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anonymous
 5 years ago
how to parameterize x^3 + y^3 = 1?
anonymous
 5 years ago
how to parameterize x^3 + y^3 = 1?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you solve for y in terms of x?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x = cos^3(t) y= sin^3(t) z = 0 right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I was close then :) I dont know how they got ^(2/3) either... id have to go back many years to remember ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its a cubic equation that crosses the x and y axis 3 times.... i see that much

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i tried to use http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtI.aspx to help but :( cant seem to figure it out

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lets see if I can raeson this..... or muddy the waters :) x = rcos ; y = rsin ; 1 = sin^2 + cos^2 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0r^3 (cos^3 + sin^3) = sin^2 + cos^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I see where the ^1/3 is coming from.... just how to get it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it seems as though thats not too far away

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do 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.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, the solution being correct is the easy part ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01 is also the "radius" or vector measure from the origin.... we could try that route.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0are there any trig identitied we could utilize for this?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ermm, not that i can think of

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0r^3 (cos^3 + sin^3) = r^2 ; r^2 = 1 so r^3 = 1 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ye, i think r^3 must have to be 1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0cos^3 + sin^3 = cos^2 + sin^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0cos^3  cos^2 = sin^3  sin^3 ??

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0typoed it...but you can see that lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ye, i see lol...what do you have to do to a square power to make it ^2/3? do you cube root it?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yes; ^2/3 means cbrt(^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it'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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, i think i get it, thanks so much for the help!
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