Results 1 to 6 of 6

Math Help - Quick Problem

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    3

    Quick Problem

    I'm not entirely sure if this belongs in this forum, but it seemed too difficult for the traditional Calculus forums. Is there any way to integrate cos(x)=c*d2x/dt2, where x is a function of time and c is some arbitrary constant. Please and thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Nov 2008
    Posts
    36
    >>cos(x)=c*d2x/dt2

    I assume its more like :

    cos(x)=c*d^2(x)/dt^2

    => 1/c * cos(x) = d^2(x)/dt^2

    why not just integrate twice?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Matter_Math View Post
    >>cos(x)=c*d2x/dt2

    I assume its more like :

    cos(x)=c*d^2(x)/dt^2

    => 1/c * cos(x) = d^2(x)/dt^2

    why not just integrate twice?
    If the question was \frac{d^2 t}{dx^2} = \frac{1}{c} \cos (x) then yes, you could ntegrate twice.

    But that's not the question. The question is \frac{d^2 x}{dt^2} = \frac{1}{c} \cos (x), which is very different.

    The usual method is to substitute \frac{d^2 x}{dt^2} = \frac{d}{dx} \left(\frac{v^2}{2}\right) where v = \frac{dx}{dt}. Now integrate with respect to x to get v and then use v to get x.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2010
    Posts
    3
    The problem with that is that puts the cos(x) with the dt^2 for integration. It would be more like 1/cos(x)*d^2x=1/c*dt^2 which I can easily integrate once, but it becomes quite problematic the second time.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by FluidOfJustice View Post
    The problem with that is that puts the cos(x) with the dt^2 for integration. It would be more like 1/cos(x)*d^2x=1/c*dt^2 which I can easily integrate once, but it becomes quite problematic the second time.
    Sorry but what you've posted makes absolutely no sense. I have told you how to solve the DE. Follow my advice, show your working and say where you're stuck if you still need help.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Feb 2010
    Posts
    3
    I was referring to the poster above you. Your solution helped enormously and I was able to solve the problem.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: April 13th 2009, 05:42 PM
  2. One Quick Mean Value Problem!!!
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 30th 2009, 05:53 PM
  3. Quick Derivative problem
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 4th 2009, 05:08 PM
  4. Quick problem
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: February 6th 2008, 01:12 PM
  5. Quick Problem.
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: October 22nd 2007, 01:16 PM

Search Tags


/mathhelpforum @mathhelpforum