Notes:Geometry of curves and surfaces

Curves

Put stuff here about regular curves, singular points, reparametrisations, so forth.

• Arc-length: For a curve $\gamma:(\alpha,\beta)\rightarrow\mathbb{R}^m$ it is easy to see that to approximate length between $t$ and $t+\delta t$ we can use:
  • $$\Vert \gamma(t+\delta t)-\gamma(t)\Vert\eq\big\vert \delta t\big\vert\cdot\left\Vert \frac{\gamma(t+\delta t)-\gamma(t)}{\delta t}\right\Vert$$ $$\approx \delta t\left\Vert \frac{\mathrm{d}\gamma}{\mathrm{d}t}\Big\vert_{t} \right\Vert$$
  • As $\delta t\longrightarrow 0$ we see: $$\Vert \gamma(t+\delta t)-\gamma(t)\Vert\approx\left\Vert\frac{d\gamma}{d t}\right\Vert dt$$
  • Suppose we want the length between $t\eq a$ and $t\eq b$, then:
    • Let $n$ be the number of segments we approximate by, and $\delta t:\eq \frac{b-a}{n}$:
      • $$L:\eq\sum_{i\eq 1}^n \Big\Vert \gamma(a+i\delta t)-\gamma(a+\delta t(i+1)) \Big\Vert \approx\sum^n_{i\eq 1}\delta t\left\Vert \frac{d\gamma}{dt}\Big\vert_{a+i\delta t} \right\Vert$$ $$\eq\delta t\sum^n_{i\eq 1}\Vert\dot{\gamma}(a+i\delta t)\big\Vert$$
      • Then as $n\longrightarrow\infty$ (so $\delta t\longrightarrow 0$) we see:
        • $$L\longrightarrow\int^b_a \big\Vert \dot{\gamma}(t)\big\Vert \mathrm{d}t$$
  • Thus $s(t)$ - a function returning the arc-length of the curve from $t_0$ to $t$ is $$s(t):\eq\int^t_{t_0}\big\Vert \dot{\gamma}(u)\big\Vert du$$
  • By the fundamental theorem of calculus: $$\dot{s}(t)\eq\frac{d}{dt}\left[\int^t_{t_0}\big\Vert \dot{\gamma}(u)\big\Vert du\right]\eq \big\Vert\dot{\gamma}(t)\big\Vert$$ - explicitly: $$\frac{ds}{dt}\eq\left\Vert \frac{d\gamma}{dt}\right\Vert$$
• If $\gamma$ is regular then $s(t)$ is a diffeomorphism
• If $\gamma$ is regular then $\bar{\gamma}:\eq\gamma\circ s^{-1}$ is a unit speed reparametrisation, and also regular
  • Note $$\frac{d\bar{\gamma} }{ds'}\eq\frac{d\gamma}{dt}\cdot\frac{d (s^{-1})}{ds'}$$ where $s$ is the length of $\gamma$ and $s'$ the parameter of $\bar{\gamma}$
    • Notice that $t:\eq s^{-1}(s')$ as $\bar{\gamma}(s'):\eq \gamma(s^{-1}(s'))$ and $\gamma(t)$ is the underlying function here, so:
    • $$\frac{d\bar{\gamma} }{ds'}\eq\frac{d\gamma}{dt}\cdot\frac{dt}{ds'}$$ thus $$\left\Vert\frac{d\bar{\gamma} }{ds'}\right\Vert\eq\left\Vert\frac{d\gamma}{dt}\right\Vert\cdot\left\vert\frac{d t}{ds'}\right\vert \eq\left\Vert\frac{d\gamma}{dt}\right\Vert\cdot\frac{1}{\left\vert\frac{ds}{dt}\right\vert}\eq \left\Vert\frac{d\gamma}{dt}\right\Vert\cdot\frac{1}{\left\Vert\frac{d\gamma}{dt}\right\Vert}\eq 1$$ - thus unit speed!

Planar curves

• Consider $\gamma:(\alpha,\beta)\rightarrow\mathbb{R}^2$ to be a curve in the plane of unit speed then for a fixed $q\in(\alpha,\beta)$ we can talk about $t:\eq\dot{\gamma}(q)$ - the tangent vector, which will be a normal vector by hypothesis
  • We want to build "coordinates" of sort. Intrinsic properties of the curve that define it.
  • Consider a primitive normal $n_p$ given by rotating $t$ anticlockwise by $\frac{\pi}{2}$, we see: $n_p\eq(-t_2,t_1)$
  • We also know that $\dot{t}\eq\ddot{\gamma}(q)$ is perpendicular to $t$, hence (in the plane) parallel to $n_p$
    • So for some $\kappa\in\mathbb{R}$ we see: $\ddot{\gamma}(q)\eq \kappa n_p$
    • $\kappa$ is the signed curvature of $\gamma$.
• Suppose $\gamma$ is regular now but not unit speed. Then:
  • $$t:\eq\frac{\frac{d\gamma}{dt} }{\frac{ds}{dt} } \eq \frac{\frac{d\gamma}{dt} }{\left\Vert\frac{d\gamma}{dt}\right\Vert}$$ - basically just by normalising $$\frac{d\gamma}{dt}$$
  • $n_p$ is obtained by rotating this anticlockwise by $\frac{\pi}{2}$ as before
  • Also $$\ddot{\gamma}(q)\eq\frac{d\mathbf{t} }{dt}\eq\frac{d\mathbf{t} }{ds}\cdot\frac{ds}{dt} \eq \kappa \frac{ds}{dt} n_p\eq\kappa \left\Vert\frac{ds}{dt}\right\Vert n_p$$