Question

Achar o comprimento das seguintes curvas: a) y=e^x entre x=0 e x=1 ; b) y=cosht entre t=-1 e t=1?

Answer 1

Comprimento do arco:

$\boxed{\begin{array}{c} L=\displaystyle\int_{x_1}^{x_2}{\sqrt{1+\left(\dfrac{dy}{dx} \right )^{\!\!2}}\,dx} \end{array}}$


a) $y=e^x$

Derivando em relação a $x,$

$\dfrac{dy}{dx}=e^x~~\Rightarrow~~\left(\dfrac{dy}{dx}\right)^{\!2}=e^{2x}\\\\\\ 1+\left(\dfrac{dy}{dx}\right)^{\!2}=1+e^{2x}\\\\\\ \sqrt{1+\left(\dfrac{dy}{dx}\right)^{\!2}}=\sqrt{1+e^{2x}}$


Portanto, o comprimento de arco é dado por

$L=\displaystyle\int_0^1\sqrt{1+e^{2x}}\,dx~~~~~~\mathbf{(i)}$


Fazendo a substituição:

$t=e^x~~\Rightarrow~~\left\{ \!\begin{array}{l} x=\mathrm{\ell n}\,t\\\\ dx=\dfrac{1}{t}\,dt \end{array} \right.$


Mudando os extremos de integração:

$\text{Quando }x=0~~\Rightarrow~~t=1\\\\ \text{Quando }x=1~~\Rightarrow~~t=e$


Substituindo, a integral $\mathbf{(i)}$ fica

$=\displaystyle\int_1^e \sqrt{1+t^2}\cdot \dfrac{1}{t}\,dt\\\\\\ =\int_1^e \dfrac{\sqrt{1+t^2}}{t}\,dt~~~~~~\mathbf{(ii)}$

___________

Fazendo outra substituição. Desta vez, uma substituição trigonométrica:

$t=\mathrm{tg\,}\theta~~\Rightarrow~~\left\{ \!\begin{array}{l} dt=\sec^2\theta\,d\theta\\\\ \theta=\mathrm{arctg\,}t \end{array} \right.$


Além disso,

$\sqrt{1+t^2}\\\\ =\sqrt{1+\mathrm{tg^2\,}\theta}\\\\ =\sqrt{\sec^2\theta}\\\\ =\left|\sec \theta\right|$


Mas pelo intervalo de integração, concluímos que

$\mathrm{arctg\,1}\le \theta\le \mathrm{arctg\,}e<\dfrac{\pi}{2}\\\\\\ \dfrac{\pi}{4}\le \theta\le \mathrm{arctg\,}e<\dfrac{\pi}{2}$


Logo, $\theta$ varia apenas no primeiro quadrante, onde a sua secante é sempre $\ge 1.$ De forma que

$\left|\sec\theta\right|=\sec\theta\\\\ \therefore~~\sqrt{1+t^2}=\sec\theta$

___________

Mudando os extremos de integração, temos que

$\text{Quando }t=1~~\Rightarrow~~\theta=\dfrac{\pi}{4}\\\\\\ \text{Quando }t=e~~\Rightarrow~~\theta=\mathrm{arctg\,}e$

(ver triângulo retângulo em anexo para o extremo $\theta=\mathrm{arctg\,}e$ )


Substituindo em $\mathbf{(ii)},$ a integral fica

$=\displaystyle\int_{\pi/4}^{\mathrm{arctg\,}e}\;\dfrac{\sec \theta}{\mathrm{tg\,}\theta}\cdot \sec^2\theta\,d\theta\\\\\\ =\int_{\pi/4}^{\mathrm{arctg\,}e}\;\dfrac{\sec^3 \theta}{\mathrm{tg\,}\theta}\,d\theta\\\\\\ =\int_{\pi/4}^{\mathrm{arctg\,}e}\;\sec^3\theta\cdot \dfrac{1}{\mathrm{tg\,}\theta}\,d\theta\\\\\\ =\int_{\pi/4}^{\mathrm{arctg\,}e}\;\sec^3\theta\cdot \mathrm{cotg\,}\theta\,d\theta\\\\\\ =\int_{\pi/4}^{\mathrm{arctg\,}e}\;\dfrac{1}{\cos^3 \theta}\cdot \dfrac{\cos \theta}{\mathrm{sen\,}\theta}\,d\theta$

