Calculus Calculus Math Forum

 February 26th, 2015, 11:31 AM #1 Newbie   Joined: Jan 2015 From: West Virginia Posts: 11 Thanks: 0 Arc Length f(x) = (x^5/6)+(1/10x^3) I get that f'(x) = (5x^4/6)-(3/10x^4). But when I square f'(x) Add plug it into the formula I have no idea what to do with the integral. Thanks. February 26th, 2015, 11:49 AM #2 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs I take it from your differentiation that the function is: $\displaystyle f(x)=\frac{x^5}{6}+\frac{1}{10x^3}$ And so we have: $\displaystyle f'(x)=\frac{5x^4}{6}-\frac{3}{10x^4}$ Hence: $\displaystyle 1+\left(f'(x)\right)^2=1+\frac{25x^8}{36}-\frac{1}{2}+\frac{9}{100x^8}=\left(\frac{5}{6}x^4+ \frac{3}{10}x^{-4}\right)^2$ Thanks from greg1313 and joshbeldon February 26th, 2015, 12:21 PM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Hello, joshbeldon!

Arc-length problems always require a lot of Algebra.
We must be VERY careful!

Quote: February 26th, 2015, 12:58 PM #4 Newbie   Joined: Jan 2015 From: West Virginia Posts: 11 Thanks: 0 Its seems that I need to work on my reverse FOIL-ing. Thanks a lot! Tags arc, calculus 2, length Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Jet1045 Calculus 22 September 15th, 2013 10:16 AM Albert.Teng Algebra 14 January 15th, 2013 07:34 AM amin7905 Algebra 4 July 19th, 2012 02:08 PM FreaKariDunk Calculus 3 October 7th, 2011 11:15 AM future_prodegy Calculus 1 April 12th, 2009 07:24 AM

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