves at a constant linear speed of 10 m/sec around a circle of radius 4 m. How large of a
central angle does it sweep out in 3.1 seconds?
Solution: Here we have t = 3.1 sec, ν = 10 m/sec, and r = 4 m. Thus, the angle θ is given by
s
ν t
(10 m/sec) (3.1 sec)
s = r θ
⇒
θ =
=
=
= 7.75 rad .
r
r
4 m
In many physical applications angular speed is given in revolutions per minute, abbrevi-
ated as rpm. To convert from rpm to, say, radians per second, notice that since there are 2 π
radians in one revolution and 60 seconds in one minute, we can convert N rpm to radians
per second by “canceling the units” as follows:
✘✘
rev
2 π rad
1 ✟✟
min
N · 2 π
N rpm = N
·
·
=
rad/sec
✟✟
min
1 ✘✘
rev
60 sec
60
This works because all we did was multiply by 1 twice. Converting to other units for angular
speed works in a similar way. Going in the opposite direction, say, from rad/sec to rpm, gives:
N · 60
N rad/sec =
rpm
2 π
102
Chapter 4 • Radian Measure
§4.4
Example 4.17
A gear with an outer radius of r 1 = 5 cm moves in the clockwise direction, causing an interlocking
gear with an outer radius of r 2 = 4 cm to move in the counterclockwise direction at an angular speed
of ω 2 = 25 rpm. What is the angular speed ω 1 of the larger gear?
Solution: Imagine a particle on the outer radius of each gear.
r 1 = 5 cm
After the gears have rotated for a period of time t > 0, the cir-
cular displacement of each particle will be the same. In other
ω 2 = 25 rpm
r 2
words, s
= 4 cm
1 = s 2, where s 1 and s 2 are the distances traveled by
the particles on the gears with radii r 1 and r 2, respectively.
But s 1 = ν 1 t and s 2 = ν 2 t, where ν 1 and ν 2 are the linear
speeds of the gears with radii r 1 and r 2, respectively. Thus,
ν 1 t = ν 2 t
⇒
ν 1 = ν 2 ,
so by formula (4.10) we get the fundamental relation between
the two gears:
Figure 4.4.2
ω 1 r 1 = ω 2 r 2
(4.11)
Note that this holds for any two gears. So in our case, we have
ω 1 (5) = (25)(4)
⇒
ω 1 = 20 rpm .
Exercises
For Exercises 1-6, assume that a particle moves along a circle of radius r for a period of time t. Given
either the arc length s or the central angle θ swept out by the particle, find the linear and angular
speed of the particle.
1. r = 4 m, t = 2 sec, θ = 3 rad
2. r = 8 m, t = 2 sec, θ = 3 rad
3. r = 7 m, t = 3.2 sec, θ = 172◦
4. r = 1 m, t = 1.6 sec, s = 3 m
5. r = 2 m, t = 1.6 sec, s = 6 m
6. r = 1.5 ft, t = 0.3 sec, s = 4 in
7. An object moves at a constant linear speed of 6 m/sec around a circle of radius 3.2 m. How large
of a central angle does it sweep out in 1.8 seconds?
8. Two interlocking gears have outer radii of 6 cm and 9 cm, respectively. If the smaller gear rotates
at 40 rpm, how fast does the larger gear rotate?
9. Three interlocking gears have outer radii of 2 cm, 3 cm, and 4 cm, respectively. If the largest gear
rotates at 16 rpm, how fast do the other gears rotate?
10. In Example 4.17, does equation (4.11) still hold if the radii r 1 and r 2 are replaced by the number of teeth N 1 and N 2, respectively, of the two gears as shown in Figure 4.4.2?
11. A 78 rpm music record has a diameter of 10 inches. What is the linear speed of a speck of dust
on the outer edge of the record in inches per second?
12. The centripetal acceleration α of an object moving along a circle of radius r with a linear speed ν
is defined as α = ν 2 . Show that α
r
= ω 2 r, where ω is the angular speed.
5 Graphing and Inverse Functions
The trigonometric functions can be graphed just like any other function, as we will now
show. In the graphs we will always use radians for the angle measure.
5.1 Graphing the Trigonometric Functions
The first function we will graph is the sine func-
y
tion. We will describe a geometrical way to create the
x 2 + y 2 = 1
( x, y) = (cos θ,sin θ)
graph, using the unit circle. This is the circle of radius
1
1 in the x y-plane consisting of all points ( x, y) which
s = r θ = θ
satisfy the equation x 2 + y 2 = 1.
θ
x
We see in Figure 5.1.1 that any point on the unit
0
1
circle has coordinates ( x, y) = (cos θ,sin θ), where θ
is the angle that the line segment from the origin to
( x, y) makes with the positive x-axis (by definition of
sine and cosine). So as the point ( x, y) goes around the
Figure 5.1.1
circle, its y-coordinate is sin θ.
We thus get a correspondence between the y-coordinates of points on the unit circle and
the values f ( θ) = sin θ, as shown by the horizontal lines from the unit circle to the graph of
f ( θ) = sin θ in Figure 5.1.2 for the angles θ = 0, π , π , π .
6
3
2
f ( θ)
1
1
π
2
π
f ( θ) = sin θ
3
π
6
θ
0
θ
1
0
π
π
π
2 π
5 π
π
x 2
6
3
2
+ y 2 = 1
3
6
Figure 5.1.2
Graph of sine function based on y-coordinate of points on unit circle
We can extend the above picture to include angles from 0 to 2 π radians, as in Figure 5.1.3.
This illustrates what is sometimes called the unit circle definition of the sine function.
103
104
Chapter 5 • Graphing and Inverse Functions
§5.1
y
f ( θ)
f ( θ) = sin θ
1
θ
θ
x
1
0
π
π
π
2 π
5 π
π
5 π
3 π
7 π
2 π
6
3
2
3
6
4
2
4
−1
x 2 + y 2 = 1
Figure 5.1.3
Unit circle definition of the sine function
Since the trigonometric functions repeat every 2 π radians (360◦), we get, for example, the
following graph of the function y = sin x for x in the interval [−2 π,2 π]:
y
y = sin x
1
x
0
π
π
−2 π
− π
3 π
π
5 π
3 π
7 π
2 π
− 7 π
4
− 3 π
2
− 5 π
4
− 3 π − π 2
− π 4
4
2
4
4
4
2
4
−1
Figure 5.1.4
Graph of y = sin x
To graph the cosine function, we could again use the unit circle idea (using the x-coordinate
of a point that moves around the circle), but there is an easier way. R