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Consider a curve $\gamma: I \subset \mathbb{R} \to A \subset \mathbb{R}^2$ and a surface $r: D \subset \mathbb{R}^2 \to \mathbb{R}^3$ with $A \subset D$. Suppose that $\gamma$ is a regular curve and $r$ is a regular surface.

Can I state that $$r \circ \gamma: I \subset \mathbb{R} \to \mathbb{R}^3$$

is a regular curve? How can I prove it?


Attempt:

I know that $\gamma'(t) \neq \bar{0} \,\, \forall t \in I$ and that $r_{u} (u,v) \times r_{v}(u,v)\neq \bar{0} \,\, \forall (u,v) \in D$ ($r_{u}$ and $r_{v}$ are the derivatives of $r$ with respect to $u$ and $v$).

I tried to calculate $(r \circ \gamma)'(t)$ but it is made of a quit complex expression and I don't see how to deduce directly that $(r \circ \gamma)'(t) \neq \bar{0} \,\,\, \forall t \in I$

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Note that the linear transformation $Dr=[r_u\ r_v]$ has trivial kernel (since, $r_u$ and $r_v$ forms a two dimensional basis) and $\gamma'$ is not zero, thus $$(r\circ \gamma)'(t)=Dr(\gamma(t))(\gamma'(t))\neq\vec{0},\forall t\in I$$

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  • $\begingroup$ Thanks for the answer! May I ask what does $Dr=[r_{u} r_{v}]$ stand for? Is it the cross product of the two derivative of $r$ with respect to $u$ and $v$? $\endgroup$ – Gianolepo Dec 8 '16 at 23:28
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    $\begingroup$ @Gianolepo is Just a matrix sucht that the collumns are the derivative of r. $\endgroup$ – DiegoMath Dec 10 '16 at 22:24
  • $\begingroup$ Thanks again! I would like to ask: suppose that $\gamma$ is piecewise defined, that is $\gamma=\cup_{i=1}^{n} \gamma_i$ and each "piece" (each curve $\gamma_i$) is regular (but still $\gamma$ can be not regular in the intersection points between the curves $\gamma_i$, since the derivative can be undefined there); suppose also that $r \in C^{2}(D)$ and it is not necessarily regular: is the composition $r \circ \gamma$ still a curve "piecewise regular" (that is, $\gamma=\cup_{i=1}^{n} r \circ \gamma_i$ with $ r \circ \gamma_i$ regular, but potentially not regular in the intersection points)? $\endgroup$ – Gianolepo Dec 20 '16 at 0:12

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