Question

example:
find the amount in an account with $2,000 invested at 5% interest compounded continuously
for 10 years.
$a = pe^{rt}=2000e^{0.05(10)} = 2000e^{0.5}approx2000(1.6487)=$3297.40$

guided practice problems

1. find the amount in an account with $1,200 invested at 3% interest compounded

continuously for 10 years.
$a =$

1. $p = $750, r = 6.5\\%, t = 5$ years → compounded continuously

$a =$

Explanation:

Step1: Identify the formula

The formula for continuous compounding is \(A = Pe^{rt}\), where \(P\) is the principal amount, \(r\) is the annual interest rate (in decimal), and \(t\) is the time in years.

Step2: Solve problem 1

Given \(P = 1200\), \(r=0.03\), \(t = 10\).
Substitute into the formula: \(A=1200e^{0.03\times10}=1200e^{0.3}\).
Using \(e^{0.3}\approx1.34986\), then \(A = 1200\times1.34986=\$1619.83\).

Step3: Solve problem 2

Given \(P = 750\), \(r = 0.065\), \(t=5\).
Substitute into the formula: \(A=750e^{0.065\times5}=750e^{0.325}\).
Using \(e^{0.325}\approx1.383\), then \(A=750\times1.383=\$1037.25\).

Answer:

1. \(A=\$1619.83\)
2. \(A=\$1037.25\)