En coordenadas rectangulares la longitud de arco de una curva parametrizada la calculamos con la integral $$\underset{t_0}{\overset{t_1}{\int }}\Vert {\alpha }^{\prime }(t)\Vert dt$$

Si $\alpha (t)=(x(t),y(t))$, y ${\alpha }^{\prime }(t)=(x^{\prime }(t),y^{\prime }(t))$, entonces $$\underset{t_0}{\overset{t_1}{\int }}\sqrt{(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2}dt$$

¿Qué integral habría que calcular si la curva está en otras coordenadas?

Por ejemplo: en coordenadas polares, es decir, si conocemos $r(t)$ y $\theta (t)$

Entonces

$x(t)=r(t)\cos (\theta (t))$

$y(t)=r(t)\sin (\theta (t))$

Derivando

$x^{\prime }(t)=r^{\prime }(t)\cos (\theta (t))–r(t){\theta }^{\prime }(t)\sin (\theta (t))$

$y^{\prime }(t)=r^{\prime }(t)\sin (\theta (t))+r(t){\theta }^{\prime }(t)\cos (\theta (t))$

Luego

$\begin{array}{rl}(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2& =(r^{\prime }(t)\cos (\theta (t))–r(t){\theta }^{\prime }(t)\sin (\theta (t)){)}^2+(r^{\prime }(t)\sin (\theta (t))+r(t){\theta }^{\prime }(t)\cos (\theta (t)){)}^2\\ & ={r^{\prime }}^2{\cos }^2\theta (t)–\cancel{2r^{\prime }(t)r(t){\theta }^{\prime }(t)\cos \theta (t)\sin \theta (t)}+r^2(t){{\theta }^{\prime }}^2(t){\sin }^2\theta (t)\\ & +{r^{\prime }}^2{\cos }^2\theta (t)+\cancel{2r^{\prime }(t)r(t){\theta }^{\prime }(t)\cos \theta (t)\sin \theta (t)}+r^2(t){{\theta }^{\prime }}^2(t){\sin }^2\theta (t)\\ & =(r^{\prime }(t){)}^2+r^2(t)({\theta }^{\prime }(t){)}^2\end{array}$

Entonces $$\underset{t_0}{\overset{t_1}{\int }}\sqrt{(r^{\prime }(t){)}^2+r^2(t)({\theta }^{\prime }(t){)}^2}dt$$

$$

La «notación diferencial»

$$d{s}^2=d{x}^2+d{y}^2⟶({\displaystyle \frac{ds}{dt}}{)}^2=({\displaystyle \frac{dx}{dt}}{)}^2+({\displaystyle \frac{dy}{dt}}{)}^2$$

Entonces $$\int ds=\int {\displaystyle \frac{ds}{dt}}dt$$

En coordenadas polares

$d{s}^2=d{r}^2+r^{2}d{\theta }^2⟷({\displaystyle \frac{ds}{dt}}{)}^2=({\displaystyle \frac{dr}{dt}}{)}^2+r^2({\displaystyle \frac{d\theta }{dt}}{)}^2$

Queremos que $To\beta =\alpha$

$T(r,\theta )=(x,y)$

$x=r\cos \theta$

$y=r\sin \theta$

$x(t)=r(t)\cos \theta (t)$

$y(t)=r(t)\sin \theta (t)$

La «diferencial de T» ( o derivada de $T$)

$$DT=\left(\begin{array}{ccc}{\displaystyle \frac{\partial x}{\partial r}}& & {\displaystyle \frac{\partial x}{\partial \theta }}\\ \\ {\displaystyle \frac{\partial y}{\partial r}}& & {\displaystyle \frac{\partial y}{\partial \theta }}\end{array}\right)$$

Esta matriz es la transformación lineal que asocia vectores tangentes en el plano $r\theta$ con vectores tangentes en el plano $xy.$

Luego $DT\cdot {\beta }^{\prime }={\alpha }^{\prime }$

Entonces $\left(\begin{array}{ccc}{\displaystyle \frac{\partial x}{\partial r}}=\cos \theta & & {\displaystyle \frac{\partial x}{\partial \theta }}=–r\sin \theta \\ \\ {\displaystyle \frac{\partial y}{\partial r}}=\sin \theta & & {\displaystyle \frac{\partial y}{\partial \theta }}=r\cos \theta \end{array}\right)$

Entonces $$\left(\begin{array}{ccc}\cos \theta & & –r\sin \theta \\ \\ \sin \theta & & r\cos \theta \end{array}\right)\left(\begin{array}{c}{r}^{\prime }\\ \\ {\theta }^{\prime }\end{array}\right)=\left(\begin{array}{c}{x}^{\prime }\\ \\ y^{\prime }\end{array}\right)$$

Luego

$x^{\prime }=r^{\prime }\cos \theta –r{\theta }^{\prime }\sin \theta$

$y^{\prime }=r^{\prime }\sin \theta +r{\theta }^{\prime }\cos \theta$

Para pedir la $\Vert {\alpha }^{\prime }t\Vert$ usamos el producto punto

${\alpha }^{\prime }\cdot {\alpha }^{\prime }=\Vert {\alpha }^{\prime }{\Vert }^2$

$\sqrt{{\alpha }^{\prime }\cdot {\alpha }^{\prime }}=\Vert {\alpha }^{\prime }\Vert$

