Fórmulas para curvatura y torsión cuando la curva no está parametrizada por longitud de arco.

(*) Expresar la curvatura y la torsión como determinantes.

(*) Usar la regla de la cadena para derivar composiciones.

CASO $n=2$ curva plana

Dada $\alpha (t)=(x(t),y(t))$ y $t=h(s)$ con $s=$ longitud de arco; entonces $$\gamma (s)=\alpha (h(s))=(u(s),v(s))$$ $${\gamma }^{\prime }(s)=(u^{\prime }(s),v^{\prime }(s))$$

Curvatura de $\gamma$ en el punto $s$ es el número $\mathcal{K}$ para el cual se cumple la ecuación $${\gamma }^{\prime \prime }(s)=\mathcal{K}(–v^{\prime }(s),u^{\prime }(s))$$

Así definida $\mathcal{K}(s)$ puede ser $\mathcal{K}(s)>0$, $\mathcal{K}(s)<0$ o incluso $\mathcal{K}(s)=0.$

Se sigue cumpliendo que $$|\mathcal{K}|=\Vert {\gamma }^{\prime \prime }(s)\Vert$$

Si $\gamma (s)$ está dada por longitud de arco, entonces $$\Vert {\gamma }^{\prime }(s)\Vert =1⟺⟨{\gamma }^{\prime }(s),{\gamma }^{\prime }(s)⟩=1$$

Derivando $$\begin{array}{rl}⟨{\gamma }^{\prime \prime }(s),{\gamma }^{\prime }(s)⟩+⟨{\gamma }^{\prime }(s),{\gamma }^{\prime \prime }(s)⟩& \equiv 0\\ ⟨{\gamma }^{\prime }(s),{\gamma }^{\prime \prime }(s)⟩& \equiv 0\end{array}$$

$$\therefore {\gamma }^{\prime \prime }(s)\text{ es ortogonal a }{\gamma }^{\prime }(s)$$

Luego, ${\gamma }^{\prime \prime }(s)$ está en la recta ortogonal a ${\gamma }^{\prime }(s).$

¿Cuál es el signo de $\mathcal{K}(s)$?

Observemos el producto punto de ${\gamma }^{\prime \prime }(s)$ con $(-v(s),u(s)).$

Signo de $\mathcal{K}(s)=$ signo $|\begin{array}{cc}{u}^{\prime }& u^{\prime \prime }\\ v^{\prime }& v^{\prime \prime }\end{array}|$

Por lo que el signo de $\mathcal{K}(s)$ está dado por el signo del determinante de $({\gamma }^{\prime }|{\gamma }^{\prime \prime }).$

Luego

$\begin{array}{rl}{u}^{\prime \prime }& =–\mathcal{K}{v}^{\prime }\\ v^{\prime \prime }& =\mathcal{K}{u}^{\prime }\end{array}$

Por lo que

$\begin{array}{rl}|\begin{array}{cc}{u}^{\prime }& u^{\prime \prime }\\ v^{\prime }& v^{\prime \prime }\end{array}|& =|\begin{array}{cc}{u}^{\prime }& –\mathcal{K}{v}^{\prime }\\ v^{\prime }& \mathcal{K}{u}^{\prime }\end{array}|\\ & =u^{\prime }{u}^{\prime }\mathcal{K}–v^{\prime }{v}^{\prime }\mathcal{K}\\ & =\mathcal{K}((u^{\prime }{)}^2–(v^{\prime }{)}^2)\\ & =\mathcal{K}\end{array}$$

Entonces $$|{\gamma }^{\prime }|=|(u^{\prime },v^{\prime })|=1$$

1° paso: La curvatura se puede ver como

$$\mathcal{K}(s)=|\begin{array}{cc}{u}^{\prime }(s)& u^{\prime \prime }(s)\\ v^{\prime }(s)& v^{\prime \prime }(s)\end{array}|$$

2° paso: Ahora consideremos la relación con cambio de variable.

