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.166.03 Textbook: Calculus by Varberg, Purcell, Rigdon, eight edition. April 11 12.7 Polar Graphing p548: 1-4,(8),10,(16),17,18,(22),27-30,33,(38) Main Idea: Key Words: polar coordinates, limacon, cardioid, hyperbolic spiral, spiral Archimedes, rose, lemniscate, circle Goal: Learn polar graphing. ------------------------------------------------------------------ Previous assignment: Page 539 Problem 11 Eliminate the cross-product term by a suitable rotation of axes and then, if necessary, translate axes (complete the squares) to put the equation in standard form. Finally, graph the equation showing the rotated axes. 2 2 5 x - 3 x y + y + 65 x - 25 y + 203 = 0 A-C 5-1 Ctn[2t] = --- = ----- = -4/3 B -3 |\ | \ 5 3| \ | | \ | | 2t\| -----------+-------------- -4 | | Sin[2t] = 3/5 Cos[2t] = -4/5 2 1-Cos[2t] 1+4/5 Sin[t] = ----------- = -------- = 9/10 2 2 2 1+Cos[2t] 1-4/5 Cos[t] = ------------ = -------- = 1/10 2 2 Sin[t] = 3/Sqrt[10] Cos[t] = 1/Sqrt[10] x = u Cos[t] - v Sin[t] = u/Sqrt[10] - 3 v/Sqrt[10] y = u Sin[t] + v Cos[t] = 3u/Sqrt[10] + v/Sqrt[10] 5 x^2 - 3 x y + y^2 + 65 x - 25 y + 203 = 0 5 (u/Sqrt[10]-3 v/Sqrt[10])^2 - 3 (u/Sqrt[10] - 3 v/Sqrt[10])(3u/Sqrt[10] + v/Sqrt[10]) + (3u/Sqrt[10] + v/Sqrt[10])^2 + 65 (u/Sqrt[10] - 3 v/Sqrt[10]) - 25(3u/Sqrt[10] + v/Sqrt[10]) + 203 --------------------------------------------------------- 2 2 u 11 v 203 - Sqrt[10] u + -- - 22 Sqrt[10] v + ----- 2 2 2 2 u 11 v 203 - Sqrt[10] u + -- - 22 Sqrt[10] v + ----- = 0 2 2 u^2 11 v^2 --- - Sqrt[10] u + ------ -22 Sqrt[10] v = -203 2 2 u^2 - 2 Sqrt[10] u + 11 v^2 - 44 Sqrt[10 v = -406 (u^2 - 2 Sqrt[10] u ) + 11 (v^2 - 4 Sqrt[10 v ) = -406 u^2 - 2 Sqrt[10] u + 10 + 11 (v^2 - 4 Sqrt[10] v + 40 ) = -406 + 10 + 440 (u-Sqrt[10)^2 + 11(v-2 Sqrt[10])^2 = 44 2 2 (u-Sqrt[10) (v-2 Sqrt[10]) ------------ + ---------------- = 1 44 4 x = u/Sqrt[10]-3 v/Sqrt[10]; y = 3 u/Sqrt[10] + v/Sqrt[10]; when x= 0 we have u/Sqrt[10] - 3 v/Sqrt[10 = 0 so u = 3 v; (y Axis) when y = 0 we have 3 u/Sqrt[10] + v/Sqrt[10] = 0 so 3u + v = 0. (x Axis) ----------------------------------------------------- f[u_] := Sqrt[4( 1- u^2/44)]; g[u_] := -Sqrt[4( 1- u^2/44)]; x = 1/Sqrt[10]-3 v/Sqrt[10]; y = 3 u/Sqrt[10] + v/Sqrt[10]; a = Sqrt[44]; b = 2; c = Sqrt[40]; h = b^2/a; offset = {Sqrt[10],2 Sqrt[10]}; p1 = ParametricPlot[ offset+ {u, f[u]} ,{u,-a,a}]; p2 = ParametricPlot[ offset+ {u, g[u]} ,{u,-a,a}]; p3 = ListPlot[{offset+{ 0, 0}, offset+{ a, 0}, offset+{-a, 0}, offset+{ 0, b}, offset+{ 0,-b}, offset+{ c, 0}, offset+{ c, h}, offset+{ c,-h}, offset+{-c, h}, offset+{-c,-h}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.02]}]; p4 = ParametricPlot[{u,-3u},{u,-10,10},PlotStyle->{RGBColor[0,1,0]}]; p5 = ParametricPlot[{u,1/3u},{u,-10,10},PlotStyle->{RGBColor[0,0,1]}]; p6 = Show[p1,p2,p3,p4,p5,AspectRatio->Automatic,PlotLabel->"green = X, blue = Y", AspectRatio->Automatic]; Display["11.ps",p6]; ------------------------------------------------------------- Page 544 Problem 30 Name the curve with the given polar equation. If it is a conic, give its eccentricity. Sketch the graph. 4 r = ----------------- 1 + 2 Sin[t] Method 1. 