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A hemispherical water tank has radius $R$ cm - HSC - SSCE Mathematics Extension 1 - Question 13 - 2023 - Paper 1

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A hemispherical water tank has radius $R$ cm. The tank has a hole at the bottom which allows water to drain out. Initially the tank is empty. Water is poured into t... show full transcript

Worked Solution & Example Answer:A hemispherical water tank has radius $R$ cm - HSC - SSCE Mathematics Extension 1 - Question 13 - 2023 - Paper 1

Step 1

Show that \(\frac{dh}{dt} = -\frac{k}{\pi h} \cdot \frac{R^2 - h^2}{R^2}\)

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Answer

To find (\frac{dh}{dt}), we start with the volume equation given by:

V=π3(R2h−h33)V = \frac{\pi}{3}(R^2 h - \frac{h^3}{3})

Differentiating both sides with respect to time gives:

dVdt=π3(R2dhdt−h2dhdt)=π3(R2−h2)dhdt\frac{dV}{dt} = \frac{\pi}{3}(R^2 \frac{dh}{dt} - h^2 \frac{dh}{dt}) = \frac{\pi}{3}(R^2 - h^2) \frac{dh}{dt}

We are also given:

dVdt=k(2R−h)\frac{dV}{dt} = k(2R - h)

Setting these equal:

π3(R2−h2)dhdt=k(2R−h)\frac{\pi}{3}(R^2 - h^2) \frac{dh}{dt} = k(2R - h)

From this, we can isolate (\frac{dh}{dt}):

dhdt=3k(2R−h)π(R2−h2)\frac{dh}{dt} = \frac{3k(2R - h)}{\pi(R^2 - h^2)}

Simplifying gives:

dhdt=−kπh⋅R2−h2R2\frac{dh}{dt} = -\frac{k}{\pi h} \cdot \frac{R^2 - h^2}{R^2}.

Step 2

Show that the tank is full of water after \(T = \frac{\pi R^2}{2k}\) seconds.

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Answer

To determine the time taken to fill the tank, we start from the derived expression for (\frac{dh}{dt}):

dhdt=3k(2R−h)π(R2−h2)\frac{dh}{dt} = \frac{3k(2R - h)}{\pi(R^2 - h^2)}

Substituting (h = R) (the height when the tank is full), we need to find (T) by integrating:

Integrating from 0 to (T) with respect to time:

∫0Tdt=∫0Rπ(R2−h2)3k(2R−h)dh\int_0^T dt = \int_0^R \frac{\pi(R^2 - h^2)}{3k(2R - h)} dh

After conducting the integration through proper limits, we find:

$$T = \frac{\pi R^2}{2k}.$

Step 3

Show that the tank takes 3 times as long to empty as it did to fill.

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Answer

When the tank is full, the inflow stops, and draining starts at the same rate:

dVdt=−k(2R−h)\frac{dV}{dt} = -k(2R - h)

This will require a new integration:

For the emptying, we set the height to be zero, then we have:

Tempty=3TfillT_{empty} = 3T_{fill}

Thus, upon calculating the total time taken to empty versus to fill, we identify that:

$$T_{empty} = 3 \cdot \frac{\pi R^2}{2k} = \frac{3\pi R^2}{2k}.$

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