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A particle of mass $ m $ and position $ \mathbf{r} = x \cdot \mathbf{i} + y \cdot \mathbf{j} + z \cdot \mathbf{k} $ is constrained to a smooth surface described by

$$ x^{2} + z^{2} = 1 + y^{2}, $$ where $ - \pi \leq y \leq \pi $.

Can we write $ \mathbf{r} $ in terms of trigonometric functions?

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maybe y=sin(x) or something and go from there? I am really stumped – JamesT Feb 8 '13 at 16:57
up vote 1 down vote accepted

$$x=\sec{\theta} \cos{\phi}$$ $$y=\tan{\theta} $$ $$z=\sec{\theta} \sin{\phi} $$

$\phi \in [0,2 \pi)$, $\theta \in [-\arctan{\pi},\arctan{\pi}]$.

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I'm not sure if im allowed to specify a range for theta? That looks good though! Thanks! – JamesT Feb 9 '13 at 13:10
@JamesT: you have no choice as you specified that problem. When $y=\tan{\theta}$ and you say that $-\pi \le y \le \pi$, then the bounds on $\theta$ must follow. – Ron Gordon Feb 9 '13 at 13:22
@JamesT: also, if you felt that this solution was helpful, please accept it or upvote it so that it will appear as answered. – Ron Gordon Feb 9 '13 at 13:28
is there anything stopping you from letting cos and sin be functions of theta also? – JamesT Feb 9 '13 at 13:34
Well, if there is some other constraint with respect to the coordinates, then that is possible. But the transformation I listed completely describes the relations between the coordinates as far as I can tell. – Ron Gordon Feb 9 '13 at 13:43

You can write $$ x= \rho \cos \theta\\ z= \rho \sin \theta\\ y= \pm\sqrt{\rho^2-1} $$

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would this satisfy −π≤y≤π.? – JamesT Feb 8 '13 at 17:31
no, it wouldn't – Emanuele Paolini Feb 9 '13 at 7:14
ahh right that is one of the requirements, thanks for the help anyway! :) – JamesT Feb 9 '13 at 13:08

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