# Calculus

A man standing 9 feet from the base of a lamppost casts a shadow 6 feet long. If the man is 6
feet tall and walks away from the lamppost at a speed of 30 feet per minute, at what rate, in feet per

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1. If the man's shadow has length s when his distance from the pole of height h is x, then using similar triangles,

(x+s)/h = s/6
6x+6s = hs
s(h-6) = 6x
so,

(h-6) ds/dt = 6 dx/dt
ds/dt = 6*30/(h-6) = 180/(h-6)

Now, we were given
15/h = 6/6, so h=15

Thus, ds/dt = 180/9 = 20 ft/min

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2. I dont get where you pulled the 9 from, 5 years late

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