How do you convert #(0, -6)# to polar form?

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Answer 1 · 2016-06-29T17:26:43

When Cartesian coordinates #(x,y)# are converted into polar coordinates #(r,theta)#, we have the relation

#x=rcostheta# and #y=rsintheta# and hence

#r=sqrt(x^2+y^2)#, #costheta=x/r# and #sintheta=y/r#.

hence for #(0,-6)#

#r=sqrt(0^2+(-6)^2)=sqrt(0+36)=sqrt36=6#

and as #costheta=0/6=0# and #sintheta=-6/6=-1#,

we have #theta=-pi/2#

Hence polar coordinates are #(6,-pi/2)#.

Answer 2 · 2016-06-29T17:40:57

Polar conversion is #(6,-pi/2).#

Cartesian #(x,y)# in polar is #(r,theta)#, where, #x=rcostheta, y=rsintheta, theta in(-pi,pi], x^2+y^2=r^2.#

Clearly, #r=6.#

Now #x=rcostheta rArr 0=6costheta rArr costheta = 0.#
#y=rsintheta rArr -6=6sintheta rArr sintheta =-1.#

We conclude that #theta=-pi/2.#

Hence, polar conversion is #(6,-pi/2).#