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Curvilinear coordinate system: Determine the standardized base vectors
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[QUOTE="pasmith, post: 6387878, member: 415692"] The position vector is always [itex]\mathbf{r}(u,v,w) = (x(u,v,w),y(u,v,w),z(u,v,w))[/itex]. Here you are given [itex]u(x,y,z)[/itex] etc., and you would have to invert this to get [itex](x,y,z)[/itex] in terms of [itex](u,v,w)[/itex], but you don't have to. The basis vector [itex]\mathbf{e}_{u_i}[/itex] points in the direction of increasing [itex]u_i[/itex] and is normal to surfaces of constant [itex]u_i[/itex]. The vector [itex]\nabla u_i[/itex] does exactly this. Now it turns out that if the [itex]\nabla u_i[/itex] are orthogonal then [itex]\nabla u_i[/itex] and [itex]\frac{\partial \mathbf{r}}{\partial u_i}[/itex] are parallel. [/QUOTE]
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Curvilinear coordinate system: Determine the standardized base vectors
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