If you have a curve in parametric form, and you have a point on that curve, how would you calculate the directional vector that has its origin in that point and its direction in the direction of the curve?
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$\begingroup$If $\overline{x}(t) = [x(t), y(t), z(t)]$ is the curve and the point is at $t_0$, then the direction vector is $[x'(t_0); y'(t_0); z'(t_0)]/\|\overline{x}(t_0)\|$, where $ \|\overline{x}(t_0)\| = \sqrt{x'(t_0)^2 + y'(t_0)^2 + z'(t_0)^2}$.
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