A man 6ft tall walks at 5ft/second away from a light that is 15ft above the ground. When he is 10ft from the base of the light, at what rate (ft/second) is the tip of his shadow moving?
Draw an horizontal line from the head of the man to the standard lamp.
You see that it forms a rectangle triangle. One of the cathetes is 9 in length and say the other is x. This is the distance from the standard lamp. This triangle is similar to the triangle formed by the man and his shadow.
This shadow is y in length.
You can find the following relation $\displaystyle *y = \frac{6}{9}x$*
Then you must differentiate it $\displaystyle \frac{dy}{dx} = \frac{2}{3} $
To have the change rate with respect to time we multiply by $\displaystyle \frac{dx}{dt} $ to get
$\displaystyle \frac{dy}{dx}\frac{dx}{dt} = \frac{2}{3}\frac{dx}{dt} $
You are given that $\displaystyle \frac{dx}{dt} = 5 $