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Finding Arc Length

  1. Jun 23, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the arc length of the graph, on the interval [1/2, 2], of

    [itex]y = \frac{x^3}{6} + \frac{1}{2x}[/itex]


    2. Relevant equations

    [itex]s = \int^b_a \sqrt{1 + [f'(x)]^2}dx[/itex]

    3. The attempt at a solution

    I began with [itex]s = \int_{1/2}^2 \sqrt{1 + (\frac{x^2}{2} - \frac{1}{2x^2})^2}dx[/itex]

    and became stuck at the step [itex]s = \int_{1/2}^2 \sqrt{1/2 + \frac{x^4}{4} - \frac{1}{4x^4}}dx[/itex]

    I can't a integration technique that would suffice in solving this problem.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 23, 2013 #2

    SteamKing

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    Check your expansion of the squared term under the square root. You have made an algebra mistake.
    After that, add up the fractions under the square root after getting a common denominator. Plug and chug.
     
  4. Jun 23, 2013 #3
    I redid the work, and came up with [itex]\frac{x^4}{2} +\frac{1}{2x^4} +\frac{1}{2}[/itex]

    Is there some way of factoring this?
     
  5. Jun 23, 2013 #4

    SteamKing

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    Nope, not in this form. Like I said, find a common denominator and add these three fractions together.
     
  6. Jun 24, 2013 #5
    Okay, I found a common denominator, and then placed the new expression under the square symbol:

    [itex]\frac{\sqrt{3x^4 + 2}}{2x^2}[/itex] Is there where you wanted me to lead to?
     
  7. Jun 24, 2013 #6

    ehild

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    Try again to expand and simplify the expression

    [itex] 1 +(\frac{x^2}{2} - \frac{1}{2x^2})^2[/itex]

    ehild
     
  8. Jun 24, 2013 #7
    ehild, if you could please look at post #6 and see if I correctly simplified the expression.
     
  9. Jun 24, 2013 #8

    ehild

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    It is wrong, that was I asked you to re-do it.

    ehild
     
  10. Jun 24, 2013 #9

    Ray Vickson

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    Hint: your integral is difficult, but the correct integral is easy.
     
  11. Jun 24, 2013 #10

    SteamKing

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    (a - b)^2 = a^2 - 2ab + b^2
     
  12. Jun 24, 2013 #11

    HallsofIvy

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    One of the things teachers work hard on is setting up problems that are easy for their students.

    Your function is [itex]x^3/6+ x^{-1}/2[/itex]. The derivative is [itex]x^2/2- x^{-2}/2[/itex]. Notice that the exponents are 2 and -2. The square is [itex]x^4/4- 1/2+ x^{-4}/4[/itex]. The "middle" term is a constant, because, of course, 2+ (-2)= 0. Now adding 1 gives [itex]x^4/4+ 1/2+ x^{-4}/4[/itex]. Do you not see that this is exactly the same as the previous formula except that "-1/2" has become "+1/2"? Do you not see that this is exactly "[itex](x^2/2+ x^{-2})^2[/itex]"?
     
  13. Jun 24, 2013 #12

    ehild

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    Not quite, [itex](x^2/2+ x^{-2}/2)^2[/itex] instead.

    ehild
     
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