Question #8a89a

1 Answer
Dec 21, 2016

Use the equations, #r = sqrt(x^2 + y^2)# and #theta = pi + tan^-1(y/x)#
#(-3sqrt(3), -3) to (6, (7pi)/6)#

Explanation:

Given the point:

Use the equation:

#r = sqrt(x^2 + y^2#

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

#r = sqrt(27 + 9)#

#r = 6#

There are many equations for the angle, #theta#; which one you use depends on x and y:

#"[1] "theta = tan^-1(y/x); x > 0 and y ge 0#
#"[2] "theta = pi/2; x = 0 and y > 0#
#"[3] " theta = pi + tan^-1(y/x); x < 0#
#"[4] "theta = (3pi)/2; x = 0 and y < 0#
#"[5] "theta = 2pi + tan^-1(y/x); x > 0 and y < 0#

Because #x = -3sqrt(3)#, then we do not care about the value of y and we use equation [3}:

#theta = pi + tan^-1((-3)/(-3sqrt(3))); x < 0#

#theta = (7pi)/6#