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Length of Curve

  1. Feb 8, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the length of the curve. [tex]y = x^{3/2}[/tex] from x = 0 to x = 4.


    2. Relevant equations
    [tex]L = \int^{b}_{a} \sqrt{1 + (dy/dx)^{2}} dx[/tex]


    3. The attempt at a solution
    [tex]L = \int^{b}_{a} \sqrt{1 + (dy/dx)^{2}} dx[/tex]
    [tex]L = \int^{4}_{0} \sqrt{1 + (3x^{1/2}/2)^{2}} dx[/tex]
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    I used substitution rule
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    .
    .
    [tex]L = 64/27[/tex]

    Is this correct?

    Thanks
     
  2. jcsd
  3. Feb 8, 2010 #2

    http://www.wolframalpha.com/input/?i=integrate+sqrt[+1+++[+(3/2)+x^(1/2)+]^2+]+from+x=0+to+x=4
    Its better to post all of your work.
     
  4. Feb 8, 2010 #3
  5. Feb 8, 2010 #4
    Yeah Its wrong.

    You will face: [tex]\int_0^4 \sqrt{ 1 + \frac{9}{4}x } \;\ dx[/tex]
    What did you do for it?
     
  6. Feb 8, 2010 #5

    CompuChip

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    Science Advisor
    Homework Helper

    Actually I found [itex]c(7 \sqrt{7} - 1)[/itex],
    where c is a numerical factor (I got 2/3) which I'm not sure of because I did the calculation sloppily.
    So I suggest you show the rest of the calculation as well.

    [edit] Too slow, you already have several replies. [/edit]
     
  7. Feb 8, 2010 #6
    That's exactly what I got, but when I used the substitution rule, to integrate, I did not change the limits of integration to in terms of u.
     
  8. Feb 8, 2010 #7
    Ohhh.
    BTW, Its a famous mistake. :)
     
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