Question

In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs. 100 double itself in 10 years. ${\log _e}2 = 0.6931$

Answer 1

Hint: Let us find the equation of rate of increment of principal amount per year and then integrate both sides of that equation to get the required value of r.

Complete step-by-step answer:
As we know that principal amount is increasing by r% per year.
And according to the formula of rate of increment if A be the initial amount and it is increasing by r% per year. Then it can be written as,
$\dfrac{{{\text{dA}}}}{{{\text{dt}}}}$ = r% of A, where t is in years and $\dfrac{{{\text{dA}}}}{{{\text{dt}}}}$ is the rate of increment of A per year and this is equal to r% of A.
Let the principal amount be denoted as P.
Then, according to the question.
$\dfrac{{{\text{dP}}}}{{{\text{dt}}}}$ = r% of P (1)
As we know that according to the percentage formula x% of B is written as $\dfrac{{\text{x}}}{{100}} \times B$.
So, above equation 1 can also be written as,
$\dfrac{{{\text{dP}}}}{{{\text{dt}}}}$ = $\dfrac{{\text{r}}}{{100}} \times {\text{P}}$
Now we had to find P in terms of t. So, that we can get a principal amount after t years.
So, let's multiply both sides of the above equation by $\dfrac{{{\text{dt}}}}{{\text{P}}}$. We get,
$\dfrac{{{\text{dP}}}}{{\text{P}}} = \dfrac{{{\text{r}} \times {\text{dt}}}}{{100}}$
Now to get P in terms of t we integrate both sides of the above equation .
So, $\int {\dfrac{{{\text{dP}}}}{{\text{P}}}} = \int {\dfrac{{{\text{r}} \times {\text{dt}}}}{{100}}}$ (2)
As we know that when t is equal to 0 (initially) then P is equal to Rs. 100.
And when t is equal to 10 years then P is equal to Rs. 200.
So, putting appropriate limits in the above integration. We get,
$\int\limits_{100}^{200} {\dfrac{{{\text{dP}}}}{{\text{P}}}} = \int\limits_0^{10} {\dfrac{{{\text{r}} \times {\text{dt}}}}{{100}}}$ (3)
As we know that $\dfrac{{\text{r}}}{{100}}$ is constant and it does not depend on time. It is the same for each year.
So, according to the integration conditions, if we have to find the integration of $\int {{\text{Adx}}}$ and A does not depend on t. then this integration can also be written as ${\text{A}}\int {{\text{dx}}}$.
Equation 3 can be written as,
$\int\limits_{100}^{200} {\dfrac{{{\text{dP}}}}{{\text{P}}}} = \dfrac{{\text{r}}}{{100}}\int\limits_0^{10} {{\text{dt}}}$ (4)
Now to solve the above equation we had to solve the above equation to find the value of r.
As we know that $\int\limits_{\text{A}}^{\text{B}} {{\text{dx}}} = \left[ {\text{x}} \right]_{\text{B}}^{\text{A}} = {\text{A}} - {\text{B}}$ and $\int\limits_{\text{A}}^{\text{B}} {\dfrac{{{\text{dx}}}}{{\text{x}}}} = \left[ {{{\log }_e}{\text{x}}} \right]_{\text{A}}^{\text{B}} = {\log _e}{\text{A}} - {\log _e}{\text{B}}$.
So, equation 4 can be written as,
$\left[ {{{\log }_e}{\text{P}}} \right]_{100}^{200} = \dfrac{{\text{r}}}{{100}}\left[ {\text{t}} \right]_0^{10}$. Now putting the limits in the above equation. We get,
$\left[ {{{\log }_e}200 - {{\log }_e}100} \right] = \dfrac{{\text{r}}}{{100}}\left[ {10 - 0} \right]$
As we know that according to logarithmic identities ${\log _e}{\text{xy}} = {\log _e}{\text{x}} + {\log _e}{\text{y}}$.
So, above equation becomes,
$\left[ {{{\log }_e}2 + {{\log }_e}100 - {{\log }_e}100} \right] = \dfrac{{\text{r}}}{{10}}$
${\log _e}2 = \dfrac{{\text{r}}}{{10}}$
As we are given in the question that ${\log _e}2 = 0.6931$.
So, $0.6931 = \dfrac{{\text{r}}}{{10}}$
r = 6.931
Hence, the value of r will be 6.931%.

Note: Whenever we come up with this type of question then first, we write rate of increment of principle as $\dfrac{{{\text{dP}}}}{{{\text{dt}}}}$ and the equate that with r% of P. After that we have to manipulate this equation such that we get P in terms of t. So, we multiply both sides of the equation by $\dfrac{{{\text{dt}}}}{{\text{P}}}$. And then apply integration to both sides of the integration with limits such that when t = 0 then P = 100 and when t = 10 then P = 200. After that using the integration formula we will get the required value of r. This will be the efficient and easiest way to find the solution of the problem.