To calculate the monthly payments for a loan, you can use the formula for an amortizing loan:
\[ M = P \times \frac{r(1 + r)^n}{(1 + r)^n - 1} \]
Where:
- \( M \) is the monthly payment
- \( P \) is the loan amount (principal)
- \( r \) is the monthly interest rate (annual interest rate divided by 12)
- \( n \) is the number of payments (loan term in months)
In this case:
- \( P = 38,179.00 \)
- Annual interest rate = 5.17%, so the monthly interest rate \( r = \frac{5.17}{100 \times 12} \approx 0.00430833 \)
- The loan term is 6 years, so \( n = 6 \times 12 = 72 \)
Now, let's substitute these values into the formula:
-
Calculate the monthly interest rate: \[ r = \frac{5.17}{100} \div 12 \approx 0.00430833 \]
-
Calculate \( (1 + r)^n \): \[ (1 + r)^{72} = (1 + 0.00430833)^{72} \]
Calculating this gives approximately: \[ (1 + 0.00430833)^{72} \approx 1.348835395 \]
- Substitute the values into the mortgage payment formula:
\[ M = 38,179 \times \frac{0.00430833 \times 1.348835395}{1.348835395 - 1} \]
Calculating the numerator:
\[ 0.00430833 \times 1.348835395 \approx 0.0058170496 \]
And calculating the denominator:
\[ 1.348835395 - 1 \approx 0.348835395 \]
Now substitute these values into the formula for \( M \):
\[ M \approx 38,179 \times \frac{0.0058170496}{0.348835395} \]
Calculating this gives:
\[ M \approx 38,179 \times 0.01666889076 \approx 636.71 \]
Thus, the monthly payment for Derek’s car loan will be approximately $636.71.