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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 find ( \frac{dh}{dt} ), we start with the expression for the volume of water in the tank:

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

Using the chain rule, we differentiate with respect to time:

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

From the problem, we know:

dVdt=k(2R−h)\frac{dV}{dt} = k(2R - h)

Setting these two expressions equal gives:

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

To isolate ( \frac{dh}{dt} ), rearranging yields:

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

Next, we note that (R^2 - h^2 = (R + h)(R - h)), simplifying further,

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

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 taken to fill the tank, we need to use the rate of inflow and the relationship we've derived.

Integrating ( \frac{dh}{dt} = -\frac{k}{\pi h} ), we separate variables and integrate:

∫h dh=−kπ∫dt\int h \, dh = -\frac{k}{\pi} \int dt

This results in:

h22=−kπt+C\frac{h^2}{2} = -\frac{k}{\pi}t + C

When the tank is full, (h = R), at time (T), we find:

R22=−kπT+C\frac{R^2}{2} = -\frac{k}{\pi}T + C

At the start, the tank is empty, so initially, C = 0. Thus,

R22=−kπT\frac{R^2}{2} = -\frac{k}{\pi}T

Solving for T gives:

T=πR22kT = \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 the outflow continues. The outflow rate is given by:

Q=k(2R−h)Q = k(2R - h)

When the tank is full, (h = R), so:

Q=k(2R−R)=kRQ = k(2R - R) = kR

The volume of the tank is:

V=π(R2R−R33)=2πR33V = \pi \left( R^2 R - \frac{R^3}{3} \right) = \frac{2\pi R^3}{3}

The time taken to empty is given by:

tempty=VQ=2πR33kR=2πR23kt_{empty} = \frac{V}{Q} = \frac{\frac{2\pi R^3}{3}}{kR} = \frac{2\pi R^2}{3k}

Thus, since (T = \frac{\pi R^2}{2k}):

tempty=3Tt_{empty} = 3T

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

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