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Hi I have a small doubt on Discrete Geometry, more specifically the derivation of Jacobian.

Say we have a function $x:\mathbb{R}^2\supseteq U \rightarrow \mathbb{R}^3 : (u,v) \mapsto x(u,v)$

Also now we have a parametric curve defined as $ \gamma(t) = x(u_0 + tw_1,v_0+tw_2)$

Then in the text I am referring to, it is written that $\gamma'(0)=w_1x_u + w_2x_v$
where $ x_u = {\delta x \over \delta u}$ and $x_v = {\delta x \over \delta v} $

My question is how is this formula for $ \gamma'(0) $ arrived at? I am sure its some elementary maths which I am missing out. Also if you need more information please ask for the same.

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It is the chain rule. If $h(x) = f(g(x))$, and $f,g$ are suitably differentiable, then $Dh(x) = Df (g(x)) Dg(x)$.

In your case, let $g(t) = (u_0+t w_1, v_0+t w_2)$, then $\gamma(t) = x(g(t))$, and $D \gamma(t) = Dx(g(t)) Dg(t)$.

We have $Dg(t) = \binom{w_1}{w_2}$, and $Dx(p) = (\frac{\partial x(p)}{\partial u}, \frac{\partial x(p)}{\partial v})$, hence $\gamma'(t) = D \gamma(t) = \frac{\partial x(u_0+t w_1, v_0+t w_2)}{\partial u} w_1 + \frac{\partial x(u_0+t w_1, v_0+t w_2p)}{\partial v} w_2$. Setting $t=0$ gives the desired answer.

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Thank you very much.. Sometimes the elementary concepts gets muddled up. – rajaditya_m Jan 30 '13 at 5:39
You are very welcome. – copper.hat Jan 30 '13 at 5:54

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