Chapter 10 Trigonometric functions - Ugrad.math.ubc.ca

Math 102 Notes Chapter 10
This implies (by the chain rule) that
Performing the required calculations, we have
y(t) = sin(θ(t)) = sin(ωt).
dx
dt = d cos(θ) dθ
dθ dt ,
dy
dt = d sin(θ) dθ
dθ dt .
dx
dt = − sin(θ)ω,
dy
dt = cos(θ)ω.
We will see some interesting consequences of this in a later section.
Example 2: runners on a circular track
Two runners start at the same position (call it x = 0) on a circular race track of length 400 meters.
Joe Runner takes 50 sec, while Michael Johnson takes 43.18 sec to complete the 400 meter race.
Determine the rate of change of the angle formed between the two runners and the center of the
track, assuming that the runners are running at a constant rate.
Solution:
The track is 400 meters in length (total). Joe completes one cycle around the track (2π radians) in
50 sec, while Michael completes a cycle in 43.18 sec. (This means that Joe has period of T = 50 sec,
and a frequency of ω 1 = 2π/T = 2π/50 radians per sec. Similarly, Michael’s period is T = 43.18
sec and his frequency is ω 2 = 2π/T = 2π/43.18 radians per sec. From this, we find that
dθ J
dt = 2π
50
dθ M
dt
= 0.125 radians per sec
= 2π = 0.145 radians per sec
43.18
Thus the angle between the runners, θ M − θ J changes at the rate
d(θ M − θ J )
dt
= 0.145 − 0.125 = 0.02 radians per sec
Example 3: simple law of cosines
Consider the triangle as shown in Figure 10.9. Suppose that the angle θ increases at a constant
rate, i.e. dθ/dt = k. If the sides a = 3, b = 4, are of constant length, determine the rate of change
of the length c opposite this angle at the instant that c = 5.
v.2005.1 - September 4, 2009 16