Question

I have been thinking about the walking surveyor, and here is a less complicated solution (also, correct, as it happens - the other one was faulty, as I'm sure you discovered.)
Suppose we have the following diagram:
O is the center of a circle of radius 2a
P and Q are the ends of a diameter
S is a point on the circle
θ is the measure of angle QPS
r is the distance PS
Now, think of the circle in polar coordinates, where P is the origin. Then we have
r = 2a cosθ
Now we have our surveyor walking along the diameter perpendicular to PQ, at a rate of 4 m/s. In the problem, she is approaching the center O, but it's easier to think of her as walking away from O. Then at time t, she is a distance 4t from O. We want to find the speed of her shadow (at S) when she is a distance a/2 from O.
Consider the arc length QS along the circle. We want to find ds/dt
ds = √(r^2 + r'^2) dθ = 2a dθ
If x is the distance from the surveyor to O, then
x = 4t
tanθ = x/a = 4t/a
cosθ = a/√(x^2 + a^2) = a^2/√(16t^2 + a^4)
ds/dt = 2a dθ/dt
But sec^2θ dθ/dt = 4/a so dθ/dt = 4/a cos^2θ
ds/dt = 2a * 4/a (a^2/√(16t^2 + a^4))^2 = 8a^4/(16t^2 + a^4)
Now, when x = a/2, t = a/8, so
ds/dt = 32a^2/(4a^2+1)
Your problem did not specify the radius of the circle, so just fill that in and you're done.

>whew< wasn't that simpler?

Answer 1

oops. I made a typo. I forgot to include the 4/a in the squared part. So we should have
ds/dt = 2a * (4/a * a^2/√(16t^2 + a^4))^2 = 32a^3/(16t^2 + a^4)
at t = a/8, that gives
ds/dt = 128a/(4a^2 + 1)

Answer 2

nope. the 4/a was not supposed to be squared.