Chapter 9, Problem 33SGA

To determine

To express: the equation in polar form and round $\varphi$ to the nearest degree.

Answer to Problem 33SGA

  $\frac{2\sqrt{10}}{5}=r\cos \left(\theta -18.43°\right)$

Explanation of Solution

Given information:

  $y=-3x-4$

Calculation:

Rewrite the equation in standard form i.e. $\left(Ax+By+C=0\right)$ as shown below.

  $\begin{array}{c}y=-3x-4\\ 3x+y+4=0\left[\text{Add }\left(3x+4\right)\text{ on each side}\right]\end{array}$

The value of $A,B$ and $C$ is 3, 1 and 4 respectively.

Now evaluate $\sqrt{A^2+B^2}$ as shown below.

  $\begin{array}{c}\sqrt{A^2+B^2}=\sqrt{3^2+1^2}\\ =\sqrt{9+1}\\ =\sqrt{10}\end{array}$

Since $C$ is positive divide the standard equation by $-\sqrt{A^2+B^2}$ to get the normal form of the equation i.e. $\left(\cos \varphi \right)x+\left(\sin \varphi \right)y-p=0$ .

  $\begin{array}{c}\frac{3}{-\sqrt{10}}x+\frac{1}{-\sqrt{10}}y+\frac{4}{-\sqrt{10}}=0\\ -\frac{3\sqrt{10}}{10}x-\frac{\sqrt{10}}{10}y-\frac{2\sqrt{10}}{5}=0\left[\text{Multiply and divide by }\sqrt{10}\right]\end{array}$

Thus $\cos \varphi =-\frac{3\sqrt{10}}{10}$ , $\sin \varphi =-\frac{\sqrt{10}}{10}$ and $p=\frac{2\sqrt{10}}{5}$ .

Now evaluate $\varphi$ as shown below.

  $\begin{array}{c}\tan \varphi =\frac{\sin \varphi }{\cos \varphi }\\ =\frac{-\frac{\sqrt{10}}{10}}{-\frac{3\sqrt{10}}{10}}\left[\text{Put the value of cosine and sine}\right]\\ =\frac{1}{3}\\ \varphi ={\tan }^{-1}\left(\frac{1}{3}\right)\\ =18.43°\end{array}$

The polar form of the equation is represented as follows:

  $\begin{array}{c}p=r\cos \left(\theta -\varphi \right)\\ \frac{2\sqrt{10}}{5}=r\cos \left(\theta -18.43°\right)\left[\text{Put the value of }p\text{ and }\varphi \right]\end{array}$

Thus the polar form of given equation is $\frac{2\sqrt{10}}{5}=r\cos \left(\theta -18.43°\right)$ .