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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    Can you solve for y in terms of x?

  2. anonymous
    • 5 years ago
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    z=0 by the way

  3. amistre64
    • 5 years ago
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    x = cos^3(t) y= sin^3(t) z = 0 right?

  4. anonymous
    • 5 years ago
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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

  5. amistre64
    • 5 years ago
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    I was close then :) I dont know how they got ^(2/3) either... id have to go back many years to remember ;)

  6. amistre64
    • 5 years ago
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    its a cubic equation that crosses the x and y axis 3 times.... i see that much

  7. anonymous
    • 5 years ago
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    i tried to use http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtI.aspx to help but :( cant seem to figure it out

  8. amistre64
    • 5 years ago
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    lets see if I can raeson this..... or muddy the waters :) x = rcos ; y = rsin ; 1 = sin^2 + cos^2 right?

  9. amistre64
    • 5 years ago
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    r^3 (cos^3 + sin^3) = sin^2 + cos^2

  10. anonymous
    • 5 years ago
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    yeah

  11. amistre64
    • 5 years ago
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    I see where the ^1/3 is coming from.... just how to get it :)

  12. anonymous
    • 5 years ago
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    it seems as though thats not too far away

  13. anonymous
    • 5 years ago
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    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.

  14. amistre64
    • 5 years ago
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    yeah, the solution being correct is the easy part ;)

  15. amistre64
    • 5 years ago
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    1 is also the "radius" or vector measure from the origin.... we could try that route.

  16. amistre64
    • 5 years ago
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    are there any trig identitied we could utilize for this?

  17. anonymous
    • 5 years ago
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    ermm, not that i can think of

  18. amistre64
    • 5 years ago
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    r^3 (cos^3 + sin^3) = r^2 ; r^2 = 1 so r^3 = 1 ?

  19. anonymous
    • 5 years ago
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    ye, i think r^3 must have to be 1

  20. amistre64
    • 5 years ago
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    cos^3 + sin^3 = cos^2 + sin^2

  21. amistre64
    • 5 years ago
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    cos^3 - cos^2 = sin^3 - sin^3 ??

  22. amistre64
    • 5 years ago
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    typoed it...but you can see that lol

  23. anonymous
    • 5 years ago
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    ye, i see lol...what do you have to do to a square power to make it ^2/3? do you cube root it?

  24. amistre64
    • 5 years ago
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    yes; ^2/3 means cbrt(^2)

  25. anonymous
    • 5 years ago
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    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

  26. anonymous
    • 5 years ago
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    ok, i think i get it, thanks so much for the help!

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