Calculus

A light hangs 15 ft. directly above a straight walk on which a man 6 ft. tall is walking. How fast is the end of the man's shadow travelling when he is walking away from the light at a rate of 3 miles per hour?

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1. did you make your sketch ??
let the length of the man's shadow be x ft
let the distance he is from the lightpost be y
by ratios:
6/x = 15(x+y)
6x + 6y = 15x
6y = 9x
2y = 3x
2dy/dt = 3dx/dt
but dy/dt = 3
so 6 = 3 dx/dt
dx/dt = 2

His shadow is lengthening at 2 ft/s
but he is also moving at 3 ft/s
so d(x+y)/dt = 5 ft/s

In this kind of question we have to be careful what is asked:

at what rate is the shadow increasing ---- 2 ft/s
at what rate is the shadow moving ----- 5 ft/s

an analogy would be a man walking in a car of a moving train:
if the man is walking in the car at 3 ft/s
but the train is moving at 50 ft/s, to an observer looking in would see the man moving at 53 ft/s

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