Question

independent practice / application

1. $5000 is invested at 3% interest, compounded continuously. how much will it be worth after 12 years?
2. a bank offers 4% compounded monthly, while another offers 3.9% compounded continuously. which yields a higher amount after 10 years for a $1,000 investment?
3. you invest $10,000 at a 5% annual interest rate for 10 years. how much money would you have in your account if interest was compounded weekly? how much money would you have in your account if interest was compounded continuously?

Explanation:

1.

Step1: Recall the continuous - compounding formula

The formula for continuous compounding is \(A = Pe^{rt}\), where \(P\) is the principal amount, \(r\) is the annual interest rate (in decimal form), and \(t\) is the time in years.
Given \(P=\$5000\), \(r = 0.03\), and \(t = 12\).

Step2: Substitute the values into the formula

\(A=5000\times e^{0.03\times12}\)
First, calculate \(0.03\times12 = 0.36\). Then \(A = 5000e^{0.36}\).
Since \(e^{0.36}\approx1.4333\) (using a calculator, \(e^x\) can be calculated as \(e^x=\sum_{n = 0}^{\infty}\frac{x^n}{n!}\approx1 + x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots\), for \(x = 0.36\)), \(A=5000\times1.4333=\$7166.5\)

2.

Step1: Use the compound - interest formula for the first bank

The compound - interest formula is \(A=P(1+\frac{r}{n})^{nt}\). For the first bank, \(P = 1000\), \(r=0.04\), \(n = 12\) (compounded monthly), and \(t = 10\).
\(A_1=1000(1+\frac{0.04}{12})^{12\times10}\)
\(1+\frac{0.04}{12}=1+\frac{1}{300}\approx1.00333\)
\(A_1=1000\times(1.00333)^{120}\)
Using a calculator, \((1.00333)^{120}\approx1.4908\), so \(A_1 = 1000\times1.4908=\$1490.8\)

Step2: Use the continuous - compounding formula for the second bank

For the second bank, \(P = 1000\), \(r=0.039\), \(t = 10\)
Using \(A = Pe^{rt}\), \(A_2=1000\times e^{0.039\times10}\)
\(0.039\times10=0.39\)
\(e^{0.39}\approx1.4765\) (using a calculator), so \(A_2=1000\times1.4765=\$1476.5\)

3.

Step3: Use the compound - interest formula for weekly compounding

The compound - interest formula is \(A=P(1+\frac{r}{n})^{nt}\). For weekly compounding, \(n = 52\), \(P=10000\), \(r = 0.05\), \(t = 10\)
\(A_1=10000(1+\frac{0.05}{52})^{52\times10}\)
\(1+\frac{0.05}{52}\approx1.00096\)
\(A_1=10000\times(1.00096)^{520}\)
Using a calculator, \((1.00096)^{520}\approx1.6487\), so \(A_1 = 10000\times1.6487=\$16487\)

Step4: Use the continuous - compounding formula

Using \(A = Pe^{rt}\), with \(P = 10000\), \(r=0.05\), \(t = 10\)
\(A_2=10000\times e^{0.05\times10}\)
\(0.05\times10 = 0.5\)
\(e^{0.5}\approx1.6487\) (using a calculator), so \(A_2=10000\times1.6487=\$16487\)

Answer:

1. \(\$7166.5\)
2. The bank with \(4\%\) compounded monthly yields a higher amount.
3. For weekly compounding: \(\$16487\); for continuous compounding: \(\$16487\)