User Name Remember Me? Password

 Calculus Calculus Math Forum

 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 = 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: 7,963 Thanks: 1148 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. Tags curve, 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 luketapis Calculus 3 March 5th, 2012 06:59 PM snopy Calculus 2 January 7th, 2011 08:22 AM snopy Calculus 5 December 13th, 2010 07:21 AM Ahplym Calculus 2 December 5th, 2010 09:50 AM Haytham Real Analysis 0 July 20th, 2009 09:01 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      