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anonymous

  • one year ago

Find the arc length of y=x^(2/3) for 0<x<1

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  1. anonymous
    • one year ago
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  2. amistre64
    • one year ago
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    \[\int_{}^{}ds\implies \int_{}^{}\sqrt{(x')^2+(y')^2}~dt\]

  3. amistre64
    • one year ago
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    of instead of dt, we can do with dx, and let x'=1

  4. amistre64
    • one year ago
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    what are you getting for your setup?

  5. anonymous
    • one year ago
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    im getting this \[\int\limits_{0}^{1} \sqrt{(1)^2 + (x^\frac{ 2 }{ 3 })^2} dx\]

  6. anonymous
    • one year ago
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    oh wait

  7. anonymous
    • one year ago
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    |dw:1442357936251:dw|

  8. amistre64
    • one year ago
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    thats better yeah

  9. anonymous
    • one year ago
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    and then

  10. anonymous
    • one year ago
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    |dw:1442358056890:dw|

  11. anonymous
    • one year ago
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    then you find the integral?

  12. amistre64
    • one year ago
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    then you find the integral ... trying to see if there is a nice form to it tho

  13. amistre64
    • one year ago
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    the wolfs integral suggests its a fairly simple u-sub

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  14. amistre64
    • one year ago
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    or not ...

  15. anonymous
    • one year ago
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    how do you get to that? im trying to integrate

  16. amistre64
    • one year ago
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    http://www.wolframalpha.com/input/?i=integrate+sqrt%28+1+%2B+4%2F9+x^%28-2%2F3%29+%29

  17. anonymous
    • one year ago
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    |dw:1442358829336:dw|

  18. amistre64
    • one year ago
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    i just ask the wolf to integrate it ...

  19. anonymous
    • one year ago
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    would that be the integral?

  20. anonymous
    • one year ago
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    im sure its that just trying to figure out how to get to that point

  21. amistre64
    • one year ago
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    how are you sure that is it? i have to wonder since the wolfs assessment is different

  22. amistre64
    • one year ago
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    one thing i am thing of is trying to work out a parametric like x = t^3, y=t^2 this way, t = x^(1/3), letting y = x^(2/3) by eliminating the parameter

  23. amistre64
    • one year ago
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    t=0 to 1 satisfies the substitution x' = 3t^2, y' =2t sqrt(9t^4 + 4t^2) t sqrt(9t^2 + 4) dt

  24. anonymous
    • one year ago
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    |dw:1442359448247:dw|

  25. anonymous
    • one year ago
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    would u sub work here?

  26. amistre64
    • one year ago
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    i already tried working it like that ... i was getting nowhere. but i believe if we split it into parameters it works out better for us

  27. amistre64
    • one year ago
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    t sqrt(9t^2 + 4) dt let u = 9t^2+4 du = 18t dt t sqrt(u) du/18t u^(1/2) du/18 2/3(18) u^(3/2) 1/3(9) u^(3/2) 1/27 u^(3/2)

  28. amistre64
    • one year ago
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    u = 4 to 13 when t=0 to 1

  29. amistre64
    • one year ago
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    1/27 ((13)^(3/2)-(4^(3/2)), is looking sweet on my end

  30. amistre64
    • one year ago
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    i cant make it work if we keep y as a function of x as its defined. so i wondered if parameters would make life easier, and it does

  31. anonymous
    • one year ago
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    |dw:1442360116489:dw|

  32. anonymous
    • one year ago
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    is that what you mean?

  33. amistre64
    • one year ago
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    \[y = x^{2/3}~:~x=(0,1)\] \[x^{1/3}=t~:~x=t^3~:~t=(0,1)\] \[y=t^2\] \[x'=3t^2~:~y'=2t\] \[\int_{0}^{1}\sqrt{(x')^2+(y')^2}~dt\]

  34. amistre64
    • one year ago
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    that simplifies to: \[\int_{0}^{1}t\sqrt{9t^2+4}~dt\] yes

  35. anonymous
    • one year ago
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    aaah i see what you did there

  36. anonymous
    • one year ago
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    |dw:1442360353998:dw|

  37. anonymous
    • one year ago
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    |dw:1442360476915:dw|

  38. anonymous
    • one year ago
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    and then you insert u? and then plug in the 1? and that gets you the answer

  39. amistre64
    • one year ago
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    by substitution u = 9t^2 + 4, as t goes from 0 to 1, u goes from 9(0)+4 to 9(1)+4

  40. amistre64
    • one year ago
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    so the limits for du stuff is 4 to 13

  41. amistre64
    • one year ago
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    youcould replace u witn 9t^2 + 4 if you want, but then the 13 and 4 get calculated in there inways

  42. anonymous
    • one year ago
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    umm im a bit lost but i'll try to see where it is that im confused

  43. amistre64
    • one year ago
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    k, and dont worry about being lost, my approach was prolly not 'standard' it is just what worked out in my head.

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