A water tank has the shape of an inverted circular cone with a base radius 2 m and a height 4 m. If the water is being pumped into the tank at rate 2m^3/min , find the rate at which the water level is rising 3m.

help please =/?

The volume of the cone is given by:

V = \frac13 \pi r^2 h

Also, from the height and radius, you know that:

\frac{h}{r} = \frac42

What you know are:

\frac{dV}{dt} = 2

You are requested to give:

\frac{dh}{dt}

Use the chain rule:

\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}

Hence, find V in terms of h, find the derivative of the volume of cone formula, replace in the equation with h = 3 m.

What do you get? :)