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Length of a Plane Curve

  1. Feb 22, 2004 #1
    Okay I'm understanding this for the most part but am having trouble with what I believe is an algebra step.


    Find the length of an arc curve y=x^(3/2) from (1,1) to (2,2sqrt2)
    For second part I need to do it in terms of x.

    Okay..

    Rewritten,

    x = g(y) = y^(2/3)
    ___________________

    So,

    L= int{ctod} sqrt(1 + [g'(y)]^2) dy

    = int{1 to 2sqrt2} sqrt(1 + [2/3*y^(-1/3)]^2) dy

    = int{1 to 2sqrt2} sqrt(1 + [4/9*y^(-2/3)] dy

    ___________________
    Because,

    g'(y)= 2/3*y^(-1/3)

    [g'(y)]^2 = 4/9*y^(-2/3)
    ___________________

    Now,
    For inside the square root

    1 + [g'(y)^2] = 1 + 4/9*y^(-2/3) = 1 + 4/9y^(2/3)

    = 9y^(2/3) + 4 / 9y^(2/3)

    ___________________

    Back to the Formula,

    L = int{1 to 2sqrt2) sqrt(9y^(2/3)+ 4 / 9y^(2/3))

    = int {1 to 2sqrt2} (1/3)(1/y^(1/3)) sqrt(9y^(2/3) + 4) dy
    ___________________

    Where and how do you get (1/3)(1/y^(1/3)) from the statement before?

    I know I have to use substitution but I'm confused on that step...
     
    Last edited: Feb 22, 2004
  2. jcsd
  3. Feb 22, 2004 #2
    There is no substitution.

    [tex]
    \int^1_{2\sqrt{2}}\sqrt{\frac{9y^{2/3}+4}{9y^{2/3}}}\,dy
    [/tex]
    [tex]
    \int^1_{2\sqrt{2}}\frac{\sqrt{9y^{2/3}+4}}{\sqrt{9y^{2/3}}}\,dy
    [/tex]
    [tex]
    \int^1_{2\sqrt{2}}\frac{\sqrt{9y^{2/3}+4}}{3y^{1/3}}}\,dy
    [/tex]
    [tex]
    \int^1_{2\sqrt{2}}\frac{1}{3}y^{-1/3}\sqrt{9y^{2/3}+4}\,dy
    [/tex]

    Let's hope I got all the LaTeX right...

    cookiemonster
     
  4. Feb 22, 2004 #3
    Oh okay... I see it now thanks! :)
     
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