Chapter 10.3, Problem 67E

(a)

To determine

To plot: The graph of $r_2$ for $\alpha =\frac{\pi }{6}$ , $\frac{\pi }{4}$ , $\frac{\pi }{3}$ and $\frac{\pi }{2}$ . Also, compare with the graph of $r_1$ .

Answer to Problem 67E

The graph of $r_2$ is shown in Figure (1). The graph of $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ is rotated about the origin by the angle $\alpha$ in counterclockwise direction to form the graph of $r_2\left(\theta \right)$ for $\alpha =\frac{\pi }{6}$ , $\frac{\pi }{4}$ , $\frac{\pi }{3}$ and $\frac{\pi }{2}$ .

Explanation of Solution

Given information: The given equations of curve are $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ and $r_2\left(\theta \right)=r_1\left(\theta -\alpha \right)$ .

Calculation:

Consider the given equation $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ .

Write the equation for $r_2\left(\theta \right)$ .

  $r_2\left(\theta \right)=3\left(1-\cos \left(\theta -\alpha \right)\right)$

Draw the graph of $r_2\left(\theta \right)$ for $\alpha =\frac{\pi }{6}$ , $\frac{\pi }{4}$ , $\frac{\pi }{3}$ and $\frac{\pi }{2}$ . Also, draw the graph of $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ on the same plane.

  

Figure (1)

From the graph it can be observed that the graph of $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ is rotated about the origin by the angle $\alpha$ in counterclockwise direction to form the graph of $r_2\left(\theta \right)$ for $\alpha =\frac{\pi }{6}$ , $\frac{\pi }{4}$ , $\frac{\pi }{3}$ and $\frac{\pi }{2}$ .

Therefore, the graphs of $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ and $r_2\left(\theta \right)=r_1\left(\theta -\alpha \right)$ is shown in Figure (1).

(b)

To determine

To plot: The graph of $r_2$ for $\alpha =-\frac{\pi }{6}$ , $-\frac{\pi }{4}$ , $-\frac{\pi }{3}$ and $-\frac{\pi }{2}$ . Also, compare with the graph of $r_1$ .

Answer to Problem 67E

The graph of $r_2$ is shown in Figure (1). The graph of $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ is rotated about the origin by the angle $-\alpha$ in clockwise direction to form the graph of $r_2\left(\theta \right)$ for $\alpha =-\frac{\pi }{6}$ , $-\frac{\pi }{4}$ , $-\frac{\pi }{3}$ and $-\frac{\pi }{2}$ .

Explanation of Solution

Given information: The given equations of curve are $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ and $r_2\left(\theta \right)=r_1\left(\theta -\alpha \right)$ .

Calculation:

Consider the given equation $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ .

Write the equation for $r_2\left(\theta \right)$ .

  $r_2\left(\theta \right)=3\left(1-\cos \left(\theta -\alpha \right)\right)$

Draw the graph of $r_2\left(\theta \right)$ for $\alpha =-\frac{\pi }{6}$ , $-\frac{\pi }{4}$ , $-\frac{\pi }{3}$ and $-\frac{\pi }{2}$ . Also, draw the graph of $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ on the same plane.

  

Figure (1)

From the graph it can be observed that the graph of $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ is rotated about the origin by the angle $-\alpha$ in clockwise direction to form the graph of $r_2\left(\theta \right)$ for $\alpha =-\frac{\pi }{6}$ , $-\frac{\pi }{4}$ , $-\frac{\pi }{3}$ and $-\frac{\pi }{2}$ .

Therefore, the graphs of $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ and $r_2\left(\theta \right)=r_1\left(\theta -\alpha \right)$ is shown in Figure (1).

(c)

To determine

To describe: The relationship between the graphs $r_1=f\left(\theta \right)$ and $r_2=f\left(\theta -\alpha \right)$ .

Answer to Problem 67E

The graph of $r_2\left(\theta \right)=3\left(1-\cos \left(\theta -\alpha \right)\right)$ is the graph of $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ rotated about the origin in counterclockwise direction for $\alpha =\pm \frac{\pi }{6}$ , $\pm \frac{\pi }{4}$ , $\pm \frac{\pi }{3}$ and $\pm \frac{\pi }{2}$ .

Explanation of Solution

Given information: The given equations of curve are $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ and $r_2\left(\theta \right)=r_1\left(\theta -\alpha \right)$ .

Calculation:

Consider the given equation $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ .

Write the equation for $r_2\left(\theta \right)$ .

  $r_2\left(\theta \right)=3\left(1-\cos \left(\theta -\alpha \right)\right)$

From part (a), the graph of $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ is rotated about the origin by the angle $\alpha$ in counterclockwise direction to form the graph of $r_2\left(\theta \right)$ for $\alpha =\frac{\pi }{6}$ , $\frac{\pi }{4}$ , $\frac{\pi }{3}$ and $\frac{\pi }{2}$ .

From part (b), the graph of $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ is rotated about the origin by the angle $-\alpha$ in clockwise direction to form the graph of $r_2\left(\theta \right)$ for $\alpha =-\frac{\pi }{6}$ , $-\frac{\pi }{4}$ , $-\frac{\pi }{3}$ and $-\frac{\pi }{2}$ .

From both parts it can be concluded that the graph of $r_2\left(\theta \right)=3\left(1-\cos \left(\theta -\alpha \right)\right)$ is the graph of $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ rotated about the origin in counterclockwise direction for $\alpha =\pm \frac{\pi }{6}$ , $\pm \frac{\pi }{4}$ , $\pm \frac{\pi }{3}$ and $\pm \frac{\pi }{2}$ .