Posted by **Amanda** on Saturday, August 27, 2011 at 2:36am.

A man 5.50 ft tall approaches a street light 18.0 ft above the ground at a rate of 4.0 ft/s. at what rate is the end of the man's shadow moving when he is 9.0 ft from the base of the light?

- Calculus -
**Reiny**, Saturday, August 27, 2011 at 7:51am
Make a sketch.

Let the distance between the man and the post be x ft

let the length of his shadow be y ft

by similar triangles:

5.5/y = 18/(x+y)

18y = 5.5x + 5.5y

12.5y = 5.5x

12.5dy/dt = 5.5dx/dt

given: dx/dt = 4 ft/s

then dy/dt = 5.5/12.5 or .44 ft/s

So the shadow is lengthening at .44 ft/s

and the shadow is moving at 4+.44 or 4.44 ft/sec

(The end of the shadow is both lengthening and being moved along by the movement of the man.

Similar to somebody walking in a moving train.

Suppose a train is going at 60 mph, but somebody is walking inside the train at 4 mph in the same direction as the train. To an observer at a crossing, the person would be moving at 64 mph )

- Calculus -
**wrongg**, Wednesday, February 20, 2013 at 9:11pm
wrong answer

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