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qwelly [4]
3 years ago
10

A certain college graduate borrows 7864 dollars to buy a car. The lender charges interest at an annual rate of 13%. Assuming tha

t interest is compounded continuously and that the borrower makes payments continuously at a constant annual rate k dollars per year, determine the payment rate that is required to pay off the loan in 3 years. Also determine how much interest is paid during the 3-year period.
Mathematics
1 answer:
White raven [17]3 years ago
4 0

Answer:

Therefore rate of payment = $ 3145.72

Therefore the rate of interest = =$1573.17

Step-by-step explanation:

Consider A represent the balance at time t.

A(0)=$ 7864.

r=13 % =0.13

Rate payment = $k

The balance rate increases by interest (product of interest rate and current balance) and payment rate.

\frac{dB}{dt} = rB-k

\Rightarrow \frac{dB}{dt} - rB=-k.......(1)

To solve the equation ,we have to find out the integrating factor.

Here p(t)= the coefficient of B =-r

The integrating factor =e^{\int p(t) dt

                                     =e^{\int (-r)dt

                                     =e^{-rt}

Multiplying the integrating factor the both sides of equation (1)

e^{-rt}\frac{dB}{dt} -e^{-rt}rB=-ke^{-rt}

\Rightarrow  e^{-rt}dB - e^{-rt}rBdt=-ke^{-rt}dt

Integrating both sides

\Rightarrow \int e^{-rt}dB -\int e^{-rt}rBdt=\int-ke^{-rt}dt

\Rightarrow e^{-rt}B=\frac{-ke^{-rt}}{-r} +C        [ where C arbitrary constant]

\Rightarrow B(t)=\frac{k}{r} +Ce^{rt}

Initial condition B=7864 when t =0

\therefore 7864= \frac{k}{r} - Ce^0

\Rightarrow  C= \frac{k}{r} -7864

Then the general solution is

B(t)=\frac{k}{r}-( \frac{k}{r}-7864)e^{rt}

To determine the payment rate, we have to put the value of B(3), r and t in the general solution.

Here B(3)=0, r=0.13 and t=3

B(3)=0=\frac{k}{0.13}-( \frac{k}{0.13}-7864)e^{0.13\times 3}

\Rightarrow- 0.48\frac{k}{0.13} +11614.98=0

⇒k≈3145.72

Therefore rate of payment = $ 3145.72

Therefore the rate of interest = ${(3145.72×3)-7864}

                                                 =$1573.17

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