Chapter 7.4, Problem 18E

a.

To determine

To find : the amount due to end of $3$ years if interest is compound annually, quarterly, monthly, weekly, daily, hourly, and continuously

Explanation of Solution

Given information : The question is

If $\$1000$ is borrowed at $8\%$ interest,

Formula used : $A_0{\left(1+\frac{r}{n}\right)}^{nt}$

Calculation :

Annual $=1000{\left(1+\frac{0.8}{1}\right)}^{1\times 3}\Rightarrow 1259.71$

Quarterly $=1000{\left(1+\frac{0.8}{1}\right)}^{4\times 3}\Rightarrow 1268.24$

Monthly $=1000{\left(1+\frac{0.8}{1}\right)}^{12\times 3}\Rightarrow 1270.24$

Weekly $=1000{\left(1+\frac{0.8}{1}\right)}^{52\times 3}\Rightarrow 1271.01$

Daily $=1000{\left(1+\frac{0.8}{1}\right)}^{365\times 3}\Rightarrow 1271.22$

Hourly $=1000{\left(1+\frac{0.8}{365\times 24}\right)}^{\left(365\times 24\right)\times 3}\Rightarrow 1271.25$

Continuously $=1000e^{0.8\times 3}\Rightarrow 1271.25$

b.

To determine

To sketch : the graph

Explanation of Solution

Given information : the equation is $A\left(t\right)=1000e^{0.06t}$ , $A\left(t\right)=1000e^{0.08t}$ , & $A\left(t\right)=1000e^{0.1t}$

Graph : Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expression $A\left(t\right)=1000e^{0.06t}$ , $A\left(t\right)=1000e^{0.08t}$ , & $A\left(t\right)=1000e^{0.1t}$

This is required to graph

Step 4: Press $\boxed{\text{GRAPH}}$ button to graph the function.

The graph is obtained as:

  

Interpretation :

When the interest rate is $6\%$ per annum, $A\left(t\right)=1000e^{0.06t}$ this curve is shown in graph in green color

When the interest rate is $8\%$ per annum, $A\left(t\right)=1000e^{0.08t}$ this curve is shown in graph in pink color

When the interest rate is $10\%$ per annum, $A\left(t\right)=1000e^{0.1t}$ this curve is shown in graph in blue color