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

1 Answer
Nov 17, 2016

Please see the explanation.

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)