Chapter 7.6, Problem 35E

a.

To determine

To find:the normal form of the line.

Answer to Problem 35E

  $\frac{5}{13}x+\frac{12}{13}y-3=0$

Explanation of Solution

Given information:

  $5x+12y-39=0$

Calculation:

In the given equation, the value of $A,B\text{ and }C$ is 5, 12, and $-3$ . Since $C$ is negative, so divide each terms of equation by $\sqrt{A^2+B^2}$ as shown below.

  $\begin{array}{c}\frac{5}{\sqrt{5^2+{12}^2}}x+\frac{12}{\sqrt{5^2+{12}^2}}y-\frac{39}{\sqrt{5^2+{12}^2}}=0\\ \frac{5}{\sqrt{169}}x+\frac{12}{\sqrt{169}}y-\frac{39}{\sqrt{169}}=0\\ \frac{5}{13}x+\frac{12}{13}y-3=0\end{array}$

Thus, the normal for of line is $\frac{5}{13}x+\frac{12}{13}y-3=0$ .

b.

To determine

To find:the value of $\phi$ and graph the result.

Answer to Problem 35E

  $-\frac{12}{13}x+\frac{5}{13}y-3=0$

Explanation of Solution

Given information:

Replace $\varphi$ by $\varphi +90°$ .

The equation evaluated in part a. is $\frac{5}{13}x+\frac{12}{13}y-3=0$ .

Calculation:

Since the normal equation of line is $x\cos \varphi +y\sin \varphi -p=0$ .The value $\varphi$ of can be evaluated as follows:

  $\begin{array}{c}\cos \varphi =\frac{5}{13}\\ \varphi ={\cos }^{-1}\frac{5}{13}\\ =67°\end{array}$

Now replace $\varphi$ by $\varphi +90°$ and evaluate $\cos \left(\varphi +90°\right)\text{ and }\sin \left(\varphi +90°\right)$ as shown below.

  $\begin{array}{c}\cos \left(\varphi +90°\right)=\cos \left(67°+90°\right)\\ =\cos \left(157°\right)\\ =-\frac{12}{13}\end{array}$

  $\begin{array}{c}\sin \left(\varphi +90°\right)=\sin \left(67°+90°\right)\\ =\sin \left(157°\right)\\ =\frac{5}{13}\end{array}$

Now replace the value of $\cos \varphi \text{ and sin}\varphi$ by $\cos \left(\varphi +90°\right)\text{ and }\sin \left(\varphi +90°\right)$ respectively in the given equation. Thus, the normal for of new line is $-\frac{12}{13}x+\frac{5}{13}y-3=0$ . Use the graphing calculator to graph the result obtained as follows:

  

The new line is the normal line that is rotated by $90°$ counterclockwise about the origin.

c.

To determine

To test:the Armando’s conjecture by selecting any value of $\delta$ .

Answer to Problem 35E

The new line is the normal line that is rotated by $\left(\delta \right)45°$ counterclockwise about the origin.

Explanation of Solution

Given information:

  $\delta$ not divisible by $90°$ .

Armando tries to rotate the graph of line $\delta °$ about the origin. He hypothesized the following would be an equation of new line.

  $x\cos \left(\varphi +\delta \right)+y\sin \left(\varphi +\delta \right)-p=0$

The value of $\varphi$ evaluated in part b. is $67°$ .

The value of $p$ is 3.

Calculation:

Let $\delta$ be $45°$ . Now evaluate the normal of a new line equation given by Armando as follows:

  $\begin{array}{c}x\cos \left(\varphi +\delta \right)+y\sin \left(\varphi +\delta \right)-p=0\\ x\cos \left(67°+45°\right)+y\sin \left(67°+45°\right)-3=0\left[\text{Put the value of }\varphi ,\delta \text{ and }p\right]\\ x\cos \left(112°\right)+y\sin \left(112°\right)-3=0\\ -0.381x+0.925y-3=0\end{array}$

The normal form of new line equation is $-0.381x+0.925y-3=0$ .

Now sketch a graph for new line equation with the help of graphing calculator.

  

Thus, the new line is the normal line that is rotated by $45°$ counterclockwise about the origin.

d.

To determine

To explain:whether Armando is correct by including graph in your result.

Answer to Problem 35E

Yes,Armando’s hypothesis is correct.

Explanation of Solution

Given information:

Armando tries to rotate the graph of line $\delta °$ about the origin. He hypothesized the following would be an equation of new line.

  $x\cos \left(\varphi +\delta \right)+y\sin \left(\varphi +\delta \right)-p=0$

Calculation:

The results obtained in part b. and c., the line with normal form $x\cos \varphi +y\sin \varphi -p=0$ makes and angle $\varphi$ with the positive $x$ -axis and has the normal of length $p$ . The graph of Armando’s equation where $\varphi$ is replace by $\varphi +\delta$ and makes an angle of $\varphi +\delta$ with the positive $x$ -axis and has the normal of length $p$ . Therefore, the graph of Armando’s equation shown below is the graph of the original line that is rotated by $\delta °$ counterclockwise. So, he is correct.