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Math Help - differentiate sec^2 psi

  1. #1
    ain
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    differentiate sec^2 psi

    Dear all,

    I already calculate the differential below.. and i want to check it whether the answer that I got is correct or not.

    X= [ x xdot psi psidot y ydot z zdot]

    solve d/dX (m1/m2*psidot/cos^2 psi )

    = [0 0 (2*m1/m2*sec^2 psi *tan psi* psidot) (m1/m2*sec^2 psi) 0 0 0 0]
    = (m1/m2)*sec^2 psi [ 0 0 (2 *tan psi* psidot) 1 0 0 0 0]


    Thank you very much.
    Last edited by ain; November 23rd 2011 at 08:15 PM.
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  2. #2
    Master Of Puppets
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    Re: differentiate sec^2 psi

    I'm really struggling with this, please be more concise.
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  3. #3
    A Plied Mathematician
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    Re: differentiate sec^2 psi

    Quote Originally Posted by ain View Post
    Dear all,

    I already calculate the differential below.. and i want to check it whether the answer that I got is correct or not.

    X= [ x xdot psi psidot y ydot z zdot]

    solve d/dX (m1/m2*psidot/cos^2 psi )

    = [0 0 (2*m1/m2*sec^2 psi *tan psi* psidot) (m1/m2*sec^2 psi) 0 0 0 0]
    = (m1/m2)*sec^2 psi [ 0 0 (2 *tan psi* psidot) 1 0 0 0 0]


    Thank you very much.
    You appear to be attempting to solve a vector differential equation. However, the ambiguities in your notation are too great to decipher. I would recommend using LaTeX to write your DE. For example, the code

    [TEX]\frac{d}{dt}\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}\dot{x}\\ \dot{y}\\ \dot{z}\end{array}\right].[/TEX]

    produces the output

    \frac{d}{dt}\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}\dot{x}\\ \dot{y}\\ \dot{z}\end{array}\right].
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