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 $(r \circ \gamma)'(t) \neq \bar{0} \,\,\, \forall t \in I$
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 $$(r\circ \gamma)'(t)=Dr(\gamma(t))(\gamma'(t))\neq\vec{0},\forall t\in I$$