Boyd Rose
02/05/2024 · Senior High School
(2,0) Calcular o comprimento de arco da curva \( y=1-\ln (\operatorname{sen} x) \) do ponto onde \( x=\frac{7}{7} \) ao ponto onde \( x=\frac{\pi}{4} \)
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Para calcular o comprimento de arco da curva \( y = 1 - \ln(\sin x) \) entre os pontos \( x = 1 \) (que é \( \frac{7}{7} \)) e \( x = \frac{\pi}{4} \), utilizamos a fórmula do comprimento de arco:
\[
L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx
\]
Primeiro, precisamos encontrar a derivada \( \frac{dy}{dx} \).
A função é:
\[
y = 1 - \ln(\sin x)
\]
Calculando a derivada:
\[
\frac{dy}{dx} = -\frac{1}{\sin x} \cdot \cos x = -\cot x
\]
Agora, substituímos \( \frac{dy}{dx} \) na fórmula do comprimento de arco:
\[
L = \int_{1}^{\frac{\pi}{4}} \sqrt{1 + \left(-\cot x\right)^2} \, dx
\]
Sabemos que \( \cot^2 x = \frac{\cos^2 x}{\sin^2 x} \), então:
\[
1 + \cot^2 x = 1 + \frac{\cos^2 x}{\sin^2 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x} = \frac{1}{\sin^2 x}
\]
Portanto, temos:
\[
\sqrt{1 + \cot^2 x} = \sqrt{\frac{1}{\sin^2 x}} = \frac{1}{\sin x}
\]
Assim, a integral do comprimento de arco se torna:
\[
L = \int_{1}^{\frac{\pi}{4}} \frac{1}{\sin x} \, dx
\]
A integral de \( \frac{1}{\sin x} \) é \( -\ln(\cot x + \csc x) \). Portanto, calculamos:
\[
L = \left[-\ln(\cot x + \csc x)\right]_{1}^{\frac{\pi}{4}}
\]
Calculando os limites:
1. Para \( x = \frac{\pi}{4} \):
\[
\cot\left(\frac{\pi}{4}\right) = 1 \quad \text{e} \quad \csc\left(\frac{\pi}{4}\right) = \sqrt{2}
\]
Portanto:
\[
\cot\left(\frac{\pi}{4}\right) + \csc\left(\frac{\pi}{4}\right) = 1 + \sqrt{2}
\]
2. Para \( x = 1 \):
\[
\cot(1) = \frac{\cos(1)}{\sin(1)} \quad \text{e} \quad \csc(1) = \frac{1}{\sin(1)}
\]
Assim:
\[
\cot(1) + \csc(1) = \frac{\cos(1)}{\sin(1)} + \frac{1}{\sin(1)} = \frac{\cos(1) + 1}{\sin(1)}
\]
Portanto, a integral se torna:
\[
L = -\ln(1 + \sqrt{2}) + \ln\left(\frac{\cos(1) + 1}{\sin(1)}\right)
\]
Simplificando, temos:
\[
L = \ln\left(\frac{\cos(1) + 1}{\sin(1)(1 + \sqrt{2})}\right)
\]
Assim, o comprimento de arco da curva \( y = 1 - \ln(\sin x) \) do ponto \( x = 1 \) ao ponto \( x = \frac{\pi}{4} \) é:
\[
L = \ln\left(\frac{\cos(1) + 1}{\sin(1)(1 + \sqrt{2})}\right)
\]
Quick Answer
O comprimento de arco da curva \( y = 1 - \ln(\sin x) \) entre \( x = 1 \) e \( x = \frac{\pi}{4} \) é \( L = \ln\left(\frac{\cos(1) + 1}{\sin(1)(1 + \sqrt{2})}\right) \).
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