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 $\alpha (t) = \big( x(t), y(t) \big)$ y $t = h (s)$ con $s = $ longitud de arco; entonces $$\gamma (s) = \alpha ( h ( s)) = ( u (s) , v (s) )$$ $${\gamma}’ (s) = ( u’ (s) , v’ (s) )$$

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

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 $$\big|\mathcal{K}\big| = \big\| {\gamma}^{\prime \prime} (s) \big\|$$

Si $\gamma (s)$ está dada por longitud de arco, entonces $$\big\| {\gamma}’ (s) \big\| = 1 \iff \langle{\gamma}’ (s) , {\gamma}’ (s) \rangle = 1$$

Derivando $$\begin{align*} \langle{\gamma}^{\prime \prime} (s) , {\gamma}’ (s) \rangle + \langle{\gamma}’ (s) , {\gamma}^{\prime \prime} (s) \rangle &\equiv 0 \\ \langle{\gamma}’ (s) , {\gamma}^{\prime \prime} (s) \rangle &\equiv 0 \end{align*}$$

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

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

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

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

Signo de $\mathcal{K} (s) = $ signo $\Bigl| \begin{smallmatrix} u’ & {u}^{\prime \prime} \\ v’ & {v}^{\prime \prime} \end{smallmatrix} \Bigr|$

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

Luego

$\begin{align*} {u}^{\prime \prime} &= – \, \mathcal{K} v’ \\ {v}^{\prime \prime} &= \mathcal{K} u’ \end{align*}$

Por lo que

$\begin{align*} \Bigl| \begin{smallmatrix} u’ & {u}^{\prime \prime} \\ v’ & {v}^{\prime \prime} \end{smallmatrix} \Bigr| &= \Bigl| \begin{smallmatrix} u’ & – \mathcal{K}{v’} \\ v’ & \mathcal{K} {u’} \end{smallmatrix} \Bigr| \\ &= u’ u’ \mathcal{K} \, – \, v’ v’ \mathcal{K} \\ &= \mathcal{K} \big( (u’)^2 – (v’)^2 \big) \\ &= \mathcal{K}\end{align*}$$

Entonces $$\big| {{\gamma}’} \big| = \big| (u’ , v’) \big| = 1 $$

1° paso: La curvatura se puede ver como

$$ \mathcal{K} (s) = \Bigl| \begin{smallmatrix} u’ (s) & {u}^{\prime \prime} (s) \\ v’ (s) & {v}^{\prime \prime} (s) \end{smallmatrix} \Bigr|$$

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

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

Derivando

$u’ (s) = \dfrac{du}{ds} = \dfrac{dx}{dt} \dfrac{dt}{ds} = \dot{x} (t) \dfrac{dt}{ds}$

${u}^{\prime \prime} (s) = \dfrac{d}{ds} \Big( \dfrac{dx}{dt} \dfrac{dt}{ds} \Big) $

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

${u}^{\prime \prime} (s) = \dfrac{d^2x}{dt^2} \dfrac{dt}{ds} \dfrac{dt}{ds} + \dfrac{dx}{dt} \dfrac{d^2t}{ds^2} {u’}’ (s)$

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

Entonces la curvatura está dada por:

$\mathcal{K} (s) = \begin{vmatrix} u’ (s) & {u}^{\prime \prime} (s) \\ {} \\ v’ (s) & {v}^{\prime \prime} (s) \end{vmatrix}$

${}$

$\mathcal{K} (s) = \begin{vmatrix} \dfrac{du}{ds} & \dfrac{d^2u}{ds^2} \\ {}\\ \dfrac{dv}{ds} & \dfrac{d^2v}{ds^2} \end{vmatrix}$

${}$

$\mathcal{K} (s) = \begin{vmatrix} \dfrac{dx}{dt} \dfrac{dt}{ds} & {} & {} & \dfrac{d^2x}{dt^2} \Big( \dfrac{dt}{ds}\Big)^2 + \dfrac{dx}{dt} \dfrac{d^2t}{ds^2} \\ {} \\ \dfrac{dy}{dt} \dfrac{dt}{ds} & {} & {} & \dfrac{d^2y}{dt^2} \Big( \dfrac{dt}{ds}\Big)^2 + \dfrac{dy}{dt} \dfrac{d^2t}{ds^2} \end{vmatrix}$

