Results 1 to 9 of 9

Math Help - arc length - stuck with intergration

  1. #1
    Super Member
    Joined
    Aug 2009
    Posts
    639

    arc length - stuck with intergration

    how do you solve these questions:

    1. find the arc length of Y^3= 8X^2 from x=1 and x=8.

    my working: ( i know how to form the equation for intergration but i do not know how to carry out the intergration)

    differenciating the qns, i got 3y^2 (dy/dx) = 3/4 X ^(1/3)


    then i would like to intergrate ( 1 + (9/16 X ^(2/3)))^(1/2) dx from x=8 to x=1.

    but im stuck now..how do i perform intergration on this equation?

    2.find the arc length of 6xy = x^4 + 3

    by implicit differentiation, i got
    (dy/dx)^2 = ( 5/6x^2 + 1/ (2x^2))^2

    and then i would have to perform intergration on

    (1+ (dy/dx)^2 ) ^(1/2) from x= 2 to x= 1...

    but i am stuck as to how i should carry out the intergration.

    thank you!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    ok


    \int_{8}^{1} \left( 1+ \frac{9}{16} x^{\frac{2}{3}} \right)^{\frac{1}{2}} \cdot dx

    let y = x^{\frac{1}{3}}

    y^3 = x \Rightarrow 3y^2 dy = dx

    \int \left(1+ \frac{9}{16} y^2 \right) 3y^2 dy

    let \tan u = \frac{3y}{4}

    \tan ^2u = \frac{9y^2}{16} \Rightarrow y^2 = \frac{16\tan ^2u}{9}

    and \sec u = \sqrt{1 + \frac{9 y^2 }{16}}

    \sec^2 u\cdot du = \frac{3}{4} dy

    \int \sec u \cdot 3\left(\frac{16\tan ^2u}{9}\right) \left(\frac{4\sec ^2u}{3} \right) du

    factorize the constant

    \int \sec ^3u \tan ^2 u \cdot du

    \int sec^4 u \cdot du  - \int sec u\cdot du

    can you handle this
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,366
    Thanks
    41
    Quote Originally Posted by alexandrabel90 View Post
    how do you solve these questions:

    1. find the arc length of Y^3= 8X^2 from x=1 and x=8.

    my working: ( i know how to form the equation for intergration but i do not know how to carry out the intergration)

    differenciating the qns, i got 3y^2 (dy/dx) = 3/4 X ^(1/3)


    then i would like to intergrate ( 1 + (9/16 X ^(2/3)))^(1/2) dx from x=8 to x=1.

    but im stuck now..how do i perform intergration on this equation?

    2.find the arc length of 6xy = x^4 + 3

    by implicit differentiation, i got
    (dy/dx)^2 = ( 5/6x^2 + 1/ (2x^2))^2

    and then i would have to perform intergration on

    (1+ (dy/dx)^2 ) ^(1/2) from x= 2 to x= 1...

    but i am stuck as to how i should carry out the intergration.

    thank you!
    For the first one, it might be easier if you introduce parametric equation for your curve

    x = t^3, y = 2 t^2; t = 1 \to 2

    and use \int \limits_c \sqrt{\dot{x}^2 + \dot{y}^2}\,dt.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Aug 2009
    Posts
    639
    how did you think of forming this parametric equation?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Aug 2009
    Posts
    639
    may i also ask how to solve the second question?
    thank you!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,736
    Thanks
    642
    Hello, alexandrabel90!

    Here's my approach to the first one . . .


    1. Find the arc length of y^3\:=\: 8x^2 . from x=1 to x=8.

    We have: ./ y \:=\:2x^{\frac{2}{3}}

    Then: . \frac{dy}{dx} \:=\:\tfrac{4}{3}x^{-\frac{1}{3}}\quad\Rightarrow\quad \left(\frac{dy}{dx}\right)^2 \:=\:\frac{16}{9}x^{-\frac{2}{3}} \quad\Rightarrow\quad 1 + \left(\frac{dy}{dx}\right)^2 \:=\:1 + \frac{16}{9x^{\frac{2}{3}}}

    . . 1 + \left(\frac{dy}{dx}\right)^2 \:=\:\frac{9x^{\frac{2}{3}} + 16}{9x^{\frac{2}{3}}}  \quad\Rightarrow\quad  \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \:=\:\frac{\sqrt{9x^{\frac{2}{3}} + 16}}{3x^{\frac{1}{3}}}