$$u(s)=x(h(s))=x(t)$$

Derivando

$u^{\prime }(s)={\displaystyle \frac{du}{ds}}={\displaystyle \frac{dx}{dt}}{\displaystyle \frac{dt}{ds}}=\dot{x}(t){\displaystyle \frac{dt}{ds}}$

$u^{\prime \prime }(s)={\displaystyle \frac{d}{ds}}({\displaystyle \frac{dx}{dt}}{\displaystyle \frac{dt}{ds}})$

$u^{\prime \prime }(s)=({\displaystyle \frac{d}{ds}}{\displaystyle \frac{dx}{dt}}{\displaystyle \frac{dt}{ds}}+{\displaystyle \frac{dx}{dt}}({\displaystyle \frac{d}{ds}}{\displaystyle \frac{dt}{ds}})$

$u^{\prime \prime }(s)={\displaystyle \frac{d^{2}x}{d{t}^2}}{\displaystyle \frac{dt}{ds}}{\displaystyle \frac{dt}{ds}}+{\displaystyle \frac{dx}{dt}}{\displaystyle \frac{d^{2}t}{d{s}^2}}{u^{\prime }}^{\prime }(s)$

$u^{\prime \prime }(s)=\ddot{x}(t)({\displaystyle \frac{dt}{ds}}{)}^2+\dot{x}(t){\displaystyle \frac{d^{2}t}{d{s}^2}}$

Entonces la curvatura está dada por:

$\mathcal{K}(s)=\left|\begin{array}{cc}{u}^{\prime }(s)& u^{\prime \prime }(s)\\ \\ v^{\prime }(s)& v^{\prime \prime }(s)\end{array}\right|$

$$

$\mathcal{K}(s)=\left|\begin{array}{cc}{\displaystyle \frac{du}{ds}}& {\displaystyle \frac{d^{2}u}{d{s}^2}}\\ \\ {\displaystyle \frac{dv}{ds}}& {\displaystyle \frac{d^{2}v}{d{s}^2}}\end{array}\right|$

$$

$\mathcal{K}(s)=\left|\begin{array}{cccc}{\displaystyle \frac{dx}{dt}}{\displaystyle \frac{dt}{ds}}& & & {\displaystyle \frac{d^{2}x}{d{t}^2}}({\displaystyle \frac{dt}{ds}}{)}^2+{\displaystyle \frac{dx}{dt}}{\displaystyle \frac{d^{2}t}{d{s}^2}}\\ \\ {\displaystyle \frac{dy}{dt}}{\displaystyle \frac{dt}{ds}}& & & {\displaystyle \frac{d^{2}y}{d{t}^2}}({\displaystyle \frac{dt}{ds}}{)}^2+{\displaystyle \frac{dy}{dt}}{\displaystyle \frac{d^{2}t}{d{s}^2}}\end{array}\right|$

$$

$\mathcal{K}(s)=\left|\begin{array}{cccc}{\displaystyle \frac{dx}{dt}}{\displaystyle \frac{dt}{ds}}& & & {\displaystyle \frac{d^{2}x}{d{t}^2}}({\displaystyle \frac{dt}{ds}}{)}^2\\ \\ {\displaystyle \frac{dy}{dt}}{\displaystyle \frac{dt}{ds}}& & & {\displaystyle \frac{d^{2}y}{d{t}^2}}({\displaystyle \frac{dt}{ds}}{)}^2\end{array}\right|+\left|\begin{array}{cccc}{\displaystyle \frac{dx}{dt}}{\displaystyle \frac{dt}{ds}}& & & {\displaystyle \frac{dx}{dt}}{\displaystyle \frac{d^{2}t}{d{s}^2}}\\ \\ {\displaystyle \frac{dy}{dt}}{\displaystyle \frac{dt}{ds}}& & & {\displaystyle \frac{dy}{dt}}{\displaystyle \frac{d^{2}t}{d{s}^2}}\end{array}\right|$

