Question

Water pours into a tank at the rate of 2000 cm3/min. The tank is cylindrical with radius 2 meters. How fast is the height of water in the tank changing when the height of the water is 50 cm?

Answer 1

Volume of water in the tank:

$V=\pi (2\,\mathrm m)^2h=\pi(200\,\mathrm{cm})^2h$

Differentiate both sides with respect to time t :

$\dfrac{\mathrm dV}{\mathrm dt}=\pi(200\,\mathrm{cm})^2\dfrac{\mathrm dh}{\mathrm dt}$

V changes at a rate of 2000 cc/min (cubic cm per minute); use this to solve for dh/dt :

$2000\dfrac{\mathrm{cm}^3}{\rm min}=\pi(40,000\,\mathrm{cm}^2)\dfrac{\mathrm dh}{\mathrm dt}$

$\dfrac{\mathrm dh}{\mathrm dt}=\dfrac{2000}{40,000\pi}\dfrac{\rm cm}{\rm min}=\dfrac1{20\pi}\dfrac{\rm cm}{\rm min}$

(The question asks how the height changes at the exact moment the height is 50 cm, but this info is a red herring because the rate of change is constant.)