How do you convert $(0, 2)$ $(0,2)$ from cartesian to polar coordinates?

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How do you convert $(0, 2)$ $(0,2)$ from cartesian to polar coordinates?

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Answer 1 · 2017-07-10T14:08:40

$(2, \frac{π}{2})$ $(2, pi/2)$

Explanation:

A polar coordinate is in the form $(r, θ)$ $(r, theta)$ , where $r$ $r$ is the distance from the origin and $θ$ $theta$ is the corresponding angle. We can see here that $r = 2$ $r=2$ and $θ = \frac{π}{2}$ $theta=pi/2$ . However, we can also use the following formulas:

$r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$

$tan θ = \frac{y}{x}$ $tan theta = y/x$

$r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$
$r = \sqrt{x^{2} + y^{2}}$ $r=sqrt(x^2+y^2)$
$r = \sqrt{0^{2} + 2^{2}}$ $r=sqrt(0^2+2^2)$
$r = 2$ $r=2$

$tan θ = \frac{2}{0}$ $tan theta = 2/0$
This is undefined, but $tan (\frac{π}{2})$ $tan (pi/2)$ is undefined anyways.

The polar coordinate is $(2, \frac{π}{2})$ $(2, pi/2)$ .

Answer 2 · 2017-07-10T14:13:06

$(2, \frac{π}{2})$ $(2,pi/2)$

Explanation:

$to convert from cartesian to polar$ $"to convert from "color(blue)"cartesian to polar"$

$that is (x, y) \to (r, θ) where$ $"that is " (x,y)to(r,theta)" where"$

$∙ x r = \sqrt{x^{2} + y^{2}}$ $•color(white)(x)r=sqrt(x^2+y^2)$

$∙ x θ = {tan}^{- 1} (\frac{y}{x}) x; - π < θ \leq π$ $•color(white)(x)theta=tan^-1(y/x)color(white)(x);-pi< theta<= pi$

$here x = 0 and y = 2$ $"here " x=0" and " y=2$

$\Rightarrow r = \sqrt{0^{2} + 2^{2}} = 2$ $rArrr=sqrt(0^2+2^2)=2$

$θ = {tan}^{- 1} (\frac{2}{0}) \leftarrow undefined$ $theta=tan^-1(2/0)larrcolor(red)" undefined"$

$\Rightarrow θ = \frac{π}{2}$ $rArrtheta=pi/2$

$\Rightarrow (0, 2) \to (2, \frac{π}{2})$ $rArr(0,2)to(2,pi/2)$

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How do you convert $(0, 2)$ $(0,2)$ from cartesian to polar coordinates?

2 Answers

Explanation:

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