distance s r
θ
time t = 0 time t > 0
Figure 4.4.1 Radian measure and arc length can be applied to the study
of circular motion. In physics the average speed of an object is defined as:
average speed = distance traveled time elapsed
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 4.4.1. Then it makes sense to define the (average) linear speedνof the object as:
ν = s
t (4.8)
Let θ be the angle swept out by the object in that period of time. Then we define the (average) angular speedωof the object as:
ω = θ
t (4.9)
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 angleθin a circle of radius r is s = rθ, we see that
ν = s t = rθ
t = θ t · r ,
so that we get the following relation between linear and angular speed:
ν = ωr (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 ofπ3 radians in 0.5 seconds as it moves along a circle of radius 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 =π3 rad. So the angular speed ω is
ω = θ t =
π 3 rad
0.5 sec ⇒ ω = 2π
3 rad/sec , and thus the linear speed ν is
ν = ω r = µ2π
3 rad/sec
¶
(3 m) ⇒ ν = 2π m/sec .
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 ν = s
t = 35 feet
2.7 sec ⇒ ν = 12.96 ft/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 = r θ ⇒ θ = s
r = ν t
r = (10 m/sec) (3.1 sec)
4 m = 7.75 rad .
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 2π 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:
Nrpm = N ✘✘rev
✟✟✟
min · 2πrad
1✘✘rev · 1✟✟✟ min
60 sec = N · 2π
60 rad/sec
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 rad/sec = N · 60 2π rpm
102 Chapter 4 • Radian Measure §4.4
Example 4.17
A gear with an outer radius of r1= 5 cm moves in the clockwise direction, causing an interlocking gear with an outer radius of r2= 4 cm to move in the counterclockwise direction at an angular speed of ω2= 25 rpm. What is the angular speed ω1of the larger gear?
r1= 5 cm
r2= 4 cm ω2= 25 rpm
Figure 4.4.2 Solution: Imagine a particle on the outer radius of each gear.
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 words, s1= s2, where s1and s2are the distances traveled by the particles on the gears with radii r1and r2, respectively.
But s1= ν1t and s2= ν2t, where ν1 and ν2 are the linear speeds of the gears with radii r1and r2, respectively. Thus,
ν1t = ν2t ⇒ ν1= ν2,
so by formula (4.10) we get the fundamental relation between the two gears:
ω1r1 = ω2r2 (4.11)
Note that this holds for any two gears. So in our case, we have ω1(5) = (25)(4) ⇒ ω1 = 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 r1and r2are replaced by the number of teeth N1and N2, 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 =νr2. Show that α = ω2r, 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
x The first function we will graph is the sine
func-tion. We will describe a geometrical way to create the graph, using the unit circle. This is the circle of radius 1 in the x y-plane consisting of all points (x, y) which satisfy the equation x2+ y2= 1.
We see in Figure 5.1.1 that any point on the unit 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 circle, its y-coordinate is sinθ.
We thus get a correspondence between the y-coordinates of points on the unit circle and the values f (θ) = sinθ, as shown by the horizontal lines from the unit circle to the graph of
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 2πradians, 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
Figure 5.1.3 Unit circle definition of the sine function
Since the trigonometric functions repeat every 2πradians (360◦), we get, for example, the following graph of the function y = sin x for x in the interval [−2π, 2π]:
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. Recall from Section 1.5 that cos x = sin (x + 90◦) for all x. So cos 0◦ has the same value as sin 90◦, cos 90◦ has the same value as sin 180◦, cos 180◦has the same value as sin 270◦, and so on. In other words, the graph of the cosine function is just the graph of the sine function shifted to the left by 90◦=π/2 radians, as in Figure 5.1.5:
To graph the tangent function, use tan x =sin xcos x to get the following graph:
Graphing the Trigonometric Functions • Section 5.1 105
x y
0
−8
−6
−4
−2 2 4 6 8
π
4 π
2 3π
4
π 5π 4
3π 2
7π
4 2π
−π4
−π2
−3π4
−5π4 −π
−3π2
−7π4
−2π
y = tan x
Figure 5.1.6 Graph of y = tan x
Recall that the tangent is positive for angles in QI and QIII, and is negative in QII and QIV, and that is indeed what the graph in Figure 5.1.6 shows. We know that tan x is not defined when cos x = 0, i.e. at odd multiples of π2: x = ±π2, ±3π2 , ±5π2 , etc. We can figure out what happens near those angles by looking at the sine and cosine functions. For example, for x in QI near π2, sin x and cos x are both positive, with sin x very close to 1 and cos x very close to 0, so the quotient tan x =sin xcos x is a positive number that is very large. And the closer xgets toπ2, the larger tan x gets. Thus, x =π2 is a vertical asymptote of the graph of y = tan x.
