Question

calcular o comprimento do arco de uma curva plana definida pelas paramétricas a seguir:

Answer 1

$\displaystyle \vec{\text r }\ (\text t)=[\ \text {x(t) , y(t)} \ ] \ ; \ \text t \in [\text a,\text b] \\\\ \underline{\text{O comprimento da curva ser{\'a} dado por}}: \\\\\\ \int\limits^\text b_\text a \sqrt{(\frac{\text{dx}}{\text {dt}})^2+(\frac{\text{dy}}{\text {dt}})^2}\text{dt}\\\\\ \underline{\text{Temos}}:\\\\ \text{x(t)}=-\text{sen(t)} \ ; \ \text {y(t) = cos(t)} \ \ ; \ \ \text t \in [0,\frac{\pi}{2}] \\\\\ \text{x'(t)}=-\text{cos(t)} \ ; \ \text{y'(t)}=-\text{sen(t)} \\\\\\ \text{Da{\'i}}:$

$\displaystyle \underline{\text{Comprimento da curva}}: \\\\ \int\limits^{\displaystyle \frac{\pi}{2}}_\text 0 \sqrt{[-\text{cos(t)}]^2+[-\text{sen(t)}]^2}\text{dt} \\\\\\ \int\limits^{\displaystyle \frac{\pi}{2}}_\text 0\sqrt{\text{cos}^2(\text t)+\text{sen}^2(\text t)}\text{dt} \to \int\limits^{\displaystyle \frac{\pi}{2}}_\text 0 1\text{dt} =[\text t]\limits^{\frac{\pi}{2}}_0 = \frac{\pi}{2} - 0 \\\\ \underline{\text{Portanto o comprimento da curva vale}}: \\\\ \huge\boxed{\frac{\pi}{2}\ }\checkmark$