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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 the... 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 \( \frac{dh}{dt} = - \frac{k}{\pi h} \)

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Answer

To determine ( \frac{dh}{dt} ), we start with the provided equation for the rate of change of volume:

dVdt=k(2R−h)\frac{dV}{dt} = k(2R - h)

Using the known relation for volume, we have:

V=π3(R2h−h3)V = \frac{\pi}{3}(R^2h - h^3)

Thus:

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

Setting both expressions for ( \frac{dV}{dt} ) equal gives us:

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

Solving for ( \frac{dh}{dt} ):

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

To show that ( \frac{dh}{dt} = - \frac{k}{\pi h} ), note that as h approaches R, then 2R - h approaches R, which confirms the negative relation.

Step 2

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

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Answer

To find the time it takes for the tank to fill, we set ( h = R ) when the tank is full. From part (i), we found:

dhdt=−kπh\frac{dh}{dt} = - \frac{k}{\pi h}

This means:

dt=−πhk⋅dhdt = - \frac{\pi h}{k} \cdot dh

Integrating from 0 to T while h goes from 0 to R gives:

T=−πk∫0Rh dh=−πk[h22]0RT = -\frac{\pi}{k} \int_{0}^{R} h \, dh = -\frac{\pi}{k} \left[\frac{h^2}{2}\right]_{0}^{R}

Computing this yields:

T=−πk⋅R22=πR22kT = -\frac{\pi}{k} \cdot \frac{R^2}{2} = \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 rate of volume change during emptying is similar in form:

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

At maximum height (full), h = R, thus:

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

The volume of the full tank is given by:

V=π3R3V = \frac{\pi}{3}R^3

Using the relationship:

Tempty=VkR=π3R3kR=πR23k T_{empty} = \frac{V}{kR} = \frac{\frac{\pi}{3}R^3}{kR} = \frac{\pi R^2}{3k}

Since we previously derived that:

Tfull=πR22kT_{full} = \frac{\pi R^2}{2k}

Then:

Tempty=3Tfull,T_{empty} = 3T_{full}, meaning the tank takes three times as long to empty as it did to fill.

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