# Thread: Calc I: Arc length

1. ## Calc I: Arc length

I am having some trouble with finding the length of the curve
ln(secx), 0≤ X ≤π/4 (the last limit is pi over 4, the pi doesnt look right)

I am just not sure how to approach taking the derivative of this problem to later incorporate into the arc length forumla:

Any help would be great, and please include a description.
Thanks

2. Hello, euchresucks!

Find the length of the curve: .$\displaystyle f(x) \:=\:\ln(\sec x),\qquad 0 \leq x \leq\frac{\pi}{4}$

We have: .$\displaystyle f(x) \:=\:\ln(\sec x)$

Then: .$\displaystyle f'(x) \;=\;\frac{1}{\sec x}(\sec x\tan x) \;=\;\tan x$

And: .$\displaystyle 1 + \left[f'(x)\right]^2 \;=\;1 + \tan^2\!x \;=\;\sec^2\!x$

Hence: .$\displaystyle \sqrt{1 + \left[f'(x)\right]^2} \;=\;\sqrt{\sec^2\!x} \;=\;\sec x$

Therefore: .$\displaystyle S \;=\;\int^{\frac{\pi}{4}}_0 \sec x\,dx\quad \hdots\;$ Got it?

3. Thanks so much, i feel like such a goon but you nailed my problem right on the head. I had it in my head that the derivative of secx was sec^2(x). I don't know why but thank you so much.