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Find the relative rate of change of volume to rate of change of height in a water tank

by RoRi

Consider a water tank shaped like a hemisphere with a radius of 10 feet. At time t let,

    \begin{align*} h &= \text{ depth of the water in the tank}, \\ r &= \text{ radius of the surface of the water}, \\ V &= \text{ volume of water in the tank}. \end{align*}

Find the rate of change of the volume relative to the rate of change of the height (\frac{dV}{dh}) when h = 5.
If water is flowing into the tank at a constant rate of 5 \sqrt{3} cubic feet per second, find \frac{dr}{dt} when h = 5.

For this problem we will consider the graph of the following function (the hemisphere tank and water will then be obtained as solids of revolution of this graph about the y-axis):

Rendered by QuickLaTeX.com

First, we find a formula for the volume of the water in the tank as a function of h. We do this by considering the solid of revolution (for a review of solids of revolution see these exercises) of \sqrt{20y-y^2} about the y-axis:

    \begin{align*} V & = \pi \int_0^h \left( \sqrt{20y - y^2} \right)^2 \, dy \\ &= \pi \int_0^h (20y -y^2) \, dy \\ &= \pi \left( 10y^2 - \frac{y^3}{3} \right)\Bigr \rvert_0^h \\ &= 10 \pi h^2 - \frac{\pi h^3}{3}. \end{align*}

Differentiating with respect to h we then have,

    \[ \frac{dV}{dh} = 20 \pi h - \pi h^2. \]

So, when h = 5 feet we have

    \[ \frac{dV}{dh} = 75 \pi. \]

Next, we are given \frac{dV}{dt} = 5 \sqrt{3} cubic feet per second. We know from above, \frac{dV}{dh} = 20 \pi h - \pi h^2. So,

    \[ \frac{dV}{dh} \cdot \frac{dh}{dt} = \frac{dV}{dt} \quad \implies \quad \frac{dh}{dt}= \frac{5 \sqrt{3}}{20 \pi h - \pi h^2}. \]

Then, to get r in terms of h we evaluate,

    \begin{align*} r^2 + (10-h)^2 = 100 \quad &\implies \quad r = \sqrt{20h - h^2} \\ &\implies \frac{dr}{dh} = \frac{10-h}{\sqrt{20h - h^2}}. \end{align*}

Thus,

    \[ \frac{dr}{dt} = \left( \frac{dh}{dt} \right) \left( \frac{dr}{dh} \right) = \left( \frac{5 \sqrt{3}}{20 \pi h - \pi h^2} \right) \left( \frac{10-h}{\sqrt{20h - h^2}} \right). \]

So, if h = 5, we have

    \[ \frac{dr}{dt} = \left( \frac{\sqrt{3}}{15 \pi} \right) \left( \frac{5}{\sqrt{75}} \right) = \frac{1}{15 \pi} \text{ feet per second}. \]

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