${}$

$\mathcal{K} (s) = \begin{vmatrix} \dfrac{dx}{dt} \dfrac{dt}{ds} & {} & {} & \dfrac{d^2x}{dt^2} \Big( \dfrac{dt}{ds}\Big)^2 \\ {} \\ \dfrac{dy}{dt} \dfrac{dt}{ds} & {} & {} & \dfrac{d^2y}{dt^2} \Big( \dfrac{dt}{ds}\Big)^2 \end{vmatrix} + \begin{vmatrix} \dfrac{dx}{dt} \dfrac{dt}{ds} & {} & {} & \dfrac{dx}{dt} \dfrac{d^2t}{ds^2} \\ {} \\ \dfrac{dy}{dt} \dfrac{dt}{ds} & {} & {} & \dfrac{dy}{dt} \dfrac{d^2t}{ds^2} \end{vmatrix}$

${}$

$\mathcal{K} (s) =\Big( \dfrac{dt}{ds}\Big)^3 \begin{vmatrix} \dfrac{dx}{dt} & {} & {} & \dfrac{d^2x}{dt^2} \\ {} \\ \dfrac{dy}{dt} & {} & {} & \dfrac{d^2y}{dt^2} \end{vmatrix} + \Big(\dfrac{dt}{ds}\Big) \dfrac{d^2t}{ds^2} \begin{vmatrix} \dfrac{dx}{dt} & {} & {} & \dfrac{dx}{dt} \\ {} \\ \dfrac{dy}{dt} & {} & {} & \dfrac{dy}{dt} \end{vmatrix}$

${}$

$\mathcal{K} (s) =\Big( \dfrac{dt}{ds}\Big)^3 \begin{vmatrix} \dfrac{dx}{dt} & {} & {} & \dfrac{d^2x}{dt^2} \\ {} \\ \dfrac{dy}{dt} & {} & {} & \dfrac{d^2y}{dt^2} \end{vmatrix}$

${}$

Pero $s = \int\|{\alpha}’ \| (t) dt$ entonces $\dfrac{ds}{dt} = \|{\alpha}’ (t) \| = \sqrt{ \Big(\dfrac{dx}{dt}\Big)^2 + \Big(\dfrac{dy}{dt}\Big)^2}$

Por lo tanto

$\mathcal{K} (s) = \dfrac{ \begin{vmatrix} \dfrac{dx}{dt} & {} & {} & \dfrac{d^2x}{dt^2} \\ {} \\ \dfrac{dy}{dt} & {} & {} & \dfrac{d^2y}{dt^2} \end{vmatrix}}{\Bigg(\Big(\dfrac{dx}{dt}\Big)^2 + \Big(\dfrac{dy}{dt}\Big)^2 \Bigg)^{\frac{3}{2}}}$

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

${}$

CASO $n = 3$

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

$$T (s) = {\gamma}’ (s)$$ $$N (s) = \dfrac{{\gamma}^{\prime \prime} (s)}{\| {\gamma}^{\prime \prime} (s)\|}$$

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

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

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

${{\gamma}’} (s) \times {\gamma}^{\prime \prime} (s) = \mathcal{K} (s) \big( T (s) \times N (s) \big) = \mathcal{K} (s) B(s)$

