Results 1 to 3 of 3

Math Help - Find the length of the curve ⃗r(t) = ⟨ 1/3*t^3, t^2, 1⟩, 0 ≤ t ≤ 2.

  1. #1
    Member
    Joined
    Sep 2010
    Posts
    151
    Thanks
    3

    Find the length of the curve ⃗r(t) = ⟨ 1/3*t^3, t^2, 1⟩, 0 ≤ t ≤ 2.

    r'(t) = <t^2, 2t,0>

    |r'(t)| = sqrt( t^4 + 4t^2)

    Arc length is integral of |r'(t)|, but how can i simplify it so i can integrate it?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2011
    From
    Crna Gora
    Posts
    420
    Thanks
    64

    Re: Find the length of the curve ⃗r(t) = ⟨ 1/3*t^3, t^2, 1⟩, 0 ≤ t ≤ 2.

    Quote Originally Posted by linalg123 View Post
    r'(t) = <t^2, 2t,0>

    |r'(t)| = sqrt( t^4 + 4t^2)

    Arc length is integral of |r'(t)|, but how can i simplify it so i can integrate it?
    Integral , click "show steps"
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2008
    Posts
    106

    Re: Find the length of the curve ⃗r(t) = ⟨ 1/3*t^3, t^2, 1⟩, 0 ≤ t ≤ 2.

    \int_0^2 \sqrt{t^4+4t^2}dt  = \int_0^2 t \sqrt{t^2+4} dt

    make the substitution  u = t^2 + 4
    and the integral should be easy.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: March 18th 2012, 02:11 PM
  2. [SOLVED] Given that -5≤x≤5 and -10≤y≤-2, find
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 2nd 2011, 06:12 AM
  3. Replies: 2
    Last Post: March 19th 2011, 02:38 AM
  4. Replies: 10
    Last Post: November 15th 2009, 07:04 PM
  5. Replies: 1
    Last Post: May 27th 2008, 08:56 AM

Search Tags


/mathhelpforum @mathhelpforum