Question

area-calculo ajda pf

Answer 1

(a) Temos um arco de cicloide dado pelas seguintes equações paramétricas:

$\begin{array}{cc} \left\{ \begin{array}{l} x(t)=a(t-\mathrm{sen\,}t)\\ \\ y(t)=a(1-\cos t) \end{array} \right.&~~~~0\leq t\leq 2\pi. \end{array}$


Sabemos que á área sob a curva parametrizada é dada por

$A=\displaystyle\int\limits_{0}^{2\pi}{y(t)\cdot \dfrac{dx}{dt}\,dt}~~~~~~\mathbf{(i)}$


Encontrando a derivada de $x$ em relação a $t:$

$\dfrac{dx}{dt}=a(1-\cos t)=y(t)$


Substituindo em $\mathbf{(i)},$ a área sob o arco de cicloide é dada por

$A=\displaystyle\int\limits_{0}^{2\pi}{y(t)\cdot y(t)\,dt}\\ \\ \\ =\int\limits_{0}^{2\pi}{y^{2}(t)\,dt}\\ \\ \\ =\int\limits_{0}^{2\pi}{\left[a(1-\cos t) \right ]^{2}\,dt}\\ \\ \\ =\int\limits_{0}^{2\pi}{a^{2}(1-\cos t)^{2}\,dt}\\ \\ \\ =a^{2}\int\limits_{0}^{2\pi}{(1-\cos t)^{2}\,dt}$

$=a^{2}\displaystyle \int\limits_{0}^{2\pi}{(1-2 \cos t+\cos^{2}t)\,dt}\\ \\ \\ =a^{2}\int\limits_{0}^{2\pi}{\left[1-2 \cos t+\left(\dfrac{1}{2}+\dfrac{1}{2}\cos 2t \right )\right]\!dt}\\ \\ \\ =a^{2}\int\limits_{0}^{2\pi}{\left[\dfrac{3}{2}-2 \cos t+\dfrac{1}{2}\cos 2t \right]\!dt}\\ \\ \\ =a^{2}\int\limits_{0}^{2\pi}{\left(\dfrac{3}{2}-2 \cos t \right)}+a^{2}\int\limits_{0}^{2\pi}{\dfrac{1}{2}\cos 2t\,dt}\\ \\ \\ =a^{2}\cdot\left.\left(\dfrac{3t}{2}-2\,\mathrm{sen\,}t \right)\right|_{0}^{2\pi}+a^{2}\cdot \left.\left(\dfrac{1}{4}\,\mathrm{sen\,}2t \right )\right|_{0}^{2\pi}\\ \\ \\ =a^{2}\cdot \left(\dfrac{3\cdot (2\pi)}{2} \right )\\ \\ \\ =3\pi a^{2}.$

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(b) O comprimento de um arco da cicloide é dado por

$L=\displaystyle\int\limits_{0}^{2\pi}{\sqrt{\left(\dfrac{dx}{dt} \right )^{\!\!2}+\left(\dfrac{dy}{dt} \right )^{\!\!2}}\,dt}\\\\\\ =\int\limits_{0}^{2\pi}{\sqrt{[a(1-\cos t)]^{2}+(a\,\mathrm{sen\,}t)^{2}}\,dt}\\\\\\ =\int\limits_{0}^{2\pi}{\sqrt{a^{2}(1-\cos t)^{2}+a^{2}\,\mathrm{sen^{2}\,}t}\,dt}\\\\\\ =\int\limits_{0}^{2\pi}{\sqrt{a^{2}[(1-\cos t)^{2}+\mathrm{sen^{2}\,}t]}\,dt}\\\\\\ =\int\limits_{0}^{2\pi}{a\sqrt{1-2\,\cos t+\cos^{2}\,t+\mathrm{sen^{2}\,}t}\,dt}\\\\\\ =a\int\limits_{0}^{2\pi}{\sqrt{1-2\,\cos t+1}\,dt}\\\\\\ =a\int\limits_{0}^{2\pi}{\sqrt{2-2\cos t}\,dt}$

$=a\displaystyle\int\limits_{0}^{2\pi}{\sqrt{2(1-\cos t)}\,dt}\\\\\\ =a\int\limits_{0}^{2\pi}{\sqrt{4\cdot \left(\dfrac{1-\cos t}{2} \right)}\,dt}\\\\\\ =a\int\limits_{0}^{2\pi}{\sqrt{4\cdot \mathrm{sen^{2}}\!\left( \frac{t}{2} \right)}\,dt}\\\\\\ =a\int\limits_{0}^{2\pi}{2\,\left|\,\mathrm{sen}\!\left(\frac{t}{2} \right )\right|dt}$


No intervalo de integração, $\mathrm{sen}\!\left(\dfrac{t}{2} \right )\geq 0.$ Portanto, podemos dispensar o módulo, e ficamos com

$=a\displaystyle\int\limits_{0}^{2\pi}{2\,\mathrm{sen}\!\left(\frac{t}{2} \right )dt}\\\\\\ =a\cdot \left[-4\cos\!\left(\dfrac{t}{2} \right ) \right ]_{0}^{2\pi}\\\\\\ =a\cdot \left[-4\cos\!\left(\dfrac{2\pi}{2} \right )+4\cos\!\left(\dfrac{0}{2} \right ) \right ]\\\\\\ =a\cdot \left[-4\cos \pi+4\cos 0 \right ]\\\\ =a\cdot \left[-4\cdot (-1)+4\cdot 1 \right ]\\\\ =a\cdot \left[4+4 \right ]\\\\ =8a.$