$$

$\mathcal{K}(s)=({\displaystyle \frac{dt}{ds}}{)}^3\left|\begin{array}{cccc}{\displaystyle \frac{dx}{dt}}& & & {\displaystyle \frac{d^{2}x}{d{t}^2}}\\ \\ {\displaystyle \frac{dy}{dt}}& & & {\displaystyle \frac{d^{2}y}{d{t}^2}}\end{array}\right|+({\displaystyle \frac{dt}{ds}}){\displaystyle \frac{d^{2}t}{d{s}^2}}\left|\begin{array}{cccc}{\displaystyle \frac{dx}{dt}}& & & {\displaystyle \frac{dx}{dt}}\\ \\ {\displaystyle \frac{dy}{dt}}& & & {\displaystyle \frac{dy}{dt}}\end{array}\right|$

$$

$\mathcal{K}(s)=({\displaystyle \frac{dt}{ds}}{)}^3\left|\begin{array}{cccc}{\displaystyle \frac{dx}{dt}}& & & {\displaystyle \frac{d^{2}x}{d{t}^2}}\\ \\ {\displaystyle \frac{dy}{dt}}& & & {\displaystyle \frac{d^{2}y}{d{t}^2}}\end{array}\right|$

$$

Pero $s=\int \Vert {\alpha }^{\prime }\Vert (t)dt$ entonces ${\displaystyle \frac{ds}{dt}}=\Vert {\alpha }^{\prime }(t)\Vert =\sqrt{({\displaystyle \frac{dx}{dt}}{)}^2+({\displaystyle \frac{dy}{dt}}{)}^2}$

Por lo tanto

$\mathcal{K}(s)={\displaystyle \frac{\left|\begin{array}{cccc}{\displaystyle \frac{dx}{dt}}& & & {\displaystyle \frac{d^{2}x}{d{t}^2}}\\ \\ {\displaystyle \frac{dy}{dt}}& & & {\displaystyle \frac{d^{2}y}{d{t}^2}}\end{array}\right|}{(({\displaystyle \frac{dx}{dt}}{)}^2+({\displaystyle \frac{dy}{dt}}{)}^2{)}^{\frac{3}{2}}}}$

Luego, cuando $n=2$ $$\mathcal{K}(t)={\displaystyle \frac{\dot{x}\ddot{y}–\dot{y}\ddot{x}}{((\dot{x}{)}^2+(\dot{y}{)}^2{)}^{\frac{3}{2}}}}$$

$$

CASO $n=3$

1° paso: Expresar la curvatura y la torsión en términos de determinantes.

$$T(s)={\gamma }^{\prime }(s)$$ $$N(s)={\displaystyle \frac{{\gamma }^{\prime \prime }(s)}{\Vert {\gamma }^{\prime \prime }(s)\Vert }}$$

Luego $${\gamma }^{\prime \prime }(s)=\mathcal{K}(s)N(s)$$

y $B(s)=T(s)\times N(s)$

entonces ${\gamma }^{\prime }(s)\times {\gamma }^{\prime \prime }(s)=T(s)\times \mathcal{K}(s)N(s)$

${\gamma }^{\prime }(s)\times {\gamma }^{\prime \prime }(s)=\mathcal{K}(s)(T(s)\times N(s))=\mathcal{K}(s)B(s)$

$\Vert {\gamma }^{\prime }(s)\times {\gamma }^{\prime \prime }(s)\Vert =\mathcal{K}$ con el parámetro longitud de arco, ya que $\Vert B(s)\Vert =1.$

$$

Cuando no está parametrizada por longitud de arco, $$\gamma (s)=(x(s),y(s),z(s))$$ $${\gamma }^{\prime }(s)=({\displaystyle \frac{dx}{ds}},{\displaystyle \frac{dy}{ds}},{\displaystyle \frac{dz}{ds}})$$ $$\gamma (s)=({\displaystyle \frac{d^{2}x}{d{s}^2}},{\displaystyle \frac{d^{2}y}{d{s}^2}},{\displaystyle \frac{d^{2}z}{d{s}^2}})$$

