Question

convert to polar equation : (x ^2+y^2)^3= 4x^2y^2

Answer 1

wherever you se an "x", make it "r cos(t)"

wherever you see a "y"; make it "r sin(t)"

Answer 2

(rcos(t)^2+rsin(t)^2)^3= 4rcos(t)^2 sin(t)^2

Answer 3

cos^2 + sin^2 = 1 therefore:

[r(1)]^3 = r^3

r^3= 4(rcos(t))^2 (rsin(t))^2

divide thru by r^2

r = 4cos(t)sin(t)

Answer 4

minus the typos of courser

r = 4 cos(t)^2 sin(t)^2

r = [2cos(t)sin(t)]^2

r = (sin(2t))^2

Answer 5

i think you are missing r^2 a few steps up...
i.e.\[(x^2+y^2)^3\]is \[\left((r\cos t)^2+(r\sin t)^2\right)^3\]=\[(r^2\cos^2t+r^2\sin^2t)^3\]=\[(r^2(\cos^2t+\sin^2t))^3\]=\[r^6\]

Answer 6

yeah...youre right :)