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Calculus/rates of change

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Expert: - 12/1/2008


Question
A man who is 6 feet tall is walking at a range of 84 ft/min toward a street lamp that is 20 feet tall. At what rate is th elength of his shadow changing?

Answer
I believe the problem is this:
A man is 6 feet tall,
he is moving at 84 ft/min,
the lamp is 20 feet tall, and
he is x feet away from the lamp.

There is a difference in height of 14 feet between the man and the lamp.  When he is x feet away from the lamp, then 14/x = 6/y, where y is the distance on the ground.

Multiply by xy and you get 14y = 6x, or y = (4/7)x

It can be said dy = (4/7) dx.  dx was given as 84 (or -84 if we're looking at the distance left to the lamp).  Put that into the equation to get dy.

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