What is the polar form of #( 5,9 )#?
1 Answer
Jan 13, 2016
Explanation:
To convert from cartesian (x , y ) to polar (
use the following relationships :
#r^2 = x^2 + y^2 rArr r = sqrt(x^2 + y^2 ) , theta = tan^-1(y/x ) #
using (5 , 9 ):
# r = sqrt(5^2 + 9^2 ) = sqrt(25 + 81) = sqrt106 # and
#theta = tan^-1(9/5) = 1.06 #
#rArr (5 , 9 ) → ( sqrt106 , 1.06 )#
