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Unformatted text preview: Math 497C Sep 17, 2004 1 Curves and Surfaces Fall 2004, PSU Lecture Notes 4 1.9 Curves of Constant Curvature Here we show that the only curves in the plane with constant curvature are lines and circles. The case of lines occurs precisely when the curvature is zero: Exercise 1. Show that the only curves with constant zero curvature in R n are straight lines. ( Hint : We may assume that our curve, : I R n has unit speed. Then = k k . So zero curvature implies that = 0. Integrating the last expression twice yields the desired result.) So it remains to consider the case where we have a planar curve whose curvature is equal to some nonzero constant c . We claim that in this case the curve has to be a circle of radius 1 /c . To this end we introduce the following definition. If a curve : I R n has nonzero curvature, the principal normal vector field of is defined as N ( t ) := T ( t ) k T ( t ) k , where T ( t ) := ( t ) / k ( t ) k is the tantrix of as we had defined earlier. Thus the principal normal is the tantrix of the tantrix. Exercise 2. Show that T ( t ) and N ( t ) are orthogonal. ( Hint : Differentiate both sides of the expression h T ( t ) , T ( t ) i = 1). So, if is a planar curve, { T ( t ) , N ( t ) } form a moving frame for R 2 , i.e., any element of R 2 may be written as a linear combination of T ( t ) and N ( t ) for any choice of t . In particular, we may express the derivatives of T and 1 Last revised: September 20, 2004 1 N in terms of this frame. The definition of N already yields that, when is parametrized by arclength, T ( t ) = ( t ) N ( t ) . To get the corresponding formula for N , first observe that N ( t ) = aT ( t ) + bN ( t ) . for some a and b . To find a note that, since h T, N i = 0, h T , N i = h T, N i . Thus = h N ( t ) , T ( t ) i = h T ( t ) , N ( t ) i = ( t ) . Exercise 3. Show that b = 0. ( Hint: Differentiate h N ( t ) , N ( t ) i = 1). So we conclude that N ( t ) = ( t ) T ( t ) , where we still assume that t is the arclength parameter. The formulas for the derivative may be expressed in the matrix notation as T ( t ) N ( t ) = ( t ) ( t ) T ( t ) N ( t ) . Now recall that our main aim here is to classify curves of constant curva ture in the plane. To this end define the center of the osculating circle of as p ( t ) := ( t ) + 1 ( t ) N ( t ) . The circle which is centered at p ( t ) and has radius of 1 / ( t ) is called the osculating circle of at time t . This is the circle which best approximates up to the second order: Exercise 4. Check that the osculating circle of is tangent to at ( t ) and has the same curvature as at time t ....
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This note was uploaded on 08/25/2011 for the course MATH 6456 taught by Professor Staff during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 Staff
 Math, Geometry

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