Results 1 to 10 of 10

Math Help - length of the curve: where is my mistake?

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    71

    length of the curve: where is my mistake?

    Ok, so let me start with the information that is given. I need to find the length of the curve.
    x=t^3
    y=3t^2/2
    0≤ t ≥ square root (3)
    So that's the information that is given. Now I believe the correct equation to use is this Mathwords: Arc Length of a Curve (parametric form)
    (sorry for the link, but i have no idea how to type something like that)
    Ok. so first I take the derivative of each of the functions, which gives:
    y'=3t^2
    x'=3t
    then i square each of them, which gives:
    9t^4
    9t^2
    So putting that into the fuction, it gives
    (9t^4 +9t^2)^ 1/2

    then taking the integral of that, i get
    2(9t^4 +9t^2)^ 3/2/ 3(36t^3 +18t)

    and evalutating at square root 3, i get 3.42857....
    but the answer in the book says 7.
    can anyone help me out here and point out what i did wrong? i may be way off, as i'm just learning this, but i need to know how to do this.










    oh for the love of god none of my notation worked. i will edit this and make it better.
    ok, that should be be better
    Last edited by isuckatcalc; January 25th 2009 at 02:23 PM. Reason: confusing notation
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,110
    Thanks
    976
    your integral is incorrect.

    note ...

    \sqrt{9t^4 + 9t^2} = \sqrt{9t^2(t^2 + 1)} = 3t\sqrt{t^2+1}

    S = \frac{3}{2} \int_0^{\sqrt{3}} 2t\sqrt{t^2+1} \, dt

    using u = t^2 + 1 ...

    \frac{3}{2} \int_1^4 \sqrt{u} \, du

    now integrate.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2009
    Posts
    71
    Thanks for the help. Could you possibly explain this a bit though?
    So you factor out 3/2, and get 2t. i'm confused as to how you changed the integral to be from 1 to 4, and what happened to the 2t?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by isuckatcalc View Post
    Thanks for the help. Could you possibly explain this a bit though?
    So you factor out 3/2, and get 2t. i'm confused as to how you changed the integral to be from 1 to 4, and what happened to the 2t?
    skeeter has made the substitution u = t^2 + 1. Are you familiar with solving integrals by using the technique often refered to as a "u-substitution" ....?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2009
    Posts
    71
    Quote Originally Posted by mr fantastic View Post
    skeeter has made the substitution u = t^2 + 1. Are you familiar with solving integrals by using the technique often refered to as a "u-substitution" ....?
    well... somewhat. if you could explain it though, that would be great, i apparently don't know it well enough, haha.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by isuckatcalc View Post
    well... somewhat. if you could explain it though, that would be great, i apparently don't know it well enough, haha.
    Then perhaps you should go back and revise integration.

    u-substitution - Google Search
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jan 2009
    Posts
    71
    so i integrate and get 2/3U^3/2
    substitute t^2+1
    2/3 (t^2+1) 3/2

    but i'm still not getting 7 when solving for 4 and 1.
    Last edited by isuckatcalc; January 25th 2009 at 04:24 PM. Reason: .
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by isuckatcalc View Post
    so i integrate and get 2/3U^3/2
    substitute t^2+1
    2/3 (t^2+1) 3/2

    but i'm still not getting 7 when solving for 4 and 1.
    When you have a definite integral there is absolutely no need to go back to the old variable!

    Just solve \frac{3}{2} \int_1^4 \sqrt{u} \, du = \frac{3}{2} \cdot \frac{2}{3} [u^{3/2}]_1^4 = etc.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,110
    Thanks
    976
    4 and 1 are u-values ... you do not back substitute t^2+1 for u

    \frac{3}{2} \cdot \frac{2}{3}\left[u^{\frac{3}{2}}\right]_1^4 = 4^{\frac{3}{2}} - 1^{\frac{3}{2}} = 8 - 1 = 7
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Jan 2009
    Posts
    71
    nevermind, thanks guys
    Last edited by isuckatcalc; January 25th 2009 at 04:33 PM. Reason: .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Length of a curve
    Posted in the Calculus Forum
    Replies: 6
    Last Post: December 6th 2010, 12:24 PM
  2. Length of a Curve
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 9th 2009, 06:10 PM
  3. arc length and parameterized curve length
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 5th 2008, 03:33 AM
  4. length of the curve
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 25th 2007, 09:47 PM
  5. Length of curve
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 9th 2007, 10:36 PM

Search Tags


/mathhelpforum @mathhelpforum