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Suppose that a surface patch $\tilde{\sigma}(\tilde u,\tilde v)$ is a reparametrization of a surface patch $\sigma (u,v)$.

Let $$\tilde E d\tilde u^2+ 2\tilde F d\tilde ud\tilde v+\tilde G d\tilde v^2$$ and $$ Edu^2+2Fdudv+Gdv^2$$ be their first fundamental form.

Show that $$du=\frac{\partial u}{\partial \tilde u}d\tilde u+\frac{\partial u}{\partial \tilde v}d\tilde v$$

$$dv=\frac{\partial v}{\partial \tilde u}d\tilde u+\frac{\partial v}{\partial \tilde v}d\tilde v$$


Firstly I do not understand the following statement:

a surface patch $\tilde{\sigma}(\tilde u,\tilde v)$ is a reparametrization of a surface patch $\sigma (u,v)$.

How to show this mathematically? And I cannot solve the question. Please can explain this. Thanks a lot.

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Let $\sigma: U\subseteq \mathbb R^2\rightarrow \mathbb R^3$ $(u,v)\mapsto \sigma(u,v)=(x(u,v),y(u,v),z(u,v))$ parametrize a 2 dimensional surface in $\mathbb R^3$ by coordinates $(u,v)$.

Let $\varphi$ be a bijective $C^1$-map $\varphi:V\subseteq\mathbb R^2\rightarrow U $ with

$$(\tilde{u},\tilde{v})\mapsto \varphi(\tilde{u},\tilde{v}):=(u,v). $$

In other words, the coordinates $(u,v)$ are functions of the coordinates $(\tilde{u},\tilde{v})$ via $\varphi$. The composition

$$\tilde{\sigma}:=\sigma\circ \varphi $$

is a reparametrization of the surface patch, with $\tilde{\sigma}(\tilde{u},\tilde{v}):= \sigma(\varphi(\tilde{u},\tilde{v}))=\sigma(u,v).$

The "question" in the OP is a straightforward application of the chain rule for the compositions $u=u(\tilde{u},\tilde{v})$, $v=v(\tilde{u},\tilde{v})$.

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  • $\begingroup$ But I have a question. In The book's solution, $\tilde{\sigma}_{\tilde u}=\sigma_u +\frac{\partial u}{\partial \sigma u}+ \sigma_v\frac{\partial v}{\partial \tilde u}$ how to be written these equations? I am confused. $\endgroup$
    – 1190
    Commented Nov 29, 2013 at 21:15
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    $\begingroup$ $\frac{\partial \tilde{\sigma}}{\partial \tilde{u}}(\tilde{u},\tilde{v})=\frac{\partial \sigma\circ \varphi}{\partial \tilde{u}}(\tilde{u},\tilde{v})$: can you apply the chain rule to the composition $\sigma\circ \varphi$? Remember that $\varphi(\tilde{u},\tilde{v})=(u,v)$ by definition. $\endgroup$
    – Avitus
    Commented Nov 29, 2013 at 21:20

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