How do you find the polar coordinates given $(-2, -2sqrt3)$ ?

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Answer 1 · 2016-11-17T22:30:41

Please see the explanation.

Polar coordinates are an order pair of $(r, theta)$

The conversion from $(x, y)$ to r is:

$r = sqrt(x^2 + y^2)$

The conversion from $(x, y)$ to $theta$ is more complicated:

If $x > 0 and y >= 0$ , then $theta = tan^-1(y/x)" [1]"$

If $x = 0 and y > 0$ , then $theta = pi/2" [2]"$

If $x = 0 and y < 0$ , then $theta = (3pi)/2" [3]"$

If $x < 0$ , then $theta = pi + tan^-1(y/x)" [4]"$

If $x > 0 and y < 0$ , then $theta = 2pi + tan^-1(y/x)" [5]"$

For the given point $(-2, -2sqrt(3))$

$r = sqrt((-2)^2 + (-2sqrt(3))^2)$

$r = sqrt(4 + 12)$

$r = sqrt(16)$

$r = 4$

Because the x coordinate is less than zero, we use equation [4]:

$theta = pi + tan^-1((-2sqrt(3))/(-2))$

$theta = pi + pi/3$

$theta = (4pi)/3$

The polar point is $(4, (4pi)/3)$

How do you find the polar coordinates given (-2, -2sqrt3)(-2, -2sqrt3)?