Question

A wire filament is shaped to match the graph of the function of $y=\ln \left(2\cos x\right)$from $x=-\frac{\mathrm{\pi }}{3}$to $x=\frac{\mathrm{\pi }}{3}$, measured in inches, as shown in the figure. Given that the length of a curve y = f(x) from x = a to x = b can be calculated

with the formula${\int }_a^b\sqrt{1+{\left(f\text{'}\left(x\right)\right)}^2}dx$find the length of the filament.

Short answer

Answer is$2.634$units.

Step-by-step solution

Step 1. Given information

Function is$y=\ln \left(2\cos x\right)$

Step 2. Explanation

Length of the curve $={\int }_a^b\sqrt{1+{\left[f\text{'}\left(x\right)\right]}^2}dx$

$$\begin{gathered} ={\int }_{\frac{-\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{3}}\sqrt{1+{\left(-\tan x\right)}^2}dx \\ ={\int }_{\frac{-\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{3}}\sqrt{1+{\tan }^{2}x}dx \\ ={\int }_{\frac{-\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{3}}secxdx \\ ={\left[\ln \left(\tan x+secx\right)\right]}_{\frac{-\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{3}} \\ =\ln \left(\sqrt{3}+2\right)-\ln \left(-\sqrt{3}+2\right) \end{gathered}$$

$\approx 2.634$units