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1.5: Un Compendio de Fórmula Curva

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A continuación$$\vecs{r} (t)=\big(x(t)\,,\,y(t)\,,\,z(t)\big)$$ se presenta una parametrización de alguna curva. Los vectores$$\hat{\textbf{T}}(t),\ \hat{\textbf{N}}(t),\$$ y$$\\hat{\textbf{B}}\$$ son los vectores tangentes unitarios, normales y binormales, respectivamente, en$$\vecs{r} (t)\text{.}$$ El vector tangente apunta en la dirección de desplazamiento (es decir, dirección de incremento$$t$$) y el vector normal apunta hacia el centro de curvatura. La longitud del arco de vez$$0$$ en cuando$$t$$ se denota$$s(t)\text{.}$$ El binormal$$\ \hat{\textbf{B}}(t)=\hat{\textbf{T}} (t)\times \hat{\textbf{N}}\$$ es perpendicular al plano que mejor se ajusta a la curva en$$\vecs{r} (t)\text{.}$$ Algunas fórmulas utilizan una parametrización de longitud de arco, que se denota$$\vecs{r} (s)\text{.}$$

 la velocidad $$\displaystyle \vecs{v} (t)=\dfrac{d\vecs{r} }{dt}(t)=\dfrac{ds}{dt}(t)\,\hat{\textbf{T}}(t)$$ el vector tangente unitario $$\hat{\textbf{T}}(t)=\frac{\vecs{v} (t)}{|\vecs{v} (t)|}$$(parametrización general) $$\hat{\textbf{T}}(s)=\dfrac{d\vecs{r} }{ds}(s)$$(parametrización de longitud de arco) la aceleración $$\displaystyle \textbf{a}(t)=\frac{\mathrm{d}^{2}\vecs{r}}{\mathrm{d}t^{2}}(t)=\frac{\mathrm{d}^{2}s}{\mathrm{d}t^{2}}(t)\,\hat{\textbf{T}}(t) +\kappa(t)\big(\dfrac{ds}{dt}(t)\big)^2\hat{\textbf{N}}(t)$$ la velocidad $$\displaystyle \dfrac{ds}{dt}(t) = |\vecs{v} (t)| = \big|\dfrac{d\vecs{r} }{dt}(t)\big|$$ la longitud del arco $$\displaystyle s(T) = \int_0^T\! \dfrac{ds}{dt}(t)\,\text{d}t = \int_0^T\! \sqrt{x'(t)^2\!+\!y'(t)^2\!+\!z'(t)^2}\,\text{d}t$$ la curvatura $$\kappa(t) = \big|\dfrac{d\hat{\textbf{T}}}{dt}(t)\big|/\dfrac{ds}{dt}(t) =\displaystyle{ \frac{|\vecs{v} (t)\times\textbf{a}(t)|}{(\dfrac{ds}{dt}(t))^3} }$$ $$\kappa(s) = \big|\dfrac{d\phi}{ds}(s)\big| = \big|\dfrac{d\hat{\textbf{T}}}{ds}(s)\big|$$ el vector normal de la unidad $$\displaystyle \hat{\textbf{N}}(t) = \dfrac{d\hat{\textbf{T}}}{dt}(t)/\big|\dfrac{d\hat{\textbf{T}}}{dt}(t)\big| \qquad \hat{\textbf{N}}(s) = \dfrac{d\hat{\textbf{T}}}{ds}(s)/\kappa(s)$$ el radio de curvatura $$\displaystyle \rho(t)=\frac{1}{\kappa(t)}$$ el centro de curvatura $$\displaystyle \vecs{r} (t)+\rho(t)\hat{\textbf{N}}(t)$$ la torsión $$\displaystyle \displaystyle \tau(t)=\frac{\big(\vecs{v} (t)\times\textbf{a}(t)\big) \cdot \dfrac{d\textbf{a}}{dt}(t)} {|\vecs{v} (t)\times\textbf{a}(t)|^2}$$ el binormal $$\displaystyle \displaystyle \hat{\textbf{B}}(t)=\hat{\textbf{T}}(t)\times \hat{\textbf{N}}(t)=\frac{\vecs{v} (t)\times\textbf{a}(t)}{|\vecs{v} (t)\times\textbf{a}(t)|}$$

Bajo parametrización de longitud de arco (es decir, si$$t=s$$) tenemos$$\hat{\textbf{T}}(s)=\frac{d\vecs{r} }{ds}(s)$$ y las fórmulas de Frenet-Serret

\begin{align*} \dfrac{d\hat{\textbf{T}}}{ds}(s)&=\phantom{-}\kappa(s)\ \hat{\textbf{N}}(s)\cr \dfrac{d\hat{\textbf{N}}}{ds}(s)&=\phantom{-}\tau(s)\ \hat{\textbf{B}}(s)-\kappa(s)\ \hat{\textbf{T}} (s)\cr \dfrac{d\hat{\textbf{B}}}{ds}(s)&=-\tau(s)\ \hat{\textbf{N}}(s)\cr \end{align*}

que en forma de matriz es

\begin{align*} \dfrac{d}{ds} \left[ \begin{matrix}\hat{\textbf{T}}(s) \\ \hat{\textbf{N}}(s)\\ \hat{\textbf{B}}(s)\end{matrix} \right] =\left[\begin{matrix} 0 & \kappa(s) & 0 \\ -\kappa(s) & 0 &\tau(s) \\ 0 &-\tau(s) & 0 \end{matrix}\right] \left[\begin{matrix}\hat{\textbf{T}} (s) \\ \hat{\textbf{N}}(s)\\ \hat{\textbf{B}}(s)\end{matrix}\right] \end{align*}

Cuando la curva se encuentra completamente en el$$xy$$ plano, la curvatura viene dada por

$\begin{gather*} \kappa(t) =\frac{\big| \dfrac{dx}{dt}(t)\ \frac{\mathrm{d}^{2}y}{\mathrm{d}t^{2}}(t)-\dfrac{dy}{dt}(t)\ \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}(t) \big|}{\Big[\big(\dfrac{dx}{dt}(t)\big)^2 +\big(\dfrac{dy}{dt}(t)\big)^2\Big]^{3/2}} \end{gather*}$

Cuando la curva se encuentra completamente en el$$xy$$ plano y la$$y$$ coordenada -se da como una función,$$y(x)\text{,}$$ de la$$x$$ coordenada, la curvatura es

$\begin{gather*} \kappa(x) =\frac{\big|\frac{\mathrm{d}^{2}y}{\mathrm{d}t^{2}}(x)\big|} {\Big[1+\big(\dfrac{dy}{dx}(x)\big)^2\Big]^{3/2}} \end{gather*}$

Observe que esto se desprende de la fórmula anterior desde$$\dfrac{dx}{dx}=1$$ y$$\frac{\mathrm{d}^{2}x}{\mathrm{d}x^{2}}=0\text{.}$$

This page titled 1.5: Un Compendio de Fórmula Curva is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform.