Calc1231W7T4 - Simple and compound and continuous compound interest

Friday, 7 September 2018

6:46 PM

Machine generated alternative text: Suppose that Jethro deposits $2000 into an account which pays simple interest of 4% per annum; which we express as the rate r = 0.04. Simple interest means Jethro only earns interest on the principle $2000 investment After one year the total in the account is $ 2080 O . After 5 years the total in the account is $ 2400 Suppose that his friend Marlene deposits $1000 into an account which pays an interest rate of 3% per annum, which we express as the rate r = 0.03 , but compounded monthly Note that this 3% per annum translates into the rate 0.25% per month. After 1 month Marlene's account contains $ 1000*1.0025 O , and after 12 months (1 year) it would contain $ O After three years it would contain $ 1000M 0025" 1000*1 0025* K) 35 Note: you can leave complicated expressions like 2000 x 1 + for MapleTA to evaluate. If you invest an amount A at an annual interest rate of r per annum, compounded n times per year, how much do you have at the end of the year? Let's derive this important formula. In first compounding period of — of a year, you earn -h A in interest So after the first period; you would have a total of A(l + In the second period you also earn interest on your interest, so you would now have a total of A(l + -L)2 _ And so after one year (i.e after all n periods) you would have a total of O + r)r/n.

 

Machine generated alternative text: Obviously having your interest compounded is preferable to having just a fixed interest But how much better? In the last question you hopefully deduced that an amount O invested at an annual rate r times a year would yield after one year O (which is a number between 0 and 1, with 1 representing 100% interest), but compounded n Suppose we invest $1000 for 5 years Using a calculator; Maple, or the GeoGebra App at the bottom of the page, determine what is better: O Simple interest of 20% for 5 years @ Compound interest of 16% for 5 years (compounded monthly) 12x5 Note: this amounts to finding what is bigger 1000(1 + 0.2 x 5) or 1000 1 + Which is better? @ Simple interest of 10% for 5 years O Compound interest of 6% for 5 years (compounded monthly) Which is better (note: since these numbers are so small the geogebra app might not be helpful here) O Simple interest of 3% for 5 years @ Compound interest of 2.9% for 5 years (compounded monthly)

 

Machine generated alternative text: We have seen that ifwe invest an amount A at an annual interest rate r compounded n times in a year then after one year we would have + -L) n. What is the approximate rate of change of our money? If we consider the first l/n period of a year in which our money goes from A to A (1 + -L) then the change in money is while the length of the time period (in years) is At So the rate of change of money per time is the quotient of these two; which is At We conclude that if the time interval is small, then approximately This is a familiar differential equation, with solution Cert Cert/ n Crt Cre t lin dt for some constant C. In fact this would correspond to the interest being compounded continuously" In this case C represents tne initial amount

 

Machine generated alternative text: Suppose a young graduate is depositing money into a bank account continuously at the rate of about $5, 000 per year, and the account earns interest of 4% annually (compounded continuously). To start with lets say they have $18, 000 in the account Assuming they dont make any withdrawals; how much money is in the account after six years? In this case the change in money has two contributions: Cl tax refunds the savings that are added the interest earned Cl donations to charity Since the interest is being compounded continuously at a annual rate of r .04 O the change in money per unit time (a year) is defined by the equation = 0.04A + 5000. dt This will determine A the amount in the account, as a function oftime t. Happily this is a first order linear ODE ' 10 which we know how to solve. The key is rearrange the equation into the form - 0.04A = 5000 and then multiply both sides by an integrating factor, which is and integrating, not forgetting a constant of integration. Then use the initial condition A (0) 18000 O to find that constant of integration Motoring through this, we get the answer so that after 4 years they ought to have, to the nearest dollar; 42812 -125000 + O dollars in the account

 

Machine generated alternative text: Suppose that you took out a college loan totalling $40, 000 with interest of 7.5% that kicks in once you start work. At that point you set up an online payment plan which continuously deducts money from your pay at a rate which comes out to $9, 000 per year. How much Will you owe after four years? This also involves a first-order linear differential equation; with the balance of the loan L(t) going up as interest is accumulated, and going down as payments are made. So the rate of change of L(t) Will involve a positive contribution for the interest; and a negative contribution for the payments: it will be 0.075 dt Using our deep knowledge of solving first order linear ODEs, we find that O L— 9000 So after four years you will owe, to the nearest dollar, 12011 dollars o

 

 

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