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Use related rates to find the rate at which we need to add water to a leaky tank

by RoRi

Given a water tank shaped like a right circular cone with radius 15 feet at its base and altitude 10 feet. Water leaks from the tank at a constant rate of 1 cubic foot per second, and water is added to the tank at a constant rate of c cubic feet per second. Find the value of c so that the water level will be rising at a rate of 4 feet per second when the depth of the water is 2 feet.

The following diagram illustrates the situation:

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First, we compute the radius of the water level in terms of the height of the water,

    \[ r = \frac{15}{10} h = \frac{3}{2} h \quad \implies \quad \frac{dr}{dt} = \frac{3}{2} \frac{dh}{dt}. \]

Furthermore, the radius at the water line is 3 feet when the height of the water is 2 feet. So, computing the volume of water in terms of r and h we have,

    \begin{align*} V &= \frac{1}{3} \pi r^2 h \\ \implies \frac{dV}{dt} &= \frac{2}{3} \pi r h \frac{dr}{dt} + \frac{1}{3} \pi r^2 \frac{dh}{dt}. \end{align*}

(As in the previous exercise, we are careful to use the product rule when evaluating this derivative.)
The problem then gives us the rate of change of volume of the water is

    \[ \frac{dV}{dt} = (c-1) \text{ cubic feet per second}. \]

(Since we are adding c cubic feet per second, and 1 cubic foot per second is leaking out.) So we set \frac{dh}{dt} = 4 and solve for c,

    \begin{align*} (c-1) &= \frac{2}{3} \pi (3)(2) \left( \frac{3}{2} \right) (4) + \frac{1}{3} \pi (9)(4) = 36 \pi \\ \implies c&= 36 \pi + 1 \text{ cubic feet per second}. \end{align*}

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