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 February 21st, 2009, 04:02 PM #1 Member   Joined: Feb 2009 Posts: 76 Thanks: 0 Length of the given curve r(t)= 6t i + 3?2 t^2 j + 2t^3 k, o <= t <= 1 The answer is: 8 my work: r(t) = 0 i + 6?2 t j + 6t^2k |r'(t)) = ? [ 0^2 + (6?2 t j)^2 + (6t^2)^2 ] = 6?2 t + 6t^2 L = $$\int_{0}^{1}|r'(t)|dt=\int_{0}^{1} (6\sqrt{2}t+6t^2)dt=\left [6\sqrt{2}t^2+2t^3 \right ]_{0}^{1}=(6\sqrt{2}+2)-0$$ what did i do wrong? why isnt the answer 8
 February 21st, 2009, 06:12 PM #2 Member   Joined: Feb 2009 Posts: 34 Thanks: 0 Re: Length of the given curve everytime is a problem with integration but this time is simple one...almost everything go out so good luck
 February 22nd, 2009, 07:06 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,943 Thanks: 1132 Math Focus: Elementary mathematics and beyond Re: Length of the given curve Pardon me, but what are i, j and k?
 February 22nd, 2009, 07:35 AM #4 Member   Joined: Feb 2009 Posts: 34 Thanks: 0 Re: Length of the given curve here you have the parametric form of a position vector r that points to the every point of a curve so r is a position vector with components for each axis so "r" is in a form x*i+y*j+z*k to find length of the curve for each point you have to find the derivative of that position vector (witch is also a vector tangent to "r") and after that make a sum (witch become an integral for infinitesimal parts) in our case 0 to 1.

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