how to parameterize x^3 + y^3 = 1?

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how to parameterize x^3 + y^3 = 1?

Mathematics
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Can you solve for y in terms of x?
z=0 by the way
x = cos^3(t) y= sin^3(t) z = 0 right?

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the answer is given as the same as what you put, but the powers are (2/3), not 3 and i'm unsure how they got that
I was close then :) I dont know how they got ^(2/3) either... id have to go back many years to remember ;)
its a cubic equation that crosses the x and y axis 3 times.... i see that much
i tried to use http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtI.aspx to help but :( cant seem to figure it out
lets see if I can raeson this..... or muddy the waters :) x = rcos ; y = rsin ; 1 = sin^2 + cos^2 right?
r^3 (cos^3 + sin^3) = sin^2 + cos^2
yeah
I see where the ^1/3 is coming from.... just how to get it :)
it seems as though thats not too far away
Do you see that if you plug in x = (cos theta)^{2/3) and y=(sin theta)^(2/3) you get the original equation? That's why the answer is correct, but doesn't give much insight into how you were supposed to guess it.
yeah, the solution being correct is the easy part ;)
1 is also the "radius" or vector measure from the origin.... we could try that route.
are there any trig identitied we could utilize for this?
ermm, not that i can think of
r^3 (cos^3 + sin^3) = r^2 ; r^2 = 1 so r^3 = 1 ?
ye, i think r^3 must have to be 1
cos^3 + sin^3 = cos^2 + sin^2
cos^3 - cos^2 = sin^3 - sin^3 ??
typoed it...but you can see that lol
ye, i see lol...what do you have to do to a square power to make it ^2/3? do you cube root it?
yes; ^2/3 means cbrt(^2)
it's because you want cosx^2+sinx^2=1. So if x=cost^(2/3) when you plug it in, the square will distrbute in to leave cos^2
ok, i think i get it, thanks so much for the help!

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