Chapter 8.2, Problem 2E

(a)

To determine

To fill: The condition to plot all the points $\left(r,\theta \right)$ for a polar graph $r=f\left(\theta \right)$.

Answer to Problem 2E

To graph the polar equation $r=f\left(\theta \right)$ all the points $\left(r,\theta \right)$ are plotted that satisfy the equation.

Explanation of Solution

The graph of polar equation $r=f\left(\theta \right)$ consists of all points $P$ that have at least one polar representation $\left(r,\theta \right)$ whose coordinates satisfy the equation.

Thus, to graph the polar equation $r=f\left(\theta \right)$ all the points $\left(r,\theta \right)$ are plotted that satisfy the equation.

(b)

To determine

To evaluate: The graph and the description of the graph of the polar equations $r=3$ and $\theta =\frac{\pi }{4}$.

Answer to Problem 2E

The graph of the equation $r=3$ is a circle with radius $3$ and centered at the origin.

The graph of the equation $\theta =\frac{\pi }{4}$ is a line passing through the origin with slope $1$.

Explanation of Solution

Calculation:

Section 1:

The given polar equation is,

$r=3$

Simplify the above equation taking square to each side,

$\begin{array}{l}{r}^2=9\\ x^2+y^2=9\left(\because x^2+y^2=r^2\right)\end{array}$

This is the equation of a circle centered at the origin with radius 3.

Draw the graph of the equation $r=3$.

Figure (1)

From Figure (1), it is noticed that it is a circle centered at the origin with radius 3.

Thus, the graph of the equation $r=3$ is a circle with radius $3$ and centered at the origin.

Section 2:

The given polar equation is,

$\theta =\frac{\pi }{4}$

Simplify the above equation using the formula to convert in rectangular coordinate,

$\frac{y}{x}=\tan \theta$

Substitute $\frac{\pi }{4}$ for $\theta$ in above formula,

$\begin{array}{l}\frac{y}{x}=\tan \left(\frac{\pi }{4}\right)\\ \frac{y}{x}=\left(1\right)\left(\because \tan \left(\frac{\pi }{4}\right)=1\right)\\ y=x\end{array}$

this is a equation of line passing through the origin with slope $1$.

Draw the graph of the equation $\theta =\frac{\pi }{4}$.

Figure (2)

From Figure (1), it is noticed that it is a line passing through the origin with slope $1$.

Thus, the graph of the equation $\theta =\frac{\pi }{4}$ is a line passing through the origin with slope $1$.