Question

A water tank in the shape of a right circular cone has a height of 10 feet. The top rim of the tank is a circle with a radius of 4 feet. If water is pumped into the tank at the rate of 2 cubic feet per minute, then the rate of change of the water depth, in feet per minute, when the depth is 5 feet, is

• A. $\frac{1}{\pi}$
• B. $\frac{3}{2 \pi}$
• C. $\frac{2}{\pi}$
• D. $\frac{1}{2 \pi}$

Answer 1

Answer: (D)

$\frac{ dV }{ dt }=2$ cubic feet per minute; $\frac{ dh }{ dt }=$ ?
$V =\frac{1}{3} \pi x ^2 h $
$\tan \theta=\frac{ x }{ h }=\frac{ r }{10}=\frac{4}{10}$
$ \Rightarrow x =\frac{2 h }{5} $
$V =\frac{1}{3} \cdot \pi \cdot \frac{4 h ^2}{25} \cdot h =\frac{4 \pi}{75} \cdot h ^3$
$\frac{ dV }{ dt }=2=\frac{4 \pi}{75} 3 h ^2 \frac{ dh }{ dt } ; \frac{ dh }{ dt }=\frac{150}{12 \pi h ^2}=\frac{150}{12 \cdot \pi \cdot 25}=\frac{6}{12 \pi}=\frac{1}{2 \pi}$