My Math Forum Arc Length

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 February 26th, 2015, 10:31 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, 10:49 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 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, 11:21 AM   #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:
 $y \;=\;\frac{1}{6}x^{^5}\,+\,\frac{1}{10x^{^{-3}}$

$y\;=\;\frac{1}{6}x^{^5}\,+\,\frac{1}{10}x^{^{-3}}$

$y' \;=\;\frac{5}{6}x^{^4}\,-\,\frac{3}{10}x^{^{-4}}$

$(y')^2 \;=\;\left(\frac{5}{6}x^{^4}\,-\,\frac{3}{10}x^{^{-4}}\right)^2 \;=\;\frac{25}{36}x^{^8}\,-\,\frac{1}{2}\,+\,\frac{9}{100}x^{^{-8}}$

$(y')^2\,+\,1 \;=\;\frac{25}{36}x^{^8}\,-\,\frac{1}{2}\,+\,\frac{9}{100}x^{^{-8}} \,+\,1\;=\;\frac{25}{36}x^{^8}\,+\,\frac{1}{2}\,+ \, \frac{9}{100}x^{^{-8}} \;=\; \left(\frac{5}{6}x^{^4}\,+\,\frac{3}{10}x^{^{-4}}\right)^2$

$\sqrt{(y')^2\,+\,1}\;=\;\sqrt{\left(\frac{5}{6}x^{ ^4}\,+\,\frac{3}{10}x^{^{-4}}\right)^2 } \;=\;\frac{5}{6}x^{^4}\,+\,\frac{3}{10}x^{^{-4}}$

$\text{Therefore: }\:L \;=\;\int^{\;\;\;b}_a\left(\frac{5}{6}x^{^4}\,+\, \frac{3}{10}x^{^{-4}}\right)\,dx$

 February 26th, 2015, 11:58 AM #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!

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