Asked 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?
Answers
Answered by
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)
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)
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