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February 21st, 2009, 04:02 PM   #1
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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
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February 21st, 2009, 06:12 PM   #2
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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
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February 22nd, 2009, 07:06 AM   #3
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Re: Length of the given curve

Pardon me, but what are i, j and k?
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February 22nd, 2009, 07:35 AM   #4
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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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