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anonymous

  • 4 years ago

**Calc 2 Help Needed** Find the length of the arc formed by y = (1/8) * (0.5x^2 -16ln(x)) from x=2 to x=6.

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  1. Zarkon
    • 4 years ago
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    what seems to be the problem?

  2. Zarkon
    • 4 years ago
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    I assume you know the formula \[\int\limits_{a}^{b}\sqrt{1+(f'(x))^2}dx\]

  3. anonymous
    • 4 years ago
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    oh yes.. I do.. I can't figure out the derivative.

  4. anonymous
    • 4 years ago
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    of y = (1/8) * (0.5x^2 -16ln(x))

  5. Zarkon
    • 4 years ago
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    \[\frac{dy}{dx}=\frac{1}{8}\left(x-\frac{16}{x}\right)\]

  6. anonymous
    • 4 years ago
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    Ok so now it is sqrt(1+ 1/8 * (x-16/x)

  7. Zarkon
    • 4 years ago
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    no...look at the formula again.

  8. anonymous
    • 4 years ago
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    Ahh forgot ^2

  9. Zarkon
    • 4 years ago
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    correct

  10. anonymous
    • 4 years ago
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    so 1+ ((x-16/8)^2

  11. Zarkon
    • 4 years ago
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    no

  12. anonymous
    • 4 years ago
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    I can't bring the 1/8 in? Or shouldn't rather.

  13. Zarkon
    • 4 years ago
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    you lost an x

  14. anonymous
    • 4 years ago
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    ohhh the ^2 wouldn't allow it.

  15. Zarkon
    • 4 years ago
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    \[\frac{1}{8}\left(x-\frac{16}{x}\right)\] \[=\frac{1}{8}\left(\frac{x^2-16}{x}\right)\] \[=\left(\frac{x^2-16}{8x}\right)\] then 1+square it \[=1+\left(\frac{x^2-16}{8x}\right)^2\]

  16. anonymous
    • 4 years ago
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    so how did the top x-16 go to x^2-16?

  17. anonymous
    • 4 years ago
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    oh common denominator

  18. Zarkon
    • 4 years ago
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    yes...\[x=\frac{x^2}{x}\] for x not equal to zero

  19. anonymous
    • 4 years ago
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    ok so after squaring I have 1+ ((x^2 - 16)^2/64x^2.. Well wolfram did.

  20. Zarkon
    • 4 years ago
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    make into one fraction

  21. anonymous
    • 4 years ago
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    with the one?

  22. Zarkon
    • 4 years ago
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    yes

  23. anonymous
    • 4 years ago
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    ok so now I have 64x^2 - (x^2 - 16)^2 all over 64x^2

  24. Zarkon
    • 4 years ago
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    why the minus sign

  25. anonymous
    • 4 years ago
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    + I mean.. sorry

  26. Zarkon
    • 4 years ago
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    multiply out the numerator...then factor

  27. anonymous
    • 4 years ago
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    so now:( x^4 + 32x^2 + 256) / 64x^2

  28. Zarkon
    • 4 years ago
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    yes...now factor the numerator

  29. anonymous
    • 4 years ago
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    the top would factor to (x^2 + 16)^2

  30. Zarkon
    • 4 years ago
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    yes...now take the square root of the whole thing

  31. anonymous
    • 4 years ago
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    the sqrt would cancel the ^2

  32. Zarkon
    • 4 years ago
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    yep

  33. anonymous
    • 4 years ago
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    so (x+16)/64

  34. Zarkon
    • 4 years ago
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    you lost an x again.

  35. Zarkon
    • 4 years ago
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    also it is not 64 anymore

  36. anonymous
    • 4 years ago
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    Whoops.. (x+16)/8x?

  37. Zarkon
    • 4 years ago
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    close \[\frac{x^2+16}{8x}\]

  38. anonymous
    • 4 years ago
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    Ohhh that one.. I am the king of mistakes tonight.

  39. Zarkon
    • 4 years ago
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    write as \[\frac{x^2+16}{8x}=\frac{x^2}{8x}+\frac{16}{8x}=\frac{x}{8}+\frac{2}{x}\] and integrate from 2 to 6

  40. anonymous
    • 4 years ago
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    2.5

  41. Zarkon
    • 4 years ago
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    no

  42. anonymous
    • 4 years ago
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    OMG why did I do that.. ok let me try again.

  43. anonymous
    • 4 years ago
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    I keep getting a negative.

  44. Zarkon
    • 4 years ago
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    what did you get for the integral before evaluating the limits?

  45. anonymous
    • 4 years ago
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    Oh wow, I need to integrate..

  46. Zarkon
    • 4 years ago
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    :)

  47. anonymous
    • 4 years ago
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    Sorry, you probably think I am absolutely dumb.

  48. Zarkon
    • 4 years ago
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    no

  49. anonymous
    • 4 years ago
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    1.33333

  50. Zarkon
    • 4 years ago
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    no

  51. anonymous
    • 4 years ago
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    After integrating I had x^2/16 + 4/2x

  52. Zarkon
    • 4 years ago
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    \[\int\left(\frac{x}{8}+\frac{2}{x}\right)dx\] \[=\frac{x^2}{16}+2\ln|x|+c\]

  53. anonymous
    • 4 years ago
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    ok after wolfram it says the second part should contain a log

  54. anonymous
    • 4 years ago
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    4.19721 with a little confidence.

  55. Zarkon
    • 4 years ago
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    yes...which equals \[2(\ln(3)+1)\]

  56. anonymous
    • 4 years ago
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    YESSS!!!!!! I can't believe that simple problem turned so nasty!

  57. Zarkon
    • 4 years ago
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    ;)

  58. anonymous
    • 4 years ago
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    Would you care to help me solve another?

  59. anonymous
    • 4 years ago
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    I'll start a new problem of course.

  60. Zarkon
    • 4 years ago
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    hmmm...don't know if I have that much time. post it in a new thread...someone will help...If I have time I'll look

  61. anonymous
    • 4 years ago
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    Okay thanks so much.. Were you by any chance a math major?

  62. Zarkon
    • 4 years ago
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    I'm a Math professor

  63. anonymous
    • 4 years ago
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    Cool, where at?

  64. Zarkon
    • 4 years ago
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    so yes...I was a math major

  65. Zarkon
    • 4 years ago
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    I don't give out that info :)

  66. anonymous
    • 4 years ago
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    Ahhh.. haha, I understand. Well I'm a student at a SEC school. Hopefully you aren't my teacher.

  67. Zarkon
    • 4 years ago
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    I'm not in the SEC

  68. anonymous
    • 4 years ago
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    Haha okay.. Well thanks again for the help. This stuff is hard to understand in a lecture hall.

  69. Zarkon
    • 4 years ago
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    good luck

  70. anonymous
    • 4 years ago
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    Thank you. Have a great night.

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