How do you find the arc length of the curve #y=lncosx# over the interval [0, pi/3]?

Question

Aurora Alvarado · Accepted Answer

#2*arc coth(sqrt(3))# #f'(x)=-tan(x)# so we have #int_0^(pi/3)sqrt(1+tan^2(x))dx=2arc coth(sqrt(3))# Note that #1+tan^2(x)=(sin^2(x)+cos^2(x))/cos^2(x)=1/cos^2(x)#

Charles Anctil · Answer

#L=ln(2+sqrt3)# units.

#y=ln(cosx)# #y'=-tanx# Arc length is given by: #L=int_0^(pi/3)sqrt(1+tan^2x)dx# Simplify: #L=int_0^(pi/3)secxdx# Integrate directly: #L=[ln|secx+tanx|]_0^(pi/3)# Insert the limits of integration: #L=ln(2+sqrt3)#

Scarlett Allison · Answer

undefined To find the arc length of the curve $y = \ln(\cos x)$ over the interval $[0, \frac{\pi}{3}]$, you use the arc length formula for a function $y = f(x)$ over an interval $[a, b]$:

$$ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$

First, find $\frac{dy}{dx}$:

$$ \frac{dy}{dx} = \frac{d}{dx}[\ln(\cos x)] = \frac{-\sin x}{\cos x} = -\tan x $$

Now, square it:

$$ \left(\frac{dy}{dx}\right)^2 = \tan^2 x $$

Now, substitute into the arc length formula and integrate over the given interval:

$$ L = \int_{0}^{\frac{\pi}{3}} \sqrt{1 + \tan^2 x} \, dx $$

$$ = \int_{0}^{\frac{\pi}{3}} \sqrt{1 + \tan^2 x} \, dx $$

$$ = \int_{0}^{\frac{\pi}{3}} \sqrt{\sec^2 x} \, dx $$

$$ = \int_{0}^{\frac{\pi}{3}} \sec x \, dx $$

$$ = \ln|\sec x + \tan x| \bigg|_{0}^{\frac{\pi}{3}} $$

$$ = \ln|\sec(\frac{\pi}{3}) + \tan(\frac{\pi}{3})| - \ln|\sec(0) + \tan(0)| $$

$$ = \ln|\sqrt{3} + \frac{\sqrt{3}}{3}| - \ln|1 + 0| $$

$$ = \ln\left(\frac{4\sqrt{3}}{3}\right) $$

So, the arc length of the curve $y = \ln(\cos x)$ over the interval $[0, \frac{\pi}{3}]$ is $\ln\left(\frac{4\sqrt{3}}{3}\right)$.