Question

A bank pays interest by continuous compounding, that is, by treating the interest rate as the instantaneous rate of change of principal. Suppose in an account interest accrues at 8% per year, compounded continuously. Calculate the percentage increase in such an account over one year.

Answer 1

Let P₀ be the initial amount and P be the amount at any time t.
We have,
$$\begin{gathered} \frac{dP}{dt}=\frac{8P}{100} \\ \Rightarrow \frac{dP}{dt}=\frac{2P}{25} \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{dP}{P}=\frac{2}{25}dt \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}t,\mathrm{we}\mathrm{get} \\ \log P=\frac{2}{25}t+C.....\left(1\right) \\ \mathrm{Now}, \\ P=P_0\mathrm{at}t=0 \\ \therefore \log P_0=0+C \\ \Rightarrow C=\log P_0 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \log P=\frac{2}{25}t+\log P_0 \\ \Rightarrow \log \frac{P}{P_0}=\frac{2}{25}t \\ \Rightarrow e^{\frac{2}{25}t}=\frac{P}{P_0} \\ \mathrm{To}\mathrm{find}\mathrm{the}\mathrm{amount}\mathrm{after}1\mathrm{year},\mathrm{we}\mathrm{have} \\ {e}^{\frac{2}{25}}=\frac{P}{P_0} \\ \Rightarrow e^{0.08}=\frac{P}{P_0} \\ \Rightarrow 1.0833=\frac{P}{P_0} \\ \Rightarrow P=1.0833P_0 \\ \mathrm{Percentage}\mathrm{increase}=\left(\frac{P-P_0}{P_0}\right)\times 100\% \\ =\left(\frac{1.0833P_0-P_0}{P_0}\right)\times 100\% \\ =0.0833\times 100\% \\ =8.33\% \end{gathered}$$