Q.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 $(\log_e 2 = 0.6931)$.

The key idea is Exponential Growth Rate: when a quantity increases continuously at a fixed percentage rate, it follows the law $\frac{dP}{dt} = \frac{r}{100} P$.

Step 1 – Set up the differential equation

Let $P$ be the principal. Continuous growth at $r\%$ per year gives:

Step 2 – Solve

Separate variables and integrate:

So $P = P_0 e^{rt/100}$, where $P_0$ is the initial principal.

Step 3 – Apply given data …

The key idea here is that "principal increases continuously at the rate of $r\%$ per year" means we are dealing with exponential growth — the same mathematics that governs population growth, radioactive decay, and compound interest compounded every instant. Unlike simple interest or annual compounding, continuous growth means the money is growing at every moment, and the growth itself is proportional to the current amount.

This is a classic differential equation situation: if $P(t)$ is the principal at time $t$ (in years), then the statement "increases continuously at the rate of $r\%$ per year" translates to:

The factor $\frac{r}{100}$ converts the percentage rate into a decimal. The solution to this is $P(t) = P_0 e^{rt/100}$, where $P_0$ is the initial principal.

Now let's work through the problem step by step.

1. Set up the continuous growth equation. Let $P_0 = 100$ (Rs). After $t = 10$ years, the amount doubles to $P = 200$. The continuous growth formula is:

Substituting the known values:

2. Simplify the equation. Divide both sides by 100:

3. Solve for $r$ using natural logarithms. Take the natural log of both sides:

Therefore:

4. Plug in the given value of $\ln 2$. …

Method: Continuous growth ("rate proportional to amount")

Use this for interest/population problems where a quantity grows continuously at a rate proportional to itself, and you must find the rate constant.

Steps

Step 1: Write the proportional-rate equation

"Increases continuously at rate $r\%$" means

Step 2: Separate and integrate to an exponential

$\dfrac{dP}{P}=\dfrac{r}{100}\,dt$ gives $P=P_0\,e^{rt/100}$.

Step 3: Use the given condition and take logs …

Common Mistakes

Mistake 1: Writing the rate as $r$ instead of $\dfrac{r}{100}$

Why it's wrong: "$r\%$ per year" means $\frac{dP}{dt}=\frac{r}{100}P$; using $r$ directly gives an answer off by a factor of 100. Correct approach: convert the percentage to $\frac{r}{100}$.

Mistake 2: Not taking logarithms to release the exponent …