Two regular surfaces [MATH] S [/MATH] and [MATH] \overline{S} [/MATH], in [MATH] \mathbb{R}^{3} [/MATH], which have a point [MATH] p [/MATH] in common, are said to have contact of orden [MATH] \geq 1 [/MATH] at [MATH] p [/MATH] if there exists parametrizations with the same domain [MATH] \mathbf{x}(u,v) [/MATH], [MATH] \overline{\mathbf{x}}(u,v) [/MATH] at of [MATH] S [/MATH] and [MATH] \overline{S} [/MATH], respectively, such that [MATH] \mathbf{x}_{u} = \overline{\mathbf{x}}_{u} [/MATH] and [MATH] \mathbf{x}_{v} = \overline{\mathbf{x}}_{v} [/MATH] at [MATH] p [/MATH].
Prove that two regular surfaces have contact of order [MATH] \geq 1 [/MATH] if and only if they have a common tangent plane at [MATH] p [/MATH], i.e., they are tangent at [MATH] p [/MATH].
My attempt:
Let [MATH] S [/MATH] and [MATH] \overline{S} [/MATH] be regular surfaces and their tangent planes at [MATH] p [/MATH] are [MATH] T_{p}(S) [/MATH] and [MATH] T_{p}(\overline{S}) [/MATH] respectively.
[MATH] (\Rightarrow) [/MATH] Suposse that [MATH] S [/MATH] and [MATH] \overline{S} [/MATH] have contact of order [MATH] \geq 1 [/MATH], then if [MATH] \mathbf{x}(u,v) [/MATH] and [MATH] \overline{\mathbf{x}}(u,v) [/MATH] are parametrizatons with the same domain of [MATH] S [/MATH] and [MATH] \overline{S} [/MATH], respectively; it is true that, by definition, [MATH] \mathbf{x}_{u} = \overline{\mathbf{x}}_{u} [/MATH] and [MATH] \mathbf{x}_{v} = \overline{\mathbf{x}}_{v} [/MATH] at [MATH] p [/MATH]. Thus, plane tangents are generated by the same two vectors; therefore, [MATH] T_{p}(S) = T_{p}(\overline{S}) [/MATH].
My question: I'm stuck with the [MATH] (\Leftarrow) [/MATH] part, can you please help me?
Prove that two regular surfaces have contact of order [MATH] \geq 1 [/MATH] if and only if they have a common tangent plane at [MATH] p [/MATH], i.e., they are tangent at [MATH] p [/MATH].
My attempt:
Let [MATH] S [/MATH] and [MATH] \overline{S} [/MATH] be regular surfaces and their tangent planes at [MATH] p [/MATH] are [MATH] T_{p}(S) [/MATH] and [MATH] T_{p}(\overline{S}) [/MATH] respectively.
[MATH] (\Rightarrow) [/MATH] Suposse that [MATH] S [/MATH] and [MATH] \overline{S} [/MATH] have contact of order [MATH] \geq 1 [/MATH], then if [MATH] \mathbf{x}(u,v) [/MATH] and [MATH] \overline{\mathbf{x}}(u,v) [/MATH] are parametrizatons with the same domain of [MATH] S [/MATH] and [MATH] \overline{S} [/MATH], respectively; it is true that, by definition, [MATH] \mathbf{x}_{u} = \overline{\mathbf{x}}_{u} [/MATH] and [MATH] \mathbf{x}_{v} = \overline{\mathbf{x}}_{v} [/MATH] at [MATH] p [/MATH]. Thus, plane tangents are generated by the same two vectors; therefore, [MATH] T_{p}(S) = T_{p}(\overline{S}) [/MATH].
My question: I'm stuck with the [MATH] (\Leftarrow) [/MATH] part, can you please help me?
