Results 1 to 5 of 5

Math Help - Length of a curve

  1. #1
    Newbie
    Joined
    Dec 2010
    Posts
    8

    Length of a curve

    I calculate the length of a curve. Task but I can not modify (derivate, cube) that I could substitute in the formula

    Could anyone help me please?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Ahplym View Post
    I calculate the length of a curve. Task but I can not modify (derivate, cube) that I could substitute in the formula

    Could anyone help me please?

    \displaystyle{y^{2/3}+x^{2/3}=a^{2/3}\Longrightarrow y=\sqrt{\left(a^{2/3}-x^{2/3}\right)^3}} , and now you can calculate y'.

    Note: Don't forget to take into account the domain of definition of y !

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2010
    Posts
    8
    Thank you. I got derivate but I can not potentiate.

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,909
    Thanks
    771
    Hello, Ahplym!

    tonio told you what has to be done.
    But it takes some Olympic-level gymnastics.


    \text{Calculate the length of the curve: }x^{\frac{2}{3}} + y^{\frac{2}{3}} \:=\:a^{\frac{2}{3}}

    \displaystyle \text{But I cannot modify it to substitute in the formula: }\: \int^b_a\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx

    We have: . y^{\frac{2}{3}} \;=\;\ a^{\frac{2}{3}} - x^{\frac{2}{3}} \quad\Rightarrow\quad y \;=\;\left(a^{\frac{2}{3}} - x^{\frac{2}{3}}\right)^{\frac{3}{2}}
    Differentiate: . \dfrac{dy}{dx} \;=\;\frac{3}{2}\left(a^{\frac{2}{3}}-x^{\frac{2}{3}}\right)^{\frac{1}{2}}\left(\text{-}\frac{2}{3}x^{-\frac{1}{3}}\right) \;=\;\dfrac{-\left(a^{\frac{2}{3}} - x^{\frac{2}{3}}\right)^{\frac{1}{2}}} {x^{\frac{1}{3}}}

    Square: . \left(\dfrac{dy}{dx}\right)^2 \;=\;\dfrac{a^{\frac{2}{3}} - x^{\frac{2}{3}}}{x^{\frac{2}{3}}}

    \displaystyle \text{Add 1: }\;1 + \left(\dfrac{dy}{dx}\right)^2 \;=\;1 + \dfrac{a^{\frac{2}{3}} - x^{\frac{2}{3}}} {x^{\frac{2}{3}}} \;=\; \frac{a^{\frac{2}{3}}}{x^{\frac{2}{3}}}

    Take the square root: . \displaystyle \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;=\;\sqrt{\frac{a^{\frac{2}{3}}}{x^{\frac{2}{3}}}  } \;=\;\frac{a^{\frac{1}{3}}}{x^{\frac{1}{3}}}


    We have: . \displaystyle{L \;=\;\int^b_a \frac{a^{\frac{1}{3}}}{x^{\frac{1}{3}}}\,dx \;=\;a^{\frac{1}{3}}\!\!\int^b_a \!\!x^{\frac{1}{3}}\,dx



    The graph is an astroid (a hypocycloid of four cusps).
    It is symmetric to both axes and the Origin.

    We can integrate from 0 to \,a and multiply by 4.


    . . \displaystyle L \;=\;4a^{\frac{1}{3}}\!\!\int^a_0\!\! x^{-\frac{1}{3}}\,dx \;=\;4a^{\frac{1}{3}}\cdot\tfrac{3}{2}x^{\frac{2}{  3}}\,\bigg]^a_0

    . . . . =\;6a^{\frac{1}{3}}\bigg[a^{\frac{2}{3}} - 0^{\frac{2}{3}}\bigg] \;=\;6a^{\frac{1}{3}}\cdot a^{\frac{2}{3}} \;=\;6a

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2010
    Posts
    8
    Thank you so much.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Length of a curve
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 21st 2010, 07:03 PM
  2. arc length and parameterized curve length
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 5th 2008, 03:33 AM
  3. length of curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 3rd 2008, 12:58 PM
  4. Length Of the curve
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 29th 2007, 08:44 AM
  5. length of curve
    Posted in the Calculus Forum
    Replies: 8
    Last Post: May 23rd 2007, 07:32 PM

Search Tags


/mathhelpforum @mathhelpforum