Chapter 9.5, Problem 94E

To determine

To calculate:

The rectangular equation of the following polar equation and names its graph equation:

  $r=\cos \theta +3\sin \theta$

Answer to Problem 94E

The rectangular equation is $x^2+y^2-x-3y=0$ , on rearranging get the circle equation in rectangular form.

Explanation of Solution

Given information:

  $r=\cos \theta +3\sin \theta$

Formula used:

The relation between polar coordinates and rectangular coordinates is given by the following:

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

Here,

  $r=\sqrt{x^2+y^2}$

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

Calculation:

Form the given relation between polar coordinates and rectangular coordinates:

  $r=\sqrt{x^2+y^2}$

And

  $\begin{array}{l}x=r\cos \theta \\ \cos \theta =\frac{x}{r}\\ \cos \theta =\frac{x}{\sqrt{x^2+y^2}}\end{array}$$\begin{array}{l}y=r\sin \theta \\ \sin \theta =\frac{y}{r}\\ \sin \theta =\frac{y}{\sqrt{x^2+y^2}}\end{array}$

The given equation can be rewrite as:

  $\begin{array}{l}r=\cos \theta +3\sin \theta \\ \sqrt{x^2+y^2}=\frac{x}{\sqrt{x^2+y^2}}+\frac{3y}{\sqrt{x^2+y^2}}\\ \sqrt{x^2+y^2}=\frac{x+3y}{\sqrt{x^2+y^2}}\\ \left(\sqrt{x^2+y^2}\right)\left(\sqrt{x^2+y^2}\right)=x+3y\\ x^2+y^2-x-3y=0\end{array}$

It is the rectangular form.

On rearranging:

  $\begin{array}{l}{x}^2+y^2-x-3y=0\\ x^2-2x\left(\frac{1}{2}\right)+{\left(\frac{1}{2}\right)}^2+y^2-2y\left(\frac{3}{2}\right)+{\left(\frac{3}{2}\right)}^2={\left(\frac{1}{2}\right)}^2+{\left(\frac{3}{2}\right)}^2\\ {\left(x-\frac{1}{2}\right)}^2+{\left(y-\frac{3}{2}\right)}^2=\frac{5}{2}\end{array}$

This is equation of circle in the rectangular form on comparing it ${\left(x-h\right)}^2+{\left(y-k\right)}^2=R^2$ .