   Chapter 6.5, Problem 20E

Chapter
Section
Textbook Problem

(a) It $1000 is borrowed at 8% interest, find the amounts due at the end of 3 years if the interest is compounded (i) annually, (ii) quarterly, (iii) monthly, (iv) weekly, (v) daily, (vi) hourly, and (vii) continuously.(b) Suppose$1000 is borrowed and the interest is compounded continuously. If A(t) is the amount due after t years, where 0 ≤ ≤ 3, graph A(t) for each of the interest rates 6%, 8%, and 10% on a common screen.

(a)

To determine

To find: The amounts due to the end of 3 years if the interest compounded (i) annually, (ii) quarterly, (iii) monthly, (iv) weekly, (v) daily, (vi) hourly, and (vii) continuously.

Explanation

Given:

Initial amount is \$1000.

Interest rate is 8%.

Number of years is 3.

Formula used:

The value of the investment is A(t)=A0(1+rn)nt.

Where,

A0 is the initial investment.

r is the interest rate.

n is the number of times the interest is compounded per year.

t is the number of years.

Calculation:

Section (i)

If n=1, the amount of A(3) is computed as follows,

Substitute A0=1000,r=0.08 and t=3 in A(t)=A0(1+rn)nt

A(3)=(1000)(1+0.081)1(3)=(1000)(1+0.08)3=(1000)(1.08)3=1259.71

Thus, the amount is A(3)=1259.71.

Section (ii)

If n=4, the amount of A(3) is computed as follows,

Substitute A0=1000,r=0.08 and t=3 in A(t)=A0(1+rn)nt

A(3)=(1000)(1+0.084)4(3)=(1000)(1+0.02)12=(1000)(1.02)12=1268.24

Thus, the amount is A(3)=1268.24.

Section (iii)

If n=12, the amount of A(3) is computed as follows,

Substitute A0=1000,r=0.08 and t=3 in A(t)=A0(1+rn)nt

A(3)=(1000)(1+0.0812)12(3)=(1000)(1+0.0066666)36=(1000)(1.0066666)36=1270.24

Thus, the amount is A(3)=1270.24.

Section (iv)

If n=52, the amount of A(3) is computed as follows,

Substitute A0=1000,r=0.08 and t=3 in A(t)=A0(1+rn)nt

A(3)=(1000)(1+0

(b)

To determine

To sketch: The graph of amount A(t) due after t years for each of the interest rates 6%,8% and 10%.

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