Chapter 12.2, Problem 52E

To determine

To calculate:

Given equation in polar form.

Answer to Problem 52E

Given equation $3x-5y+5=0$ in polar form is $\frac{5}{\sqrt{34}}=r\cos \left(\theta +1.03\right)$

Explanation of Solution

Given information:

Given equation: $3x-5y+5=0$

Calculation:

Consider,

  $3x-5y+5=\mathrm{0...}\left(1\right)$

It is in the form,

  $Ax+By+C=0$

For converting the normal form will find,

  $\begin{array}{l}\pm \sqrt{A^2+B^2}\\ \pm \sqrt{3^2+{\left(-5\right)}^2}\\ \pm \sqrt{34}\end{array}$

From eq. $\left(1\right)$

  $C$ is positive.

The normal form of the equation is,

  $\frac{3x}{-\sqrt{34}}-\frac{5y}{-\sqrt{34}}+\frac{5}{-\sqrt{34}}=0$

The normal form is,

  $x\cos \varphi +y\sin \varphi -p=0$

Thus, $\cos \varphi =\frac{-3}{\sqrt{34}},\sin \varphi =\frac{5}{\sqrt{34}},p=\frac{5}{\sqrt{34}}$

Now,

  $\begin{array}{l}\tan \varphi =\frac{\sin \varphi }{\cos \varphi }\\ \tan \varphi =\left(\frac{5}{\sqrt{34}}\right)\left(\frac{\sqrt{34}}{-3}\right)\\ \tan \varphi =-\frac{5}{3}\\ \varphi ={\tan }^{-1}\left(-\frac{5}{3}\right)\\ \varphi =-1.03\end{array}$

Now substitute $\varphi$ in polar equation,

  $\begin{array}{l}p=r\cos \left(\theta -\varphi \right)\\ \frac{5}{\sqrt{34}}=r\cos \left(\theta +1.03\right)\end{array}$

Hence, Given equation in polar form is $\frac{5}{\sqrt{34}}=r\cos \left(\theta +1.03\right)$