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: $\displaystyle 1+\left(f'(x)\right)^2=1+\frac{25x^8}{36}\frac{1}{2}+\frac{9}{100x^8}=\left(\frac{5}{6}x^4+ \frac{3}{10}x^{4}\right)^2$ 
February 26th, 2015, 11:21 AM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, joshbeldon! Arclength problems always require a lot of Algebra. We must be VERY careful! Quote:  
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 FOILing. Thanks a lot!


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arc, calculus 2, length 
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