Chapter 9.4, Problem 5CFU

To determine

To write: The given equation in polar form and round $\phi$ to the nearest degree

Answer to Problem 5CFU

The polar form of the equation is $2=r\cos \left(\theta -{307}^{\circ }\right)$ and the nearest degree to the value of $\phi \text{ is }{307}^{\circ }$

Explanation of Solution

Given information:

The equation is $3x-4y-10=0$

Calculation:

The standard form of the equation is $3x-4y-10=0$ . First, write the equation in normal form to find the values of $p\text{ and }\phi$ . To convert to normal form, it is necessary to find the value of $\pm \sqrt{A^2+B^2}$

  $\begin{array}{l}\pm \sqrt{A^2+B^2}\\ \pm \sqrt{3^2+{\left(-4\right)}^2}\\ \pm \sqrt{9+16}\\ \pm 5\end{array}$

Since $C$ is negative, use $+5$

The normal form of the equation is,

  $\begin{array}{l}\frac{3}{5}x-\frac{4}{5}y-\frac{10}{5}=0\\ \frac{3}{5}x-\frac{4}{5}y-2=0\end{array}$

The normal form is $x\cos \phi -y\sin \phi -p=0$

So, it can be seen from the normal form that $\cos \phi =\frac{3}{5},\sin \phi =-\frac{4}{5}\text{ and }p=2$

Since $\cos \phi >0\text{ and }\sin \phi <0$ , so the normal lies in the fourth quadrant

  $\begin{array}{l}\tan \phi =\frac{\sin \phi }{\cos \phi }\\ \tan \phi =\frac{\left(-\frac{4}{5}\right)}{\left(\frac{3}{5}\right)}\\ \tan \phi =-\frac{4}{3}\\ \phi \approx {307}^{\circ }\left[\text{add 360 to the arctangent value}\right]\end{array}$

Substitute the values for $p\text{ and }\phi$ into the polar form

  $\begin{array}{l}p=r\cos \left(\theta -\phi \right)\\ \Rightarrow 2=r\cos \left(\theta -{307}^{\circ }\right)\end{array}$

Therefore, the polar form of the equation is $2=r\cos \left(\theta -{307}^{\circ }\right)$ and the nearest degree to the value of $\phi \text{ is }{307}^{\circ }$