Entonces $${\gamma }^{\prime }(s)\times {\gamma }^{\prime \prime }(s)=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \\ {\displaystyle \frac{dx}{ds}}& {\displaystyle \frac{dy}{ds}}& {\displaystyle \frac{dz}{ds}}\\ \\ {\displaystyle \frac{d^{2}x}{d{s}^2}}& {\displaystyle \frac{d^{2}y}{d{s}^2}}& {\displaystyle \frac{d^{2}z}{d{s}^2}}\end{array}\right|$$

$${\gamma }^{\prime }(s)\times {\gamma }^{\prime \prime }(s)=\left|\begin{array}{ccccc}\hat{i}& & \hat{j}& & \hat{k}\\ \\ {\displaystyle \frac{dx}{dt}}{\displaystyle \frac{dt}{ds}}& & {\displaystyle \frac{dy}{dt}}{\displaystyle \frac{dt}{ds}}& & {\displaystyle \frac{dz}{dt}}{\displaystyle \frac{dt}{ds}}\\ \\ {\displaystyle \frac{d^{2}x}{d{t}^2}}({\displaystyle \frac{dt}{ds}}{)}^2+{\displaystyle \frac{dx}{dt}}{\displaystyle \frac{d^{2}t}{d{s}^2}}& & {\displaystyle \frac{d^{2}y}{d{t}^2}}({\displaystyle \frac{dt}{ds}}{)}^2+{\displaystyle \frac{dy}{dt}}{\displaystyle \frac{d^{2}t}{d{s}^2}}& & {\displaystyle \frac{d^{2}z}{d{t}^2}}({\displaystyle \frac{dt}{ds}}{)}^2+{\displaystyle \frac{dz}{dt}}{\displaystyle \frac{d^{2}t}{d{s}^2}}\end{array}\right|$$

Por lo tanto, para $n=3$ se tiene que $$\mathcal{K}(t)={\displaystyle \frac{\Vert \dot{\gamma }(t)\times \ddot{\gamma }(t)\Vert }{\Vert \dot{\gamma }(t){\Vert }^3}}$$

$$

Torsión

$$T^{\prime }(s)=\mathcal{K}(s)N(s)$$

$$N^{\prime }(s)=–\mathcal{K}(s)T(s)–\tau (s)B(s)$$

$$B^{\prime }(s)=\tau (s)N(s)$$

Tenemos que:

${\gamma }^{\prime }=T(s)$

${\gamma }^{\prime \prime }=T^{\prime }(s)=\mathcal{K}N$

${\gamma }^{\prime \prime \prime }={\displaystyle \frac{d}{ds}}(\mathcal{K}N={\mathcal{K}}^{\prime }N+\mathcal{K}{N}^{\prime }={\mathcal{K}}^{\prime }N+\mathcal{K}(–\mathcal{K}(s)T(s)–\tau (s)B(s))={\mathcal{K}}^{\prime }N-{\mathcal{K}}^2(s)T(s)–\mathcal{K}\tau (s)B(s)$

Resolviendo el triple producto escalar de $({\gamma }^{\prime },{\gamma }^{\prime \prime },{\gamma }^{\prime \prime \prime })$ se tiene que:

$\begin{array}{rl}({\gamma }^{\prime },{\gamma }^{\prime \prime },{\gamma }^{\prime \prime \prime })& =(T,\mathcal{K}N,-{\mathcal{K}}^{2}T+{\mathcal{K}}^{\prime }N–\mathcal{K}\tau B\\ & =\mathcal{K}(T,N,-{\mathcal{K}}^{2}T+{\mathcal{K}}^{\prime }N–\mathcal{K}\tau B)\\ & =\mathcal{K}(T,N,-{\mathcal{K}}^{2}T)+(T,N,{\mathcal{K}}^{\prime }N)+(T,N,-\mathcal{K}\tau B))\\ & =–{\mathcal{K}}^3(T,N,T)+\mathcal{K}{\mathcal{K}}^{\prime }(T,N,N)–{\mathcal{K}}^2\tau (T,N,B)\end{array}$

