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.