Question

dS dt
=1000r 100 e(r/100)
The rate of change of the value of an investment, S, with respect to time, t ≥ 0, is given by
where r is the annual interest rate (assumed constant) and the principal of the investment is S(0) = 1 000
1.(a) Find an expression for S(t), that is, the value of the investment at time t.

Answer 1

dS/dt = 1000($\frac {r} {100}$ )$e^{r/100}$

dS = 10r$e^{r/100}$ dt

Integrating with respect to t

$\int$ dS = $\int$ 10r$e^{r/100}$ dt +C , where C is integration constant.

S = 10r$e^{r/100}$ t + C

When t = 0, S =1000

So 1000 = C

Therefore S = 10r$e^{r/100}$ t + 1000

The value of investment at time t is

S(t) = 10r$e^{r/100}$ t + 1000

$\mathbb { NOTE}$

As per growth rate formula if a correction is made as

$\frac {dS}{dt}$ = 1000($\frac {r}{100}$ ) $e^{rt/100}$

$\int$ dS = $\int$ 10r$e^{rt/100}$ dt

=> S = 10r$e^{rt/100}$ .$\frac {100}{r}$ + C

=> S = 1000$e^{rt/100}$ +C

When t= 0 , S = 1000

=> 1000 = 1000 + C

=> C = 0

So S = 1000$e^{rt/100}$

The value of investment at time t is

S(t) = 1000$e^{rt/100}$