Por lo tanto $$({\gamma }^{\prime },{\gamma }^{\prime \prime },{\gamma }^{\prime \prime \prime })=–{\mathcal{K}}^2\tau$$

Por lo tanto, para curvas parametrizadas por longitud de arco $$\tau (s)={\displaystyle \frac{–({\gamma }^{\prime },{\gamma }^{\prime \prime },{\gamma }^{\prime \prime \prime })}{{\mathcal{K}}^2(s)}}$$

Partiendo de $\tau (s)={\displaystyle \frac{–({\gamma }^{\prime },{\gamma }^{\prime \prime },{\gamma }^{\prime \prime \prime })}{{\mathcal{K}}^2(s)}}$

Buscamos una expresión en términos de $t$ y las derivadas con respecto a $t$.

1° paso:

${\gamma }^{\prime }=\dot{\gamma }(t){\displaystyle \frac{dt}{ds}}$

${\gamma }^{\prime \prime }=\ddot{\gamma }(t)({\displaystyle \frac{dt}{ds}}{)}^2+\dot{\gamma }(t){\displaystyle \frac{d^{2}t}{d{s}^2}}$

${\gamma }^{\prime \prime \prime }=\stackrel{⃛}{\gamma }(t)({\displaystyle \frac{dt}{ds}}{)}^3+\ddot{\gamma }(t)2{\displaystyle \frac{dt}{ds}}{\displaystyle \frac{d^{2}t}{d{s}^2}}+\ddot{\gamma }(t){\displaystyle \frac{dt}{ds}}{\displaystyle \frac{d^{2}t}{d{s}^2}}+\dot{\gamma }(t){\displaystyle \frac{d^{3}t}{d{s}^3}}$

Entonces ${\gamma }^{\prime \prime \prime }=\stackrel{⃛}{\gamma }(t)({\displaystyle \frac{dt}{ds}}{)}^3+3\ddot{\gamma }(t){\displaystyle \frac{dt}{ds}}{\displaystyle \frac{d^{2}t}{d{s}^2}}+\dot{\gamma }(t){\displaystyle \frac{d^{3}t}{d{s}^3}}$

Luego $$\begin{array}{rl}({\gamma }^{\prime },{\gamma }^{\prime \prime },{\gamma }^{\prime \prime \prime })& =(\dot{\gamma }(t){\displaystyle \frac{dt}{ds}},\ddot{\gamma }(t)({\displaystyle \frac{dt}{ds}}{)}^2+\dot{\gamma }(t){\displaystyle \frac{d^{2}t}{d{s}^2}},\stackrel{⃛}{\gamma }(t)({\displaystyle \frac{dt}{ds}}{)}^3+3\ddot{\gamma }(t){\displaystyle \frac{dt}{ds}}{\displaystyle \frac{d^{2}t}{d{s}^2}}+\dot{\gamma }(t){\displaystyle \frac{d^{3}t}{d{s}^3}})\\ & =(\dot{\gamma }(t){\displaystyle \frac{dt}{ds}},\ddot{\gamma }(t)({\displaystyle \frac{dt}{ds}}{)}^2,\stackrel{⃛}{\gamma }(t)({\displaystyle \frac{dt}{ds}}{)}^3)+(\dot{\gamma }(t){\displaystyle \frac{dt}{ds}},\ddot{\gamma }(t)({\displaystyle \frac{dt}{ds}}{)}^2,3\ddot{\gamma }(t){\displaystyle \frac{dt}{ds}}{\displaystyle \frac{d^{2}t}{d{s}^2}})+\\ & (\dot{\gamma }(t){\displaystyle \frac{dt}{ds}},\ddot{\gamma }(t)({\displaystyle \frac{dt}{ds}}{)}^2,\dot{\gamma }(t){\displaystyle \frac{d^{3}t}{d{s}^3}}))+(\dot{\gamma }(t){\displaystyle \frac{dt}{ds}},\dot{\gamma }(t){\displaystyle \frac{d^{2}t}{d{s}^2}},\stackrel{⃛}{\gamma }(t)({\displaystyle \frac{dt}{ds}}{)}^3+3\ddot{\gamma }(t){\displaystyle \frac{dt}{ds}}{\displaystyle \frac{d^{2}t}{d{s}^2}}+\dot{\gamma }(t){\displaystyle \frac{d^{3}t}{d{s}^3}})\end{array}$$

