2022 H2 Mathematics Paper 2 Question 2

Differential Equations (DEs)

Answers

(a)

k=252h.{k = \frac{25}{2} h.}

(b)

t=12 s.{t=12 \textrm{ s}.}

(c)

h=1011 m.{h=\frac{10}{11} \textrm{ m}.}

Full solutions

(a)

V=0.9h m2=900h litres\begin{align*} V &= 0.9h \textrm{ m}^2 \\ &= 900h \textrm{ litres} \end{align*}
Since the container is filled in 72{72} seconds,
72k=900hk=12.5h  ■\begin{align*} 72k &= 900h \\ k &= 12.5h \; \blacksquare \end{align*}

(b)

dVdt=ktV=∫kt  dt=k2t2+c\begin{align*} \frac{\mathrm{d}V}{\mathrm{d}t} &= kt \\ V &= \int kt \; \mathrm{d}t \\ &= \frac{k}{2} t^2 + c \end{align*}
When t=0,{t = 0, } V=0.{V=0.} Hence c=0.{c = 0.}
V=k2t212.5h2t2\begin{align*} V &= \frac{k}{2} t^2 \\ \frac{12.5h}{2} t^2 \end{align*}
When the container is full, V=900h.{V = 900h.}
900h=12.5h2t2t2=144t=12 s  ■\begin{align*} 900h &= \frac{12.5h}{2} t^2 \\ t^2 &= 144 \\ t &= 12 \textrm{ s} \; \blacksquare \end{align*}

(c)

dVdt=kt+25V=∫kt+25  dt=k2t2+25t+C\begin{align*} \frac{\mathrm{d}V}{\mathrm{d}t} &= kt + 25 \\ V &= \int kt + 25 \; \mathrm{d}t \\ &= \frac{k}{2} t^2 + 25t + C \end{align*}
When t=0,{t = 0, } V=0.{V=0.} Hence C=0.{C = 0.}
V=k2t2+25t=12.5h2t2+25t\begin{align*} V &= \frac{k}{2} t^2 + 25t \\ &= \frac{12.5h}{2} t^2 + 25 t \end{align*}
When the container is full, V=900h,{V = 900h,} t=10.{t=10.}
900h=12.5h2(10)2+25(10)275h=250h=1011 m  ■\begin{align*} 900h &= \frac{12.5h}{2} (10)^2 + 25(10) \\ 275 h &= 250 \\ h &= \frac{10}{11} \textrm{ m} \; \blacksquare \end{align*}

Question Commentary

It is interesting that differential equations came out again given that it has come up in paper 1 (I believe this is the first time we have seen this topic come out twice).

Once we identify the underlying topic the integration required here is much easier (and does not require the separation of variables). All in all this question feels more like a general application of mathematics to real life context.