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,
\[\frac{dP}{dt} = \frac{8P}{100}\]
\[ \Rightarrow \frac{dP}{dt} = \frac{2P}{25}\]
\[\Rightarrow \frac{dP}{P} = \frac{2}{25}dt\]
Integrating both sides with respect to t, we get
\[\log P = \frac{2}{25}t + C . . . . . \left( 1 \right)\]
Now,
\[P = P_0\text{ at }t = 0 \]
\[ \therefore \log P_0 = 0 + C\]
\[ \Rightarrow C = \log P_0 \]
\[\text{ Putting the value of C in }\left( 1 \right),\text{ we get }\]
\[\log P = \frac{2}{25}t + \log P_0 \]
\[ \Rightarrow \log\frac{P}{P_0} = \frac{2}{25}t\]
\[ \Rightarrow e^{\frac{2}{25}t} = \frac{P}{P_0}\]
To find the amount after 1 year, we have
\[ e^\frac{2}{25} = \frac{P}{P_0}\]
\[ \Rightarrow e^{0 . 08} = \frac{P}{P_0}\]
\[ \Rightarrow 1 . 0833 = \frac{P}{P_0}\]
\[ \Rightarrow P = 1 . 0833 P_0 \]
\[\text{ Percentage increase }= \left( \frac{P - P_0}{P_0} \right) \times 100 % \]
\[ = \left( \frac{1 . 0833 P_0 - P_0}{P_0} \right) \times 100 % \]
\[ = 0 . 0833 \times 100 % \]
\[ = 8 . 33 %\]