Question

If the interest is compounded continuously at 6% per annum, how much worth Rs 1000 will be after 10 years? How long will it take to double Rs 1000?

Answer 1

Let P₀ be the initial amount and P be the amount at any time t. Then,
$$\frac{dP}{dt}=\frac{6P}{100}$$
$$\Rightarrow \frac{dP}{dt}=0.06P$$
$$\Rightarrow \frac{dP}{P}=0.06dt$$
Integrating both sides with respect to t, we get
$$\log P=0.06t+C$$
Now,
$$P=P_0\text{ at }t=0$$
$$\therefore \log P_0=0+C$$
$$\Rightarrow C=\log P_0$$
Putting the value of C, we get
$$\log P=0.06t+\log P_0$$
$$\Rightarrow \log \frac{P}{P_0}=0.06t$$
$$\Rightarrow e^{0.06t}=\frac{P}{P_0}$$
To find the amount after 10 years, we get
$$\Rightarrow e^{0.06\times 10}=\frac{P}{P_0}$$
$$\Rightarrow e^{0.6}=\frac{P}{P_0}$$
$$\Rightarrow 1.822=\frac{P}{P_0}$$
$$\Rightarrow P=1.822P_0$$
$$\Rightarrow P=1.822\times 1000=\text{ Rs. }1822$$
To find the time after which the amount will double, we have
$$P=2P_0$$
$$\therefore \log \frac{2P_0}{P_0}=0.06t$$
$$\Rightarrow \log 2=0.06t$$
$$\Rightarrow t=\frac{0.6931}{0.06}=11.55\text{ years }$$