related rates

Question

A 6-foot-tall woman walks at 6 ft/s toward a street light that is 18 ft above the ground. What is the rate of change of the length of her shadow when she is 16 ft from the street light? At what rate is the tip of her shadow moving?

Answer 1

Set x as woman-to-post distance.

Set y as (tip of shadow behind woman)-to-post distance.

By similar triangles, obtain 18/6 = y/(y − x) or 3y − 3x = y

which goes to 2y = 3x.

Differentiate through 2y = 3x to gain 2(dy/dt) = 3(dx/dt).

Given that dx/dt is -6 feet per second (negative because

x is decreasing), one reaches 2(dy/dt) = 3(-6) and

dy/dt is -9.

The tip of the shadow is moving at 9 feet per

second toward the post of the streetlight.

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At x = 16, the length of the woman's shadow is y − 16.

The principle of similar triangles gives 18/6 = y/(y − 16) or 3y − 48 = y

or 2y − 48 = 0 or distance y (from tip of shadow behind girl to streetlight)

is 24 and length of shadow is (24 − 16) or 8 feet.

Also, the rate of change in shadow length is d(y − x)/dt or (dy/dt − dx/dt) or

(-9) − (-6) or -3 feet per second (that is, 3 feet decrease in shadow length every second).