# differentiate sec^2 psi

• Nov 23rd 2011, 06:11 PM
ain
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.
• Nov 23rd 2011, 08:30 PM
pickslides
Re: differentiate sec^2 psi
I'm really struggling with this, please be more concise.
• Nov 24th 2011, 09:09 AM
Ackbeet
Re: differentiate sec^2 psi
Quote:

Originally Posted by ain
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

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