2023HSCShort answer

Q13 (a) (i)

2 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).

(i) Show that

\frac{dh}{dt}=\frac{k}{\pi h}.

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

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	2
Obtains $\frac{dV}{dh}$ , or equivalent merit	1

Sample answer:

\begin{aligned} V&=\pi\left(Rh^2-\frac{h^3}{3}\right)\\ \frac{dV}{dt}&=\frac{dV}{dh}\times\frac{dh}{dt}\\ \frac{dV}{dt}&=\pi(2Rh-h^2)\frac{dh}{dt}=\pi h(2R-h)\frac{dh}{dt} \end{aligned}

But we are given

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

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

\pi h\frac{dh}{dt}=k\quad\text{as }0\le h\le R\text{ so }2R-h\ne0

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

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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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