Money Flow  The rate of a continuous money flow starts at $1000 and increases exponentially at 5% per year for 4 years. Find the present value and accumulated amount if interest earned is 3.5% compounded continuously.

Respuesta :

Answer:

Present value =  $4,122.4

Accumulated amount = $4,742

Step-by-step explanation:

Data provided in the question:

Amount at the Start of money flow = $1,000

Increase in amount is exponentially at the rate of 5% per year

Time = 4 years

Interest rate = 3.5%  compounded continuously

Now,

Accumulated Value of the money flow = [tex]1000e^{0.05t}[/tex]

The present value of the money flow = [tex]\int\limits^4_0 {1000e^{0.05t}(e^{-0.035t})} \, dt[/tex]

= [tex]1000\int\limits^4_0 {e^{0.015t}} \, dt[/tex]

= [tex]1000\left [\frac{e^{0.015t}}{0.015} \right ]_0^4[/tex]

= [tex]1000\times\left [\frac{e^{0.015(4)}}{0.015} -\frac{e^{0.015(0)}}{0.015} \right][/tex]

= 1000 × [70.7891 - 66.6667]

= $4,122.4

Accumulated interest = [tex]e^{rt}\int\limits^4_0 {1000e^{0.05t}(e^{-0.035t}} \, dt[/tex]

= [tex]e^{0.035\times4}\times4,122.4[/tex]

= $4,742

The present value and interest accumulated would be as follows:

Present Value = $ 4,122.4

Interest Accumulated = $ 4742

Given that,

Principal at the beginning of money flow = $1,000

Exponential interest rate = 5% per year

Time Period = 4 years

So,

The accumulated money flow's worth = [tex]1000e^{0.05t}[/tex]

The current value of the money can be determined by [tex]\int\limits^4_0 1000e^{0.05t}(e^{-0.035t}) {} \, dt[/tex]

On solving, we get

The present value = $ 4,122.4

Interest Accumulated = $4,742

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