Question

Poderia me ajudar nessa pergunta?

Answer 1

i) el vector posición es: $\vec r(t)=(e^t\cos t,e^t\sint)$
el vector tangente: $\vec r\,'(t)=((e^t\cos t)',(e^t\sin t)')=\left(e^t(\cos t-\sin t),e^t(\sin t+\cos t)\right)$

La norma 

       $\|\vec r\,'(t)\|=\left\|\left(e^t(\cos t-\sin t),e^t(\sin t+\cos t)\right)\right\|\\ \\ \|\vec r\,'(t)\|=e^t\left\|\left(\cos t-\sin t,\sin t+\cos t\right)\right\|\\ \\ \|\vec r\,'(t)\|=e^t\sqrt{(\cos t-\sin t)^2+(\sin t+\cos t)^2}\\ \\ \boxed{\|\vec r\,'(t)\|=e^t\sqrt{2}}$

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Longitud de la curva

               $\displaystyle L=\int_{0}^1\|\vec r\,'(t)\|dt\\ \\ L=\sqrt{2}\int_{0}^1e^t dt\\ \\ \\ \boxed{L=\sqrt{2}(e-1)}$