Answer
$\ln(\sqrt 2+1)$
Work Step by Step
Given: $r(t)=\cos ti+\sin tj+ln(\cos t)k$ ; $0 \leq t \leq \frac{\pi}{4}$
To calculate the length of the curve we will have to use the formula:
$L=\int_a^b |r'(t)| dt$
Thus,
$r'(t)=\lt -\sin t, \cos t, \dfrac{1}{\cos t}(-\sin t)\gt=\lt -\sin t, \cos t, -\tan t\gt$
and $|r'(t)|=\sqrt {(\sin t)^2+( \cos t)^2+(-\tan t)^2}dt$
$=\sqrt{ 1+\tan^2 t}$
$=\sec t$
$L=\int_{0}^\frac{\pi}{4} \sec tdt=ln(\sec t+ \tan t)_{0}^\frac{\pi}{4}$
$=ln(\sec(\frac{\pi}{4})+ \tan (\frac{\pi}{4}))-ln(\sec (0)+ \tan (0))$
$L=\ln(\sqrt 2+1)$
Hence, $L=\ln(\sqrt 2+1)$
