Chapter 12.8, Problem 40PPS

a.

To determine

Explain the maximum and minimum points mean in the context of the situation.

Answer to Problem 40PPS

The maximum point means that the rope is the highest at $1.25$ second and that the height is $68$ inches.

The minimum point means that the rope is the lowest at $2.75$ second and that the height is $2$ inches.

Explanation of Solution

Given:

It is given in the question that the graph shows the approximates the height of a jump rope h in inches as a function of time t in seconds. A Maximum point on the graph is $(1.25,68)$ and a minimum point is $(2.75,2)$ .

  

Concept Used:

In this, use the concept of study the graph and describe the maximum and minimum points.

Calculation:

Since, from the graph

The x − coordinate represents the time.

The y − coordinate represents the height.

The maximum point means that the rope is the highest at $1.25$ second and that the height is $68$ inches.

The minimum point means that the rope is the lowest at $2.75$ second and that the height is $2$ inches.

Conclusion:

The maximum point means that the rope is the highest at $1.25$ second and that the height is $68$ inches.

The minimum point means that the rope is the lowest at $2.75$ second and that the height is $2$ inches.

b.

To determine

Calculate the equation for the midline, the amplitude and the period of the function.

Answer to Problem 40PPS

The midline, amplitude and the period are $35,33\&\frac{2\pi }{3}$ .

Explanation of Solution

Given:

It is given in the question that the graph shows the approximates the height of a jump rope h in inches as a function of time t in seconds. A Maximum point on the graph is $(1.25,68)$ and a minimum point is $(2.75,2)$ .

  

Concept Used:

In this, use the concept of general equation of sine $y=a\sin b(t)+d$ , where a is the amplitude, $\frac{2\pi }{\left|b\right|}$ is the period and d is the midline and it is the average between the maximum and minimum value.

Calculation: Let the equation be $y=a\sin b(t)+d$ .

Firstly find the midline, d :

  $\begin{array}{l}\frac{68+2}{2}=\frac{70}{2}=35\\ d=35\end{array}$

Now, the amplitude is the difference between the midline and maximum value:

  $\begin{array}{l}68-35=33\\ a=33\end{array}$

Since, it takes $1.5$ seconds from maximum to minimum which is a half of the period.

Now, it is known that the period is $3$ .

Let’s find b:

  $\begin{array}{l}\frac{2\pi }{\left|b\right|}=3\\ b=\frac{2\pi }{3}\end{array}$

Conclusion:

The midline, amplitude and the period are $35,33\&\frac{2\pi }{3}$ .

c.

To determine

Find the equation for the function.

Answer to Problem 40PPS

The equation for the function is $y=33\sin \frac{2\pi }{3}\theta +35$ .

Explanation of Solution

Given: It is given in the question that amplitude, period and midline are $35,33\&\frac{2\pi }{3}$ .

Concept Used:

In this, use the general equation of the sine function $y=a\sin b(t)+d$ .

Calculation: Let the function be $y=a\sin b(t)+d$ .

Now,

  $\begin{array}{l}a=33\\ b=\frac{2\pi }{3}\\ d=35\end{array}$

Put the above value in the equation,

The equation is $y=33\sin \frac{2\pi }{3}\theta +35$ .

Conclusion:

The equation is $y=33\sin \frac{2\pi }{3}\theta +35$