Question

6. For the applications that follow, you must use a variation of N=N0e^(kt) when necessary.

A. A local bank advertises that it compounds interest continuously and that $2200 will become $3500 in 3 years. What is the annual interest rate?

B. The growth rate of the demand for electricity in a certain foreign country is 12.5% per year. If the demand grows exponentially, when will the demand be one and a half times that of 2019

C. Joanne wants to have $5000 available in 5 years to pay for her driveway to be repaved. Interest is currently 1.4% compounded continuously. How much money should be invested now>

Answer 1

Ans. A) The expression for the final amount which is compounded anually:

	=	final amount
	=	initial principal balance
	=	interest rate
	=	number of times interest applied per time period
	=	number of time periods elapsed

Here, A= $3500 , P = $2200 , n=1 ( as compounded annually) , t= 3 yrs

% (Ans.)

Ans. B) Given, Demand grows exponentially.

Let Demand at year 2019 D₀.

Let Demand at certain time in future be D.

so expression for exponential growth is:

Here, k = 12.5%

Let time at which D becomes 1.5 times D₀be t.

1.5D₀₌D_0.

^(Ans)

Ans. C) The expression for continuous compounding is

	=	value at time t
	=	original principal sum
	=	nominal annual interest rate
	=	length of time the interest is applied

Here, t=5 yrs, P(5) = $5000, r = 1.4%

Let amount to be invested is P₀

So,  

P₀  = 5000/

P₀  = $4661.9691 (Ans.)