2023 H2 Mathematics Paper 1 Question 10

Differential Equations (DEs)

Answers

{{\frac{\operatorname{d}\!M}{\operatorname{d}\!t} = k \left( C - 30M \right).}}

{{3300.}}

{{ M = 88 + 22 \operatorname{e}^{-30kt}.}}

{{51. }}

{{0 < C < 2400.}}

Since the rate of change of mass is proportional to the difference between energy intake and energy expenditure, we have

\begin{align*} \frac{\operatorname{d}\!M}{\operatorname{d}\!t} &= k \left( \text{intake} - \text{expenditure} \right) \\ &= k \left( C - 30M \right) \; \blacksquare \end{align*}

To maintain his mass,

\begin{align*} \frac{\operatorname{d}\!M}{\operatorname{d}\!t} &= 0 \\ k \left( C - 30M \right) &= 0 \\ C &= 30 M \\ &= 30(110) \\ &= 3300 \; \blacksquare \end{align*}

\begin{gather*} \frac{\operatorname{d}\!M}{\operatorname{d}\!t} = k \left( 0.8C - 30M \right) \\ \frac{\operatorname{d}\!M}{\operatorname{d}\!t} = k \left( 2640 - 30M \right) \\ \int \frac{1}{88 - M} \, \mathrm{d}M = \int 30 k \, \mathrm{d}t \\ - \ln | 88 - M | = 30kt + c \\ 88 - M = A \operatorname{e}^{-30kt} \end{gather*}

When ${{t = 0,}}$ ${{M = 110,}}$

\begin{gather*} 88-110 = A \operatorname{e}^0 \\ A = - 22 \end{gather*}

\begin{gather*} 88 - M = - 22 \operatorname{e}^{-30kt} \\ M = 88 + 22 \operatorname{e}^{-30kt} \; \blacksquare \end{gather*}

When ${{t = 75,}}$ ${{M = 100,}}$

\begin{align*} 100 &= 88 + 22 \operatorname{e}^{-30(75)k} \\ \operatorname{e}^{-2250k} &= \frac{12}{22} \\ k &= \frac{\ln \frac{6}{11}}{-2250} \\ &= 0.00026939 \end{align*}

When his mass falls below ${{96,}}$

\begin{align*} 88 + 22 \operatorname{e}^{-30kt} &< 96 \\ \operatorname{e}^{-30kt} &< \frac{8}{22} \\ -30kt &< \ln \frac{8}{22} \\ t &> \frac{\ln \frac{8}{22}}{-30k} \\ t &> 125.17 \end{align*}

\begin{align*} & \text{Additional days required} \\ & = 126 - 75 \\ & = 51 \text{ days} \; \blacksquare \end{align*}

(ei)

Since ${{ \operatorname{e}^{-30kt} > 0}}$ for all real values of ${{t,}}$

\begin{align*} M &= 88 + 22 \operatorname{e}^{-30kt} \\ &> 88 \end{align*}

Hence Andrew cannot achieve a mass of ${{80 \text{ kg} \; \blacksquare}}$

(eii)

For Andrew to achieve a mass of ${{80 \text{ kg},}}$

\begin{gather*} C - 30M &< 0 \\ C &< 30M \\ C &< 30(80) \\ C &< 2400 \end{gather*}

{ 0 < C < 2400 \; \blacksquare }