   Chapter 6.5, Problem 20E

Chapter
Section
Textbook Problem

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

To determine

(a)

To calculate:

The amount at the end if the interest is compounded

i. annually

ii. quarterly

iii. monthly

iv. weekly

v. daily

vi. hourly, and

vii. continuously

Explanation

Given:

P =$1000, r = 8% per annum and t = 3 years. P = Principal amount, A = amount at the end, r = rate of interest per annum and t = time (in years Formula used: 1. If the interest is compounded n-times in a year then the amount at the end is A=P(1+r100n)nt 2. If the interest is compounded continuously then the amount at the end is A(t)=Per100t Calculation: (i) If the interest is compounded annually: In this case the interest is compounded one time in a year, therefore, the amount at the end is: A=P(1+r100n)nt =1000(1+8100)3 =1259.71 Therefore, the amount at the end is$1259.71 if the interest is compounded annually.

(ii) If the interest is compounded quarterly:

In this case the interest is compounded 4 times in a year, therefore, the amount at the end is:

A=P(1+r100n)nt    =1000(1+84×100)3×4    =1268.24

Therefore, the amount at the end is \$1268.24 if the interest is compounded quarterly.

(iii) If the interest is compounded monthly:

In this case the interest is compounded 12 times in a year, therefore, the amount at the end is:

A=P(1+r100n)nt    =1000(1+812×100)3×12    =1270

To determine

(b):

To Graph

A(t) for different interests.

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