Question

Utilize a fórmula:
$s(x) =\int_a^x \sqrt{1+\frac{d}{dt} f(t)^2}dt$
para mostrar que o perímetro de uma circunferência de raio R é ~2πR~​

Answer 1

$\boxed{\begin{array}{l}\underline{\rm Observe~a~figura~que~eu~anexei}\\\sf a~equac_{\!\!,}\tilde ao~da~circunfer\hat encia\\\sf que~tem~centro~na~origem~e~raio~R\\\sf \acute e~x^2+y^2=R^2\\\sf y^2=R^2-x^2\\\sf y(x)=\sqrt{R^2-x^2}\\\sf y(x)=(R^2-x^2)^{\frac{1}{2}} \longrightarrow semicircunfer\hat encia~de~-R~a~R.\\\sf para~obter~o~per\acute imetro,basta~calcular~o~comprimento\\\sf de~\dfrac{1}{4}~da~circunfer\hat encia~e~multiplicar~por~4.\\\displaystyle\sf isto~\acute e,usamos~a~equac_{\!\!,}\tilde ao ~L=s(x)=\int_a^b\sqrt{1+\bigg(\dfrac{d}{dt}f(t)\bigg)^2}dt\\\sf onde~a=0~e~b=R.\end{array}}$

$\boxed{\begin{array}{l}\sf y(x)=\sqrt{R^2-x^2}^{\frac{1}{2}}\\\sf\dfrac{d}{dx} y(x)=\dfrac{1}{\diagup\!\!\!2}(R^2-x^2)^{-\frac{1}{2}}\cdot(-\diagup\!\!\!2x)\\\sf\dfrac{d}{dx}y(x)=-\dfrac{x}{(R^2-x^2)^{\frac{1}{2}}}\\\sf\bigg(\dfrac{d}{dx}y(x)\bigg)^2+1=\dfrac{x^2}{R^2-x^2}+1\\\\\sf\bigg(\dfrac{d}{dx}y(x)\bigg)^2+1=\dfrac{R^2+\diagup\!\!\!\!\!x^2-\diagup\!\!\!\!\!x^2}{R^2-x^2}=\dfrac{R^2}{R^2-x^2}\\\\\sf\sqrt{\bigg(\dfrac{d}{dx}y(x)\bigg)^2+1}=\dfrac{R}{\sqrt{R^2-x^2}}\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf L_1=\int_0^R\sqrt{\bigg(\dfrac{d}{dx}y(x)\bigg)^2+1}dx\\\\\displaystyle\sf L_1=R\int_0^R \dfrac{dx}{\sqrt{R^2-x^2}}=R\bigg[arcsen\bigg(\dfrac{x}{R}\bigg)\bigg]_0^R\\\sf L_1=R~arcsen\bigg(\dfrac{R}{R}-\dfrac{0}{R}\bigg)\\\sf L_1=R~arc sen(1)=\dfrac{\pi R}{2}\\\sf L=s(x)=4\cdot L_1\\\sf s(x)=\diagup\!\!\!4^2\cdot\dfrac{\pi R}{\diagup\!\!\!2}\\\huge\boxed{\boxed{\boxed{\boxed{\sf s(x)=2\pi R}}}} \end{array}}$