How do you convert #(2, -3)# into polar form?

1 Answer
Jim G.
Aug 23, 2016

#(2,-3)to(sqrt13,-0.983)#

Explanation:

To convert from #color(blue)"cartesian to polar form"#

That is #(x,y)to(r,theta)#

#color(orange)"Reminders"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(r=sqrt(x^2+y^2))color(white)(a/a)|))) " and " color(red)(|bar(ul(color(white)(a/a)color(black)(theta=tan^-1(y/x))color(white)(a/a)|)))#

here x = 2 and y = -3

#rArrr=sqrt(2^2+(-3)^2)=sqrt13#

now (2 ,-3) is a point in the 4th quadrant so we must ensure that #theta# is in the 4th quadrant.

#theta=tan^-1(-3/2)=-0.983" in 4th quadrant "#

#rArr(2,-3)to(sqrt13,-0.983)=(sqrt13,-56.31^@)#

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