A Car Loan: Suppose the amount of principal left to be paid back on a car loan i

Question

A Car Loan: Suppose the amount of principal left to be paid back on a car loan is given by P(t) and suppose a payment of k dollars is made on the loan each month. Then the rate at which the remaining principal P(t) changes with respect to time (in exist/month) is the net result of the monthly payment of k dollars (which is a positive constant) and the interest which is proportional to the remaining principal (here the proportionality constant is the monthly continuous interest rate r). a. Write a differential equation to represent this situation using the function and constants named above. b. A student purchases a car for exist12,000 with a 3-year loan at 9% annual interest, compounded continuously. Calculate the monthly interest rate r and then use this value to rewrite the DE from pert a. Then solve this differential equation for P(t). Be sure to determine the value of C and state your model for P(t) clearly. c. Use your model for P(t) to find the monthly payment k so that the loan in paid off (P(t) = 0) in 3 years (36 months). Give your answer rounded to the nearest cent. d. Derive the general payment formula solving for k in terms of r, n (the number of months), and p 0 (the initial principal), showing all work to support it.

Explanation / Answer

solution:

a) Let P(t) = principal amount after t months

k = monthly payment

r = rate of interest

dp/dt = rate of change

rate of change = int rate * principal - monthly payment

dp/dt = rP - k

b) car loan = 20291.9 = P(0)

rate of interest = 9%

monthly rate = 9/12 = 0.75%

dp/dt = 0.0075P - k

taking log

0.0075P - k = Ce^0.0075t

putting t= 0

we get

C = 152.18 - k

Thus,

?p = ((152.18-k)(e^0.0075*t)+k)/0.0075

C) after 60 months principal amount =0

P(60)= 0

((152.18-k)(e^0.0075*60)+k)/0.0075 =0

k= 419.95

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