Question

Efetue os seguintes calculos:

Answer 1

$\frac{dx}{d\theta} =a*(-sen(\theta)+sen(\theta)+\theta*cos(\theta))\\\\ \boxed{\boxed{\frac{dx}{d\theta} =a*\theta*cos(\theta)}}\\\\\\ \frac{dy}{d\theta} =a*(cos(\theta)-[cos(\theta)+\theta*sen(\theta)]\\\\\boxed{\boxed{ \frac{dy}{d\theta}=a*\theta*sen(\theta) }}$

o comprimento da curva

$S= \int\limits_0^\pi \sqrt{(a*\theta*cos(\theta))^2+(a*\theta*sen(\theta)^2}\; d\theta\\\\S= \int\limits_0^\pi \sqrt{(a*\theta)^2*(cos^2(\theta)+sen(\theta)^2)}\\\\S=\sqrt{a^2} \int\limits_0^\pi {\theta} \, dx \\\\\boxed{\boxed{S=|a|* \frac{\pi^2}{2} }}$

b)
$x' =e^t*cos(t)-e^t*sen(t)\\\boxed{\boxed{x'=e^t*[cos(t)-sen(t)]}}\\\\\\y'=e^t*sen(x)+e^t*cos(t)\\\boxed{\boxed{y'=e^t*[cos(t)+sen(t)]}}$

o comprimento
$\int\limits_0^\pi \sqrt{(e^t*[cos(t)-sen(t)])^2+(e^t*[cos(t)+sen(t)])^2} \;dt\\\\ \int\limits_0^\pi e^{t}\sqrt{[cos(t)-sen(t)]^2+[cos(t)+sen(t)]^2]}\;dt\\\\ \int\limits_0^\pi e^{t}\sqrt{cos^2(t)-2cos(t)sen(t)+sen^2(t)+cos^2{t}+2cos(t)sen(t)+sen^2{t}}\\\\ \sqrt{2}\int\limits_0^\pi e^{t}\;dt = \sqrt{2}(e^\pi-1)$

c)
$x'=-sen(t)+ \frac{1}{tan(\frac{t}{2})} * \frac{sec^2\frac{t}{2}}{2}\\\\x'=-sen(t)+ \frac{cos(\frac{t}{2})}{sen(\frac{t}{2})} * \frac{1}{2cos^2(\frac{t}{2})} \\\\ x'=-sen(t)+ \frac{1}{2*sen(\frac{t}{2})*cos(\frac{t}{2})} \\\\x'= -sen(t)+ \frac{1}{2* sen( 2*\frac{t}{2} )* \frac{1}{2} } \\\\\boxed{\boxed{x'=-sen(t)+ \frac{1}{sen(t)} }}\\\\\\ \boxed{\boxed{y'=cos(t)}}$

logo:

$S= \int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} }\sqrt{[ \frac{1}{sen(t)}-sen(t) ]^2+[cos(t)]^2} \\\\ S= \int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} }\sqrt{ \frac{1}{sen^2(t)}-2 \frac{sen(t)}{sen(t)} +sen^2(t)+cos^2(t) }\\\\ S= \int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} }\sqrt{ \frac{1}{sen^2(t)}-1 }\\\\ S= \int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} }\sqrt{ \frac{1-sen^2(t)}{sen^2(t)} }\\\\ S= \int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} } \sqrt{ \frac{cos^2(t)}{sen^2(t)} }$

$S=\int_{ \frac{\pi}{4} }^{ \frac{3\pi}{4} } \left| \frac{cos(t)}{sen(t)} \right |$

por simetria 
no intervalo π/4 até π/2 , cos(t) e sen(t) são positivos
e no intervalo π/2 até 3π/4 , cos(t) é positivo e sen(t) negativo (a integral vai dar um resultado com o sinal trocado )

dividindo em duas integrais
$S=\int_{ \frac{\pi}{4} }^{ \frac{\pi}{2} } \frac{cos(x)}{sen(x)} + \int_{ \frac{\pi}{2} }^{ \frac{3\pi}{2} }| \frac{cos(x)}{sen(x)} |\\\\S=\int_{ \frac{\pi}{4} }^{ \frac{\pi}{2} } \frac{cos(x)}{sen(x)} - \int_{ \frac{\pi}{2} }^{ \frac{3\pi}{2} } \frac{cos(x)}{sen(x)} \\\\ S=\int_{ \frac{\pi}{4} }^{ \frac{\pi}{2} } \frac{du}{u} - \int_{ \frac{\pi}{2} }^{ \frac{3\pi}{2} } \frac{du}{u} \\\\ S= [ln[sen( \frac{\pi}{2})] -ln[sen( \frac{\pi}{4})] ] - [ln[sen( \frac{3\pi}{4})]-ln[sen( \frac{\pi}{2})] ]$

$S=ln(1)-ln( \frac{1}{\sqrt{2}} ) -ln( \frac{1}{\sqrt{2}} )+ln(1)\\\\S=-2ln( \frac{1}{\sqrt{2}} )\\\\S=-2ln(2^{ \frac{-1}{2} })\\\\S=ln(2)$