Question

A)The rate of change of the value of an investment ,S, with respect to time, t>= 0, is given by ds/dt= 1000(r/10)^e^rt/100 Where r is the annual interest rate (assumed constant) and the principal of the investment S(0)=10 000 1. Find an expression for S(t), that is, the value of the investment at time t. 2. Verify that your expression for S(t) is correct by computing S'(t) 3. Explain why S(t) is continuous for t>=O
4..How long would it take for the investment to be exactly 15 000

Answer 1

since $\frac{ds}{dt} = 1000(\frac{r}{10})(e^{\frac{rt}{100}})$

(i) Expression for S(t)

${ds} = 1000(\frac{r}{10})(e^{\frac{rt}{100}}){dt}$

Integrating both sides,

$\int_0^t{ds} = \int_0^t1000(\frac{r}{10})(e^{\frac{rt}{100}}){dt}$

Solving the integration on both sides,

$S(t) - S(0) = 10000[e^{\frac{rt}{100}}]_0^t$

$S(t) = 10000(e^{\frac{rt}{100}} )$ . . . . . . . . (i)

(ii) Differentiating equation (i) on both sides,

S'(t) = $1000(\frac{r}{10})(e^{\frac{rt}{100}})$

(iii)

Look at the graph, For t >= 0, It is continuous, there is no break in the curve at any point.

Since Function is exponential and exponential function is continuous so it is also continuous.

(iv) For calculating time,

15000 = $10000(e^{\frac{rt}{100}} )$

Solving it we get,

$t = \frac{100log(1.5)}{r} = \frac{17.609}{r} years$

If r = 1 then

t = 17.609 years = 17 years 7 months (approximately)