Instructor Hentzel Office Phone: 515-294-8141 E-mail: hentzel@iastate.edu Math Department Fax: 515-294-5454 http://www.math.iastate.edu/hentzel/class.166.ICN Textbook: Calculus Early Vectors Preliminary Edition By Stewart. Monday, April 14 11.7 Arc Length and Curvature p705: (1),2,(8),12,13,(14),17,(18),22,23 Main Idea: Remember that there are easy ways to calculate these things. But generate these formulas from scratch using the basis formulas. T' is in the direction of the normal, and A = At T + An N. Key Words: T, N, At, An, Position Vector, Velocity Vector, Acceleration Vector, Curvature, Arc Length, Goal: Find T, N, At An Previous Assignment Fri, Apr 11 11.6 Space p696: 42,(44),47,(48),57,(60) Curves Page 696 Problem 44 Find the unit tangent vector T(t) at the point with the given value of the parameter t. r(t) = (e^(2t) Cos[t], e^(2t) Sin[t], e^(2t) ) r(t) = e^(2t) ( Cos[t], Sin[t], 1) v(t) = 2 e^(2t) (Cos[t], Sin[t], 1) + e^(2t)(-Sin[t], Cos[t],0) v(t) = e^(2t) ( 2 Cos[t]-Sin[t], 2 Sin[t]+Cos[t], 2) |v(t)| = e^(2t) Sqrt[ 4 Cos[t]^2 -4 Cos[t] Sin[t] + Sin[t]^2 4 Sin[t]^2 +4 Cos[t] Sin[t] + Cos[t]^2 +4] = e^(2t) Sqrt[9] = 3 e^(2t) T = (1/3) ( 2 Cos[t]-Sin[t], 2 Sin[t]+Cos[t], 2) T(Pi/2) = (-1/3, 2/3, 2/3) r[t_] = E^(2t) { Cos[t], Sin[t], 1}; a = ParametricPlot3D[ r[t], {t,Pi/4,3 Pi/4}]; b = ParametricPlot3D[ r[Pi/2] + x{-1,2,2},{x,0,10}]; c = Show[a,b,PlotLabel->"Page 696 Problem 44"]; Display["44.ps",c]; ------------------------------------------------------------------------- Page 696 Problem 48 Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. r(t) = ( Sin[ Pi t], Sqrt[t], Cos[ Pi t] ) at( 0,1,-1) v(t) = ( Pi Cos[ Pi t], 1/2 t^(-1/2), -Pi Sin[Pi,t] ) v(t) = ( - Pi, 1/2, 0) r[t_] := { Sin[ Pi t], Sqrt[t], Cos[Pi t]}; a = ParametricPlot3D[ r[t], {t,-3,3}]; b = ParametricPlot3D[ r[1] + x{-Pi, 1/2, 0},{x,-1,1}]; c = Show[a,b,PlotLabel->"Page 696 Problem 48"]; Display["48.ps",c]; ------------------------------------------------------------ Page 696 Problem 60 t=4 INT ( Sqrt[t], t e^(-t), t^(-2) ) t=1 | | t=4 | ( t^(3/2)/3/2, -te^(-t) - e^(-t), - t^(-1) | | | t=1 ( 16/3 - 2/3, (-5)e^(-4)+2e^(-1), -1/4+1 ) ----------------------------------------------------- New Material. Arc Length is the distance measured along the curve. To find Arc Length, integrate the speed. Find the arc length of the helix r(t) from (1,0,0) to (1,0,2 Pi) r(t) = ( Cos[t], Sin[t], t ) v(t) = ( - Sin[t], Cos[t], 1) speed(t) = ||v(t)|| = Sqrt[ Sin[t]^2 + Cos[t]^2 + 1 ] = Sqrt[2]. So the particle is traveling at a constant speed along the spiral t=2 Pi path. INT speed(t) = 2 Sqrt[2] Pi t=0 