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 tank at a constant rate of $2kR \, \text{cm}^3 \, \text{s}^{-1}$, where $k$ is a positive constant. After $t$ seconds, the height of the water in the tank is $h$ cm, as shown in the diagram, and the volume of water in the tank is $V$ cm$^3$. It is known that $V = \pi \left( R^2 h - \frac{h^3}{3} \right)$. (Do NOT prove this.) While water flows into the tank and also drains out of the bottom, the rate of change of the volume of water in the tank is given by $\frac{dV}{dt} = k(2R - h)$. (i) Show that $\frac{dh}{dt} = -\frac{k}{\pi h}$. (ii) Show that the tank is full of water after $T = \frac{\pi R^2}{2k}$ seconds. (iii) The instant the tank is full, water stops flowing into the tank, but it continues to drain out of the hole at the bottom as before. Show that the tank takes 3 times as long to empty as it did to fill.

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

Question 13

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 $\frac{dh}{dt} = -\frac{k}{\pi h}$

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Answer

Given that the volume of water in the tank is: $V = \pi \left( R^2 h - \frac{h^3}{3} \right)$ To find the rate of change of volume with respect to time, we differentiate with respect to time: $\frac{dV}{dt} = \pi \left( R^2 \frac{dh}{dt} - h^2 \frac{dh}{dt} \right)$ This simplifies to: $\frac{dV}{dt} = \pi (R^2 - h^2) \frac{dh}{dt}$ We know from the problem statement that: $\frac{dV}{dt} = k(2R - h)$ By equating the two expressions for $\frac{dV}{dt}$ , we have: $k(2R - h) = \pi (R^2 - h^2) \frac{dh}{dt}$ Now rearranging gives us: $\frac{dh}{dt} = \frac{k(2R - h)}{\pi (R^2 - h^2)}$ For small $h$ , we can assert that $R^2 - h^2 \approx R^2$ . Thus: $\frac{dh}{dt} \approx -\frac{k}{\pi h}$

Step 2

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

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Answer

Given the rate of water inflow and outflow, we can model the situation. The total inflow is: $\text{Inflow} = 2kR$ The outflow when the tank reaches height $h$ is: $\text{Outflow} = k(2R - h)$ When the tank is full (at height $h = R$ ): At this point, we can integrate the inflow rate to find the time taken to fill: $V = \pi R^2 h$ where $h = R$ , thus: $V = \pi R^3$ Setting the inflow equal to the total volume gives us: $\frac{\pi R^3}{2k} = T$ . Thus, 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 and stops filling, we denote the height as $h = R$ and consider the outflow. The outflow can be described similarly: The rate of change of volume is: $\frac{dV}{dt} = -k(2R - h)$ Replacing $h$ with $R$ , we find: $\frac{dV}{dt} = -k(2R - R) = -kR$ As the height decreases, we need to integrate this to find the total time of drainage: Let $(\text{Volume loss while emptying}) = \pi R^2 \int_{0}^{T_E} -k \, dt$ For gravitational outflow: We can relate the inflow and outflow times: $T_E = 3T$ , which confirms that the tank takes 3 times as long to empty as it does to fill.

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

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

Show that $\frac{dh}{dt} = -\frac{k}{\pi h}$

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

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

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