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(A) Let $P$ be the principal at time $t$. It is given that the principal increases continuously at the rate of $r \%$ per year.$\frac{dP}{dt} = \left(

Vedclass · Accepted Answer

$6.93$

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(A) Let $P$ be the principal at time $t$. It is given that the principal increases continuously at the rate of $r \%$ per year.$\frac{dP}{dt} = \left(\frac{r}{100}\right) P$Separating the variables,we get:$\frac{dP}{P} = \frac{r}{100} dt$Integrating both sides:$\int \frac{dP}{P} = \int \frac{r}{100} dt$$\log_{e} P = \frac{rt}{100} + C$At $t = 0$,$P = 100$. Substituting these values:$\log_{e} 100 = \frac{r(0)}{100} + C \Rightarrow C = \log_{e} 100$So,$\log_{e} P = \frac{rt}{100} + \log_{e} 100$$\log_{e} P - \log_{e} 100 = \frac{rt}{100}$$\log_{e} \left(\frac{P}{100}\right) = \frac{rt}{100}$Given that the principal doubles in $10$ years,so at $t = 10$,$P = 200$:$\log_{e} \left(\frac{200}{100}\right) = \frac{r(10)}{100}$$\log_{e} 2 = \frac{r}{10}$Given $\log_{e} 2 = 0.6931$:$0.6931 = \frac{r}{10}$$r = 6.931 \%$Thus,the value of $r$ is $6.931 \%$.

In a bank,the principal increases continuously at the rate of $r \%$ per year. Find the value of $r$ if $Rs \, 100$ doubles itself in $10$ years $\left(\log _{e} 2=0.6931\right)$. (in $\%$)

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