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

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

a)

$\int$ dS = $\int$ $100re^{rt/100} dt$

=> S(t) = 100r $\int e^{rt/100}dt$ , r is constant

=> S(t) = 100r ($e^{rt/100}$ )($\frac {100}{r}$ )+C

=> S(t) = 10000$e^{rt/100} + C$

When t =0 , S = 10000

So 10000 = 10000 + $C$

=> $C = 0$

So S(t) = 10000 $e^{rt/100}$

Investment at time t is,

S(t) = 10000$e^{rt/100}$

b)

S'(t) = 10000 $e^{rt/100}$ ($\frac {r}{100})$

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

[ Verified ]

c) S(t) is continuous for t≥0 as S(t) is

an exponential function and exponential function is continuous function everywhere.

d) $\lim_{t→∞} S(t)$

$= \lim_{t→∞} 10000e^{rt/100}$

= ∞ [as e> 1]

INTERPRETATION:

Investment grows without bound as time grows unboundedly.

e)

15000 = 10000 $e^{rt/100}$

=> $e^{rt/100}$ = 15000/10000 = 1.5

So rt/100 = ln(1.5)

=> t = $\frac {100}{r} ln(1.5)$

It will take $\frac {100}{r} ln(1.5)$ years to be 15000