Hello, I'm Sundayclose. Your recent edit(s) to the page Attention deficit hyperactivity disorder appear to have added incorrect information, so they have been reverted for now. If you believe the information was correct, please cite a reliable source or discuss your change on the article's talk page. If you would like to experiment, please use your sandbox. If you think I made a mistake, or if you have any questions, you can leave me a message on my talk page. Thank you. Sundayclose ( talk) 19:43, 9 November 2021 (UTC) reply

Hello Sundayclose, Thank you for your attention. It is a matter of spelling convention though, for acronyms < 's > and < s > are both used, and my spellchecker didn't recognize the spelling without the apostrophe, so I over-eagerly changed them. Yodo9000 ( talk) 18:00, 25 April 2022 (UTC) reply

I noticed an error regarding a position of a general point in spherical polars. Checking the history I noticed you made the 'correction' which was false. I hope to convince you that the correction you made was wrong by giving you the exercise of differentiating both position vectors with respect to time and noticing the discrepancy with the given velocity vector. In fact the unit vector in the increasing r direction actually contains all the information of the two polar angles. I hope this clears things up and avoids any back and fourth corrections. — Preceding unsigned comment added by 163.1.150.134 ( talk) 09:14, 23 April 2022 (UTC) reply

Dear anonymous, ${\displaystyle v\neq {\dot {r}}}$ in general, as ${\displaystyle v}$ is nonzero for a particle orbiting the origin at a constant distance ${\displaystyle r}$ (such that ${\displaystyle {\dot {r}}=0}$).

${\displaystyle \mathbf {r} =r\mathbf {\hat {r}} }$ implies that ${\displaystyle \varphi ,\theta }$ are zero, or undefined, and differentiating this equation will not give you ${\displaystyle \mathbf {v} ={\dot {r}}\mathbf {\hat {r}} +r\,{\dot {\theta }}\,{\hat {\boldsymbol {\theta }}}+r\,{\dot {\varphi }}\sin \theta \,\mathbf {\hat {\boldsymbol {\varphi }}} }$, where ${\displaystyle {\dot {\varphi }},{\dot {\theta }}}$ would appear out of nowhere, not being defined.

The point in spherical coordinates is written as ${\displaystyle (r,\theta ,\phi )}$, writing this as a vector gives ${\displaystyle r\mathbf {\hat {r}} +\theta {\hat {\boldsymbol {\theta }}}+\varphi \mathbf {\hat {\boldsymbol {\varphi }}} }$.

I will revert your change, and think it is best if this is continued at Talk:Spherical coordinate system if necessary. You can also leave a short message here in that case, I might miss it otherwise (or not). Yodo9000 ( talk) 17:54, 25 April 2022 (UTC) reply

As the anonymous person who reverted you're change I'm glad you finally see the error you made. Explicitly, the unit vector r-hat is a function of theta and phi and the tuple you mention is not equivalent to your idea of an arbitrary position vector. If this is still not clear I recommend you read the entire article but specifically section 5 where the unit vectors are defined in terms of the usual cartesian unit vectors. I think you may also have introduced more confusion when adding the bracketed comment about the position being better stated as a triple. In kinematics we prefer to do computations with unit vectors, so that we can differentiate easily for example. However I will not change it and leave it to you to discuss on the talk page. Artanhx ( talk) 18:29, 26 July 2022 (UTC) reply

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