2023HSCShort answer

Q13 (a) (iii)

3 marks

Question

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 $2kR\ \mathrm{cm^3\,s^{-1}}$ , 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 $V\ \mathrm{cm^3}$ .

It is known that

V=\pi\left(Rh^2-\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).

(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.

Uncropped source page 12 for question 13.a.iii; requires human review and cropping — Reviewed visual · source page 12

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	3
Finds correct expression for $\frac{dh}{dt}$ , or equivalent merit	2
Models the situation to find the new differential equation $\frac{dV}{dt}=-kh$ , or equivalent merit	1

Sample answer:

When the tank is full the inflow of water stops. This means that the rate of change in volume is only dependent on $h$ .

-kh=\frac{dV}{dt}=\frac{dV}{dh}\times\frac{dh}{dt}

-kh=\pi h(2R-h)\frac{dh}{dt}

\frac{dh}{dt}=\frac{-k}{\pi(2R-h)}

When $t=0$ , $h=R$ ; when $t=T_E$ , $h=0$ .

\begin{aligned} \int_R^0\pi(2R-h)\,dh&=\int_0^{T_E}-k\,dt\\ \left[\pi\left(2Rh-\frac{h^2}{2}\right)\right]_R^0&=-kT_E\\ -\pi\left(2R^2-\frac{R^2}{2}\right)&=-kT_E\\ \frac{3\pi R^2}{2}&=kT_E\\ T_E&=\frac{3\pi R^2}{2k}=3T \end{aligned}

So the tank takes 3 times as long to empty.

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Syllabus alignment

Calculus

Rates of ChangeUse derivatives to model and solve rates of change, related rates, and exponential growth and decay problems.

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