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