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

${}$

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

Entonces $${\gamma}’ (s) \times { \gamma}^{\prime \prime} (s) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ {} \\ \dfrac{dx}{ds} & \dfrac{dy}{ds} & \dfrac{dz}{ds} \\ {} \\ \dfrac{d^2x}{ds^2} & \dfrac{d^2y}{ds^2} & \dfrac{d^2z}{ds^2}\end{vmatrix}$$

$${\gamma}’ (s) \times { \gamma}^{\prime \prime} (s) = \begin{vmatrix} \hat{i} & {} & \hat{j} & {} & \hat{k} \\ {} \\ \dfrac{dx}{dt} \dfrac{dt}{ds} & {} & \dfrac{dy}{dt} \dfrac{dt}{ds} & {} &\dfrac{dz}{dt} \dfrac{dt}{ds} \\ {} \\ \dfrac{d^2x}{dt^2}\Big(\dfrac{dt}{ds}\Big)^2 + \dfrac{dx}{dt}\dfrac{d^2t}{ds^2} & {} & \dfrac{d^2y}{dt^2}\Big(\dfrac{dt}{ds}\Big)^2 + \dfrac{dy}{dt}\dfrac{d^2t}{ds^2} & {} & \dfrac{d^2z}{dt^2}\Big(\dfrac{dt}{ds}\Big)^2 + \dfrac{dz}{dt}\dfrac{d^2t}{ds^2} \end{vmatrix}$$

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

${}$

Torsión

$$T’ (s) = \mathcal{K} (s) N (s)$$

$$N’ (s) = – \, \mathcal{K} (s) T (s) – \, \tau (s) B (s)$$

$${B \, }’ (s) = \tau (s) N (s)$$

Tenemos que:

${\gamma}’ = T (s)$

${\gamma}^{\prime \prime} = T’ (s) = \mathcal{K} N$

$ {\gamma}^{\prime \prime \prime} = \dfrac{d}{ds}(\mathcal{K}N = \mathcal{K}’ N + \mathcal{K} N’ = \mathcal{K}’ N + \mathcal{K}( – \, \mathcal{K} (s) T (s) – \, \tau (s) B (s)) = \mathcal{K}’ N- \, \mathcal{K}^2 (s) T (s) – \, \mathcal{K} \tau (s) B (s)$

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

$\begin{align*}({\gamma}’, {\gamma}^{\prime \prime}, {\gamma}^{\prime \prime \prime}) &= (T, \mathcal{K}N, -\, \mathcal{K}^2T + \mathcal{K}’N \, – \, \mathcal{K} \tau B \\ &= \mathcal{K} (T, N, -\, \mathcal{K}^2 T + \mathcal{K}’ N \, – \, \mathcal{K} \tau B) \\ &= \mathcal{K} \Big(T, N, -\, \mathcal{K}^2 T) + (T, N, \mathcal{K}’ N) + (T, N, -\, \mathcal{K} \tau B) \Big) \\ &= – \, \mathcal{K}^3 (T, N, T) + \mathcal{K} \mathcal{K}’ (T, N, N) \, – \, \mathcal{K}^2 \tau (T, N, B)\end{align*}$

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

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

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

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

1° paso:

${\gamma}’ = \dot{\gamma} (t) \dfrac{dt}{ds}$

${\gamma}^{\prime \prime } = \ddot{\gamma} (t) \Big( \dfrac{dt}{ds} \Big)^2 + \dot{\gamma} (t) \dfrac{d^2t}{ds^2}$

${\gamma}^{\prime \prime \prime} = \dddot{\gamma} (t) \Big(\dfrac{dt}{ds}\Big)^3 + \ddot{\gamma} (t) 2 \dfrac{dt}{ds} \dfrac{d^2t}{ds^2} + \ddot{\gamma} (t) \dfrac{dt}{ds} \dfrac{d^2t}{ds^2} + \dot{\gamma} (t) \dfrac{d^3t}{ds^3}$

Entonces ${\gamma}^{\prime \prime \prime} = \dddot{\gamma} (t) \Big(\dfrac{dt}{ds}\Big)^3 + 3 \ddot{\gamma} (t) \dfrac{dt}{ds} \dfrac{d^2t}{ds^2} + \dot{\gamma} (t) \dfrac{d^3t}{ds^3}$

