Question

I na bank,principal increses continusoly at the rate of r% per year.find thevalue of r if Rs 100 double itself in 10 years(log e^2=0.6931)?I na bank,principal increses continusoly at the rate of 5% per year.find thevalue of r if Rs 1000 double itself in 10 years(e^1/2= 1.648)?

Answer 1

Answer on Question #37546 – Math - Differential Calculus

Let $p, t$, and $r$ represent the principal, time, and rate of interest respectively. It is given that the principal increases continuously at the rate of $r\%$ per year. So we have

$$\frac{dp}{dt} = p \left(\frac{r}{100}\right)$$ $$\frac{dp}{p} = \frac{r}{100} dt$$

Integrate both sides

$$\int \frac{dp}{p} = \int \frac{r}{100} dt$$ $$\ln p = \frac{rt}{100} + \text{const} = \frac{rt}{100} + k$$ $$p = e^{\frac{rt}{100} + k}$$

When $t = 0, p = 100$:

$$100 = e^{0 + k} = e^{k}$$

If $t = 10$, then $p = 2 \cdot 100 = 200$:

$$200 = e^{\frac{r}{10} + k}$$ $$200 = e^{\frac{r}{10}} \cdot e^{k} = e^{\frac{r}{10}} \cdot 100$$ $$\frac{r}{10} = \ln 2$$ $$r = 0.6931 \cdot 10 = 6.931$$

The value of $r$ is 6.93%.

Consider the second case. Let $p$ and $t$ be the principal and time respectively. It is given that the principal increases continuously at the rate 5% per year. Then we have

$$\frac{dp}{dt} = p \left(\frac{5}{100}\right)$$ $$\int \frac{dp}{p} = \int \frac{1}{20} dt$$ $$\ln p = \frac{t}{20} + k$$ $$p = e^{\frac{t}{20} + k}$$

When $t = 0, p = 1000$:

$$1000 = e^{0 + k} = e^{k}$$

If $t = 10$, then:

$$p = e^{\frac{1}{2} + k}$$ $$p = e^{1/2} \cdot e^{k} = e^{\frac{1}{2}} \cdot 1000 = 1.648 \cdot 1000 = 1648$$

After 10 years the amount will worth $1648$.