Matrix Differentiation Problem

[leonmate]

Simple question really,

I'm not sure why the constant pulled out of the derivative becomes negative (-w²). I've tried looking for answers by googling but can't come up with anything.
I feel like its because the first term (1,1) is negative but I want to be sure.

Thanks

 

[gleem]

The minus sign is pulled out so that there are fewer minus signs inside the matrix is my guess.

 

[HallsofIvy]

You have
$$R(\omega t)= \begin{pmatrix}cos(\omega t) & -sin(\omega t) & 0 \\ sin(\omega t) & cos(\omega t) & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
(Rotation about the z-axis with constant angular velocity $\omega$)

I presume you know that the derivative of $sin(\omega t)$ is $\omega cos(\omega t)$ and the derivative of $cos(\omega t)$ is $-\omega sin(\omega t)$ and, further, that you differentiate a matrix by differentiating its components separately.

So you should see, easily, that
$$\frac{dR}{dt}= \begin{pmatrix} -\omega sin(\omega t) & -\omega cos(\omega t) & 0 \\ \omega cos(\omega t) & -\omega sin(\omega t) & 0 \\ 0 & 0 & 0 \end{pmatrix}$$
and, of course, you can factor "$\omega$" out of the matrix.

To find the second derivative, differentiate again:
$$\frac{d^2R}{dt^2}= \begin{pmatrix} -\omega^2 cos(\omega t) & \omega^2 sin(\omega t) & 0 \\ -\omega^2 sin(\omega t) & -\omega^2 cos(\omega t) & 0 \\ 0 & 0 & 0 \end{pmatrix}$$
and factoring $-\omega^2$ out of that gives
$$-\omega^2 \begin{pmatrix} cos(\omega t) & - sin(\omega t) & 0 \\ sin(\omega t) & cos(\omega t) & 0 \\ 0 & 0 & 0 \end{pmatrix}= -\omega^2 R$$

 

[Fredrik]

If $f(t)=\cos(\omega t)$ and $g(t)=\sin(\omega t)$, what are $f''(t)$ and $g''(t)$?