How do you convert $(- 3, 3)$ $(-3, 3)$ into polar coordinates?

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How do you convert $(- 3, 3)$ $(-3, 3)$ into polar coordinates?

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Answer 1 · 2018-04-20T09:22:41

$(- 3, 3) \to (3 \sqrt{2}, \frac{3 π}{4})$ $(-3,3)to(3sqrt2,(3pi)/4)$

Explanation:

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

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

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

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

$here x = - 3 and y = 3$ $"here "x=-3" and y=3$

$\Rightarrow r = \sqrt{{(- 3)}^{2} + 3^{2}} = \sqrt{18} = 3 \sqrt{2}$ $rArrr=sqrt((-3)^2+3^2)=sqrt18=3sqrt2$

$(- 3, 3) is in the second quadrant so θ must be in the$ $(-3,3)" is in the second quadrant so "theta" must be in the "$
$second quadrant$ $"second quadrant"$

$θ = {tan}^{- 1} (1) = \frac{π}{4} \leftarrow related acute angle$ $theta=tan^-1(1)=pi/4larrcolor(red)"related acute angle"$

$\Rightarrow θ = π - \frac{π}{4} = \frac{3 π}{4} \leftarrow in second quadrant$ $rArrtheta=pi-pi/4=(3pi)/4larrcolor(red)"in second quadrant"$

$\Rightarrow (- 3, 3) \to (3 \sqrt{2}, \frac{3 π}{4})$ $rArr(-3,3)to(3sqrt2,(3pi)/4)$

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How do you convert $(- 3, 3)$ $(-3, 3)$ into polar coordinates?

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