# Find two polar equivalents of the Cartesian point (-3 \sqrt{3}, 3)

## Question:

Find two polar equivalents of the Cartesian point {eq}\displaystyle (-3 \sqrt{3}, 3) {/eq}

## Polar coordinates

Polar coordinates of a Cartesian point, {eq}\displaystyle (x,y){/eq} is a pair of two coordinates, a real number {eq}\displaystyle r {/eq} and an angle {eq}\displaystyle \theta, {/eq} i.e. {eq}\displaystyle (r,\theta). {/eq}

The real number {eq}\displaystyle r {/eq} is the signed distance between the pole and the point, while the angle {eq}\displaystyle \theta {/eq} is the angle between the polar axis and the line going through the pole and the point.

A Cartesian point can be written in four different polar forms: {eq}\displaystyle (r,\theta), (r, -\theta), (-r, \theta), \text{ or } (-r, -\theta) {/eq}

The angle {eq}\displaystyle \theta {/eq} can be measured negatively if the rotation is done clockwise from the polar axis, while {eq}\displaystyle r {/eq} is negative if the angle {eq}\displaystyle \theta {/eq} is in the opposite quadrant form the quadrant containing the point.

To convert from Cartesian to polar coordinates, we use the following formulae

{eq}\displaystyle r=\sqrt{x^2+y^2}, \\ \displaystyle \theta=\arctan\left(\frac{y}{x}\right), \text { if the point in in the Quadrant I or IV, or} \\ \displaystyle \theta=\arctan\left(\frac{y}{x}\right) +\pi, \text { if the point in in the Quadrant II or III}. {/eq}

The above formulae will give only one form of polar coordinate, the positive one {eq}\displaystyle (r,\theta). {/eq} For the other three forms, we just need to read the point negatively.

## Answer and Explanation:

The Cartesian point {eq}\displaystyle (-3 \sqrt{3}, 3) {/eq} is in the Quadrant II, therefore we will add {eq}\displaystyle \pi {/eq} to the inverse...

See full answer below.

Become a Study.com member to unlock this answer! Create your account

View this answerThe Cartesian point {eq}\displaystyle (-3 \sqrt{3}, 3) {/eq} is in the Quadrant II, therefore we will add {eq}\displaystyle \pi {/eq} to the inverse tangent.

Using the conversion formulae, we obtain

{eq}\displaystyle r=\sqrt{(-3\sqrt{3})^2+3^2}=\sqrt{9\cdot 3+9}=\sqrt{9\cdot4}=6, \\ \displaystyle \theta=\arctan\left(\frac{3}{-3\sqrt{3}}\right) +\pi =\arctan\left(-\frac{\sqrt{3}}{3}\right)+\pi=-\frac{\pi}{6}+\pi=\frac{5\pi}{6}. {/eq}

So, one polar form is {eq}\displaystyle \boxed{\left(6, \frac{5\pi}{6}\right)}. {/eq}

#### Learn more about this topic:

from Calculus: Help and Review

Chapter 1 / Lesson 16