Answer
Water is being pumped into the tank at a rate of $289,253~cm^3/min$
Work Step by Step
We can use similar triangles to write an expression for the radius $r$:
$\frac{r}{h} = \frac{2}{6}$
$r = \frac{h}{3}$
We can differentiate both sides of the equation for volume with respect to $t$:
$V = \frac{1}{3}\pi~r^2~h$
$V = \frac{1}{27}\pi~h^3$
$\frac{dV}{dt} = \frac{1}{9}\pi~h^2~\frac{dh}{dt}$
$\frac{dV}{dt} = \frac{1}{9}\pi~(200~cm)^2~(20~cm/min)$
$\frac{dV}{dt} = 279,253~cm^3/min$
We can find the rate $P$ at which water is being pumped into the tank:
$P-10,000~cm^3/min = 279,253~cm^3/min$
$P = 289,253~cm^3/min$
Water is being pumped into the tank at a rate of $289,253~cm^3/min$
