What is the polar form of #( -23,-3 )#?

1 Answer
Dec 5, 2015

The polar form of #(-23, -3)# is

#(sqrt(538), tan^(-1)(3/23)+pi) ~~ (23.195, 3.271)#

Explanation:

This question has a list of equations used when converting between rectangular and polar coordinates.

In this case, we will be using
#{(r^2 = x^2 + y^2), (tan(theta)=y/x):}#

#=>{(r = sqrt(x^2 + y^2)), (theta = tan^(-1)(y/x)):}#

#r = sqrt((-23)^2 + (-3)^2)#
#theta = tan^(-1)((-3)/(-23)))^(color(red)("*"))#

#" "^color(red)("*")#(While calculating #theta#, the #-3# and #-23# cancel negatives, causing the resulting angle puts us in quadrant #I# when we want quadrant #III#. To fix this, all we need to do is add or subtract #pi# from the angle to put us in the correct quadrant.)

#=>{(r = sqrt(538)), (theta = tan^(-1)(3/23)+pi):}#

Thus we get the polar form of #(-23, -3)# to be

#(sqrt(538), tan^(-1)(3/23)+pi) ~~ (23.195, 3.271)#