anonymous
  • anonymous
convert to polar equation : (x ^2+y^2)^3= 4x^2y^2
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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amistre64
  • amistre64
wherever you se an "x", make it "r cos(t)" wherever you see a "y"; make it "r sin(t)"
amistre64
  • amistre64
(rcos(t)^2+rsin(t)^2)^3= 4rcos(t)^2 sin(t)^2
amistre64
  • amistre64
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)

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amistre64
  • amistre64
minus the typos of courser r = 4 cos(t)^2 sin(t)^2 r = [2cos(t)sin(t)]^2 r = (sin(2t))^2
anonymous
  • anonymous
i think you are missing r^2 a few steps up... i.e.\[(x^2+y^2)^3\]is \[\left((r\cos t)^2+(r\sin t)^2\right)^3\]=\[(r^2\cos^2t+r^2\sin^2t)^3\]=\[(r^2(\cos^2t+\sin^2t))^3\]=\[r^6\]
amistre64
  • amistre64
yeah...youre right :)

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