curvatureexample0

# curvatureexample0 - MA 261 ARCLENGTH AND CURVATURE TURKAY...

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MA 261 ARCLENGTH AND CURVATURE T ¨ URKAY YOLCU 1. formulas 1.1. The unit tangent and unit normal vectors. Given a curve r ( t ) , we know that r 0 ( t ) is a tangent vector at the point indicated with the vector r ( t ) . Then the normalized vector T ( t ) = r 0 ( t ) | r 0 ( t ) | is called the unit tangent vector. Since T is a unit vector, | T ( t ) | = 1 for each t , and so T 0 ( t ) is perpendicular to T ( t ) and the normalized vector N ( t ) = T 0 ( t ) | T 0 ( t ) | is called the unit normal vector. 1.2. Arclength. The length L of r ( t ) from t = a to t = b is given by L = Z b a | r 0 ( u ) | du Forexample, if r ( t ) = h f ( t ) , g ( t ) , h ( t ) i then | r 0 ( t ) | = p ( f 0 ( t )) 2 + ( g 0 ( t )) 2 + ( h 0 ( t )) 2 . 1.3. Arclength parametrization. Suppose that the length of the curve r ( t ) start- ing from t = a is denoted by s ( t ) . Then s ( t ) = Z t a | r 0 ( u ) | du. Therefore, by the fundamental theorem of calculus, we have ds dt = | r 0 ( t ) | . Moreover, if we solve s = s ( t ) for t, we get t as a function of s, say t ( s ) . Using this, we obtain another parametrization r 2 ( s ) of r ( t ) by setting r 2 ( s ) = r ( t ( s )) . Here, r 2 ( s ) is called the arclength parametrization of r ( t ) . 1

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2 T ¨ URKAY YOLCU 1.4. Observations. Here r 2 ( s ) is called the unit speed parametrization, because | r 0 2 ( s ) | = 1 for each s. Indeed, recall that t ( s ) is the function for which s ( t ( s )) = s, where s ( t ) = Z t a | r 0 ( u ) | du. Observe that r 0 2 ( s ) = d ds ( r 2 ( s )) = d ds ( r ( t ( s ))) = r 0 ( t ( s )) t 0 ( s ) . Also, using ds dt = | r 0 ( t ) | = | r 0 ( t ( s )) | we obtain that t 0 ( s ) = dt ds = 1 ds dt = 1 | r 0 ( t ( s )) | . Therefore, we are left with r 0 2 ( s ) = r 0 ( t ( s )) | r 0 ( t ( s )) | is a unit vector. Thus, | r 0 2 ( s ) | = 1. 1.5. Curvature. Now suppose that r ( s ) is a curve parametrised by the arclength. Then the curvature κ is defined to be the magnitude of the rate of change of the unit tangent vector: κ = dT ds . If the curve is not parametrised by arclength, but parametrised with t, then using dT ds = dT dt ds dt = T 0 ( t ) | r 0 ( t ) | we can write the curvature as κ = T 0 ( t ) | r 0 ( t ) | = | T 0 ( t ) | | r 0 ( t ) | .
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