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# LectureNotes4U - Math 497C Curves and Surfaces Fall 2004...

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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, α : I R n has unit speed. Then κ = α . So zero curvature implies that α = 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 α : I R n has nonzero curvature, the principal normal vector field of α is defined as N ( t ) := T ( t ) T ( t ) , where T ( t ) := α ( t ) / α ( t ) is the tantrix of α 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 T ( t ) , T ( t ) = 1). So, if α 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

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N in terms of this frame. The definition of N already yields that, when α is parametrized by arclength, T ( t ) = κ ( 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 T, N = 0, T , N = T, N . Thus α = N ( t ) , T ( t ) = T ( t ) , N ( t ) = κ ( t ) . Exercise 3. Show that b = 0. ( Hint: Differentiate N ( t ) , N ( t ) = 1). So we conclude that N ( t ) = κ ( 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 ) = κ ( t ) 0 0 κ ( 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 α as p ( t ) := α ( t ) + 1 κ ( t ) N ( t ) . The circle which is centered at p ( t ) and has radius of 1 ( t ) is called the osculating circle of α at time t . This is the circle which best approximates α up to the second order: Exercise 4. Check that the osculating circle of α is tangent to α at α ( t ) and has the same curvature as α at time t . Now note that if α is a circle, then it coincides with its own osculating circle. In particular p ( t ) is a fixed point (the center of the circle) and α ( t ) p ( t ) is constant (the radius of the circle). Conversely: Exercise 5. Show that if α has constant curvature c , then (i) p ( t ) is a fixed point, and (ii) α ( t ) p ( t ) = 1 /c ( Hint : For part (i) differentiate p ( t ); part (ii) follows immediately from the definition of p ( t )).
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LectureNotes4U - Math 497C Curves and Surfaces Fall 2004...

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