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?
Yes, the solution you provided does indeed offer a simpler approach to finding the speed of the shadow of the walking surveyor. Here's a breakdown of the steps involved:
1. Given the diagram and setup, express the distance PS (r) using polar coordinates as r = 2a cosθ.
2. Consider the surveyor walking away from the center O with a constant speed of 4 m/s. At time t, the distance from O to the surveyor is x = 4t.
3. To find the speed of the shadow at S, we need to calculate ds/dt, where ds is the arc length QS along the circle. Using the formula ds = √(r^2 + r'^2) dθ (where r' denotes dr/dt), we can simplify it to ds = 2a dθ.
4. Next, we need to find dθ/dt. Since tanθ = x/a = 4t/a, we can express cosθ as a/√(x^2 + a^2) = a^2/√(16t^2 + a^4).
5. To find dθ/dt, we use the derivative of the equation tanθ = x/a with respect to t, which gives us sec^2θ dθ/dt = 4/a. Simplifying, we get dθ/dt = 4/a cos^2θ.
6. Combining the expressions for ds and dθ/dt, we have ds/dt = 2a * 4/a (a^2/√(16t^2 + a^4))^2. Simplifying further, we obtain ds/dt = 8a^4/(16t^2 + a^4).
7. Finally, when x = a/2, t = a/8. Substituting this into the expression for ds/dt, we get ds/dt = 32a^2/(4a^2+1).
8. If the radius of the circle is not specified, you can substitute the appropriate value to obtain the final answer.
So, overall, the simplified approach involves using polar coordinates and considering the distance of the surveyor from the center at a specific time to derive the speed of the shadow at a given distance from the center.
nope. the 4/a was not supposed to be squared.
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)