Chapter 9, Problem 32SGA

To determine

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

Answer to Problem 32SGA

  $\frac{3\sqrt{5}}{5}=r\cos \left(\theta -26.6°\right)$

Explanation of Solution

Given information:

  $2x+y=-3$

Calculation:

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

  $2x+y+3=0$

The value of $A,B$ and $C$ is 2, 1 and 3 respectively.

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

  $\begin{array}{c}\sqrt{A^2+B^2}=\sqrt{2^2+1^2}\\ =\sqrt{4+1}\\ =\sqrt{5}\end{array}$

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

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

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

Now evaluate $\varphi$ as shown below.

  $\begin{array}{c}\tan \varphi =\frac{\sin \varphi }{\cos \varphi }\\ =\frac{\frac{-\sqrt{5}}{5}}{\frac{-2\sqrt{5}}{5}}\left[\text{Put the value of cosine and sine}\right]\\ =\frac{1}{2}\\ \varphi ={\tan }^{-1}\left(\frac{1}{2}\right)\\ =26.6°\end{array}$

The polar form of the equation is represented as follows:

  $\begin{array}{c}p=r\cos \left(\theta -\varphi \right)\\ \frac{3\sqrt{5}}{5}=r\cos \left(\theta -26.6°\right)\left[\text{Put the value of }p\text{ and }\varphi \right]\end{array}$

Thus the polar form of given equation is $\frac{3\sqrt{5}}{5}=r\cos \left(\theta -26.6°\right)$ .