Suppose that $P$ dollars in principal is invested for $t$ years at the given interest rates with continuous compounding. Determine the amount that the investment is worth at the end of the given time period. \[ P=\$ 14,000, t=11 \text { yr } \] (a) $2 \%$ interest (b) $3 \%$ interest (c) $6.5 \%$ interest
Solution
Step 1 ：Convert the interest rate from a percentage to a decimal by dividing by 100. So, \(r = \frac{2}{100} = 0.02\).
Step 2 ：Substitute \(P = $14,000\), \(r = 0.02\), and \(t = 11\) into the formula \(A = Pe^{rt}\): \(A = 14000 * e^{(0.02*11)} = 14000 * e^{0.22}\)
Step 3 ：\(A \approx $17,080.79\)
Step 4 ：Convert the interest rate to a decimal: \(r = \frac{3}{100} = 0.03\).
Step 5 ：Substitute \(P = $14,000\), \(r = 0.03\), and \(t = 11\) into the formula: \(A = 14000 * e^{(0.03*11)} = 14000 * e^{0.33}\)
Step 6 ：\(A \approx $18,716.68\)
Step 7 ：Convert the interest rate to a decimal: \(r = \frac{6.5}{100} = 0.065\).
Step 8 ：Substitute \(P = $14,000\), \(r = 0.065\), and \(t = 11\) into the formula: \(A = 14000 * e^{(0.065*11)} = 14000 * e^{0.715}\)
Step 9 ：\(A \approx $23,982.96\)
Step 10 ：So, the investment would be worth approximately \(\boxed{$17,080.79}\) at 2% interest, \(\boxed{$18,716.68}\) at 3% interest, and \(\boxed{$23,982.96}\) at 6.5% interest.