$=\displaystyle\int_{\pi/4}^{\mathrm{arctg\,}e}\;\dfrac{1}{\cos^2 \theta\,\mathrm{sen\,}\theta}\,d\theta~~~~~~~~~(1=\cos^2\theta+\mathrm{sen^2\,}\theta)\\\\\\ =\int_{\pi/4}^{\mathrm{arctg\,}e}\;\dfrac{\cos^2\theta+\mathrm{sen^2\,}\theta}{\cos^2 \theta\,\mathrm{sen\,}\theta}\,d\theta\\\\\\ =\int_{\pi/4}^{\mathrm{arctg\,}e}\;\dfrac{\cos^2\theta}{\cos^2\theta\,\mathrm{sen\,}\theta}\,d\theta+\int_{\pi/4}^{\mathrm{arctg\,}e}\;\dfrac{\mathrm{sen^2\,}\theta}{\cos^2 \theta\,\mathrm{sen\,}\theta}\,d\theta\\\\\\ =\int_{\pi/4}^{\mathrm{arctg\,}e}\;\dfrac{1}{\mathrm{sen\,}\theta}\,d\theta+\int_{\pi/4}^{\mathrm{arctg\,}e}\;\dfrac{\mathrm{sen\,}\theta}{\cos^2 \theta}\,d\theta\\\\\\ =\int_{\pi/4}^{\mathrm{arctg\,}e}\;\mathrm{cossec\,}\theta\,d\theta+\int_{\pi/4}^{\mathrm{arctg\,}e}\;\mathrm{tg\,}\theta\,\sec\theta\,d\theta$

$=[\mathrm{\ell n}(\mathrm{cossec\,}\theta-\mathrm{cotg\,}\theta)+\sec \theta]\,|_{\pi/4}^{\mathrm{arctg\,}e}\\\\\\ =[\mathrm{\ell n}(\mathrm{cossec}(\mathrm{arctg\,}e) -\mathrm{cotg}(\mathrm{arctg\,}e))+\sec(\mathrm{arctg\,}e)]\\\\-[\mathrm{\ell n}(\mathrm{cossec\,}\frac{\pi}{4}-\mathrm{cotg\,}\frac{\pi}{4})+\sec \frac{\pi}{4}]\\\\\\ =\left[\mathrm{\ell n}\!\left(\dfrac{\sqrt{1+e^2}}{e}-\dfrac{1}{e} \right )+\sqrt{1+e^2}\right]-[\mathrm{\ell n}(\sqrt{2}-1)+\sqrt{2}]\\\\\\ =\mathrm{\ell n}\!\left(\dfrac{\sqrt{1+e^2}-1}{e} \right )+\sqrt{1+e^2}-\mathrm{\ell n}(\sqrt{2}-1)-\sqrt{2}\\\\\\ =\mathrm{\ell n}(\sqrt{1+e^2}-1)-\mathrm{\ell n}\,e+\sqrt{1+e^2}-\mathrm{\ell n}(\sqrt{2}-1)-\sqrt{2}\\\\\\ \therefore~~\boxed{\begin{array}{c} L=\mathrm{\ell n}(\sqrt{1+e^2}-1)+\sqrt{1+e^2}-\mathrm{\ell n}(\sqrt{2}-1)-\sqrt{2}-1 \end{array}}$

(se quiser usar uma caluladora, o valor de $L$ é aproximadamente igual a $2$ unidades de comprimento )