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 \(6.93\%.\)