Basic question pertaining to Polar Coordinates & how to employ them

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warhammer
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I have a question that might be considered vague or even downright idiotic but just wanted to know that once we find out the velocity & acceleration of a body in angular motion in plane polar coordinates, and are asked to integrate the expressions in order to find position at some specified time 't' , would we proceed to integrate the velocity expression (say) like we do usually with respect to time when the said expression involves r cap as well as theta cap; since the polar unit vectors are not "fixed" unlike the unit vectors of i,j,k in Cartesian system?
 
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The position vector is [itex]\mathbf{r} = r\mathbf{e}_r(\theta)[/itex]. You need both [itex]r(t)[/itex] and [itex]\theta(t)[/itex] to fully specify the position. The velocity vector is [tex]\dot{\mathbf{r}} = \dot r \mathbf{e}_r + r\dot\theta \frac{d}{d\theta}\mathbf{e}_r = \dot r \mathbf{e}_r + r\dot \theta \mathbf{e}_\theta[/tex] so if [itex]\dot{\mathbf{r}}= v_r(t)\mathbf{e}_r + v_\theta(t) \mathbf{e}_\theta[/itex] then [tex] \begin{align*}<br /> r &= \int v_r\,dt \\<br /> \theta &= \int \frac{v_\theta}{r}\,dt<br /> \end{align*}[/tex]

(You can always resolve components because [itex]\mathbf{e}_r \cdot \mathbf{e}_r = \mathbf{e}_\theta \cdot \mathbf{e}_\theta = 1[/itex] and [itex]\mathbf{e}_r \cdot \mathbf{e}_\theta = 0[/itex] are constant.)
 
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pasmith said:
The position vector is [itex]\mathbf{r} = r\mathbf{e}_r(\theta)[/itex]. You need both [itex]r(t)[/itex] and [itex]\theta(t)[/itex] to fully specify the position. The velocity vector is [tex]\dot{\mathbf{r}} = \dot r \mathbf{e}_r + r\dot\theta \frac{d}{d\theta}\mathbf{e}_r = \dot r \mathbf{e}_r + r\dot \theta \mathbf{e}_\theta[/tex] so if [itex]\dot{\mathbf{r}}= v_r(t)\mathbf{e}_r + v_\theta(t) \mathbf{e}_\theta[/itex] then [tex] \begin{align*}<br /> r &= \int v_r\,dt \\<br /> \theta &= \int \frac{v_\theta}{r}\,dt<br /> \end{align*}[/tex]

(You can always resolve components because [itex]\mathbf{e}_r \cdot \mathbf{e}_r = \mathbf{e}_\theta \cdot \mathbf{e}_\theta = 1[/itex] and [itex]\mathbf{e}_r \cdot \mathbf{e}_\theta = 0[/itex] are constant.)
Thank you so much for your help sir
 
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