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 October 15th, 2013, 02:50 AM #1 Member   Joined: Sep 2013 Posts: 32 Thanks: 0 Fundamental forms Let $f(x,y)=(x,y,h(x,y))$ be a parametrization of the graph T_h of $h:\mathbb{R}^2\to \mathbb{R}.$ Find the first fundamental form for $T_h$ and also compute the second fundamental form. For the first fundamental form, I got that $f_u= \langle 1, 0, f_u \rangle$ and $f_v \langle 0,1,f_v \rangle.$ Then $f_u \dot\ d_u= 1^2 + f_u^2, f_u \dot\ f_v = f_uf_v$ and $f_v \dot\ f_v= 1^2 + f_v^2.$ How can I complete this?
 October 15th, 2013, 04:27 AM #2 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Fundamental forms I advice that in your final answers, you use $(x, y)$ instead of $(u, v)$ \ and $h$ insted of$f$ as these were the variables given. So you've computed the first fundamental form; $I= \left( \begin{array}{cc} \langle f_x , f_x \rangle \ &\ \langle f_x , f_y \rangle \\ \langle f_y , f_x \rangle \=&\ \langle f_y , f_y \rangle \end{array} \right)= \left( \begin{array}{cc} 1 + h_x^2 \ &\ h_xh_y \\ h_xh_y \=&\ 1 + h_y^2 \end{array} \right)=$ To get the second fundamental form, you of course need the Gauss map $N$of the graph. $II= \left[\langle N , \dfrac{\partial ^2f}{\partial u_i \partial u_j}\rangle \right]_{1 \leq i,j \leq 2}$ where $u_i= x \ , \ u_2 = y$ Using the formula $N= \dfrac{f_x \times f_y}{||f_x \times f_y||}$ and $f_{xx}= (0, 0, h_{xx}) , \ f_{xy} = f_{yx} = (0, 0, h_{yx}) , \ f_{yy} = (0, 0, h_{yy})$ the rest is no stress.
 October 15th, 2013, 03:55 PM #3 Member   Joined: Sep 2013 Posts: 32 Thanks: 0 Re: Fundamental forms Thank you very much!

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