Luego $$ \begin{align*} ({\gamma}’, {\gamma}^{\prime \prime}, {\gamma}^{\prime \prime \prime}) &= \Big( \dot{\gamma} (t) \dfrac{dt}{ds}, \ddot{\gamma} (t) \Big( \dfrac{dt}{ds} \Big)^2 + \dot{\gamma} (t) \dfrac{d^2t}{ds^2}, \dddot{\gamma} (t) \Big(\dfrac{dt}{ds}\Big)^3 + 3 \ddot{\gamma} (t) \dfrac{dt}{ds} \dfrac{d^2t}{ds^2} + \dot{\gamma} (t) \dfrac{d^3t}{ds^3}\Big) \\ &= \Bigg( \dot{\gamma} (t) \dfrac{dt}{ds}, \ddot{\gamma} (t) \Big( \dfrac{dt}{ds} \Big)^2 , \dddot{\gamma} (t) \Big(\dfrac{dt}{ds}\Big)^3 \Bigg) + \Bigg( \dot{\gamma} (t) \dfrac{dt}{ds}, \ddot{\gamma} (t) \Big( \dfrac{dt}{ds} \Big)^2 , 3 \ddot{\gamma} (t) \dfrac{dt}{ds} \dfrac{d^2t}{ds^2} \Bigg) + \\ & \Bigg( \dot{\gamma} (t) \dfrac{dt}{ds}, \ddot{\gamma} (t) \Big( \dfrac{dt}{ds} \Big)^2 , \dot{\gamma} (t) \dfrac{d^3t}{ds^3}\Big) \Bigg)+ \Bigg( \dot{\gamma} (t) \dfrac{dt}{ds}, \dot{\gamma} (t) \dfrac{d^2t}{ds^2}, \dddot{\gamma} (t) \Big(\dfrac{dt}{ds}\Big)^3 + 3 \ddot{\gamma} (t) \dfrac{dt}{ds} \dfrac{d^2t}{ds^2} + \dot{\gamma} (t) \dfrac{d^3t}{ds^3}\Bigg) \end{align*}$$

${}$

Entonces

$\begin{align*} ({\gamma}’, {\gamma}^{\prime \prime}, {\gamma}^{\prime \prime \prime}) &= \Bigg( \dot{\gamma} (t) \dfrac{dt}{ds}, \ddot{\gamma} (t) \Big( \dfrac{dt}{ds} \Big)^2 , \dddot{\gamma} (t) \Big(\dfrac{dt}{ds}\Big)^3 \Bigg) \\ &= \Bigg(\dfrac{dt}{ds}\Bigg)^6 ({\gamma}’, {{\gamma}’}’, {{{\gamma}’}’}’) \end{align*}$

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

Luego $$\tau (s) = \dfrac{-\, ({\gamma}’, {\gamma}^{\prime \prime}, {\gamma}^{\prime \prime \prime})}{\mathcal{K}^2} = \dfrac{\dfrac{ ({\gamma}’, {\gamma}^{\prime \prime}, {\gamma}^{\prime \prime \prime}) }{ \|\dot{\gamma}\|^6}}{\dfrac{\|\dot{\gamma} \times \ddot{\gamma} \|^2}{\|\dot{\gamma}\|^6}}$$

$$\therefore \tau (s) = \dfrac{-\, ({\gamma}’, {\gamma}^{\prime \prime}, {\gamma}^{\prime \prime \prime})}{\|\dot{\gamma} \times \ddot{\gamma} \|^2}$$

${}$

Forma canónica local

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

Sin pérdida de generalidad, $\gamma (0) = \vec{0}$ entonces ${\gamma}’ (0)$

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

Entonces $\big( x (s), y (s), z (s)\big) = (0, 0, 0) + s (1, 0, 0) + \dfrac{s^2}{2} (0,\mathcal{K}, 0) + \dfrac{s^3}{6} (-\, \mathcal{K}, \mathcal{K}’ , – \, \mathcal{K} \tau) + \dotsc$

Por lo que $$x (s) = s + \dfrac{s^3}{6} \mathcal{K}^2 (0) + Residuo_x (s)$$ $$y (s) = \dfrac{s^2}{2} \mathcal{K} (0) + \dfrac{s^3}{6} {\mathcal{K}}’ (0) + Residuo_y (s)$$ $$z (s) = \dfrac{\mathcal{K} (0) \tau (0)}{6} s^3 + Residuo_z (s)$$