Chapter 11.3, Problem 67E

Solution

(a)

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

Given information:

The curves: $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ , $r_2\left(\theta \right)=r_1\left(\theta -\alpha \right)$

Calculation:

The given curves are $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ and $r_2\left(\theta \right)=r_1\left(\theta -\alpha \right)$ .

Rewrite the curve $r_2\left(\theta \right)=r_1\left(\theta -\alpha \right)$ as:

  $\begin{array}{l}{r}_2\left(\theta \right)=r_1\left(\theta -\alpha \right)\\ r_2\left(\theta \right)=3\left[1-\cos \left(\theta -\alpha \right)\right]\end{array}$

For the given values of $\alpha =\frac{\pi }{6},\frac{\pi }{4},\frac{\pi }{3},\frac{\pi }{2}$ ; the graph of the curve $r_2$ is shown below:

  

The graphs of the curves $r_1\left(\theta \right)$ and $r_2\left(\theta \right)$ are same. Also, the graphs rotate about the origin in anticlockwise direction.

(b)

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

Given information:

The curves: $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ , $r_2\left(\theta \right)=r_1\left(\theta -\alpha \right)$

Calculation:

The given curves are $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ and $r_2\left(\theta \right)=r_1\left(\theta -\alpha \right)$ .

Rewrite the curve $r_2\left(\theta \right)=r_1\left(\theta -\alpha \right)$ as:

  $\begin{array}{l}{r}_2\left(\theta \right)=r_1\left(\theta -\alpha \right)\\ r_2\left(\theta \right)=3\left[1-\cos \left(\theta -\alpha \right)\right]\end{array}$

For the given values of $\alpha =-\frac{\pi }{6},-\frac{\pi }{4},-\frac{\pi }{3},-\frac{\pi }{2}$ ; the graph of the curve $r_2$ is shown below:

  

The graphs of the curves $r_1\left(\theta \right)$ and $r_2\left(\theta \right)$ are same. Also, the graphs rotate about the origin in anticlockwise direction.

(c)

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

Given information:

The curves: $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ , $r_2\left(\theta \right)=r_1\left(\theta -\alpha \right)$

Calculation:

The given curves are $r_1\left(\theta \right)=3\left(1-\cos \theta \right)$ and $r_2\left(\theta \right)=r_1\left(\theta -\alpha \right)$ .

Rewrite the curve $r_2\left(\theta \right)=r_1\left(\theta -\alpha \right)$ as:

  $\begin{array}{l}{r}_2\left(\theta \right)=r_1\left(\theta -\alpha \right)\\ r_2\left(\theta \right)=3\left[1-\cos \left(\theta -\alpha \right)\right]\end{array}$

For the given values of $\alpha =\frac{\pi }{6},\frac{\pi }{4},\frac{\pi }{3},\frac{\pi }{2}$ ; the graph of the curve $r_2$ is shown below:

  

The graphs of the curves $r_1\left(\theta \right)$ and $r_2\left(\theta \right)$ are same. Also, the graphs rotate about the origin in anticlockwise direction.

For the given values of $\alpha =-\frac{\pi }{6},-\frac{\pi }{4},-\frac{\pi }{3},-\frac{\pi }{2}$ ; the graph of the curve $r_2$ is shown below:

  

The graphs of the curves $r_1\left(\theta \right)$ and $r_2\left(\theta \right)$ are same. Also, the graphs rotate about the origin in anticlockwise direction.

It is seen that the graphs of $r_1\left(\theta \right)$ and $r_2\left(\theta \right)$ rotated counterclockwise about the origin by the angle $\alpha$ .