In a bank,the principal increases continuousl | vedclass

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Question

(A) Let $P$ be the principal at any time $t$. According to the given problem,the rate of increase is $\frac{dP}{dt} = \frac{5}{100} P = \frac{P}{20}$.

Vedclass · Accepted Answer

$t = 20 \log_e 2$

Vedclass · Answer

(A) Let $P$ be the principal at any time $t$. According to the given problem,the rate of increase is $\frac{dP}{dt} = \frac{5}{100} P = \frac{P}{20}$.Separating the variables,we get $\frac{dP}{P} = \frac{dt}{20}$.Integrating both sides,we get $\int \frac{dP}{P} = \int \frac{dt}{20}$,which gives $\log_e P = \frac{t}{20} + C_1$.This implies $P = e^{\frac{t}{20} + C_1} = C e^{\frac{t}{20}}$,where $C = e^{C_1}$.At $t = 0$,$P = 1000$. Substituting these values,$1000 = C e^0$,so $C = 1000$.Thus,the equation is $P = 1000 e^{\frac{t}{20}}$.To find the time $t$ when the principal doubles,we set $P = 2000$:$2000 = 1000 e^{\frac{t}{20}}$$2 = e^{\frac{t}{20}}$Taking the natural logarithm on both sides,$\log_e 2 = \frac{t}{20}$.Therefore,$t = 20 \log_e 2$ years.

In a bank,the principal increases continuously at the rate of $5 \%$ per year. In how many years will Rs $1000$ double itself?

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