Calculus
posted by Trevor .
A man is 6ft tall and is walking at night straight toward a lighted street lamp at a rate of 5 ft/sec. If the lamp is 20 ft above the ground, find the rate at which the length of his shadow is changing.

If the man is x away from the lamp, and his shadow is s,
s/6 = (x+s)/20
s = 3/7 x
so,
ds/dt = 3/7 * dx/dt = 3/7 (5) = 15/7 ft/s 
I don't know if this is right because if you are using x for the distance the man is away from the light post then you cant set x equal to the rate at which the man is walking.

You are correct. However, dx/dt is the rate he is walking away. That is the value I used.