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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

  1. Why the vector $\mathbf{T}$ is the tangent vector for line $\Gamma$ at point $X_0$?
  2. Why the $\mathbf{J}F(X_0)$ should not equal to zero? What if it is zero?
  3. Why $x'(t_0), y'(t_0), z'(t_0)$ should not all vanish? What if all vanish?
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2. If they all vanish, you then have a so-called "singular point" where you either cannot draw a unique tangent, or you cannot draw a tangent at all; isolated points (second case) and self-intersection points (first case) are two common examples. – J. M. Nov 29 '10 at 12:47
up vote 3 down vote accepted

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.

alt text

(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.

alt text

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+1 - nice examples! – Alex B. Nov 30 '10 at 1:50
Could you tell me what tool you use to draw these pictures? – Jichao Nov 30 '10 at 14:47
@Jichao: I used Mathematica. – Mike Spivey Nov 30 '10 at 16:15
  1. The coordinates $x(t)$ and so on tell you how to trace out a curve, as $t$ changes. The vector T describes the change in the $x$, $y$ and $z$-coordinates as $t$ changes infinitesimally. Surely, that's exactly what the tangent vector to a curve should do. More algebraically, the tangent to the curve at $X_0$ is defined as $$\lim_{\delta\rightarrow 0} \frac{(x(t_0),y(t_0)z(t_0)) - (x(t_0+\delta),y(t_0+\delta)z(t_0+\delta))}{\delta},$$ whenever the limit is defined.

  2. As J.M. wrote, in that case you don't have a well-defined tangent. Again, algebraically, your calculation has shown that J$F(X_0)$ is normal to tangents to all curves on the surface passing through $X_0$, so it should be the normal vector to the surface at that point. But if the surface has a singularity, then there is no well-defined normal vector.

  3. As remarked above, the derivatives $x'(t_0)$ and so on describe, how quickly the coordinates change as you trace out your curve. In pictorial terms, you can imagine walking down the curve as the parameter $t$ slowly increases. If all these derivatives vanish, it means that as $t$ approaches $t_0$, your walk is slowing down to a grinding halt. That means that either you have chosen a bad parametrisation, i.e. you are staying in one place for some range of $t$, which you might as well leave out then, or you are approaching a singularity, which is bad news if you are computing tangents.

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