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anonymous

  • 5 years ago

Anyone know how to find the arc length of the curve: y=(((x^(4))/(4))+(1/(8x^(2))) on the interval 1less than or equal to x less than or equal to 2

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  1. amistre64
    • 5 years ago
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    ds = sqrt(dx^2 + dt^2) if I recall correctly

  2. amistre64
    • 5 years ago
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    since y is a function of "x" dx^2 = 1

  3. amistre64
    • 5 years ago
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    dt was supposed to be dy.....

  4. anonymous
    • 5 years ago
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    \[L=\int\limits_{a}^{b}\sqrt{1+(dy/dx)^2 dx}\]

  5. anonymous
    • 5 years ago
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    a=1 b=2 but what is (dy/dx)?

  6. amistre64
    • 5 years ago
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    looks right.... but I think it has a "ds" on the end, not that it matters much :) and make sure your "dx" is out of the radical

  7. amistre64
    • 5 years ago
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    dy/dx = the derivative of your function

  8. anonymous
    • 5 years ago
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    I have (dy/dx)= x^(3) - (1/(4x^(3)))

  9. amistre64
    • 5 years ago
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    the length thing is the square root of the sum off the derivatives of x and y; since y is a function of x; x'=1

  10. anonymous
    • 5 years ago
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    Is that right?

  11. amistre64
    • 5 years ago
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    id have to try to decipher the original equation you posted and maybe implicit it for y'...

  12. anonymous
    • 5 years ago
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    okay hold on let me write it again, sorry for it being sloppy hahah

  13. amistre64
    • 5 years ago
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    :)

  14. anonymous
    • 5 years ago
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    \[y=(x^{4}/4) + 1/(8x^2)\]

  15. anonymous
    • 5 years ago
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    A lil neater :D

  16. amistre64
    • 5 years ago
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    neater yes; but make sure that this is the correct equation.... double check it :)

  17. anonymous
    • 5 years ago
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    Yes, lol. And I have to find the arc length of the curve from \[1\le x \le 2\]

  18. amistre64
    • 5 years ago
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    4x^3 16x y' = ------ (-) --------- 4 (8x^2)^2 this should be y' = dy/dx

  19. amistre64
    • 5 years ago
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    can reduce as needed :)

  20. anonymous
    • 5 years ago
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    y' = x^(3) - (1/4x^(3)) reduced?

  21. amistre64
    • 5 years ago
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    yep, that looks good :)

  22. anonymous
    • 5 years ago
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    haha awesome.

  23. anonymous
    • 5 years ago
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    now I have to square that. That is when I got : x^(6) - 1/2 + 1/(16x^(6))

  24. amistre64
    • 5 years ago
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    as long as you kept track of all the stuff, ill trust you on that :)

  25. anonymous
    • 5 years ago
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    let me double check real quick, I'm not sureee..

  26. amistre64
    • 5 years ago
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    can we get that into 1 term? get like denominators be easier?

  27. anonymous
    • 5 years ago
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    I think that might make everything a bit more confusing... lol

  28. anonymous
    • 5 years ago
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    hold on a sec :)

  29. amistre64
    • 5 years ago
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    4x^6 - 1 -------- ?? 4x^3

  30. anonymous
    • 5 years ago
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    ummmm... I'm lost. Sorry! What did you do?

  31. anonymous
    • 5 years ago
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    nvm, I see what you did haha

  32. anonymous
    • 5 years ago
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    hm, maybe.. let me see real quick

  33. amistre64
    • 5 years ago
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    y' = x^(3) - (1/4x^(3)) get like denominators.... 4x^3 is common

  34. anonymous
    • 5 years ago
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    ahhhh, smart move, that helped. let me make that change real fast

  35. amistre64
    • 5 years ago
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    16x^12 -8x^6 +1 ------------------ = y'^2 16x^6

  36. anonymous
    • 5 years ago
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    Yeah! now I have to put that back into the equation for arc length..

  37. amistre64
    • 5 years ago
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    yep...

  38. amistre64
    • 5 years ago
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    reduce it if you want :)

  39. anonymous
    • 5 years ago
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    \[\int\limits_{1}^{2} x^6 + 1/16x^6 + 1/2 dx\]

  40. amistre64
    • 5 years ago
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    (S) sqrt[1 + x^6 + (1/16x^6) - (1/2)] dx

  41. anonymous
    • 5 years ago
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    oh thats not too bad

  42. amistre64
    • 5 years ago
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    dont forget its the "square root" of 1 + y'^2 ....

  43. anonymous
    • 5 years ago
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    ohhhhh, this is where I got caught up...

  44. anonymous
    • 5 years ago
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    yeah, thats where I didn't know what to do next. Should I take the square root of each term separately?

  45. amistre64
    • 5 years ago
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    cant do that; they aint multiplied together, it just a one lump sum under that radical and you cant break that up....

  46. anonymous
    • 5 years ago
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    x^(3) + 1/4x^(3) + 1/sqrt2 ?

  47. anonymous
    • 5 years ago
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    oh you cant do that :( darnnnn

  48. anonymous
    • 5 years ago
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    are you sure?

  49. amistre64
    • 5 years ago
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    watch: sqrt(4*3) = sqrt(4) * sqrt(3) = 2sqrt(3) BUT!! sqrt(4+3) NOT equal sqrt(4) + sqrt(3)

  50. anonymous
    • 5 years ago
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    ohhhhh, you're right. soooo... what do I do?

  51. anonymous
    • 5 years ago
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    :(

  52. amistre64
    • 5 years ago
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    I dont really know..... thats were I usually have to open up the book and re-read what they have in there for that section :)

  53. amistre64
    • 5 years ago
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    it looks like you might be able to do some sort of trig substitution, but i just aint confident about my abilities in that yet...

  54. anonymous
    • 5 years ago
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    ahhh, is there a quadratic function we can make out of those three terms?

  55. anonymous
    • 5 years ago
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    oh yeahhh, trig substitution... ugh

  56. amistre64
    • 5 years ago
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    sqrt(1+tan^2) = something or another.... but I always mess it up :)

  57. anonymous
    • 5 years ago
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    lol I'll check...

  58. anonymous
    • 5 years ago
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    thanks for helping me on such an exhausting problem! :D

  59. amistre64
    • 5 years ago
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    lol ..... I hope I learned something :)

  60. amistre64
    • 5 years ago
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    if y' = tan sqrt(1+tan^2) = sqrt(sec^2) = sec... (S) sec du or something...

  61. anonymous
    • 5 years ago
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    hahaha, ummm... I dont know if its a trig substitution. But if you wanted to move to another problem or anything thats fine. I dont want to bother you all night with just this problem

  62. anonymous
    • 5 years ago
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    totally your choice :)

  63. amistre64
    • 5 years ago
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    well, library is closing soon, which takes my internet with it :) So I will be heading out anywhoos :) It was fun... Ciao

  64. anonymous
    • 5 years ago
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    Thanks again :)

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