r[t_] := {Cos[t], Sin[t], t}; v[t_] = D[r[t],t]; a = ParametricPlot3D[r[t],{t,-2 Pi ,2 Pi }]; H = Table[ParametricPlot3D[r[-6+i/2] + h v[-6+i/2],{h,0,1}],{i,1,20}]; Show[a,H,PlotLabel->"Helix"]; Definition. The rate that the unit tangent vector changes direction divided by the speed is called the curvature. k = | T' |/|v|. Find the curvature of the helix. r(t) = ( Cos[t], Sin[t], t ) v(t) = ( - Sin[t], Cos[t], 1) T(t) = 1/Sqrt[2] (-Sin[t], Cos[t], 1) T' = 1/Sqrt[2] ( -Cos[t], -Sin[t], 0) k = |T'|/|v| = 1/2 Find the curvature of the twisted cubic r(t) = (t, t^2, t^3) r(t) = (t, t^2, t^3) v(t) = (1, 2t, 3t^2) T = 1/Sqrt[ 1+4t^2+9t^4 ] (1,2t, 3t^2) T' = -1/2 (1+4t^2+9t^4)^(-3/2) (8t+36 t^3) (1,2t,3t^2) + 1/Sqrt[1+4t^2+9t^4](0,2,6t) T' =(1+4t^2+9t^4)^(-3/2) (-4t-18t^3) (1, 2t, 3t^2) (1+4t^2+9t^4)^(-3/2) ( 0, 2 + 8t^2 + 18 t^4, + 6t +24 t^3 + 54 t^5 ) T' =(1+4t^2+9t^4)^(-3/2) (-4t-18t^3, -8t^2 -36t^4, -12t^3 -54t^5) (1+4t^2+9t^4)^(-3/2) ( 0, 2 +8t^2 +18 t^4, +6t+24 t^3+54 t^5 ) T' =(1+4t^2+9t^4)^(-3/2) (-4t-18t^3, 2 -18t^4, 6t +12t^3 ) 2 4 4 + 36 t + 36 t |T'|^2 = ------------------ 2 4 2 (1 + 4 t + 9 t ) | 2 4 | 2 Sqtt| 1 + 9 t + 9 t | |T'| = | | --------------------------- 2 4 (1 + 4 t + 9 t ) |T'| 2 Sqrt[1+9t^2+9t^4] k = ------ = ------------------- |v| 3/2 (1+4t^2+9t^4) Easier way: V = |v| T A = |v|' T + |v| T' A = |v|' T + |v| | T' | N A = |v|' T + |v|^2 k N |v|^2 A = |v|' T + -------- N r So remember, the component of the acceleration that that is perpendicular to the line of motion is |v^2|/r. To isolate that component, use the cross product to kill off the T. |v^2| AxT = -------- NxT r |v^2| |AxT| = -------- r v(t) = (1,2t,3t^2) a(t) = (0,2,6t) | i j k | AxV = | 0 2 6t | | 1 2t 3t^2 | AxV = (-6t^2, 6t, -2) |v|^3 |AxV| = -------- = Sqrt[36t^4 + 36t^2 + 4] = 2 Sqrt[9t^4 + 9t^2 + 1] r 2 Sqrt[9t^4 + 9t^2 + 1] k = 1/r = ----------------------- 3/2 (1+4t^2+9t^4) For the helix, decompose the acceleration into tangential and normal components. r(t) = ( Cos[t], Sin[t], t ) v(t) = ( - Sin[t], Cos[t], 1) a(t) = ( -Cos[t], -Sin[t], 0) |v|^2 A = At T + ------ N r At = A.T = ( -Cos[t], -Sin[t], 0).1/Sqrt[2] ( - Sin[t], Cos[t], 1) = 1/Sqrt[2] (+Cos[t] Sin[t]-Sin[t] Cos[t] + 0) = 0. | i j k | An = |AxT| = 1/Sqrt[2] |-Cos[t] -Sin[t] 0 | |-Sin[t] Cos[t] 1 | 1/Sqrt[2] | ( -Sin[t], Cost[t], -1 ) | = Sqrt[2]/Sqrt[2] = 1 A = 0 T + 1 N |v|^2 1 = ---- and since v = Sqrt[2], r = 2 r The radius of curvature is 2 and the curvature is 1/2. In class problem. 1. Find the curvature of the circle r(t) = a(Cos[t], Sin[t],0); 2. Find the formula for the curvature of the plane curve r(t) = (x, f(x), 0).