/ Logarithmic Exponential applications Compound interest
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The first part of this worksheet investigates the connection between the two formulas we will use for compound interest. We will justify for ourselves that the continuous compounding formula A = Pe rt works. The second part of the worksheet focuses on solving problems involving these equations. The first four are guided examples. They should provide enough examples to get through the last four on your own.
Recall the formula for compound interest is A = P 1 + r n
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)
nt
where A is the amount in an
account after you invest P dollars (the principal) for t years. The interest is compounded n times a year at an annual interest rate r (decimal form). If the account is compounded continuously, the formula is reduced to A = Pe rt where A is the amount in an account after you invest P dollars (the principal) for t years. The r is still the annual interest rate in decimal form. The e is the irrational number e; use the button on the left of your calculator.
1. They say that the A = Pe rt formula comes from the other formula when n, the number of times it compounds per year, is assumed to be really large or infinity. Well look at the similarities between the two formulas. To make it easier, well use specific numbers for P, r, and t and well look at how the two formulas relate to each other as n gets bigger. We will see that the A = Pe rt formula is a good estimate of A = P 1 + r n really big numbers for n. Copy the graphs in the given window and complete the table using the Value (or Eval) function of your calculator. (You must be in the given window.)
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nt
if we use
Suppose we invest $100 in an account that pays interest at an annual rate of 12% for 10 years compounded x times a year. (Here, were using x instead of n because our calculators require it.) Graph both y1 = 100 1 + .12 x
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)(
10 x )
and y 2 = 100e (.1210) using the window [0, 6] x [0, 400].
Change your window to [0, 1500] x [0, 400] and use the Value (or Eval on the TI85 or 86) function to complete the table. Enter the values for x at the prompt, it will tell you the (10x ) y value for y1 = 100 1 + .12 x , then arrow down and it will tell you the y value for y 2 = 100e (.1210) .
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x number of times it compounds per year 1 once a year 4 once every 3 months 12 once a month 24 once every 2 weeks 365 once a day 1460 once every 6 hours
y = 100 1 + .12
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)( x
10 x )
y = 100e(.1210 )
What is happening to the values of y = 100 1 + .12 x as x gets bigger and bigger?
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10 x )
in comparison to y = 100e (.1210)
Practice using the formulas: (The previous idea is useful for understanding but not necessary to do the following problems.) Some are partially done to guide you. 1. I invest $500 in an account that pays 5% interest, compounded monthly, for a total of 15 years. How much money will the account have in it after 15 years? (Start off with the nt formula A = P 1 + r n ; put in the information that you know. Remember to r write in decimal form. Notice we are looking for A.)
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Did you put 500(1 + .05/12) ^ (12*15) into your calculator? Did you get $1056.85?
2. I will invest a certain amount of money in an account that pays 15% interest compounded every two months (6 times a year). How much must I invest now to have nt $10,000 in ten years? (Start with the formula A = P 1 + r n ; put in the information that you know. Remember to write r in decimal form. Notice we are looking for P.)
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Did you get 10,000 = P 1 + .15 6
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6*10
? Notice the complicated stuff on the right simplifies
so that our equation is really 10,000 = P 4.3998 . Now solve for P.
Did you get $2272.83 as the amount to be invested? (You would get $2272.84 if you used exact values and did not round intermediate answers.)
3. Jody invests $1000 in an account that pays 12% interest, compounded monthly. How long must she leave the money in the account in order to have $5000? (Start with the nt formula A = P 1 + r n ; put in the information that you know. Remember to write r in decimal form. Notice we are looking for t. This is a little more difficult. Remember when we solve for a variable in an exponent, we need to use the inverse relationship between the exponential and logarithmic functions.)
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Did you get 5000 = 1000 1 + .12 12
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12 t
? Simplify the stuff inside the parentheses and
divide both sides by 1000 to get the exponential factor by itself. You should get 5 = 1.0112t . Solve it by the methods discussed in class. (I would take the natural log of both sides of the equation.)
Did you get 13.48 years?
4. Fred has $750 to invest in an account that pays 6% annual interest, compounded continuously. He wants to leave his money in the account until it has grown to $1200. How long must he wait? (Start with...