Question

In a bank,principal increases continuously at the rate of \(5\%\) per year. An amount of Rs1000 is deposited with this bank,how much will it worth after 10years\((e^{0.5}=1.648)\).

Answer 1

Let p and t be the principal and time respectively.
It is given that the principal increases continuously at the rate of \(5\%\) per year.
\(⇒\frac{dp}{dt}=(\frac{5}{100})p\)
\(⇒\frac{dp}{dt}=\frac{p}{20}\)
\(⇒\frac{dp}{p}=\frac{dt}{20}\) [Seperating variables]
Integrating both sides,we get:
\(∫\frac{dp}{p}=\frac{1}{20}∫dt\)
\(⇒log\,p=\frac{t}{20}+C\)
\(⇒p=\frac{t}{e^{20}}+C...(1)\)
Now,then t=0,p=1000.
\(⇒1000=e^c...(2)\)
At t=10,equation(1)becomes:
\(p=\frac{1}{e^2}+C\)
\(⇒p=e^{0.5}×e^C\)
\(⇒p=1.648\times1000\)
\(⇒p=1648\)
Hence,after 10years the amount will worth Rs1648.