Posted by **Trevor** on Thursday, February 28, 2013 at 12:01pm.

A man is 6ft tall and is walking at night straight toward a lighted street lamp at a rate of 5 ft/sec. If the lamp is 20 ft above the ground, find the rate at which the length of his shadow is changing.

- Calculus -
**Steve**, Thursday, February 28, 2013 at 12:40pm
If the man is x away from the lamp, and his shadow is s,

s/6 = (x+s)/20

s = 3/7 x

so,

ds/dt = 3/7 * dx/dt = 3/7 (-5) = -15/7 ft/s

- Calculus -
**Trevor**, Thursday, February 28, 2013 at 1:52pm
I don't know if this is right because if you are using x for the distance the man is away from the light post then you cant set x equal to the rate at which the man is walking.

- Calculus -
**Steve**, Thursday, February 28, 2013 at 3:21pm
You are correct. However, dx/dt *is* the rate he is walking away. That is the value I used.

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