Question

Water is draining from a cone-shaped tank at a rate of 3.5 meters^3/second. The tank is 15 meters high and its top radius is 5 meters. How fast is the water level falling when the water level is 6 meters high?

Answer 1

You know the rate of change of volume with respect to time: \(\dfrac{dV}{dt} = 3.5\text{ m}^3/\text{s}\)

You want to find the rate of change of the height with respect to time: \(\dfrac{dh}{dt}\)

You can find \(\dfrac{dh}{dt} = \dfrac{dh}{dV}*\dfrac{dV}{dt}\) if you find an expression for the height based on the volume and take its derivative. Bust out those geometry skills :-)

When you have formula for \(\dfrac{dh}{dt}\), evaluate it when \(h =6\) to get the answer.