Supplement to 9.4 Introduction This handout parallels the extra material that was covered in the lecture on 9.4. Specifically, you will learn how to use differential equations to analyze loans and plan for your retirement. Continuous Growth Rates vs. Discrete Growth Rates What is a continuous growth rate? Recall that our model of constant continuous growth isdyrydt=, where r represents the rate that y is growing. Because we are using derivatives, this must be a continuous growth rate. That is, it is the rate (expressed as a percent) that the population is growing at a particular instance. This concept is hard to get your brain around at first, but after working through the example below, it will make sense. Imagine that A and B are each opening a new bank and are trying to attract new customers. Due to certain banking regulations, they are only allowed to offer a return of 6% on savings account. Trying to drum up business, both approach a potential customer who is trying to decide where to deposit his $1,000. A: If you put your $1,000 in my bank and come back in one year, you will have $1,060 in your account. B: You will make more money by going with me. Instead of giving you 6% at the end of the year, I will give you 3% six months from now and 3% after the next six months. Notice that you have extra money after the first six months and you will earn interest on your interest during the next six months. This is how it will work: After six months, you will have $1,030 in your account [$1,000 (1.03)⋅]. After the next six months, you will have $1,060.90. [$1,030 (1.03)⋅]. In other words, after one year, you will have ()()$1,000 1.03 1.03⋅ ⋅ =$1,060.90 A: I can beat that. I will pay you 1.5% interest every 3 months. That is still 6% but now your interest earns interest, and the interest on your interest will earn interest, and that interest will earn interest too. Here is how your account will grow: After 3 months: ()$1,000 1.015 $1,015.00= After 6 months: ()$1,000 1.015 (1.015) $1,030.23= After 9 months: ( )3$1,000 1.015 $1,045.68=After 1 year: ( )4$1,000 1.015 $1,061.36= B: Oh yeah. I will compound the interest every day. At the end of the year, this fine gentleman will have ()12.0612$1,000 1 $1,061.68+ = … You can imagine A and B offering to compound interest daily, hourly and even by the second. Eventually, one will decide to take the limit of this process. Mathematically, the depositor will have ( ).06lim $1,000 1 $1061.84nnn→∞ + =  in his account at the end of the year. Using techniques you learned in Calculus 1, you know that( ).06.06lim $1,000 1 $1000nnne→∞ + = . This is what is meant by continuously compounded interest. This matches up with our original equation dyrydt=. The solution, as you know, is ( )r ty t Ce= . How do you compute a continuous growth rate? Suppose you are trying to decide which investment is better: 6% compounded annually or 5.85% compounded continuously? The best way to compare is to convert the 6% annual compounding to a continuous rate. Here is how to do this: Imagine that you are investing $1,000 at 6% compounded annually. At the end of the year, you will have $1,060. If you had an interest rate of r compounded continuously and still made $1,060 by the end of the year, r would satisfy the equation1060 1000re= . With a bit of algebra shows you see that()10601000ln .0582 5.82%r = ≈ = . So the 5.85% compounded continuously is a better deal. This technique will also work with other compounding periods. Let’s convert 6% compounded quarterly to a continuously compounded rate. Calculate what $1,000 will grow to and find the equivalent continuous rate: ( )( )4.0641061.361000$1,000 1 $1,061.36$1,061.36 $1,000ln .0596 5.96%r ter+ === ≈ =Planning Your Retirement Suppose that on the day of your retirement you have $1,000,000 in an account that earns 5% compounded continuously. You also decide that you will withdraw $60,000 per year for your expenses. How long will your money last? (Assume that the money is withdrawn continuously). Solution: Let ( )x t= the amount of money you have in the bank in year t. The interest that the bank pays you increases the amount of money you have and will be a positive contribution to the rate of change of your balance. Your withdrawals will cause your balance to decrease and will add a negative term to the rate your balance is changing. Specifically, the initial value problem is .05 60 000dxxdt= − , (0) 1000 000x=. You want to find the value of t such that( ) 0x t=. Here is how you work it ou: .05.05( 1200 000) .05 .051200 000 1200 000ln( 1200000) .05 ( ) 1200000tdx dx dxx dt dtdt x xx t C x t Ce= − → = → =− −→ − = + → = +  Using the initial condition (0) 1000 000x= we see that200000C= −, the amount of money left in the account at time t is .05( ) 200000 1200 000tx t e= − + . If 1200000.05 .052000000 200000 1200 000 35.84t te e t= − + → = → ≈ . The money will last slightly less than 36 years. Buying a House Imagine you are thinking about taking out a 30 year fixed rate mortgage. You want to borrow $500,000 and can get an interest rate of 6%. What would your monthly payment be? Solution: To use the tools of this section, we will make some simplifying assumptions. First, quoted mortgage rates are compounded monthly. We could get the continuous rate. It turns out that this does not affect the answer significantly, so we will stick with .06, which is easy to work with. Secondly, mortgage payments are made monthly. Our model assumes that payments are made continuously. Again, this will not affect the final answer by too much.Since the 6% is the yearly rate, let Y be the yearly payment. You will divide Y by 12 to get the monthly payment. As in the previous problem, let ( )x t be the amount of money you owe on the loan at year t and determine what factors increase the balance of the loan and which ones decrease the balance. Specifically, the interest will add a term of .06x to the rate of change of your balance, and your yearly payment Y will bring the amount of money you owe down. Here is the equation: .06dxx Ydt= −. Since the initial balance is$500,000, we have(0) 500 000x=. Since the loan will be paid off in 30 years, we have(30) 0x=. Armed with this information we can find the monthly payment. ( )( )( )( ).06.06.06.06.06 .06.06 .06.06 .06 .06.06