A surface is locally the graph of a function

We know that any surface is locally the graph of a function. In fact, we proved two things. First, the graph is defined in one of the three coordinate planes and, second, we know exactly what plane is. The proof says that if $p\in S$ and $X$ is a parametrization around $p$, then the rank of the derivative $dX_q$, $q=X^{-1}(p)$, is $2$. In particular, a $2\times 2$ matrix of $$dX_q=\left(\begin{array}{ccc}x_u& y_u&z_u\\ x_v& y_v&z_v\end{array}\right)$$ has non zero determinant, where $X(u,v)=(x(u,v),y(u,v),z(u,v))$. If we choose $$\left|\begin{array}{cc}x_u& y_u\\ x_v&y_v\end{array}\right|(q)\not=0,$$ then it is a graph on the $xy$-plane. 

It is clear that if $S$ is a graph on the $xy$-plane around $p$, then $T_pS$ can not be vertical, that is, $T_pS$ can not be orthogonal to the $xy$-plane. This can also proved in terms of the projection map. Let $\pi:{\mathbb R}^3\rightarrow{\mathbb R}^2$ be the projection $\pi(x,y,z)=(x,y)$ onto the $xy$-plane. This map is smooth, so when we restrict to $S$ is a differentiable map, which we denote by $\pi$ again. If $S$ is a graph on the $xy$-plane, then $\pi$ would be a local diffeomorphism. Indeed, the differential of $\pi$ is $$(d\pi)_p(v)=(\pi\circ\alpha)'(0)=(\alpha_1'(0),\alpha_2'(0))=(v_1,v_2)$$ where $\alpha$ is a curve that represents $v\in T_pS$ and $\alpha(t)=(\alpha_1(t),\alpha_2(t),\alpha_3(t))$. We ask when $(d\pi)_p$ is an isomorphism. Thus if $v\in T_pS$ where $(d\pi)_p(v)=(0,0)$ then $v_1=v_2=0$. This means that $v=(0,0,v_3)$ is a vertical vector. We conclude that $(d\pi)_p$ is an isomorphism if and only if  the tangent plane is not orthogonal to the $xy$-plane. In such a case, the inverse function theorem asserts the existence of an open set $V\subset S$ around $p\in S$ and and open set $W\subset{\mathbb R}^2$ around $\pi(p)$ such that $\pi:V\rightarrow W$ is a diffeomorphism. If $\phi=\pi^{-1}:W\rightarrow V$ and we stand for $\phi=(\phi_1,\phi_2,\phi_3)$, then for any $(x,y,z)\in V$ we have $$(x,y,z)=\phi(\pi(x,y,z))=\phi(x,y)=(\phi_1(x,y),\phi_2(x,y),\phi_3(x,y)).$$ This proves that $$V=\mbox{graph}(f)=\{(x,y,f(x,y)):(x,y)\in W\},$$ where $f=\phi_3:W\rightarrow\mathbb{R}$.

Curvature of a planar curve by reversing its direction

After the last exercise this afternoon, I will revise what is the curvature of a planar curve when we reverse its orientation. In order to simplify the notation, we will assume that the domain $I$ of the planar curve is $I=\mathbb{R}$, so we have $\alpha:\mathbb{R}\rightarrow{\mathbb R}^2$, which we suppose parametrized by the length-arc. In classroom we have said "when we reverse the orientation, then the curvature change the sign at the point". Firstly, we explanation of our intuition.

When we change the orientation on the curve, we are taking $-s$ instead of $s$ as the new parameter the curve, and as we increase $s$, we are going in the initial curve on the contrary direction. Then the result that we want to prove is that when we pass by a point of the curve, its curvature is the opposite if we go along the curve in the other direction. That is, we say that at $s_0$  the curvature of $\alpha$ when we reverse the sense of the parameter is the opposite with respect to the other direction.

The computations are clear and were made in the classroom. Define $\beta(s)=\alpha(-s)$ the curve $\alpha$ parameterized in the opposite direction. Then $T_\beta(s)=-T_\alpha(-s)$, $N_\beta(s)=JT_\beta(s)=-JT_\alpha(-s)=-N_\alpha(-s)$ and $T_\beta'(s)=T'_\alpha(-s)$. Thus $$\kappa_\beta(s)=\langle T'_\beta(s),N_\beta(s)\rangle=-\langle T'_\alpha(-s),N_\alpha(-s)\rangle=-\kappa_\alpha(-s). (*)$$

The problem appears when we utilize the words to describe the above result. The symbol $\kappa_\beta(s)$ is the curvature of $\alpha$ with the reverse direction at the point $s$. The right hand side says that it coincides with the opposite of the curvature of $\alpha$ at $-s$.

An example. Consider the curve $\alpha(t)=(t,t^2+t^3)$. This curve is not parameterized by the length-arc, but the above argument (which it involves only the curvature) holds for any curve. The curvature of $\alpha$ is $$\alpha'(t)=(1,2t+3t^2), \alpha''(t)=(0,2+6t)\Rightarrow\kappa_\alpha(t)=\frac{2+6t}{(1+(2t+3t^2)^2)^{3/2}}.$$ We fix the point $t=1$, that is, $\alpha(1)=(1,2)$. Then the curvature at $t=1$ is $\kappa_\alpha(1)=8/26^{3/2}$.  We reverse the orientation of $\alpha$ by doing $\beta(t)=(-t,t^2-t^3)$. Now the point $t=1$ for $\alpha$ corresponds with $t=-1$. The curvature of $\beta$ is $$\kappa_\beta(t)=\frac{-2+6t}{(1+(2t-3t^2)^2)^{3/2}}.$$ Then $\kappa_\beta(-1)=-8/26^{3/2}$, which corresponds with the formula  (*).

To finish, I propose a question. Suppose now that the curve is the graph of a function $y=f(x)$, where we know that the curvature of the curve is $f''(x)/(1+f'(x)^2)^{3/2}$. If we move in the opposite direction, then the curve is the graph of the function $y=f(-x)$. Using the above formula,  the curvature would be $f''(-x)/(1+f'(-x)^2)^{3/2}$ because the first derivative gives $-f'(-x)$, but the second one yields $f''(-x)$. But according to (*), this should give $-f''(-x)/(1+f'(-x)^2)^{3/2}$, the oppossite sign.

Question: what is it happening?