Question

qual o comprimento do arco y=cos^3 (t), x=sen^3 (t) entre t=0 e t=pi/4

Answer 1

Temos uma curva parametrizada:

$\gamma:~\left\{ \begin{array}{l} x(t)=\mathrm{sen^3\,}t\\\\ y(t)=\cos^3 t \end{array} \right.~~~~~~~~~~0\le t\le \dfrac{\pi}{4}$

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Encontrando o vetor tangente à curva:

$\gamma'(t)=(x'(t),\;y'(t))\\\\ \gamma'(t)=((\mathrm{sen^3\,}t)',\;(\cos^3 t)')\\\\ \gamma'(t)=(3\,\mathrm{sen^2\,}t\cdot (\mathrm{sen\,}t)',\;3\cos^2 t\cdot (\cos t)')\\\\ \gamma'(t)=(3\,\mathrm{sen^2\,}t\cdot \cos t,\;3\cos^2 t\cdot (-\mathrm{sen\,}t))\\\\ \boxed{\begin{array}{c}\gamma'(t)=(3\,\mathrm{sen^2\,}t\cos t,\;-3\cos^2 t\,\mathrm{sen\,}t) \end{array}}$

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Para o cálculo do comprimento do arco, precisamos apenas da norma (módulo) do vetor tangente:

$\|\gamma'(t)\|=\|(3\,\mathrm{sen^2\,}t\cos t,\;-3\cos^2 t\,\mathrm{sen\,}t)\|\\\\\ =\sqrt{(3\,\mathrm{sen^2\,}t\cos t)^2+(-3\cos^2 t\,\mathrm{sen\,}t)^2}\\\\ =\sqrt{9\,\mathrm{sen^4\,}t\cos^2 t+9\cos^4 t\,\mathrm{sen^2\,}t}~~~~~~~~~~(\mathrm{sen^2\,}t=1-\cos^2 t)\\\\ =\sqrt{9\,\mathrm{sen^4\,}t\cos^2 t+9\cos^4 t\,(1-\cos^2 t)}\\\\ =\sqrt{9\,\mathrm{sen^4\,}t\cos^2 t+9\cos^4 t-9\cos^4 t\cos^2 t}\\\\ =\sqrt{9\,\mathrm{sen^4\,}t\cos^2 t-9\cos^4 t\cos^2 t+9\cos^4 t}\\\\ =\sqrt{9\cos^2 t\,(\mathrm{sen^4\,}t-\cos^4 t)+9\cos^4 t}\\\\ =\sqrt{9\cos^2 t\,((\mathrm{sen^2\,}t)^2-(\cos^2 t)^2)+9\cos^4 t}$


Usando produtos notáveis (diferença entre dois quadrados)

$=\sqrt{9\cos^2 t\,(\mathrm{sen^2\,}t+\cos^2 t)\,(\mathrm{sen^2\,}t-\cos^2 t)+9\cos^4 t}\\\\ =\sqrt{9\cos^2 t\cdot 1\cdot \,(\mathrm{sen^2\,}t-\cos^2 t)+9\cos^4 t}\\\\ =\sqrt{9\cos^2 t\,\mathrm{sen^2\,}t-9\cos^2 t\cos^2 t+9\cos^4 t}\\\\ =\sqrt{9\,(\cos t\,\mathrm{sen\,}t)^2-9\cos^4 t+9\cos^4 t}\\\\ =\sqrt{9\,(\cos t\,\mathrm{sen\,}t)^2}\\\\ =3\sqrt{(\cos t\,\mathrm{sen\,}t)^2}\\\\ =3\sqrt{\left(\dfrac{1}{2}\cdot 2\cos t\,\mathrm{sen\,}t\right)^{\!2}}\\\\ =3\sqrt{\left(\dfrac{1}{2}\,\mathrm{sen\,}2t\right)^{\!2}}\\\\\\ =3\sqrt{\left(\dfrac{1}{2}\right)^2\mathrm{sen^2\,}2t}$

$=\dfrac{3}{2}\sqrt{\mathrm{sen^2\,}2t}\\\\\\ =\dfrac{3}{2}\left|\mathrm{sen\,}2t\right|\\\\\\ \therefore~~\boxed{\begin{array}{c} \|\gamma'(t)\|=\dfrac{3}{2}\left|\mathrm{sen\,}2t\right| \end{array}}$

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O comprimento do arco é dado pela seguinte integral (de linha)

$L=\displaystyle\int_\gamma d\ell=\int_{t_{1}}^{t_{2}}\|\gamma'(t)\|\,dt\\\\\\ =\int_0^{\pi/4}\dfrac{3}{2}\left|\mathrm{sen\,}2t\right|dt~~~~~~\mathbf{(i)}$


Se $0\le t\le \frac{\pi}{4}\,,$ então

$0\le 2t\le \dfrac{\pi}{2}~~\Rightarrow~~\mathrm{sen\,}2t\ge 0$


Como $\mathrm{sen\,}2t$ nunca é negativo no intervalo de integração, podemos dispensar o módulo e a integral $\mathbf{(i)}$ fica

$=\displaystyle\int_0^{\pi/4}\dfrac{3}{2}\,\mathrm{sen\,}2t\,dt\\\\\\ =\int_0^{\pi/4}\dfrac{3}{4}\cdot 2\,\mathrm{sen\,}2t\,dt\\\\\\ =\dfrac{3}{4}\int_0^{\pi/4}\mathrm{sen\,}2t\cdot 2\,dt\\\\\\ =\dfrac{3}{4}\int_0^{\pi/2}\mathrm{sen\,}u\,du~~~~~~~~~(u=2t)\\\\\\ =\dfrac{3}{4}\cdot (-\cos u)|_0^{\pi/2}\\\\\\ =\dfrac{3}{4}\cdot \left[-\cos \dfrac{\pi}{2}-(-\cos 0) \right ]\\\\\\ =\dfrac{3}{4}\cdot \left[0+1 \right ]\\\\\\ \therefore~~\boxed{\begin{array}{c} L=\dfrac{3}{4}\mathrm{~u.c.} \end{array}}$