07-30-2008, 11:27 AM
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#12 (permalink)
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Experienced Zuner
Join Date: Jul 2008
Location: Chicago
Posts: 139
Reputation: 17
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Quote:
Originally Posted by Netrix
It does not matter why he cares.
Polar Relationships:
x = r cos(t)
y = r sin(t)
x^2 + y^2 = r^2.
So:
sin(t) = x / r
cos(t) = y / r
sin(t) = x / sqrt(x^2 + y^2)
r = 2 + 2 sin(2t) (Assuming this is what is meant, as opposed to 2 + 2 sin^2(t))
sqrt(x^2 + y^2) = 2 + 2(sin(2t))
sqrt(x^2 + y^2) = 2 + 2(2 sin(t) + cos(t))
sqrt(x^2 + y^2) = 2 + 2(2(x / sqrt(x^2 + y^2)) + (y / sqrt(x^2 + y^2))
sqrt(x^2 + y^2) = 2 + 4x / sqrt(x^2 + y^2) + y / sqrt(x^2 + y^2)
Multiply by (x^2 + y^2)sqrt(x^2 + y^2)
(x^2 + y^2)^2 = 2((x^2 + y^2)sqrt(x^2 + y^2) + 4x(x^2 + y^2) + y(x^2 + y^2)
x^4 + y^4 + 2(x^2)(y^2) = (2x^2 + 2y^2)(x^2 + y^2)^(1/2) + 4x^3 + 4x(y^2) + (x^2)y + y^3
x^4 + y^4 - 4x^3 - y^3 + 2(x^2)(y^2) - (x^2)y - 4x(y^2) - (2x^2 + 2y^2)(x^2 + y^2)^(1/2) = 0
I think that is right...
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 HOLY &#$!, good work!

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