    We have: . L \;=\;\int^8_0\frac{\sqrt{9x^{\frac{2}{3}} + 16}}{3x^{\frac{1}{3}}}\,dx

    Let: . 3x^{\frac{1}{3}} \:=\:4\tan\theta \quad\Rightarrow\quad x^{\frac{1}{3}} \:=\:\tfrac{4}{3}\tan\theta \quad\Rightarrow\quad x \:=\:\tfrac{64}{27}\tan^3\!\theta
    . . . dx \:=\:\tfrac{64}{9}\tan^2\!\theta\sec^2\!\theta\,d\  theta

    Substitute: . L \;=\;\int\frac{4\sec\theta}{3\cdot\frac{4}{3}\tan\  theta}\left(\tfrac{64}{9}\tan^2\!\theta\sec^2\!\th  eta\,d\theta\right) \;=\;\tfrac{64}{9}\int \sec^3\!\theta\tan\theta\,d\theta

    \text{And we have: }\;L \;=\;\tfrac{64}{9}\int \underbrace{\sec^2\!\theta}_{u^2}\underbrace{(\sec  \theta\tan\theta\,d\theta)}_{du}

    Got it?

    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,736
    Thanks
    642
    Hello, alexandrabel90!

    Okay, here's the second one . . .


    2.Find the arc length of: . 6xy \:=\: x^4 + 3 . . What limits?

    We have: . y \:=\:\tfrac{1}{6}x^3 + \tfrac{1}{2}x^{-1}

    Then: . \frac{dy}{dx} \:=\:\tfrac{1}{2}x^2 - \tfrac{1}{2}x^{-2} \:=\:\frac{x^2-x^{-2}}{2}

    . . \left(\frac{dy}{dx}\right)^2 \;=\;\frac{(x^2 - x^{-2})^2}{4} \;=\;\frac{x^4 - 2 + x^{-4}}{4}

    . . 1 + \left(\frac{dy}{dx}\right)^2 \;=\;1 + \frac{x^4 - 2 + x^{-4}}{4} \;=\; \frac{x^4 + 2 + x^{-4}}{4} \;=\;\frac{(x^2 + x^{-2})^2}{4}

    . . \sqrt{1+\left(\frac{dy}{dx}\right)^2} \;=\;\frac{x^2 + x^{-2}}{2}


    And we have: . L \;=\;\tfrac{1}{2}\int^b_a\left(x^2 + x^{-2}\right)\,dx . . . etc.

    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,366
    Thanks
    41
    Quote Originally Posted by alexandrabel90 View Post
    how did you think of forming this parametric equation?
    When learning about arc length we learn about it when we have parametic equation. When I saw y^3 = 8 x^2 I saw

    \left(\frac{y}{2}\right)^3 = x^2 and that's when I thought of it.

    If you let \frac{y}{2} = t^p, x = t^q and pick p and q to match the powers. Here p = 2 and q = 3 work! Also note that when x = 0 then t = 1 and x = 8 then t = 2. Then it's a matter of assembling the arc length formula

    \dot{x} = 3x^2, \dot{y} = 4x so

    \int_0^1 \sqrt{9x^4 + 16x^2}\,dx = 3\int_0^1 x\sqrt{x^2 + \frac{16}{9}}\,dx

    then we can use the sub u = x^2 + \frac{16}{9}.

    Using the original formula in terms of x works (as Soroban shows) but I think is a little more difficult.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Aug 2009
    Posts
    639
    thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Stuck on Parametric Arc Length
    Posted in the Calculus Forum
    Replies: 6
    Last Post: December 10th 2009, 07:15 PM
  2. Stuck: Arc Length
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 23rd 2009, 03:43 AM
  3. Intergration problem im stuck on
    Posted in the Calculus Forum
    Replies: 9
    Last Post: October 25th 2008, 05:59 AM
  4. stuck on this intergration
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 9th 2008, 05:16 PM
  5. intergration, i m stuck.
    Posted in the Calculus Forum
    Replies: 8
    Last Post: September 9th 2008, 09:00 PM

Search Tags


/mathhelpforum @mathhelpforum