2016 H2 Mathematics Paper 2 Question 1

Differentiation II: Maxima, Minima, Rates of Change

Answers

0.0251 m/min{0.0251 \textrm{ m/min}}

Full solutions

Let V{V} and h{h} denote the volume and depth of the water respectively
tan⁡α=rhrh=0.5r=12h\begin{align*} \tan \alpha &= \frac{r}{h} \\ \frac{r}{h} &= 0.5 \\ r &= \frac{1}{2}h \end{align*}
V=13πr2h=13π(12h)2h=112πh3\begin{align*} V &= \frac{1}{3} \pi r^2 h \\ &= \frac{1}{3} \pi \left( \frac{1}{2} h \right)^2 h \\ &= \frac{1}{12} \pi h^3 \end{align*}
dVdh=14πh2\frac{\mathrm{d}V}{\mathrm{d}h} = \frac{1}{4}\pi h^2
dVdt=dVdh×dhdt0.1=14πh2×dhdtdhdt=25πh2\begin{align*} \frac{\mathrm{d}V}{\mathrm{d}t} &= \frac{\mathrm{d}V}{\mathrm{d}h} \times \frac{\mathrm{d}h}{\mathrm{d}t} \\ 0.1 &= \frac{1}{4} \pi h^2 \times \frac{\mathrm{d}h}{\mathrm{d}t} \\ \frac{\mathrm{d}h}{\mathrm{d}t} &= \frac{2}{5\pi h^2} \end{align*}
When the volume of water in the container is 3 m3,{3 \textrm{ m}^3,}
112πh3=3h=36π3\begin{align*} \frac{1}{12} \pi h^3 &= 3 \\ h &= \sqrt[3]{\frac{36}{\pi}} \end{align*}
Rate of increase of the depth of water:
dhdt=25π÷(36π3)2=0.0251 m/min (3sf)  ■\begin{align*} \frac{\mathrm{d}h}{\mathrm{d}t} &= \frac{2}{5\pi} \div \left( \sqrt[3]{\frac{36}{\pi}} \right)^2 \\ &= 0.0251 \textrm{ m/min} \textrm{ (3sf)} \; \blacksquare \end{align*}