Table of Contents
Quick Intro II
This is like a mini-lesson with an overview of the main objects of study. It will often contain a list of key words, definitions and properties – all that is new in this lesson. We will use this opportunity to make connections with other concepts. It can be also used as a review of the lesson.
A Quick Intro to Compound Interest
Key Words. Compound Interest, principal/initial amount , annual interest rate, compounding, compounded annually, compounded monthly, compounded quarterly, compounded daily, continuous compounding, final amount.
Suppose the $P$ dollars in principal is invested at an annual rate $r$.
$\bigstar$ For interest compounded $n$ times per year, the final amount obtained after $t$ years is
$$A= P\left(1+\dfrac{r}{n}\right)^{tn}.$$
$\bigstar$ For interest compounded continuously, the final amount obtained after $t$ years is
$$A = Pe^{rt}.$$
- Compounded annually means that the interest is compounded every year, so once a year and $n=1$.
- Compounded quarterly means that the interest is compounded four times a year, so $n=4$.
- Compounded monthly means that the interest is compounded every month, so twelve times a year and $n=12$.
- Compounded daily means that the interest is compounded every day, so $n=365$
Video Lesson II
Many times the mini-lesson will not be enough for you to start working on the problems. You need to see someone explaining the material to you. In the video you will find a variety of examples, solved step-by-step – starting from a simple one to a more complex one. Feel free to play them as many times as you need. Pause, rewind, replay, stop… follow your pace!
Video Lesson 2
A description of the video
This video is used in Lessons 36 and 37. For today’s lesson, watch from [5:23] to [9:16].
In this video you will the following example:
- An investment of \$1,000 earning interest at a rate of $2\%$ per year compounded annually.
Try Questions II
Now that you have read the material and watched the video, it is your turn to put in practice what you have learned. We encourage you to try the Try Questions on your own. When you are done, click on the “Show answer” tab to see if you got the correct answer.
Try Question 4
Determine the final amount in a savings account when \$700 is invested at $3\%$, compounded quarterly, for 7 years.
Show Answer 4
We use the formula $A=P\left(1+\dfrac{r}{n}\right)^{nt}$ where $P=700$, $r=0.03$, $n=4$, and $t=7$. So
\[A=700\left(1+\dfrac{0.03}{4}\right)^{4\cdot 7} \approx 862.9. \]
The final amount at the end of 7 years will be \$862.90 approximately.
WeBWorK II
You should now be ready to start working on the WeBWorK problems. Doing the homework is an essential part of learning. It will help you practice the lesson and reinforce your knowledge.
WeBWorK II
It is time to do the homework on WeBWork:
CompoundInterests
When you are done, come back to this page for the Exit Questions.
Exit Questions II
After doing the WeBWorK problems, come back to this page. The Exit Questions include vocabulary checking and conceptual questions. Knowing the vocabulary accurately is important for us to communicate. You will also find one last problem. All these questions will give you an idea as to whether or not you have mastered the material. Remember: the “Show Answer” tab is there for you to check your work!
Exit Questions
- Why are we dividing by $n$ in the interest formula?
- Why is 1 being added to $r/ n$ in the formula?
- Can we use the interest formula in the context of population growth?
$\bigstar$ How much do you have to invest today at $2\%$ compounded monthly to obtain \$50,000 in return in 3 years? What if it is compounded continuously? Which one is better?
Show Answer
$\bullet$ For the monthly compounding interest, we use the formula $A=P\left(1+\dfrac{r}{n}\right)^{nt}$. We set the equation
\[50000 = P\left(1+\dfrac{0.02}{12}\right)^{12\cdot 3}\]
and solve for $P$:
\[ P=\dfrac{50000}{\left(1+\dfrac{0.02}{12}\right)^{36} } \approx 47090.58. \]
$\bullet$ For the continuous compound interest, we use the formula $A=Pe^{rt}$. We set the equation
\[50000 = Pe^{0.02\cdot 3}\]
and solve for $P$:
\[ P=\dfrac{50000}{e^{0.06}} \approx 47088.23. \]
$\bullet$ Conclusion: The continuous compounding is slightly better.
Need more help?
Don’t wait too long to do the following.
- Watch the additional video resources.
- Talk to your instructor.
- Form a study group.
- Visit a tutor. For more information, check the tutoring page.