Likewise, for x in QII very close to π2, sin x is very close to 1 and cos x is negative and very close to 0, so the quotient tan x =cos xsin x is a negative number that is very large, and it gets larger in the negative direction the closer x gets to π2. The graph shows this. Similarly, we get vertical asymptotes at x = −π2, x = 3π2 , and x = −3π2 , as in Figure 5.1.6. Notice that the graph of the tangent function repeats everyπradians, i.e. two times faster than the graphs of sine and cosine repeat.
The graphs of the remaining trigonometric functions can be determined by looking at the graphs of their reciprocal functions. For example, using csc x =sin x1 we can just look at the graph of y = sin x and invert the values. We will get vertical asymptotes when sin x = 0, namely at multiples ofπ: x = 0, ±π, ±2π, etc. Figure 5.1.7 shows the graph of y = csc x, with the graph of y = sin x (the dashed curve) for reference.
106 Chapter 5 • Graphing and Inverse Functions §5.1
Likewise, Figure 5.1.8 shows the graph of y = sec x, with the graph of y = cos x (the dashed curve) for reference. Note the vertical asymptotes at x = ±π2, ±3π2 . Notice also that the graph is just the graph of the cosecant function shifted to the left by π2 radians.
x
Graphing the Trigonometric Functions • Section 5.1 107
The graph of y = cot x can also be determined by using cot x =cot x1 . Alternatively, we can use the relation cot x = −tan (x + 90◦) from Section 1.5, so that the graph of the cotangent function is just the graph of the tangent function shifted to the left by π2 radians and then reflected about the x-axis, as in Figure 5.1.9:
x
Solution: Multiplying a function by −1 just reflects its graph around the x-axis. So reflecting the graph of y = sin x around the x-axis gives us the graph of y = −sin x:
108 Chapter 5 • Graphing and Inverse Functions §5.1
It is worthwhile to remember the general shapes of the graphs of the six trigonometric functions, especially for sine, cosine, and tangent. In particular, the graphs of the sine and cosine functions are called sinusoidal curves. Many phenomena in nature exhibit sinusoidal behavior, so recognizing the general shape is important.
Example 5.2
Draw the graph of y = 1 + cos x for 0 ≤ x ≤ 2π.
Solution: Adding a constant to a function just moves its graph up or down by that amount, depend-ing on whether the constant is positive or negative, respectively. So adddepend-ing 1 to cos x moves the graph of y = cos x upward by 1, giving us the graph of y = 1 + cos x:
For Exercises 1-12, draw the graph of the given function for 0 ≤ x ≤ 2π.
1. y = −cos x 2. y = 1 + sin x 3. y = 2 − cos x 4. y = 2 − sin x 13. We can extend the unit circle definition of the sine and cosine
functions to all six trigonometric functions. Let P be a point in QI on the unit circle, so that the line segment OP in Figure 5.1.10 has length 1 and makes an acute angle θ with the positive x-axis.
Identify each of the six trigonometric functions of θ with exactly one of the line segments in Figure 5.1.10, keeping in mind that the radius of the circle is 1. To get you started, we have sin θ = MP (why?).
14. For Exercise 13, how would you draw the line segments in Fig-ure 5.1.10 if θ was in QII? Recall that some of the trigonometric functions are negative in QII, so you will have to come up with a convention for how to treat some of the line segment lengths as negative.
15. For any point (x, y) on the unit circle and any angle α, show that the point Rα(x, y) defined by Rα(x, y) = (x cos α − y sin α, x sin α + y cos α) is also on the unit circle. What is the geometric interpretation of Rα(x, y)? Also, show that R−α(Rα(x, y)) = (x, y) and Rβ(Rα(x, y)) = Rα+β(x, y).
Properties of Graphs of Trigonometric Functions • Section 5.2 109