how do you obtain the cartesian equation for r^2=16cos(3theata)?? I've tried looking on many website and at my notes but have had no luck.

Question

i know that r^2=x^2+y^2 and x=rcos(theta) but don't know were to go from here?

Answer 1

the messy part is the cos (3Ø)
we know cos 3Ø = 4cos^3 Ø - 3cosØ

and since you know cosØ = x/r
we get
x^2 + y^2 = 16x^3/r^3 - 3x/r
x^2 + y^2 = 16x/((x^2+y^2)√(x^2 + y^2)) - 3x/√(x^2+y^2)

see:
http://www.wolframalpha.com/input/?i=plot+x%5E2+%2B+y%5E2+%3D+16x%2F%28%28x%5E2%2By%5E2%29√%28x%5E2+%2B+y%5E2%29%29+-+3x%2F√%28x%5E2%2By%5E2%29+

Answer 2

thanks so much ill have a look

Answer 3

forgot the 4 in front of the 4cos^3 Ø

should be

x^2 + y^2 = 16x^3/(64r^3) - 3x/r
x^2 + y^2 = (1/4)x/((x^2+y^2)√(x^2 + y^2)) - 3x/√(x^2+y^2)