$$

Entonces

$\begin{array}{rl}({\gamma }^{\prime },{\gamma }^{\prime \prime },{\gamma }^{\prime \prime \prime })& =(\dot{\gamma }(t){\displaystyle \frac{dt}{ds}},\ddot{\gamma }(t)({\displaystyle \frac{dt}{ds}}{)}^2,\stackrel{⃛}{\gamma }(t)({\displaystyle \frac{dt}{ds}}{)}^3)\\ & =({\displaystyle \frac{dt}{ds}}{)}^6({\gamma }^{\prime },{{\gamma }^{\prime }}^{\prime },{{{\gamma }^{\prime }}^{\prime }}^{\prime })\end{array}$

Por lo tanto $$({\gamma }^{\prime },{\gamma }^{\prime \prime },{\gamma }^{\prime \prime \prime })\Vert \dot{\gamma }{\Vert }^6$$

Luego $$\tau (s)={\displaystyle \frac{-({\gamma }^{\prime },{\gamma }^{\prime \prime },{\gamma }^{\prime \prime \prime })}{{\mathcal{K}}^2}}={\displaystyle \frac{{\displaystyle \frac{({\gamma }^{\prime },{\gamma }^{\prime \prime },{\gamma }^{\prime \prime \prime })}{\Vert \dot{\gamma }{\Vert }^6}}}{{\displaystyle \frac{\Vert \dot{\gamma }\times \ddot{\gamma }{\Vert }^2}{\Vert \dot{\gamma }{\Vert }^6}}}}$$

$$\therefore \tau (s)={\displaystyle \frac{-({\gamma }^{\prime },{\gamma }^{\prime \prime },{\gamma }^{\prime \prime \prime })}{\Vert \dot{\gamma }\times \ddot{\gamma }{\Vert }^2}}$$

$$

Forma canónica local

Sea $\gamma (s)=(x(s),y(s),z(s))$ podemos desarrollar cada función en series de Taylor. Luego $$\gamma (s)=\gamma (0)+s{\gamma }^{\prime }(0)+{\displaystyle \frac{s^2}{2}}{\gamma }^{\prime \prime }(0)+{\displaystyle \frac{s^3}{6}}{\gamma }^{\prime \prime \prime }(0)+\dots$$

Sin pérdida de generalidad, $\gamma (0)=\overrightarrow{0}$ entonces ${\gamma }^{\prime }(0)$

Luego $$e_1=(1,0,0)=T(0)$$ $$e_2=(0,1,0)=N(0)$$ $$e_3=(0,0,1)=B(0)$$

Entonces $(x(s),y(s),z(s))=(0,0,0)+s(1,0,0)+{\displaystyle \frac{s^2}{2}}(0,\mathcal{K},0)+{\displaystyle \frac{s^3}{6}}(-\mathcal{K},{\mathcal{K}}^{\prime },–\mathcal{K}\tau )+\dots$

Por lo que $$x(s)=s+{\displaystyle \frac{s^3}{6}}{\mathcal{K}}^2(0)+Residu{o}_x(s)$$ $$y(s)={\displaystyle \frac{s^2}{2}}\mathcal{K}(0)+{\displaystyle \frac{s^3}{6}}{\mathcal{K}}^{\prime }(0)+Residu{o}_y(s)$$ $$z(s)={\displaystyle \frac{\mathcal{K}(0)\tau (0)}{6}}{s}^3+Residu{o}_z(s)$$