Question

The rate of change of the value of an investment, S, with respect to time, t ≥ 0, is given by dS dt =1000r 10ert/100, where r is the annual interest rate (assumed constant) and the principal of the investment is S(0) = 10 000. (a) Find an expression for S(t), that is, the value of the investment at time t. (6 marks) (b) Verify that your expression for S(t) is correct by computing S(t). (1 mark) (c) Explain why S(t) is continuous for t ≥ 0. (1 mark) (d) Determine, by computing lim t→∞S(t), what would happen to the value of the investment if t were to grow without bound. Interpret the result. (5 marks) (e) How long would it take for the value of the investment to be exactly 15 000? (4 marks)

Answer 1

Given $\frac{dS}{dt} = 1000r e^{\frac{rt}{100}}$ and $S(0)=10000$.

a) So, By integration of given differential equation, we get $S = 1000 r e^{\frac{rt}{100}} \frac{100}{r} +c = 100000 e^{\frac{rt}{100}} + c$

Now, $S(0) = 100000 + c = 10000 \implies c = - 90000$

Hence, $S = 100000 e^{\frac{rt}{100}}- 90000$

b) Derivative of $S$ is : $\frac{ds}{dt} = 100000 \ e^{\frac{rt}{100}} \frac{r}{100} = 1000 r e^\frac{rt}{100}$ and $S(0) = 100000$

Hence, verified.

c) Exponential function is always continuous, hence S(t) is continuous.

d) $\lim_{t\to \infin} S(t) = \lim_{t\to \infin} 1000 00 e^{\frac{rt}{100}}- 90000 \to \infin$ because $r\geq 0$.

Hence, value of investment goes to infinity as t goes without bound.

e) Find $t_1,$ such that $S(t_1)=15000$.

$\implies 100000 e^\frac{rt_1}{100}-90000 = 15000 \\ \implies e^\frac{rt_1}{100} = 1.05 \\ \implies t_1 = \frac{100 ln(1.05)}{r} \approx \frac{4.88}{r}$