Chapter 9, Problem 14RE

To determine

To write:the polar equation of the rectangular coordinates $\left(x,y\right)$ of the given variables $r\text{ and }\theta$ . Also, identify the equation and graph it.

Answer to Problem 14RE

The equation of the circle is $x^2+{\left(y-\frac{1}{6}\right)}^2=\frac{1}{36}$ .

Explanation of Solution

Given:

The givenequation is $3r=\sin \theta$ .

Calculation:

As it is known that $x=r\cos \theta y=r\sin \theta$ .

  $\begin{array}{l}x=r\cos \theta \\ y=r\sin \theta \end{array}$ .

And the given equation is $3r=\sin \theta$ .

Now, r multiply both sides,

  $\begin{array}{l}3r^2=r\sin \theta \\ 3\left(x^2+y^2\right)=y\{y=r\sin \theta \}\\ \left(x^2+y^2\right)=\frac{y}{3}\\ x^2+y^2-\frac{y}{3}=0\\ x^2+y^2-\frac{y}{3}+\frac{1}{36}=0+\frac{1}{36}\left\{\text{add }\frac{1}{36}\text{ on both the sides}\right\}\\ \therefore x^2+\left(y^2-\frac{y}{3}+\frac{1}{36}\right)=\frac{1}{36}\\ x^2+{\left(y-\frac{1}{6}\right)}^2=\frac{1}{36}\{\text{as }\left(y^2-\frac{y}{3}+\frac{1}{36}\right)={\left(y-\frac{1}{6}\right)}^2\}\end{array}$

Therefore, it is found that the above equation is the equation of circle with center $C\left(0,\frac{1}{6}\right)$ and radius $\frac{1}{6}$ . Now, draw the graph of the circle.

  

Hence, polar equation of the rectangular coordinates is $x^2+{\left(y-\frac{1}{6}\right)}^2=\frac{1}{36}$ .