Lecture 5: ARC LENGTH Math21aHOMEWORK: 10.6: 8,14,20,22PLANE CURVE~r(t) = (x(t), y(t)) position~r0(t) = (x0(t), y0(t)) velocity|~r0(t)| = (x0(t), y0(t)) speed~r00(t) = (x00(t), y00(t)) acceleration ~r000(t) =(x000(t), y000(t)) jerkSPACE CURVE~r(t) = (x(t), y(t), z(t)) position~r0(t) = (x0(t), y0(t), z0(t)) velocity|~r0(t)| = (x0(t), y0(t)) speed~r00(t) = (x00(t), y00(t), z00(t)) acceleration ~r000(t) =(x000(t), y000(t), z000(t)) jerkARC LENGTH. If t ∈ [a, b] 7→ ~r(t) with velocity ~v(t) = ~r0(t) and speed |~v(t)|, thenRba|~v(t)| dt is called the arclength of the curve. For space curves this isL =Rbapx0(t)2+ y0(t)2+ z0(t)2dtPARAMETER INDEPENDENCE. The arc length is independent of the parameterization of the curve.REASON. Changing the parameter is a change of variables (substitution) in the integration.EXAMPLE. The circle parameterized by ~r(t) = (cos(t2), sin(t2)) on t = [0,√2π] has the velocity ~r0(t) =2t(−sin(t), cos(t)) and speed 2t. The arc length isR√2π02t dt = t2|√2π0= 2π.REMARK. Often, there is no closed formula for the arc length of a curve. For example, the Lissajouxfigure ~r(t) = (cos(3t), sin(5t)) has the lengthR2π0q9 sin2(3t) + 25 cos2(5t) dt. This integral must be evaluatednumerically. If you do the Mathematica Lab, you will see how to do that with the computer.THE MATERIAL BELOW IS NOT PART OF THIS COURSE.CURVATURE.~T (t) = ~r0(t)|/|~r0(t)| unit tangent vectorκ(t) = |~T0(t)|/|~r0(t)| curvatureCURVATURE FORMULA((a, b) × (c, d) = ad − bc in 2D)κ(t) =|~r0(t)×~r00(t)||~r0(t)|3EXAMPLE. CIRCLE~r(t) = (r cos(t), r sin(t)).~r0(t) = (−r sin(t), r cos(t)).|~r0(t)| = r.~T (t) = (−sin(t), cos(t)).~r00(t) = (−r cos(t), −r sin(t)).~T0(t) = (−cos(t), −sin(t)).κ(t) = |~T0(t)|/|~r0(t)| = 1/r.EXAMPLE. HELIX~r(t) = (cos(t), sin(t), t).~r0(t) = (−sin(t), cos(t), 1).|~r0(t)| = (−sin(t), cos(t), 1) =√2.~T (t) = (−sin(t), cos(t), 1)/√2.~r00(t) = (−cos(t), −sin(t), 0).~T0(t) = (−cos(t), sin(t), 0)/√2.κ(t) = |~T0(t)|/|~r0(t)| = 1/2.INTERPRETATION.If s(t) =Rt0|~r0(t)| dt, then s0(t) = ds/dt = |~r0(t)|. Because~T0(t) = d~T /dt = d~T /ds · ds/dt, we have|d~T /ds| = |~T0(t)|/|~r0(t)| = κ(t).”The curvature is the length of the acceleration vec-tor if ~r(t) traces the curve with constant speed 1.”A large curvature at a point means that the curve is stronglybent. Unlike the acceleration or the velocity, the curvature doesnot depend on the parameterization of the curve. You ”see” thecurvature, while you ”feel” the acceleration.Smallcurvatureκ =1/r =1/2Largecurvatureκ =1/r = 2CURVATURE OF A GRAPH.The curve ~r(t) = (t, f (t)), which is the graph of a function fhas the velocity ~r0(t) = (1, f0(t)) and the unit tangent vector~T (t) = (1, f0(t))/p1 + f0(t)22and after some simplificationκ(t) = |~T0(t)|/|~r0(t)| = |f00(t)|/p1 + f0(t)23EXAMPLE. f (t) = sin(t), then κ(t) = |sin(t)|/|p1 + cos2(t)3.2 4 6 8-1-0.50.51TANGENT/NORMAL/BINORMAL.~T (t) =~r0(t)|~r0(t)|tangent vector~N(t) =~T0(t)|~T0(t)|unit normal vector~B(t) =~T (t) ×~N(t) binormal vectorBecause~T (t)·~T (t) = 1, we get after differentiation~T0(t)·~T (t) = 0and~N(t) is perpendicular to~T (t).The three vectors (~T (t),~N(t),~B(t)) are unit vectors orthogonal to each other.Note. In order that (~T (t),~N(t),~B(t)) exist, we need that ~r0(t) is not zero.WHERE IS CURVATURE NEEDED?OPTICS. If a curve ~r(t) represents a wavefront and ~n(t) is a unitvector normal to the curve at ~r(t), then ~s(t) = ~r(t) + ~n(t)/κ(t)defines a new curve called the caustic of the curve. Geometerscall that curve the evolute of the original curve.HISTORY.Aristotle: (350 BC) distinguishes between straight lines, circles and ”mixed behavior”Oresme: (14’th century): measure of twist called ”curvitas”Kepler: (15’th century): circle of curvature.Huygens: (16’th century): evolutes and involutes in connection with optics.Newton: (17’th century): circle has constant curvature inversely proportional to radius. (using infinitesimals)Simpson: (17’th century): string construction of evolutes, description using fluxions.Euler: (17’th century): first formulas of curvature using second derivatives.Gauss: (18’th century): modern description, higher dimensional versions.COMPUTING CURVATURE WITH MATHEMATICA1 2 3 4 5
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