# calculus

posted by Anonymous

A woman 5.5 ft tall walks at a rate of 6 ft/sec toward a streetlight that is 22 ft above the ground. At what rate is the length of her shadow changing when she is 15 ft from the base of the light?

1. drwls

If x is her horizontal distance from the street light, the shadow length s is given by
s = (5.5/22)*(s + x)
so
(16.5/22)s = (5.5/22) x
s = x/3

Therefore ds/dt = (1/3) dx/dt
= (2.0 ft/s)*(dx/dt)

2. Mona

-2

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