Wow. This one gets tricky, since we're dealing in 3D, rather than the usual 2D.
If the bottom of the pole is at B, and the top is at T, and the point directly across the road is at Q, then if the man is x from Q, the diagonal distance across the road (BQ) is y, then
y^2 = x^2+30^2
Now, using similar triangles, if the man's shadow has length s, then
s/6 = (s+y)/18
Now, when x=40, y=50 and s=25
Taking derivatives, we have
y dy/dt = x dx/dt
50 dy/dt = 40*5
dy/dt = 8
1/6 ds/dt = 1/18 (ds/dt + dy/dt)
3 ds/dt = ds/dt + 8
2 ds/dt = 8
ds/dt = 4
A 6 foot tall man walks at a rate of 5 feet per second along one edge of a road that is 30 feet wide. On the other edge of the road is a light atop a pole 18 feet high. How fast is the length of the man's shadow increasing when he is 40 feet beyond the point directly across the road from the pole?
...I drew a diagram but I can't figure out the equations to find the length of the shadow, I distinguished as s, in terms of the distance away from the pole (as if on the same side of the road as the man), distinguished as x
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