Chapter 3.6, Problem 62E

Solution

a.

To find: What type of function is graphed.

The function is a line, because the graph is linear.

Given information:

  $\begin{array}{c}\text{Amount, }P=\$300\\ \text{Rate of interest, }r=3\%\end{array}$

Calculation:

Substituting the value of $P\text{ and }r$ to get,

  $\begin{array}{c}p(t)=300(1+0.03t)\\ p(t)=300+9t\\ p(t)=9t+300\end{array}$

Substituting the values for $t$ to plot the graph,

$t$	$p(t)$
$0$	$300$
$2$	$318$
$4$	$336$
$6$	$354$
$8$	$372$
$10$	$390$

Graph:

  

Interpretation:

Since the graph is a line, then the given function is a linear function.

b.

To find: The slope of the lined graph.

The slope of the lined graph is $9$ .

Given information:

  $\begin{array}{c}\text{Amount }P=\$300\\ \text{Rate of interest }r=3\%\end{array}$

Calculation:

From part (a) it is known that,

  $p(t)=9t+300$ , which is a linear equation

The equation for the linear function is given by,

  $y=mx+b$

Where $m$ is the slope

By comparing the above 2 equation to get,

Slope $m=9$

Therefore, the slope of the lined graph is $9$ .

c.

To describe: The effect of a graph by increasing or decreasing the interest rates.

The slope gets increases when increasing the interest rates.

The slope gets decreases when decreasing the interest rates.

Given information:

  $\begin{array}{c}\text{Amount }P=\$300\\ \text{Rate of interest }r=3\%\end{array}$

Calculation:

From part (a) it is known that,

  $p(t)=9t+300$ , which is a linear equation

On observing from the graph,

The slope increases when the rate of interest increases. Thus, the graph becomes steeper.

The slope decreases when the rate of interest decreases. Thus, the graph becomes less steep.

d.

To find: The simple interest and compound interest and compare both to know which is greater.

The Simple interest is $p_{simple}\approx \$304.50$

The Compound interest is $p_{compound}\approx \$304.47$

The Simple interest is greater than the compound interest.

Given information:

  $\begin{array}{c}\text{Amount }P=\$300\\ \text{Rate of interest }r=3\%\\ \text{time }t=0.5\end{array}$

Calculation:

The compound interest for annually is given by,

  $P(t)=p{(1+r)}^t$

Substitute the given values in the above equation to get,

  $\begin{array}{c}$ p_{compound}=300{(1+0.03)}^{0.5}$\approx \$304.47\end{array}$

Use linear function to find the simple interest,

  $P(t)=9t+300$

Substitute $t=0.5$

  $\begin{array}{c}{p}_{simple}=9(0.5)+300\\ \approx \$304.50\end{array}$

The compound interest is greater for $t>1$ .

But $0.5<1$ the simple interest is greater.

Conclusion:

  $\begin{array}{c}{p}_{compound}\approx \$304.47\\ p_{simple}\approx \$304.50\end{array}$

The simple interest is greater compared to compound interest.