Notes:Geometry of curves and surfaces

Curves

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

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

Planar curves

• Consider [ilmath]\gamma:(\alpha,\beta)\rightarrow\mathbb{R}^2[/ilmath] to be a curve in the plane of unit speed then for a fixed [ilmath]q\in(\alpha,\beta)[/ilmath] we can talk about [ilmath]t:\eq\dot{\gamma}(q)[/ilmath] - 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 [ilmath]n_p[/ilmath] given by rotating [ilmath]t[/ilmath] anticlockwise by [ilmath]\frac{\pi}{2} [/ilmath], we see: [ilmath]n_p\eq(-t_2,t_1)[/ilmath]
  • We also know that [ilmath]\dot{t}\eq\ddot{\gamma}(q) [/ilmath] is perpendicular to [ilmath]t[/ilmath], hence (in the plane) parallel to [ilmath]n_p[/ilmath]
    • So for some [ilmath]\kappa\in\mathbb{R} [/ilmath] we see: [ilmath]\ddot{\gamma}(q)\eq \kappa n_p[/ilmath]
    • [ilmath]\kappa[/ilmath] is the signed curvature of [ilmath]\gamma[/ilmath].
• Suppose [ilmath]\gamma[/ilmath] is regular now but not unit speed. Then:
  • [math]t:\eq\frac{\frac{d\gamma}{dt} }{\frac{ds}{dt} } \eq \frac{\frac{d\gamma}{dt} }{\left\Vert\frac{d\gamma}{dt}\right\Vert} [/math] - basically just by normalising [math]\frac{d\gamma}{dt} [/math]
  • [ilmath]n_p[/ilmath] is obtained by rotating this anticlockwise by [ilmath]\frac{\pi}{2} [/ilmath] as before
  • Also [math]\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[/math]