View source for 022 Exam 2 Sample A, Problem 5

<span class="exam"> '''Set up the equation to solve.  You only need to plug in the numbers - not solve for particular values!'''

<span class="exam">How much money would I have after 6 years if I invested $3000 in a bank account that paid 4.5% interest, 
::<span class="exam">(a) compounded monthly?
::<span class="exam">(b) compounded continuously?

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations: &nbsp;
|-
|The primary purpose of this problem is to demonstrate that you understand the difference between continuous compounding and compounding on an interval of time.  When we compound on an interval, say monthly, the value in the account only changes at the end of each interval.  In other words, there is no interest accrued for a week or a day.  As a result, we use the formula
|-
|
::<math>A\,=\,P\left(1+\frac{r}{n}\right)^{nt},</math>
|-
|where <math style="vertical-align: 0%;">A</math> is the value of the account, <math style="vertical-align: 0%;">P</math> is the principal (original amount invested), <math style="vertical-align: 0%;">r</math> is the annual rate and <math style="vertical-align: 0%;">n</math> is the number of compoundings per year.  The value of <math style="vertical-align: 0%;">n</math> is <math style="vertical-align: 0%;">365</math> for compounding daily, <math style="vertical-align: 0%;">52</math> for compounding weekly, and <math style="vertical-align: -5%;">12</math> for compounding monthly.  As a result, the exponent <math style="vertical-align: 0%;">nt</math>, where <math style="vertical-align: 0%;">t</math> is the time in years, is the number of compounding periods where we actually earn interest. Similarly, <math style="vertical-align: -22%;">r/n</math> is the rate per compounding period (the annual rate divided by the number of compoundings per year).
|-
|For example, if we compound monthly for <math style="vertical-align: 0%;">7</math> years at a <math style="vertical-align: -5%;">6%</math> rate, we would compound <math style="vertical-align: -5%;">nt\,=\,12 \cdot7\,=\,84</math> times, once per month, at a rate of <math style="vertical-align: -22%;">0.06/12\,=\,0.005</math> per monthly period. Notice that we <u>'''always'''</u> use the decimal version for interest rates when using these equations.
|-
|On the other hand, interest compounded continuously earns rate in just that way - continuously.  I can have any value I want for time <math style="vertical-align: 0%;">t</math>, and the total amount in the account will change with each and every second.  Therefore, we express the account value as 
|-
|
::<math>A\,=\,Pe^{rt}.</math>
|-
|The goal in the two parts of this problem is to choose the correct equation for each.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(a): &nbsp;
|-
|We are given all the pieces required.  We begin with <math style="vertical-align: -5%;">$3000</math> of principal, and compound monthly, or <math style="vertical-align: -5%;">12</math> times per year. Using the formula in 'Foundations', the equation for the account value is 
|-
|
::<math>A\,=\,P\left(1+\frac{r}{n}\right)^{nt}\,=\,3000\left(1+\frac{0.045}{12}\right)^{12\cdot 6}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(b): &nbsp;
|-
|Again, we need to apply the formula from foundations to find 
|-
|
::<math>A\,=\,Pe^{rt}\,=\,3000e^{0.045\cdot 6}.</math>

|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer: &nbsp;
|-
|
::(a) <math style="vertical-align: -80%;">A\,=\,3000\left(1+\frac{0.045}{12}\right)^{12\cdot 6}.</math> 
|-
|
::(b) <math style="vertical-align: 0%;">A\,=\,Pe^{rt}\,=\,3000e^{0.045\cdot 6}.</math>
|}

[[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]