The OP already stated that these quantities are positive so the point is moot.

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- Jul 23rd 2010, 08:23 AMlvleph
The OP already stated that these quantities are positive so the point is moot.

- Jul 23rd 2010, 08:25 AMAckbeet
Reply to lvleph at post # 16:

Take a closer look at post #3. Stability, in this context, is dealing with the boundedness of y. If y is bounded, you've got stability. Otherwise, you don't. - Jul 23rd 2010, 08:26 AMmathfilip
While we already are on the subject of physics. I could need some help understanding the derivatives of polar coordinates:

If I take the derivative with respect to time for theta(hat) - dont know how to type with math symbols

What will I get?

I know that if i do it for the radial unit vector i get - r(hat)' = theta'*theta(hat)

I can see it for the radial unit vector but not for the angular one. Any one dare to give an explanation? - Jul 23rd 2010, 08:29 AMlvleph
for the math symbols

\hat{\theta} = $\displaystyle \hat{\theta}$

\hat{r}' = $\displaystyle \hat{r}'$

As far as the polar coordinate question, I am actually not sure what you are asking. - Jul 23rd 2010, 08:33 AMmathfilip
Ok now I really understand what you meant Ackbeet with the exponentials! Thanks once again.

- Jul 23rd 2010, 08:34 AMAckbeet
To take the derivatives of $\displaystyle \hat{r}$ and $\displaystyle \hat{\theta},$ I would simply write them as functions of $\displaystyle \hat{i}$ and $\displaystyle \hat{j}$ and go from there.