Exercise 9.3 · Q20: Continuous interest: Rs 100 doubles in 10 years, so $r = 10\log_e 2 \approx 6.93$

Basics you need

Solution

1. Name the variables and write the model

   Let $t$ be the time in years and $P$ the principal in rupees at time $t$⁠. Growth at $r\%$ per year, continuously, means

   $$\frac{dP}{dt} = \frac{r}{100}\times P.$$
   At every instant the rate of increase is $r$ per cent of the amount present at that instant. The equation has order 1 and degree 1; $r$ is an unknown number, not a variable.
   A rate of change is a derivative. Basics: e^{kt} and its rate

2. Separate the variables

   The principal is always positive, so divide by $P$⁠:

   $$\frac{dP}{P} = \frac{r}{100}\,dt.$$
   Here $h(P) = P$ and $g(t) = \frac{r}{100}$⁠, a constant. Dividing by $P$ is allowed because $P > 0$⁠.
   Section 9.4.1, with $t$ and $P$ in place of $x$ and $y$⁠. Lesson: separate, then integrate

3. Integrate both sides, with one constant

   $$\log P = \frac{rt}{100} + C.$$
   $\int\frac{dP}{P} = \log|P| = \log P$ since $P > 0$⁠, and $\int\frac{r}{100}\,dt = \frac{rt}{100}$⁠. The two constants combine into one $C$⁠.
   Section 7.2, formulas (i) and (xii). Basics: every method ends with an integral

4. Use the starting amount

   At $t = 0$⁠, $P = 100$⁠: $\log 100 = 0 + C$⁠, so $C = \log 100$ and

   $$\begin{aligned} \log P - \log 100 &= \frac{rt}{100}, \\ \text{i.e.}\quad \log\frac{P}{100} &= \frac{rt}{100}. \end{aligned}$$
   The condition fixes the one constant. Equivalently $P = 100e^{rt/100}$⁠, the particular solution.
   Section 9.3: a particular solution from a condition. Basics: one constant, one condition

5. Use the doubling to find $r$

   Rs 100 doubles in 10 years: $P = 200$ when $t = 10$⁠. Then $\frac{P}{100} = 2$ and

   $$\begin{aligned} \log_e 2 &= \frac{10r}{100} = \frac{r}{10} \\ r &= 10\log_e 2 \\ &= 10 \times 0.6931 = 6.931. \end{aligned}$$
   This one equation has the one unknown $r$⁠. The given value $\log_e 2 = 0.6931$ is used, so $r = 6.931$⁠, which is $6.93$ to two decimal places.
   $e^k = 2 \iff k = \log_e 2$⁠. Lesson: interest every instant

6. Check, and compare with yearly compounding

   With $r = 6.931$⁠: $P(10) = 100e^{0.6931} \approx 199.99$⁠, which is 200 up to the rounding of $\log_e 2$⁠. The exact rate is $r = 10\log_e 2 \approx 6.9315$⁠.

   If instead $6.931\%$ were added once a year, after 10 years Rs 100 would be $100 \times 1.06931^{10} \approx 195.45$⁠, short of 200.

   Substituting back confirms the answer. Continuous growth adds growth on the growth at every instant, so it doubles the money faster than yearly compounding at the same rate.
   $\frac{d}{dt}e^{kt} = ke^{kt}$⁠. Basics: e^{kt} is k times itself

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