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

Math 102 Notes Chapter 10
We differentiate both sides of this equation with respect to t, and obtain
d tan(θ) dθ
dθ dt = d ( ) s
dt x(t)
so that
We can use the trigonometric identity
sec 2 (θ) dθ
dt = −s 1 dx
x 2 dt
dθ
dt = −s 1
sec 2 (θ)
1 dx
x 2 dt
sec 2 (θ) = 1 + tan 2 (θ)
to express our answer in terms only of the size, s, the distance of the object, x and the speed:
so
( s 2
sec 2 x
(θ) = 1 + =
x) 2 + s 2
dθ
dt = −s x 2
x 2 + s 2 1
x 2 dx
dt =
x 2
S
x 2 + s 2v.
(Two minus signs cancelled above.) Thus, the rate of change of the visual angle is sv/(x 2 + s 2 ).
This calculation has some interesting implications for reactions to visual stimuli. We will explore
some of these implications later on.
10.9 Trigonometric functions and differential equations
In this section, we will show the following relationship between trigonometric functions and differential
equations:
The functions
x(t) = cos(ωt), y(t) = sin(ωt)
satisfy a pair of differential equations,
dx
dt = −ωy,
dy
dt = ωx.
The functions
x(t) = cos(ωt),
y(t) = sin(ωt)
also satisfy a related differential equation with a second derivative
d 2 x
dt 2 = −ω2 x.
To show that these statements are true, we return to an example explored in the previous section:
we considered a point moving around a unit circle at a constant angular rate, ω, so that
dθ
dt = ω.
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