Question

In a bank principal increases continuously at the rate of $r\%$ per year.Find the value of $r$ is Rs100 doubles itself in 10years$(log_e2=0.6931).$

Answer 1

Let $p,t,$and $r$ represent the principal,time,and rate of interest respectively.
It is given that the principal increases continuously at the rate of $r\%$ per year.
$⇒\frac{dp}{dt}=(\frac{r}{100})p$
$⇒\frac{dp}{p}=(\frac{r}{100})dt$ [Seperating Variables]
Integrating both sides,we get:
$∫\frac{dp}{p}=\frac{r}{100}∫dt$
⇒logP=\frac{rt}{100}+k$$⇒p=\frac{rt}{e^{100}}+k...(1)
It is given that when t=0,p=100.
$⇒100=e^k...(2)$
Now,if t=10,then $P=2\times100=200.$
Therefore,equation(1),becomes:
$200=\frac{r}{e^{10}}+k$
$⇒200=\frac{r}{e^{10}}.e^k$
⇒$200=\frac{r}{e^{10.100}}$ (from(2))
$⇒\frac{r}{e^{10}}=2$
$⇒\frac{r}{10}=log_e2$
$⇒\frac{r}{10}=0.6931$
$⇒r=6.931$
Hence,the value of r is $6.93\%.$