Chapter 9.6, Problem 29E

To determine

To find: A sine and cosine function for the given sinusoid.

Answer to Problem 29E

  $y=-2.5\cos \left(4t\right)+5.5$ or $y=2.5\sin \left(4t-\frac{\pi }{2}\right)+5.5$ .

Explanation of Solution

Given information:

During one cycle, a sinusoid has minimum at $\left(\frac{\pi }{2},3\right)$ and maximum at $\left(\frac{\pi }{4},8\right)$ .

Calculation:

As per the given information −

During one cycle, a sinusoid has minimum at $\left(\frac{\pi }{2},3\right)$ and maximum at $\left(\frac{\pi }{4},8\right)$ .

Step 1: Find the maximum and minimum value.

Here, maximum value is $8$ and minimum value is $3$ .

Step 2: Identify the vertical shift $k$ .

The value of $k$ is the mean of maximum and minimum value, i.e.

  $k=\frac{8+3}{2}=5.5$

Step 3: Find the amplitude and period.

  $\begin{array}{l}\left|a\right|=\frac{\text{Maximum value}-\text{Minimum value}}{2}=\frac{8-3}{2}=2.5\\ \text{Since graph is not a reflection, }a>0\\ \Rightarrow a=2.5\end{array}$

Since the function goes from maximum to minimum during period $\left(\frac{\pi }{4},\frac{\pi }{2}\right)$ . So, function again will go from minimum to maximum during the period $\left(\frac{\pi }{2},\frac{3\pi }{4}\right)$ . Thus, function is repeating itself every $\frac{\pi }{2}$ period. Therefore, period of the function would be $\frac{\pi }{2}$ .

  $\begin{array}{l}\text{i}\text{.e}\text{. }\frac{\pi }{2}=\frac{2\pi }{b}\\ \Rightarrow b=4\end{array}$

Also at $t=0$ , function would be at its minimum value. So, cosine function reflected across $x-$ axis can be taken as model.

Step $5$ : The function graph would be given as −

  $y=-2.5\cos \left(4t\right)+5.5$

Hence, desired function of the given graph is $y=-2.5\cos \left(4t\right)+5.5$ .

  

Taking sine as model function:

When sine is taken as model function then function should be shifted $\frac{\pi }{8}$ units rightward.

  $\begin{array}{l}\text{So, }y=2.5\sin \left[4\left(t-\frac{\pi }{8}\right)\right]+5.5\\ \Rightarrow y=2.5\sin \left(4t-\frac{\pi }{2}\right)+5.5\end{array}$

  

Hence, desired model functions would be $y=-2.5\cos \left(4t\right)+5.5$ or $y=2.5\sin \left(4t-\frac{\pi }{2}\right)+5.5$ .