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{\bf Question}

An appeal fund is launched with a donation of £1,000; $t$ weeks
later the fund stands at $£A$, and is growing at a rate of
$£Af(t)$, where

$$f(t)=\ds\frac{1000 t}{(t^2+125)^2}$$

models the growth and decline of enthusiasm of the sponsors. Write
down the differential equation governing the appeal, find $A$ in
terms of $t$, and the time it takes to reach the target of
$50,000$. To what value does the fund tend as $t \to \infty$?
\medskip

{\bf Answer}

Rate of growth of $A$: $\ds\frac{dA}{dt}=\ds\frac{A \cdot
1000t}{(t^2+125)^2}$

when $t=0,\ A=1000$

This is variables separable

$\ds\int\ds\frac{DA}{A}=\ds\int\ds\frac{1000t \,dt}{(t^2+125)^2}$

Use a substitution. Set $u=t^2+125,\ du=2t+dt$

$\Rightarrow \ln
A=\ds\int1000\ds\frac{du}{2}\times\ds\frac{1}{u^2}$

$\Rightarrow \ln A=500\left[-\ds\frac{1}{u}\right]+c$ where
$u=t^2+125$

Thus $\ln A=c-\ds\frac{500}{(t^2+125)}$

When $t=0,\ A=1000$

$\Rightarrow \ln(1000)=c-\ds\frac{500}{125}$

$\Rightarrow c=\ln(1000)+4$

Thus

$\ \ln A=\ln(1000)+4-\ds\frac{500}{(t^2+125)}$

$\Rightarrow
\ln\left(\ds\frac{A}{1000}\right)=\ds\frac{4t^2+500-500}{(t^2+125)}$

$\Rightarrow
\ln\left(\ds\frac{A}{1000}\right)=\ds\frac{4t^2}{(t^2+125)}$

$\Rightarrow \un{A=1000 \exp\left[\ds\frac{4t^2}{t^2+125}\right]}$

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When $A=50,000$ we have

$\
\ds\frac{50,000}{1000}=\exp\left[\ds\frac{4t^2}{t^2+125}\right]$

$\Rightarrow \ln(50)=\ds\frac{4t^2}{t^2+125}$

$\Rightarrow t^2\ln(50)+125\ln(50)=4t^2$

$\Rightarrow t^2=\ds\frac{125\ln(50)}{4-\ln(50)}$

or $\un{t=\sqrt{\ds\frac{125\ln(50)}{4-\ln(50)}}=74.6=75 weeks}$

As $t \to \infty$, $A \to 1000 \exp\left[\ds\frac{4\times
\infty}{\infty}\right]=1000e^4=£54,598 \approx £54,600$
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