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Suppose we are given the matrix $$\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}\cos(\omega t)& -\sin(\omega t)\\\sin(\omega t)& \cos(\omega t)\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$

In other words the new coordinate system is a rotating coordinate system.

Is there a matrix-based way of finding $\begin{pmatrix}\dot{x}\\\dot{y}\end{pmatrix}$ and $\dot{x}^2+\dot{y}^2$ in terms of the new coordinates?

I can, by inspection, see that the latter should be $(\dot{x'}-\omega y')^2+(\dot{y'}+\omega x')^2$ but I would appreciate a way using matrices. Also if such a matrix to "change coordinates" exist, I would appreciate an intuitive explanation of what it is doing.

Many thanks!

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$\dot x$ is the derivative wrt to $t$ and $x,y$ are supposed to depend on $t$? It's the product rule. –  user20266 Jun 9 '12 at 19:12
    
@QiaochuYuan: Sorry about the ambiguity in notation! $x'$ is a new coordinate, whereas $\dot{x}$ is the time derivative of $x$ –  George Jun 9 '12 at 19:12

2 Answers 2

up vote 3 down vote accepted

If we have $x' = A x$, then $\dot{x'} = \dot{A}x + A \dot{x} = \begin{pmatrix} \dot{A} & A\end{pmatrix}\begin{pmatrix} x \\ \dot{x}\end{pmatrix}$ and thereby $$\sum_k \left(\dot{x_k'}\right)^2 = \left(\dot{x'} \right)^T \left(\dot{x'} \right) = \begin{pmatrix} x \\ \dot{x}\end{pmatrix}^T \begin{pmatrix} \dot{A} & A\end{pmatrix}^T\begin{pmatrix} \dot{A} & A\end{pmatrix} \begin{pmatrix} x \\ \dot{x}\end{pmatrix}$$ $$\sum_k \left(\dot{x_k'}\right)^2 = \begin{pmatrix} x & \dot{x}\end{pmatrix} \begin{pmatrix} \dot{A}^T \dot{A} & \dot{A}^T A \\ A^T \dot{A} & A^TA \end{pmatrix}\begin{pmatrix} x \\ \dot{x}\end{pmatrix}$$ For $k=2$, you get the following. \begin{align} \dot{\begin{pmatrix}x'\\y'\end{pmatrix}} & =\dot{\begin{pmatrix}\cos(\omega t)& -\sin(\omega t)\\\sin(\omega t)& \cos(\omega t)\end{pmatrix}}\begin{pmatrix}x\\y\end{pmatrix} + \begin{pmatrix}\cos(\omega t)& -\sin(\omega t)\\\sin(\omega t)& \cos(\omega t)\end{pmatrix} \dot{\begin{pmatrix}x\\y\end{pmatrix}}\\ \begin{pmatrix}\dot{x'}\\\dot{y'}\end{pmatrix} & =\begin{pmatrix}-\sin(\omega t)& -\cos(\omega t)\\\cos(\omega t)& -\sin(\omega t)\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} + \begin{pmatrix}\cos(\omega t)& -\sin(\omega t)\\\sin(\omega t)& \cos(\omega t)\end{pmatrix} \begin{pmatrix}\dot{x}\\\dot{y}\end{pmatrix}\\ \begin{pmatrix}\dot{x'}\\\dot{y'}\end{pmatrix} & =\begin{pmatrix}-\sin(\omega t)& -\cos(\omega t) & \cos(\omega t)& -\sin(\omega t)\\\cos(\omega t)& -\sin(\omega t)&\sin(\omega t)& \cos(\omega t)\end{pmatrix}\begin{pmatrix}x\\y\\\dot{x}\\\dot{y}\end{pmatrix}\\ \left(\dot{x'} \right)^2 + \left(\dot{y'} \right)^2 &= \begin{pmatrix}\dot{x'}&\dot{y'} \end{pmatrix}\begin{pmatrix}\dot{x'}\\\dot{y'} \end{pmatrix}= \begin{pmatrix}\dot{x'}\\\dot{y'}\end{pmatrix}^T\begin{pmatrix}\dot{x'}\\\dot{y'}\end{pmatrix} \end{align} Now plugin the previous expression to get the expression in matrix form.

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Thanks, Marvis!! –  George Jun 9 '12 at 19:28

Isn't it better to write this as: $$z' = e^{i \omega t}z$$ where $z = x + i y$? Than, weather you mean rotating coordinates or system of differential equations, the complex form is easier to handle.

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This approach does not extend to higher (real) dimensions, though. –  user20266 Jun 9 '12 at 19:16
    
@Thomas yes, but is such generalization needed? –  qoqosz Jun 9 '12 at 19:18

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