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Let $S\subseteq \mathbb{R}^3$ be an oriented regular surface and let $N$ be a field of normal unitary vector on $S$. We consider the map $F:S\times \mathbb{R}\rightarrow \mathbb{R}^3$ defined by $F(p,t):= p+tN_p$ and we want to calculate $DF_{(p,t)}:T_pS\times\mathbb{R}\rightarrow\mathbb{R}^3$.

Can someone show me how to do this calculation?

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1 Answer 1

Let us fix $(p,t)\in S\times\mathbb R$, $(u,h)\in T_{(p,t)}(S\times\mathbb R)\cong T_pS\times\mathbb R$.
Let $\gamma:I\to S$ be a smooth curve with $\gamma(0)=p$ and $\dot\gamma(0)=u$. Then $$(D_{(p,t)}F)(u,h)=\left.\frac{d}{ds}\right|_{s=0}F(\gamma(s),t+sh)= \left.\frac{d}{ds}\right|_{s=0}\left(\gamma(s)+(t+sh)N_{\gamma(s)}\right)=u+hN_p+t(D_pN)u$$ Now we recall that the Weingarten map $W:TS\to T\mathbb S^2$ is defined as the negative derivative of the Gauss map $N:S\to\mathbb S^2$, so $$\begin{array}{cccc}(D_{(p,t)}F):&T_{p,t}(S\times\mathbb R)\cong T_pS\times\mathbb R&\to&T_{F(p,t)}\mathbb R^3\cong\mathbb R^3\\&(u,h)&\mapsto&u+hN_p-t(W_p)u\end{array} $$

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