Results 1 to 3 of 3

Thread: Find Arc Length

  1. #1
    Member FalconPUNCH!'s Avatar
    Joined
    Sep 2007
    From
    Daly City, CA
    Posts
    167

    Find Arc Length

    Find the Arc Length of the curve:

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

    1 < x < 2

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

    $\displaystyle 0 < x < \frac{\pi}{3} $
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Mar 2008
    Posts
    148
    For problem 2:
    $\displaystyle L = \int_0^\frac{\pi}{3} \sqrt{1+\left(\frac{d(\ln(\cos(x))}{dx}\right)^2} $

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

    Taking the square root gives:

    $\displaystyle L = \int_0^\frac{\pi}{3}\sec(x) = \ln\left(\frac{\cos(x/2)+\sin(x/2)}{\cos(x/2)-\sin(x/2)}\right)$ evaluated at the limits.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, FalconPUNCH!

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


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


    Find the Arc Length of the curve:

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

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

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

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

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



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

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

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

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


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

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Find length of arc
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 11th 2010, 09:06 PM
  2. Find Length a
    Posted in the Geometry Forum
    Replies: 2
    Last Post: Jan 5th 2009, 09:57 PM
  3. Find the length for PR
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: Sep 20th 2008, 04:28 AM
  4. Find the Length
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Oct 29th 2007, 12:18 AM
  5. find arc length
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Mar 18th 2007, 12:39 PM

Search Tags


/mathhelpforum @mathhelpforum