Question

The function in Exercise represents the rate of flow of money in dollars per year. Assume a 10-year period at 8% compounded continuously and find the following: (a) the present value; (b) the accumulated amount of money flow at t=10.
f(t)=2000

Answer 1

Answer:

a) $P= 56,972.5$

b) $A= 1,26,792.3$

Step-by-step explanation:

Given Data:

Interest rate=$r= 0.08$  per year

No. of years=$t=10$

Rate of continuous money flow is given by the function

$f(t)=2000$

a) to find the present value of money

$P=\int\limits^n_0 {f(t)e^{-rt} } \, dt$

Put f(t)=2000 and n=10 years and r=0.08

$P=\int\limits^n_0 {2000e^{-0.08t} } \, dt$

Now integrate

$P= {2000(\frac{e^{-0.08t}}{-0.08} )$

$P= -\frac{2000}{0.08} (e^{-0.08*10}-e^{-0.08*0})$

$P= -\frac{2000}{0.08} (e^{-0.8}-e^{0})$

$P= -\frac{2000}{0.08} (0.4493-2.7282)$

$P= -\frac{2000}{0.08} (-2.2789)$

$P= -25000(-2.2789)$

$P= 56972.5$

(b) to find the accumulated amount of money at t=10

$A=P(e^{rt} )$

Where P is the present worth already calculated in part a

$A=56972.5(e^{0.08*10} )$

$A=56972.5(e^{0.8} )$

$A=56972.5(2.2255 )$

$A= 1,26,792.3$