The compound interest formula appears in many classes. It can be confusing to students when it appears in one class as

and in another as

These are basically the same formulas, but used in a different context. However, how you solve for the different quantities in either one is the same. The Math-FAQs below demonstrate how to solve for

This week you will be graphing the function from Project 3. To find the equation for this function, you need to utilize the initial population and doubling time of the population. The goal of this post is to help you to find the rate r in the function . You will need to use the doubling time assigned to you in the project letter to do this. Continue reading “How Do I Use the Doubling Time to Find the Rate?”

A couple of you have had trouble finding the payment properly. I am not sure whether this is a calculator issue (there is a screen capture for the payment in Question 3 of Section 5.4) or a formula issue. Often I see people using the interest rate for i instead of the interest rate per period. Remember, since i is the interest rate per period you need to divide it by the number of periods in a year.

Here is an additional example of finding the payment and then using it to construct an amortization table.

After a product is released or an advertising campaign is finished, sales usually drop off. Often this decrease is modeled with an exponential function. This example shows how to find when the sales have dropped a a predetermined level. This requires us to convert from exponential to logarithm…a key skill from Section 5.2.

Problem Monthly sales of a Blue Ray player are approximately

where t is the number of months the Blue Ray player has been on the market.

a. Find the initial sales.

Solution The initial sales occur at t = 0. The corresponding sales are

or 250,000 units.

b. In how many months will sales reach 500,000 units?

Solution Set S(t) equal to 500 and solve for t.

c. Will sales ever reach 1000 thousand units?

Solution Follow steps similar to part b.

Since the logarithm of zero is not defined, sales will never be 1000 thousand units.

d. Is there a limit for sales?

Solution To help us answer this question, let’s look at a graph of S(t).

Examining the graph, it appears that the sales are getting closer and closer to 1000 units, but never quite get there (part c). So the limit for sales is 1000 thousand units or 1,000,000 units. This is due to the fact that as t increases, e^{–t} gets smaller and smaller so all that is left from S(t) is 1000.

To find the rate for a sinking fund, you needed to use a graph or the TVM Solver on a TI graphing calculator to find the interest rate on a sinking fund. Let’s look at the most basic way of doing this by graphing.

Problem Find the interest rate needed for a sinking fund to accumulate 25,600 dollars in five years with monthly payments (made at the end of the month) of 400 dollars.

Solution Since this problem indicates that payments are being made, we need to use the ordinary annuity formula. For a sinking fund, the future value is known, 25600. We also know the payment, 400, and the number of periods, 60:

This is incredibly difficult to solve for i algebraically. Instead, graph each side of the equation and look for a point of intersection. On a graphing calculator we would type y1 = 400*((1+x)^60-1)/x and y2 = 25600 in the window [0, 0.004] by [23500, 27500].

The toughest part is coming up with the window on the graph. Since the x on your calculator represents the interest rate per period, it needs to small like 0 to 0.01 initially. The vertical window is much easier to determine since the second equation is y = 25600, a horizontal line. Simply pick a horizontal window that spans this value.

The annual interest rate is 12 · 0.00217 ≈ 0.02604 or approximately 2.6%.