Question

A surveyor is walking at the rate of 4 ft/sec along the diameter of a circular courtyard. A light at one end of a diameter perpendicular to her path casts a shadow on the circular wall. How fast is the shadow moving along the wall when her distance to the center of the courtyard is r/2?

Please help.

Answer 1

Let's label a few things in our diagram.
O = center of circle
P,Q are the ends of the perpendicular diameter
T is where the surveyor is standing
S is the point on the circle where the shadow falls
θ = angle QPS
a is the radius of the circle
x is the distance TO
r is the distance PS
z is the distance PT
Now, angle dθ subtends a small arc of length ds
we want to find ds/dt: how fast s is moving along the circle
r = 2a cosθ
tanθ = (a-4t)/a, so
sec^2θ dθ/dt = -4/a
sec^2θ = 1+tan^2θ = 1 + ((a-4t)/a)^2 = 2/a^2 (8t^2-4at+a^2)
and cos^2θ = a^2/(2(8t^2-4at+a^2))
ds = r dθ
ds/dt = -8 cos^3θ = -8(a^2/(2(8t^2-4at+a^2)))^(3/2)
= -2√2 a^3 / (8t^2-4at+a^2)^(3/2)
Now, what is t when x = a/2?
t = a/8
so now we know that at that time,
ds/dt = -2√2 a^3 / (5a^2/8)^(3/2) = -64 / 5√5 ft/s
the negative sign indicates that the shadow's distance from PQ is decreasing

Better double-check my math!