notes001 (25)

# notes001 (25) - p s =& 3& s p 13 ± i& 2& s p...

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September 23, 2009 NAME________________________________ r ( t ) = 3 t i + (2 t 3) j . 1. Sketch the graph of this curve over the parameter interval 2 ± t ± 2 . 2. Compute r 0 ( t ) . 3. Compute the arc length, s , of this curve (measured from t = 0 ): s = Z t 0 j r 0 ( u ) j du = ? 4. Reparameterize this curve with respect to arc length measured from the point where t = 0 in the direction of increasing t . Solution: The graph of the curve over the parameter interval 2 ± t ± 2 is a line segment that begins at the point (6 ; 7) and ends at the point ( 6 ; 1) . Also r 0 ( t ) = 3 i + 2 j and j r 0 ( t ) j = q ( 3) 2 + (2) 2 = p 13 . Thus the arc length of the curve over the parameter interval [0 ; t ] is s = Z t 0 p 13 du = p 13 t: The reparameterization of the curve with respect to arc length is

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Unformatted text preview: p ( s ) = & 3 & s p 13 ± i + & 2 & s p 13 ± & 3 ± j = & 3 p 13 13 s i + 2 p 13 13 s & 3 ! j . 1 Please understand that the curves described by r and p are the same ! For example the portion of r that is traced out as t varies over the interval & 2 ± t ± 2 is the same as the portion of p that is traced out as s varies over the interval & 2 p 13 ± t ± 2 p 13 . What is &special±about the parameter s is that it corresponds to the actual distance travelled along the curve (beginning from the point (6 ; & 7) ) in travelling to the point whose position vector is p ( s ) . 2...
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notes001 (25) - p s =& 3& s p 13 ± i& 2& s p...

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