Answer to Question #148147 in Calculus for Sean

Question

A trough is 12 ft long and its ends are in the form of inverted isosceles triangles having an
altitude of 3 ft and a base of 3 ft. Water is flowing into the trough at the rate of 2 (ft^3/min). How fast is the water level rising when the water is 1 ft deep?

Answer 1

When the depth of water in the trough is $h$ the volume of water is

$V=\frac{1}{2}12h^2$ because the width of the trough is proportional to the

depth of the water (when width will be 3 feet when the depth will be 3 feet).

After differentiating as a function of time:

$\frac{dV}{dt}=12h\frac{dh}{dt}$

$\frac{dh}{dt}=\frac{\frac{dV}{dt}}{12h}=\frac{2}{12*1}=\frac{1}{6}(ft/min)$