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anonymous

  • 4 years ago

Challenge: Compute the are of the loop of curve \[y^2=x^2\frac{1+x}{1-x}\]

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  1. anonymous
    • 4 years ago
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    Anyone

  2. anonymous
    • 4 years ago
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    Lets do brainstorming everyone give all ideas coming to mind

  3. anonymous
    • 4 years ago
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    integrate?

  4. anonymous
    • 4 years ago
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    I got it ans is \[2\int\limits_{-1}^{0}x^2\sqrt{\frac{1+x}{1-x}}\]

  5. anonymous
    • 4 years ago
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    \[2-\frac{\pi}{2}\]

  6. TuringTest
    • 4 years ago
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    \[y=2\int_{-1}^{0}x\sqrt{\frac{1+x}{1-x}}dx\]isn't it? why is yours still x^2 squared?

  7. TuringTest
    • 4 years ago
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    A=... sorry

  8. anonymous
    • 4 years ago
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    Sorry

  9. TuringTest
    • 4 years ago
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    is that a no? is should be x^2

  10. TuringTest
    • 4 years ago
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    or should not?

  11. anonymous
    • 4 years ago
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    not x^2

  12. anonymous
    • 4 years ago
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    turing yours was correct

  13. anonymous
    • 4 years ago
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    Should be

  14. TuringTest
    • 4 years ago
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    \[A=\int_{-1}^{0}x\sqrt{\frac{1+x}{1-x}}dx\]\[u=1-x\]groovy :)

  15. TuringTest
    • 4 years ago
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    but my sub wont' work

  16. anonymous
    • 4 years ago
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    that won't work.. i don't think

  17. anonymous
    • 4 years ago
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    Easier if you rationalise miltiply sqrt(1-x) in numerator and denominator

  18. TuringTest
    • 4 years ago
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    oh yeah, ok

  19. anonymous
    • 4 years ago
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    I thought it was arc....I assume everyone else is looking at area...lol

  20. anonymous
    • 4 years ago
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    Sorry 1+x

  21. anonymous
    • 4 years ago
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    compute the area of the loop of the curve

  22. anonymous
    • 4 years ago
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    =area under curve

  23. anonymous
    • 4 years ago
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    Simplifies to \[\int\limits\limits_{-1}^{2}(\frac{x}{\sqrt{1-x^2}}-\sqrt{1-x^2}+\frac{1}{{\sqrt1-x^2}})dx\]

  24. TuringTest
    • 4 years ago
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    where did the 2 come from? I have something else

  25. anonymous
    • 4 years ago
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    Typing error i menat it to be outside

  26. anonymous
    • 4 years ago
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    which should give the anwer approx.....using winplot .4292 But of course the mathematics is more fun than letting the computer do all of the work.

  27. TuringTest
    • 4 years ago
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    ok, I still have a different expression though I'm trying to figure out if they're the same...

  28. TuringTest
    • 4 years ago
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    am I doing something wrong?\[A=\int_{-1}^{0}x\sqrt{\frac{1+x}{1-x}}\cdot\sqrt{\frac{1+x}{1+x}}dx=\int_{-1}^{0}x\frac{1+x}{\sqrt{1-x}}dx\]\[=\int_{-1}^{0}\frac{x}{\sqrt{1-x}}+\frac{x^2}{\sqrt{1-x}}dx\]because my expression looks different ...or am I just tired?

  29. TuringTest
    • 4 years ago
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    2 times all that*

  30. anonymous
    • 4 years ago
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    yeah (1-x)(1+x)=1-x^2 you look tired

  31. TuringTest
    • 4 years ago
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    ok that's just a typo though

  32. TuringTest
    • 4 years ago
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    I meant that the expression is different as far as I seeam I doing something wrong?\[A=2\int_{-1}^{0}x\sqrt{\frac{1+x}{1-x}}\cdot\sqrt{\frac{1+x}{1+x}}dx=2\int_{-1}^{0}x\frac{1+x}{\sqrt{1-x^2}}dx\]\[=2\int_{-1}^{0}\frac{x}{\sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}}dx\]is what I meant

  33. anonymous
    • 4 years ago
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    If that is the case add 1and subtract 1 from x^2 to arrive at mine

  34. TuringTest
    • 4 years ago
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    oh gotcha :P

  35. nikvist
    • 4 years ago
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  36. anonymous
    • 4 years ago
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    Nice!

  37. anonymous
    • 4 years ago
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    I feel double integrals no worth this problem it is not that difficult

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