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# Parametric Shapes, and Tangent Space?

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Dear All, I remember reading somewhere that for a parametric shape, to generate tangent space, all you had to do, was take the derivate of the parametric shape. This would make sense, as the derivate=tangent, but I just wanted to make sure, as I can''t seem to find where I read this... Jesse

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Well, been a while since I used any calculus in anger, but I remember that differentiating a curve gave you the gradient, so differentiating a function that represents a surface should give you the gradient at any point on the surface, e.g. the tangent.

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The gradiant really applies to a single function of multiple variables, i.e. z=f(x,y). A parametric surface is really three differant functions, i.e. x=f(s,t), y=g(s,t) and z=h(s,t). The gradiant of f in this case would give you the direction in terms of s and t for which x increases most rapidly. If you hold one variable constant and vary the other you get a space curve. The derivative with respect to that variable gives you a tangent vector. So if you do that with both variables you get two tangents. Their cross product is the normal to the surface and thus the tangent plane.

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