How do you convert the Cartesian coordinates (2√3, 2) to polar coordinates?

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How do you convert the Cartesian coordinates (2√3, 2) to polar coordinates?

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Answer 1 · 2015-09-05T18:33:31

$(4, \frac{π}{6})$ $(4,pi/6)$ .

Explanation:

Let $(x, y)$ $(x,y)$ be a coordinate on the Cartesian plane.

The corresponding polar coordinate is $(r, θ)$ $(r,theta)$ , where:

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

(You might notice that this is similar to the distance formula; that's not a coincidence, $r$ $r$ is the distance from the point to the pole (a.k.a. the center) )

and:

$θ = {tan}^{- 1} (\frac{y}{x})$ $theta = tan^-1(y/x)$

So, given $(2 \sqrt{3}, 2)$ $(2sqrt(3),2)$ :

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

$r = \sqrt{12 + 4}$ $r = sqrt(12+4)$

$r = \sqrt{16}$ $r = sqrt(16)$

$r = 4$ $r = 4$

$θ = {tan}^{- 1} (\frac{2}{2 \sqrt{3}})$ $theta = tan^-1(2/(2sqrt(3)))$

$θ = {tan}^{- 1} (\frac{1}{\sqrt{3}})$ $theta = tan^-1(1/sqrt(3))$

$θ = {tan}^{- 1} (\frac{\sqrt{3}}{3})$ $theta = tan^-1(sqrt(3)/3)$

$θ = \frac{π}{6}$ $theta = pi/6$

We can say that $θ = \frac{π}{6}$ $theta = pi/6$ (and not $\frac{5 π}{6}$ $(5pi)/6$ , etc.) because $x$ $x$ and $y$ $y$ are both positive, which means the point is in the first quadrant.

Thus, the point in polar coordinates is $(4, \frac{π}{6})$ $(4,pi/6)$ .

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How do you convert the Cartesian coordinates (2√3, 2) to polar coordinates?

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