# What is the polar form of ( -4,32 )?

Aug 29, 2017

$\left(r , \theta\right) = \left(4 \sqrt{65} , 1.69515132\right)$

#### Explanation:

The rectangular point $\left(- 4 , 32\right)$ is in the form $\left(x , y\right)$.

Polar points are in the form $\left(r , \theta\right)$. See the attached image for what this means: To find $r$, we effectively need to find the hypotenuse of the right triangle with legs of $x$ and $y$.

Thus, $r = \sqrt{{x}^{2} + {y}^{2}}$. Here, this becomes

$r = \sqrt{{\left(- 4\right)}^{2} + {\left(32\right)}^{2}} = \sqrt{{4}^{2} + {32}^{2}} = \sqrt{{4}^{2} + {4}^{2} \left({8}^{2}\right)} = \sqrt{{4}^{2} \left(1 + {8}^{2}\right)} = 4 \sqrt{65}$

Even though the point $\left(- 4 , 32\right)$ is in Quadrant $\text{II}$, the value of $r$ is a magnitude and is still positive.

To find $\theta$, we first need to write some statement involving $\theta$ given the information that we know.

Looking at the image, we have $\theta$, the side opposite $\theta$, and the side adjacent to $\theta$ in a right triangle. Thus, we can say:

$\tan \theta = \text{opposite"/"adjacent} = \frac{y}{x}$

Solving for $\theta$:

$\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)$

Using our known values:

$\theta = {\tan}^{-} 1 \left(\frac{32}{- 4}\right) = {\tan}^{-} 1 \left(- 8\right) = - 1.44644133$

Note, however, that this is a negative value and that $- 1.44644133 > - \frac{\pi}{2}$, so this is really an angle in Quadrant $\text{IV}$.

To find the value of this angle in Quadrant $\text{II}$, we know that it will be $\pi$ minus the magnitude of the angle we determined.

That is,

$\theta = \pi - 1.44644133 = 1.69515132$

So, our point is:

$\left(r , \theta\right) = \left(4 \sqrt{65} , 1.69515132\right)$