Question

Ache o comprimento da curva :

x=(e^t)*cos t, y=(e^t)*sen t ; Intervalo(t) = [0,pi/2]

Answer 1

O comprimento $L$ de uma curva paramétrica $\left(x\left(t \right ),\,y\left(t \right ) \right )$, no intervalo $t_{1}\leq t \leq t_{2}$ é dado por

$L=\int_{t_{1}}^{t_{2}}{\sqrt{\left(\dfrac{dx}{dt} \right )^{2}+\left(\dfrac{dy}{dt} \right )^{2}}\,dt}$


Calculando as derivadas das equações paramétricas temos

$\bullet\;\;\dfrac{dx}{dt}=\dfrac{d}{dt}\left(e^{t}\cos t \right )\\ \\ \dfrac{dx}{dt}=\dfrac{d}{dt}\left(e^{t}\right)\cdot \cos t+e^{t}\cdot \dfrac{d}{dt}\left(\cos t \right )\\ \\ \dfrac{dx}{dt}=e^{t}\cdot \cos t+e^{t}\cdot \left(-\mathrm{sen\,}t \right )\\ \\ \dfrac{dx}{dt}=e^{t}\cos t-e^{t}\mathrm{sen\,}t\\ \\ \dfrac{dx}{dt}=e^{t}\cdot \left(\cos t-\mathrm{sen\,}t \right )$


$\bullet\;\;\dfrac{dy}{dt}=\dfrac{d}{dt}\left(e^{t}\mathrm{\,sen\,} t \right )\\ \\ \dfrac{dy}{dt}=\dfrac{d}{dt}\left(e^{t}\right)\cdot \mathrm{sen\,} t+e^{t}\cdot \dfrac{d}{dt}\left(\mathrm{sen\,} t \right )\\ \\ \dfrac{dy}{dt}=e^{t}\cdot \mathrm{sen\,} t+e^{t}\cdot \cos t\\ \\ \dfrac{dy}{dt}=e^{t}\cdot \left(\mathrm{sen\,} t+\cos t \right )$


Elevando ao quadrado as derivadas acima temos

$\bullet\;\;\left(\dfrac{dx}{dt}\right)^{2}=\left[\,e^{t}\cdot \left(\cos t-\mathrm{sen\,}t \right )\, \right ]^{2}\\ \\ \left(\dfrac{dx}{dt}\right)^{2}=e^{2t}\cdot \left(\cos t-\mathrm{sen\,}t \right )^{2}\\ \\ \left(\dfrac{dx}{dt}\right)^{2}=e^{2t}\cdot \left(\cos^{2} t-2\cos t\mathrm{\,sen\,}t+\mathrm{sen}^{2\,}t \right )\\ \\ \left(\dfrac{dx}{dt}\right)^{2}=e^{2t}\cdot \left(\right.\underbrace{\cos^{2}t+\mathrm{sen}^{2\,}t}_{1}-\underbrace{2\cos t\mathrm{\,sen\,}t}_{\mathrm{sen\,}2t}\left. \right )\\ \\ \left(\dfrac{dx}{dt}\right)^{2}=e^{2t}\cdot \left(1-\mathrm{sen\,}2t \right )$


$\bullet\;\;\left(\dfrac{dy}{dt}\right)^{2}=\left[\,e^{t}\cdot \left(\mathrm{sen\,}t+\cos t \right )\, \right ]^{2}\\ \\ \left(\dfrac{dy}{dt}\right)^{2}=e^{2t}\cdot \left(\mathrm{sen\,}t+\cos t \right )^{2}\\ \\ \left(\dfrac{dy}{dt}\right)^{2}=e^{2t}\cdot \left(\mathrm{sen}^{2\,}t+2\cos t\mathrm{\,sen\,}t+\cos^{2}t \right )\\ \\ \left(\dfrac{dy}{dt}\right)^{2}=e^{2t}\cdot \left(\right.\underbrace{\mathrm{sen}^{2\,}t+\cos^{2} t}_{1}+\underbrace{2\cos t\mathrm{\,sen\,}t}_{\mathrm{sen\,}2t}\left. \right )\\ \\ \left(\dfrac{dy}{dt}\right)^{2}=e^{2t}\cdot \left(1+\mathrm{sen\,}2t \right )$


Substituindo na integral para o cálculo do comprimento do arco, temos

$L=\int_{0}^{\,^{\pi}\!\!\!\diagup\!\!_{2}}{\sqrt{e^{2t}\cdot \left(1-\mathrm{sen\,}2t \right )+e^{2t}\cdot \left(1+\mathrm{sen\,}2t \right )}\,dt}\\ \\ L=\int_{0}^{\,^{\pi}\!\!\!\diagup\!\!_{2}}{\sqrt{e^{2t}\cdot \left[\,\left(1-\mathrm{sen\,}2t \right )+\left(1+\mathrm{sen\,}2t \right )\, \right ]}\,dt}\\ \\ L=\int_{0}^{\,^{\pi}\!\!\!\diagup\!\!_{2}}{\sqrt{e^{2t}\cdot 2}\,dt}\\ \\ L=\sqrt{2}\cdot\int_{0}^{\,^{\pi}\!\!\!\diagup\!\!_{2}}{\sqrt{e^{2t}}\,dt}\\ \\ L=\sqrt{2}\cdot\int_{0}^{\,^{\pi}\!\!\!\diagup\!\!_{2}}{e^{t}\,dt}\\ \\ L=\sqrt{2}\cdot\left[\,e^{t}\, \right ]_{0}^{\,^{\pi}\!\!\!\diagup\!\!_{2}}\\ \\ L=\sqrt{2}\cdot\left[\,e^{\,^{\pi}\!\!\!\diagup\!\!_{2}}-e^{0}\, \right ]\\ \\ \boxed{L=\sqrt{2}\cdot\left[\,e^{\,^{\pi}\!\!\!\diagup\!\!_{2}}-1\, \right ]\text{ u.c.}}$