Question

Poderia me ajudar nessa pergunta?

Answer 1

i) Vector posición: $\vec r(t)=(t^2,2t,\ln t)$
Vector tangente: $\vec r\,'(t)=\left((t^2)',(2t)',(\ln t)'\right)= \left(2t,2,\dfrac{1}{t}\right)$

Norma del vector tangente: 

         $\|\vec r\,'(t)\|= \left\|\left(2t,2,\dfrac{1}{t}\right)\right\| =\sqrt{(2t)^2+2^2+(1/t)^2}\\ \\ \|\vec r\,'(t)\|= \sqrt{4t^2+4+\dfrac{1}{t^2}}\\ \\ \boxed{\|\vec r\,'(t)\|=2t+\dfrac{1}{t}}$

ii) Longitud de la curva

        $\displaystyle L=\int _{1}^2\|\vec r\,'(t)\| \;dt\\ \\ L=\int _{1}^22t+\dfrac{1}{t}\; dt\\ \\ L=\left.\left(t^2+\ln t\right)\right|_{1}^2\\ \\ \\ \boxed{L=3+\ln 2}$