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I'm looking at some notes in CAGD and there is written that for the curves:

$f(u) = (-1 + u^2, 2u - u^2, 0)$

$g(v) = (2v - v^2, 1 - v^2, 0)$ $u, v$ are in the interval $[0,1]$

the tangent vector is $(2, 0, 0$).

The problem is that I have no idea how it is found. Can you please tell me how to find it, or point me some kind of resource where I can read the algorithm?

Thank you very much in advance! :)

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    $\begingroup$ A tangent vector is typically found at a specified point (or a specified value of $u$ or $v$). Did they not give some kind of information like this? $\endgroup$
    – Muphrid
    Commented Jan 21, 2013 at 21:29

1 Answer 1

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The tangent vector of a curve is just its derivative in the given point. So, now it might be $f'(u)$ for some $u$ or $g'(v)$ for some $v$. Differentiation is done coordinatewise.

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