A woman 5 ft tall walks at the rate of 7.5 ft/sec away from a streetlight that is 10 ft above the ground. At what rate is the tip of her shadow moving?

Answer :

taskmasters
Height of the woman = 5 ft
Rate at which the woman is walking = 7.5 ft/sec
Let us assume the length of the shadow = s
Le us assume the distance of the woman's feet from the base of the streetlight = x
Then
s/5 = (s + x)/12
12s = 5s + 5x
7s = 5x
s = (5/7)x
Now let us differentiate with respect to t
ds/dt = (5/7)(dx/dt)
We already know that dx/dt = 7/2 ft/sec
Then
ds/dt = (5/7) * (7/2)
        = (5/2)
        = 2.5 ft/sec
From the above deduction, it can be easily concluded that the rate at which the tip of her shadow is moving is 2.5 ft/sec. 

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