TITLE: Help understanding the definition of tangent vector or tangent plane?
QUESTION [6 upvotes]: 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?

REPLY [5 votes]: This is intended as a complement to Alex Bartel's answer. 
(1)  Consider the 2D case given by $x = x(t)$, $y = y(t)$.  Here is a graph of a parametrized function with $(x(t_0),y(t_0))$ (in red) and $(x(t_0 + \delta), y(t_0 + \delta))$ (in yellow) marked on it.  The vector $(x(t_0 + \delta), y(t_0 + \delta)) - (x(t_0),y(t_0))$ (from the numerator of the derivative definition) is turquoise, sits almost on top of the graph of the function, and thus is very close to being tangent to the graph of the function at $(x(t_0), y(t_0))$.  Dividing by $\delta$ only changes the length of that vector, not its direction.  Now imagine letting $\delta \to 0$ and watching the yellow and turquoise vectors change.  Graphically, the turquoise vector is getting closer and closer to being a tangent vector at $t_0$.  In the limit, it should actually be a tangent vector.
 
(1) (from another perspective)  If $(x(t), y(t), z(t))$ describes the position of some object moving through space, then $(x'(t), y'(t), z'(t))$ is its velocity.  Since the direction of the velocity vector gives the direction the object is moving at an instant in time, $(x'(t_0), y'(t_0), z'(t_0))$ will be tangent to the graph of the position at $t_0$.
(3) Here's an example in the 2D case where $x'(t)$ and $y'(t)$ both vanish.  The parametrized function is $x(t) = 1+t^3$ and $y(t) = t^2$, with $x'(0) = y'(0) = 0$.  Graphically, we have a cusp, and so there is no well-defined tangent vector.  This is not the only situation that can occur (see Alex's answer and J.M.'s comment), but it does show why an assumption like this is needed.