# Thread: Find Arc Length

1. ## Find Arc Length

Find the Arc Length of the curve:

1. $x = \frac{y^4}{8} + \frac{1}{4y^(2)}$

1 < x < 2

2. $y = ln(cos(x))$

$0 < x < \frac{\pi}{3}$

2. For problem 2:
$L = \int_0^\frac{\pi}{3} \sqrt{1+\left(\frac{d(\ln(\cos(x))}{dx}\right)^2}$

$1 + \left(\frac{d(\ln(\cos(x))}{dx}\right)^2 = 1 + \frac{\sin^2(x)}{\cos^2(x)} = 1 + \tan^2(x) = \sec^2(x).$

Taking the square root gives:

$L = \int_0^\frac{\pi}{3}\sec(x) = \ln\left(\frac{\cos(x/2)+\sin(x/2)}{\cos(x/2)-\sin(x/2)}\right)$ evaluated at the limits.

3. Hello, FalconPUNCH!

Formula: . $L \;=\;\int^b_a\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx \;\;\text{ or }\;\;\int^b_a\sqrt{1 + \left(\frac{dx}{dy}\right)^2}\,dy$

I hope there is a typo in #1 . . .

Find the Arc Length of the curve:

$1)\;\;x \:= \:\frac{y^4}{8} + \frac{1}{4y^2}\qquad 1 < {\color{red}y} < 2$
We have: . $x \;=\;\frac{1}{8}y^4 + \frac{1}{4}y^{-2}$

. . . . . . . $\frac{dx}{dy} \;=\;\frac{1}{2}y^3 - \frac{1}{2}y^{-3}$

. . . . $\left(\frac{dx}{dy}\right)^2 \;=\;\left(\frac{1}{2}y^3 - \frac{1}{2}y^{-3}\right)^2 \;=\;\frac{1}{4}y^6 - \frac{1}{2} + \frac{1}{4}y^{-6}$

. . $1 + \left(\frac{dx}{dy}\right)^2 \;=\;1 + \frac{1}{4}y^6 - \frac{1}{2} + \frac{1}{4}y^{-6} \;=\;\frac{1}{4}y^6 + \frac{1}{2}+ \frac{1}{4}y^{-6} \;=\;\left(\frac{1}{2}y^3 + \frac{1}{2}y^{-3}\right)^2$

$\sqrt{1 + \left(\frac{dx}{dy}\right)^2} \;=\;\frac{1}{2}y^3 + \frac{1}{2}y^{-3} \;=\;\frac{1}{2}\left(y^3 + y^{-3}\right)$

Therefore: . $L \;=\;\frac{1}{2}\int^2_1\left(y^3 + y^{-3}\right)\,dy\quad\hdots$ Got it?

$2)\;\;y \:= \:\ln(\cos x) \qquad 0 < x < \frac{\pi}{3}$

We have: . $\frac{dy}{dx}\;=\;\frac{-\sin x}{\cos x} \;=\;-\tan x$

. . . $1 + \left(\frac{dy}{dx}\right)^2 \;=\;1 + \tan^2\!x \;=\;\sec^2\!x$

. $\sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;=\;\sqrt{\sec^2\!x} \;=\;\sec x$

Therefore: . $L \;=\;\int^{\frac{\pi}{3}}_0 \sec x\,dx\quad\hdots$ Okay?