女孩子的脸虽然都是同胚的，但是有的正则性好一点，所以这部分女孩子的脸就格外好看。
― 网络
The end of this semester’s FINAL EXAM has come, and at last I could try to make time to complete this blog. This will be my first time to write a maths blog in English. I’ll translate it into Chinese later. Sounds odd… Why don’t I write it in Chinese at very first, but force myself to use English pretending to be a native English speaker? Oh, it is said that maths majors are all somehow masochists. Most of them, anyway. You know, “No pain no gain.”
I shall start with basic definitions.
Def(3-dimensional curve) A parametrized curve in 3-dimensional Euclidean space is a continuous map $\alpha: I \to \mathbb{R}^3$ of an open interval $I = (a,b)$ of the real line $\mathbb{R}$ into $\mathbb{R}^3$. As for $\alpha = (x(t),y(t),z(t))$, $t$ is the parameter of line $\alpha$. Image set${x(t),y(t),z(t)}$ is the graph of line $\alpha$.
So, differential geometry, we want the methods of differential calculus can be applied.
Def(differential curve) A parametrized differentiable curve is a continuous differentiable map. As for $\alpha = (x(t),y(t),z(t))$, $x(t)$, $y(t)$, $z(t)$ are all $C^\infty$ functions.
One obvious simple instance of non-differential curve is absolute value function.
Those two properties above, continuity and differentiability, seem to be enough for a certain curve in Euclidean space to be analysed easily. However, all the textbooks taught me “regularity” were forgoten12. What exactly is “regularity” they mentioned here? I shall read the definition again.
Def(regular curve) A parametrized differentiable curve $\alpha: \alpha(t) = (x(t), y(t), z(t))$ is regular if $$\forall t \in (a,b), |\alpha’(t)| = \sqrt{[x’(t)]^2 +[y’(t)]^2 + [z’(t)]^2}> 0.$$
We want to insure that the curve we consider has a tangent line at every point. Frenet frame, a powerful tool for analysis, consists of three vectors orthogonal to each other: unit tangent vector $\bm{t}$, unit normal vector $\bm{n}$, unit binormal vector $\bm{b}$. Both $\bm{n}$ and $\bm{b}$ come from $\bm{t}$ parallel to the tangent line. What if the derivative of $\alpha(t_0)$ was a zero vector? Unfortunately, such tangent line might not exist at $t_0$, and dear Frenet frame couldn’t work. Frent frame is only a small example to show why WE NEED TANGENT LINEs EVERYWHERE.
e.g. Consider such a plane curve, $\alpha: \alpha(t) = (t^4, t^2, t^3)$. Of course it’s differentiable. Notice that $\alpha’(0) = (0,0,0)$, the velocity vector is zero for $t = 0$, so this curve is not regular. You can check its figure below. Pay attention to $(0,0,0)$, the curve isn’t smooth there.
Someone (like me) prefer to play devil’s advocate: how can a non-zero derivative demonstrate the existence of the tangent line to the curve? The proof is as followed.
Proof Assume $\alpha: \alpha(t) = (x(t),y(t),z(t))$ is a differential curve. According to the definition of derivative of vector function, $\forall t_0 \in (a,b)$,
$$
\frac{\mathrm{d}\alpha}{\mathrm{d}t}\Big\vert_{t_0} = (\frac{\mathrm{d}x}{\mathrm{d}t},\frac{\mathrm{d}y}{\mathrm{d}t},\frac{\mathrm{d}z}{\mathrm{d}t})\Big\vert_{t_0} = \lim_{t \to t_0} \frac{\alpha(t) - \alpha(t_0)}{t-t_0}.
$$
According to differentiability,
$$
\lim_{t \to t_0-0} \frac{\alpha(t_0) - \alpha(t)}{t-t_0} = \lim_{t \to t_0+0} \frac{\alpha(t) - \alpha(t_0)}{t-t_0},
$$
i.e. $\alpha’(t_0-0) = \alpha’(t_0+0)$.
The unit direction vector of the left tangent line is
$$
\bm{e}^-(t_0) = -\lim_{t\to t_0 - 0} \frac{\alpha(t) - \alpha(t_0)}{|\alpha(t) - \alpha(t_0)|};
$$
The unit direction vector of the right tangent line is
$$
\bm{e}^+(t_0) = +\lim_{t\to t_0 + 0} \frac{\alpha(t) - \alpha(t_0)}{|\alpha(t) - \alpha(t_0)|}.
$$
Obviously, iff. $\bm{e}^-(t_0) = \bm{e}^+(t_0) = \bm{e}$, the tangent line to $\alpha$ at $t_0$ exists, and its direction vector is just $\bm{e}$.
$\bm{e}^-(t_0) \parallel \alpha’(t_0-0)$, $\bm{e}^+(t_0) \parallel \alpha’(t_0+0)$. If $\alpha’(t_0 - 0) = \alpha’(t_0 + 0) = \alpha’(t_0) \neq \bm{0}$, $\bm{e}^-(t_0)$ and $\bm{e}^+(t_0)$ would get to the same direction, hence they are equal.
Q.E.D.
This proof tells us $\alpha’(t_0) \neq \bm{0} \Rightarrow $ tangent line exists. In contrast, can we say there’s no tangent line if $\alpha’(t_0) = \bm{0}$? Sorry, just take a look at $\alpha: \alpha(t) = (t^3, t^3)$.