Question

¿Como puedo hallar la longitud de línea de las siguientes funciones?

$r(t)= -3ti + 2tj + 4tk [0,3]$
$r(t)= ti + t^2 + 2tk [0,2]$
$r(t)= <8 cost, 8 sent,t>[0,pi/2]$
$r(t)= <2(sent-cost),2(cost+sent),t> [0,pi/2]$

Answer 1

Acordándote del teorema de Pitágoras.

(1) $\displaystyle L=\int\limits_{0}^{3}\sqrt{x_t^2+y_t^2+z_t^2}~dt\\ \\ \\ L=\int\limits_{0}^{3}\sqrt{(-3)^2+(2)^2+(4)^2}~dt\\ \\ \\ L=\int\limits_{0}^{3}\sqrt{9+4+16}~dt\\ \\ \\ L=\int\limits_{0}^{3}\sqrt{29}~dt\\ \\ \\ L=3\sqrt{29}$

(2)

$r(t)=\langle{8\cos t,8\sin t,t}\rangle\text{ donde}\\ \\ x(t)=8\cos t\to x_t=-8\sin t\\ \\ y(t)=8\sin t\to y_t=8\cos t\\ \\ z(t)=t\to z_t=1\\ \\ \\ \displaystyle L=\int\limits_{0}^{\pi/2}\sqrt{x_t^2+y_t^2+z_t^2}~dt\\ \\ \\ L=\int\limits_{0}^{\pi/2}\sqrt{(-8\sin t)^2+(8\cos t)^2+(t)^2}~dt\\ \\ \\ L=\int\limits_{0}^{\pi/2}\sqrt{64+t^2}~dt$

$L=\dfrac{1}{2}\left.\left[t\sqrt{64+t^2}+64\ln(t+\sqrt{64+t^2})\right]\right|_{0}^{\pi/2}\\ \\ \\ L=\dfrac{1}{2}\left[\dfrac{\pi}{2}\sqrt{ 64+\dfrac{\pi^2}{4} }+64\ln\left(\dfrac{\pi}{2}+\sqrt{ 64+\dfrac{\pi^2}{4} }\right)-64\ln8\right]\\ \\ \\ \boxed{L=\dfrac{\pi}{8}\sqrt{ 256+\pi^2}+32\ln\left(\pi+\sqrt{ 256+\pi^2 }\right)-32\ln4}$