Si tenemos $T:V⟶W$ transf. lineal, y tenemos una función bilineal

$b:W\times W⟶\mathbb{R}$

podemos formar otra función bilineal B, tal que $B:V\times V\to \mathbb{R}$

$B(v_1,v_2):=b(T{v}_1,T{v}_2)$

Vamos a medir el tamaño de los vectores en el plano $(r,\theta )$ no con la norma del producto punto sino con la norma de este producto escalar

$\begin{array}{rl}B(({r^{\prime }}_1,{\theta }_1^{\prime }),({r^{\prime }}_2,{\theta }_2^{\prime }))& =b(DT({r^{\prime }}_1,{\theta }_1^{\prime }),DT({r^{\prime }}_2,{\theta }_2^{\prime }))\\ & =DT\left(\begin{array}{c}{r^{\prime }}_1\\ {\theta }_1^{\prime }\end{array}\right)\cdot DT\left(\begin{array}{c}{r^{\prime }}_2\\ {\theta }_2^{\prime }\end{array}\right)\end{array}$

$\begin{array}{r}({r^{\prime }}_1\cos \theta –r{\theta }_1^{\prime }\sin \theta ,{r^{\prime }}_1\sin \theta +r{\theta }_1^{\prime }\cos \theta )\cdot ({r^{\prime }}_2\cos \theta –r{\theta }_2^{\prime }\sin \theta ,r_2^{\prime }\sin \theta +r{\theta }_2^{\prime }\cos \theta )\\ =r_1^{\prime }{r}_2^{\prime }{\cos }^2\theta –\cancel{{r^{\prime }}_{1}r{\theta }_2^{\prime }\cos \theta \sin \theta }–r_2^{\prime }{\theta }_1^{\prime }\sin \theta \cos \theta \\ +r^2{\theta }_1^{\prime }{\theta }_2^{\prime }{\sin }^2\theta +r_1^{\prime }{r}_2^{\prime }{\cos }^2\theta –\cancel{{r^{\prime }}_{1}r{\theta }_2^{\prime }\cos \theta \sin \theta }\\ –r_2^{\prime }{\theta }_1^{\prime }\sin \theta \cos \theta +r^2{\theta }_1^{\prime }{\theta }_2^{\prime }{\sin }^2\theta \\ =r_1^{\prime }{r}_2^{\prime }+r^{\prime }{\theta }_1^{\prime }{\theta }_2^{\prime }\end{array}$

Nueva norma para los vectores tangentes $(r^{\prime },{\theta }^{\prime })$ en el plano $(r,\theta )$ $$\Vert (r^{\prime },{\theta }^{\prime })\Vert :=\sqrt{{r^{\prime }}^2+r^2{{\theta }^{\prime }}^2}$$

$$

Jacobiano $=\left|\begin{array}{ccc}\cos \theta & & \sin \theta \\ \sin \theta & & -\cos \theta \end{array}\right|=r{\cos }^2\theta +r{\sin }^2\theta =r$

En general, si tenemos un cambio de coordenadas

$x=f(u,v)$

$y=g(u,v)$

Sus derivadas son

${\displaystyle \frac{dx}{dt}}={\displaystyle \frac{\partial x}{\partial u}}{\displaystyle \frac{du}{dt}}+{\displaystyle \frac{\partial x}{\partial v}}{\displaystyle \frac{dv}{dt}}$

${\displaystyle \frac{dy}{dt}}={\displaystyle \frac{\partial y}{\partial u}}{\displaystyle \frac{du}{dt}}+{\displaystyle \frac{\partial y}{\partial v}}{\displaystyle \frac{dv}{dt}}$

Entonces

$\left(\begin{array}{c}{\displaystyle \frac{dx}{dt}}\\ \\ {\displaystyle \frac{dy}{dt}}\end{array}\right)=\left(\begin{array}{ccc}{\displaystyle \frac{\partial x}{\partial u}}& & {\displaystyle \frac{\partial x}{\partial v}}\\ \\ {\displaystyle \frac{\partial y}{\partial u}}& & {\displaystyle \frac{\partial y}{\partial v}}\end{array}\right)\left(\begin{array}{c}{\displaystyle \frac{du}{dt}}\\ \\ {\displaystyle \frac{dv}{dt}}\end{array}\right)$

Luego

$\begin{array}{rl}\underset{t_0}{\overset{t_1}{\int }}\Vert {\alpha }^{\prime }(t)\Vert dt& =\underset{t_0}{\overset{t_1}{\int }}\sqrt{({\displaystyle \frac{dx}{dt}}{)}^2+({\displaystyle \frac{dy}{dt}}{)}^2}dt\\ & =\underset{t_0}{\overset{t_1}{\int }}H({\displaystyle \frac{du}{dt}},{\displaystyle \frac{dv}{dt}})dt\end{array}$

$$

Longitud de arco de una curva en ${\mathbb{R}}^3$,

(*) en coordenadas cartesianas $$d{s}^2=d{x}^2+d{y}^2+d{z}^2$$

(*) en coordenadas cilíndricas $$d{s}^2=d{r}^2+rd{\theta }^2+d{z}^2$$

(*) en coordenadas esféricas $$d{s}^2=d{r}^2+r^2(d{\theta }^2+{\sin }^2\theta d{\phi }^2)$$