Solution — 2020 Question 5 Exponential & Logarithmic Functions| Graphs| Differentiation & Applications

(a)

When $t = 0,$

\begin{matrix} 6 + c & = 10 \\ c & = 4 ∎ \end{matrix}

When $t = 1,$

\begin{matrix} 4 e^{- a} & = 2 \\ e^{- a} & = \frac{1}{2} \\ \ln e^{- a} & = \ln \frac{1}{2} \\ - a & = \ln \frac{1}{2} \\ a & = - \ln \frac{1}{2} ∎ \\ = \ln 2 ∎ \end{matrix}

(b)

\begin{matrix} P & = 6 + c e^{- a t} \\ = 6 + 4 e^{- (\ln 2) t} \\ \frac{d P}{d t} & = - 4 (\ln 2) e^{- (\ln 2) t} \end{matrix}

When $t = 2,$

\begin{matrix} \frac{d P}{d t} & = - 4 (\ln 2) e^{- (\ln 2) (2)} \\ = - 0.693 15 \end{matrix}

Hence the rate of change of the population is $- 0.693$ million/year $∎$

(c)

(d)

Approximate size of population is $6 million ∎ .$

(e)

\begin{matrix} Q & = \sqrt{t} + \frac{5}{\sqrt{t}} \\ = t^{\frac{1}{2}} + 5 t^{- \frac{1}{2}} \\ \frac{d Q}{d t} & = \frac{1}{2} t^{- \frac{1}{2}} - \frac{5}{2} t^{- \frac{3}{2}} \end{matrix}

At stationary points,

\begin{matrix} \frac{d Q}{d t} & = 0 \\ \frac{1}{2} t^{- \frac{1}{2}} - \frac{5}{2} t^{- \frac{3}{2}} & = 0 \\ \frac{1}{2} t^{- \frac{1}{2}} & = \frac{5}{2} t^{- \frac{3}{2}} \\ t^{- \frac{1}{2}} & = 5 t^{- \frac{3}{2}} \\ \frac{1}{\sqrt{t}} & = \frac{5}{t^{\frac{3}{2}}} \\ t^{\frac{3}{2}} & = 5 \sqrt{t} \\ t^{3} & = 25 t \\ t^{2} & = 25 \\ t & = 5 \end{matrix}

When $t = 5,$

\begin{matrix} Q & = \sqrt{5} + \frac{5}{\sqrt{5}} \\ = 2 \sqrt{5} \end{matrix}

Hence the stationary point is $(5, 2 \sqrt{5}) ∎$

(f)

\begin{matrix} 6 + 4 e^{- t \ln 2} & = \sqrt{t} + \frac{5}{\sqrt{t}} \end{matrix}

From GC, $t = 0.340 ∎$ and the size of each population at that time is $9.16 million ∎$