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sketch the curve (x^2+y^2)^3=4x^2y^2 in polar coordinates

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Polar Coordinates, Sketch the curve (x^2+y^2)^3=4x^2y^2

asked Dec 6, 2013 in TRIGONOMETRY by andrew Scholar

2 Answers

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Using polar coordinate transformations

x = r Cosθ

y = r sinθ

x^2+y^2 = r^2 Cos^2θ+r^2 Sin^2θ

x^2+y^2 = r^2(Cos^2θ+Sin^2θ)

 x^2+y^2 = r^2

(x^2+y^2)^3 = (r^2)3

= r^6

And 4x^2y^2 = 4r^2 Cos^2θr^2 Sin^2θ

= 4r^4Cos^2θSin^2θ

Now (x^2+y^2)^3 = 4x^2y^2

r^6 = 4r^4Cos^2θSin^2θ

Divide to each side by r^4

r^6/r^4 = (4r^4Cos^2θSin^2θ)/r^4

r^2 = 2^2Cos^2θSin^2θ

r^2 = (2CosθSinθ)^2

Cancell the squres on each side.

r = 2CosθSinθ

r = 2 SinθCosθ

r = Sin2θ

answered Feb 4, 2014 by dozey Mentor
0 votes

Graph for r = Sin(2theta) for theta in [0,2pi]

answered Feb 4, 2014 by dozey Mentor

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