Here is what my textbook told me:
Assuming the formula for surface $\Sigma$ is $$F(x,y,z) = 0$$ Suppose $X_0 = (x_0, y_0, z_0)$ is a point on the surface $\Sigma$ and we assuming F(x,y,z) is differentiable and $$\mathbf{J}F(X_0) = (\frac{\partial F(X_0)}{\partial x}, \frac{\partial F(X_0)}{\partial y}, \frac{\partial F(X_0)}{\partial z}) \neq 0$$ Draw a line $\Gamma$ in the surface $\Sigma$ passing through the point $X_0$, assuming the equations for $\Sigma$ is $$x = x(t), y = y(t), z = z(t)$$ $t = t_0$ correspond to the point $X_0$ and $x'(t_0), y'(t_0), z'(t_0)$ does not all vanish. Because of the line $\Gamma$ is on the surface $\Sigma$, so $$F(x(t), y(t), z(t)) = 0$$ So $$ \frac{dF}{dt}\mid_{t=t_0} = {F_x}'(X_0)x'(t_0) + {F_y}'(X_0)y'(t_0) + {F_z}'(X_0)z'(t_0) = 0 $$ So $$ ({F_x}'(X_0), {F_y}'(X_0), {F_z}'(X_0))\cdot(x'(t_0), y'(t_0), z'(t_0)) = 0 $$
We know the vector $\mathbf{T} = (x'(t_0), y'(t_0), z'(t_0))$ is the tangent vector for the line $\Gamma$ on the point $X_0$
My questions are
- Why the vector $\mathbf{T}$ is the tangent vector for line $\Gamma$ at point $X_0$?
- Why the $\mathbf{J}F(X_0)$ should not equal to zero? What if it is zero?
- Why $x'(t_0), y'(t_0), z'(t_0)$ should not all vanish? What if all vanish?