Calculus: Related Rates?

Question

A 6 ft. tall woman is walking at a rate of 4 ft/sec away from a street lamp that is 24 feet tall. How fast is the lenght of her shadow changing?

Ans: 4/3 ft/sec

Answer 1

If the distance from the lamp to the woman is x and the length of the shadow is s, then a sketch of the problem will easily reveal that

(6/s) = 24/(x+s)   .

or   6x+6s= 24s

6x=18s

s=x/3

differentiate wrt time gives

ds/dt = (1/3) dx/dt

but the problem gives dx/dt= 4 ft/s

so ds/dt = 4/3 ft/s