Question

Determine two pairs of polar coordinates for the point (3, -3) with 0°≤ θ < 360°

Answer 1

`\left ( 3√(2),135^(\circ) \right )\,,\,\left ( 3√(2),315^(\circ) \right )`

Explanation:

Let (x,y) be the rectangular coordinates of the point.

Here,
`(x,y)=(3,-3)`

Let polar coordinates be
`(r,\theta )` such that
`r=√(x^2+y^2)\,,\,\theta =\arctan \left ( (y)/(x) \right )`

`r=√(3^2+(-3)^2)=√(18)=3√(2)`

`\theta =\arctan \left ( (-3)/(3) \right )= \arctan (-1)`

We know that tan is negative in first and fourth quadrant, we get

`\theta =\pi-(\pi)/(4)=(3\pi)/(4)=135^(\circ)\\\theta =2\pi-(\pi)/(4)=(7\pi)/(4)=315^(\circ)`

So, polar coordinates are
`\left ( 3√(2),135^(\circ) \right )\,,\,\left ( 3√(2),315^(\circ) \right )`

Answer 2

`(3 √(2) ,135 \degree) \: \: and \: \: (3 √(2) ,315 \degree)`

Explanation:

Polar coordinates are of the form:

`(r,\theta)`

where

`r = \sqrt{ {x}^(2) + {y}^(2) }`

and

`\theta = \tan^( - 1) ( (y)/(x) )`

The given rectangular coordinates are: (3,-3).

`r = \sqrt{ {3}^(2) + {( - 3)}^(2) }`

`r = √(9 + 9) = √(18)`

`r = 3 √(2)`

`\theta = \tan^( - 1) ( ( - 3)/(3) )`

`\theta = \tan^( - 1) ( - 1 )`

`\theta =135 \degree \: \: or \: \: 315 \degree`

The two polar coordinates are:

`(3 √(2) ,135 \degree) \: \: and \: \: (3 √(2) ,315 \degree)`