Name: MATH 1113 Precalculus Eric Perkerson Date: October 12, 2014 Handout No. 5 Problem 8 v.1 If P = 500 dollars is deposited in a savings account that pays interest at a rate of 4 = 19/2% per year compounded continuously, find the balance after t = 4 years. We know that for continuously compounded interest, we have At) = P e rt we plug in A4) = 500)e 19/200)4 731.14 Problem 8 v.2 How much money, invested at an interest rate of r = 6.6% per year compounded continuously, will amount to A = 500000 dollars after t = 13 years? Round your answer to the nearest cent.) Set and solve for 500000 = P e 6.6/100)13 P = 500000e 6.6/100)13 212004.63 Problem 8 v.3 Alice invests her money at Bob s bank. Bob compounds interest quarterly at a nominal rate of 6%. Alice wants the investment to be worth $7000 after 8 years. How much money should Alice invest? Set and solve for 7000 = P 1 + 6/100 ) 48) 4 P = 7000 1 + 6/100 ) 48) 4346.95 4 Problem 9 v.1 Determine the per annum interest rate required for an investment with continuous compound interest to triple in 7.5 years. Express your answer as a percent. We know that an investment with continuous compound interest of r yields an amount of money at time t of At) = P e rt so the amount of money at t = 7.5 is A7.5) = P e r7.5) = P e r/2)
and we know that our starting amount of money is the principle, P. triple the principle is 3P. So we set 3P = P e r/2) and solve for r. Hence 3P = P e r/2) 3 = e r/2) ln3) = r/2) 2 ln3) = r r = 2 ln3) 0.146482 r 0.146482100%) = 14.6482% ) 2 ln3) Problem 9 v.2 You invest $8,800 at 4% per annum compounded continuously. Determine the time T in years) for your investment to be worth $14,350. We know that an investment of $8, 800 with continuous compound interest of 4% yields an amount of money at time t of At) = )e.04)t = )e 4/100)t so if we want this to equal $14, 350 at t = T, we set and then solve for T, thus we have 14350 = )e 4/100)T 14350 = )e 4/100)T 14350 = e4/100)t ) 14350 ln = 1/25)T ) 14350 25 ln = T T = 25 ln ) 14350 12.22
Problem 10 v.1 Determine the effective interest rate if a nominal rate of 5.5% is compounded monthly. Express your answer as a percent. We know that a nominal rate of 5.5% compounded monthly yields an amount of money at time t of At) = P 1 +.055 ) 12t = P 1 + 55 ) 12t 12 and we know that the effective interest rate, call it r, yields an amount of money at time t of At) = P 1 + r) t Because these are both the same function, these expressions for the function must be the same for all t, so in particular they are the same for t = 1. So we set and then we solve for r, the effective rate of interest: P P 1 + 55 ) 121) = P 1 + r) 1) 1 + 55 ) 121) = P 1 + r) 1) P 1 + 55 ) 12 = P 1 + r) ) 12 = 1 + r 1 + 55 1 + 55 ) 12 1 = r r = 1 + 55 ) 12 1 0.0564079 r 0.0564079100%) = 5.64079% 1 + 55 ) 12 1) Problem 9 v.3 Determine the per annum interest rate required for an investment with continuous compound interest to double in 7.5 years. Express your answer as a percent. We know that a per annum interest rate of r compounded continuously yields an amount of money at time t of At) = P e rt
so the amount of money at t = 7.5 is A7.5) = P e r7.5) = P e r/2) and we know that our starting amount of money is the principle, P. double the principle is 2P. So we set 2P = P e r/2) and solve for r. Hence 2P = P e r/2) 2 = e r/2) ln2) = r/2) 2 ln2) = r r = 2 ln2) 0.0924196 r 0.0924196100%) = 9.24196% ) 2 ln2)
Problem 10 v.2 Determine the per annum interest rate required for an investment with continuous compound interest to yield an effective rate of 5.5%. Express your answer as a percent. We know that an investment with continuous compound interest of r yields an amount of money at time t of At) = P e rt and we know that an effective interest rate of 5.5%, yields an amount of money at time t of At) = P 1 +.055) t = P 1 + 55 )t Because these are both the same function, these expressions for the function must be the same for all t, so in particular they are the same for t = 1. So we set P e r1) = P 1 + 55 )1) and then we solve for r, the per annum interest rate for the investment with continuous compound interest: P e r1) = P 1 + 55 )1) P e r = P 1 + 55 ) e r = 1 + 55 r = ln 1 + 55 ) r = ln 1 + 55 ) 0.0535408 r 0.0535408100%) = 5.35408% ln 1 + 55 )