Question

A water tank has the shape of an inverted right circular cone with base radius 3 meters and height 6 meters. Water is being pumped into the tank at the rate of 12 meters3/sec. Find the rate, in meters/sec, at which the water level is rising when the water is 2 meters deep. Give 2 decimal places for your answer.

Answer 1

Given dV/dt , find dh/dt.

Use volume formula for cone.
Define radius in terms of height.
Differentiate with respect to 't'.
Sub in given values, dV/dt = 12, h=2.  
Solve for dh/dt.

$V = \frac{\pi}{3} r^2 h \\ \\\frac{r}{h} = \frac{3}{6} \rightarrow r = \frac{h}{2} \\ \\ V = \frac{\pi}{3}(\frac{h}{2})^2 h = \frac{\pi}{12} h^3 \\ \\ \frac{dV}{dt} = \frac{\pi}{12}(3h^2) \frac{dh}{dt} \\ \\ 12 = \frac{\pi}{12}(3*2^2) \frac{dh}{dt} \\ \\ \frac{dh}{dt} = \frac{12}{\pi}$