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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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    the derivative $(x'(t),y'(t))$2017-01-14

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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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    just a small note, in the off-chance the length of the vector is $0$, then one might need to use other methods or just accept there is no direction as the parameterization stopped at that point and hovered there.2017-01-14
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    I agree. If possible one should re-parameterize the curve in this case.2017-01-15