4 Sqrt[x^2+y^2] = ---------------------- 1 + 2 y/Sqrt[x^2+y^2] Sqrt[x^2+y^2] + 2 y = 4 Sqrt[x^2 + y^2] = 4 - 2y x^2 + y^2 = 16 - 16 y + 4 y^2 x^2 - 3 y^2 + 16 y = 16 x^2 - 3 (y^2 - 16/3 y ) = 16 x^2 - 3 (y^2 - 16/3 y +64/9 ) = 16 - 64/3 x^2 - 3 (y^2 - 16/3 y +64/9 ) = -16/3 -x^2 + 3 (y^2 - 16/3 y +64/9 ) = 16/3 - x^2 (y-8/3)^2 --------- + ---------- = 1 16/3 16/9 Center (0,8/3) a = 4/3 b = 4/Sqrt[3] c = 8/3 ----------------------------- center = {0,8/3}; a = 4/3; b = 4/Sqrt[3]; c = 8/3; h = b^2/a; p1 = ListPlot[{ center+{ 0, a}, center+{ 0,-a}, center+{ 0, c}, center+{ h, c}, center+{-h, c}, center+{ 0,-c}, center+{-h,-c}, center+{ h,-c}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.02]}]; p2 = ParametricPlot[center+{t, 4/3} ,{t,-b,b}]; p3 = ParametricPlot[center+{t,-4/3} ,{t,-b,b}]; p4 = ParametricPlot[center+{ 4/Sqrt[3],t} ,{t,-a,a}]; p5 = ParametricPlot[center+{-4/Sqrt[3],t} ,{t,-a,a}]; p6 = ParametricPlot[center+{ t,a/b t},{t,-c,c}]; p7 = ParametricPlot[center+{-t,a/b t},{t,-c,c}]; r[t_] := 4/(1+2 Sin[t]); p8 = ParametricPlot[r[t]{Cos[t],Sin[t]},{t,0,2 Pi}]; p9 = Show[p1,p2,p3,p4,p5,p6,p7,p8,PlotLabel->"P544 P30 r = 4/(1+2 Sin[t]"]; Display[ "30.ps",p9]; -------------------------------------------------------------- Method 2 de r = -------- 1+e Sin[t] is a conic, e is the eccentricity, and d is the distance from the origin to the directrix The origin is the focus. | | d+rCos[t] d------------/ i | / r | r/ e | / \ c d |/ t | ---------------t-------F----------------------- r | i | x | | | | | | distence to focus e = -------------------- distance to line r e = ------------------ d+r Cos[t] e (d+r Cos[t]) = r de r = --------------------- 1 - e Cos[t] ------------------------------------------- d = 1; r[e_,t_] := d e/(1-e Cos[t]); p1 = ParametricPlot[ r[1/2,t] {Cos[t],Sin[t]},{t,0,2 Pi},PlotStyle->{RGBColor[1,0,0]}]; p2 = ParametricPlot[ r[1 ,t] {Cos[t],Sin[t]},{t,0,2 Pi},PlotStyle->{RGBColor[0,1,0]}]; p3 = ParametricPlot[ r[3/2,t] {Cos[t],Sin[t]},{t,0,2 Pi},PlotStyle->{RGBColor[0,0,1]}]; p4 = Show[p1,p2,p3,PlotLabel->"P30 P9 r=de/(1-e Cos[t]) red:e=1/2,green:e=1,blue:e=3/2"]; Display["30x.ps",p4]; ---------------------------------------------------------------------------- d e Sqrt[x^2 + y^2] = ---------------------- 1-e x/Sqrt[x^2+y^2] Sqrt[x^2 + y^2] - e x = d e x^2 + y^2 = e^2 (d+x)^2 = e^2 (x^2 + 2 dx + d^2) (1-e^2) x^2 - 2 e^2 d x + y^2 = e^2 d^2 | 2 e^2 d x e^4 d^2) | e^4 d^2 (1-e^2) | x^2 - --------- + ----------- | + y^2 = e^2 d^2 + -------- | 1-e^2 (1-e^2)^2 | 1-e^2 2 | e^2 d| 2 | e^2 | (1-e^2) | x - ------| + y = e^2 d^2 | 1 + ----- | | 1-e^2 | | 1-e^2 | 2 | e^2 d| 2 | 1 | (1-e^2) | x - ------| + y = e^2 d^2 | -------- | | 1-e^2 | | 1-e^2 | 2 | e^2 d| | x - ------| | 1-e^2 | y^2 ------------------ + ---------- = 1 e^2 d^2 e^2 d^2 ------------ ------------ (1-e^2)^2 1-e^2 If e < 1 the curve is an ellipse 2 e^2 d^2 a = ---------- (1-e^2)^2 2 e^2 d^2 b = -------------- 1-e^2 2 e^2 d^2 c = a^2-b^2 = ------------ ( 1 - 1+e^2) = e^2 a^2 (1-e^2)^2 So e = c/a as expected. a^2 d^2 If e > 1 then a^2 = --------- (1-e^2)^2 e^2 d^2 b^2 = -------------- e^2 - 1 a^2 d^2 c^2 = a^2 + b^2 = --------- ( 1 - 1+e^2) = e^2 a^2 (1-e)^2 So again e = c/a as expected. If e = 1 then (1-e^2) x^2 - 2 e^2 d x + y^2 = e^2 d^2 So -2 d x + y^2 = d^2 y^2 = (2 d x + d^2 ) = 2d( x + d/2) Then you get a parabola with p = d/2 which is he distance from the center to the directrix. ----------------------------------------------------------- Page 544 Problem 38 Prove that r = a Sin[t] + b Cos[t] represents a circle and find its center and radius. a x b y Sqrt[x^2 + y^2 ] = -------------- + ------------ Sqrt[x^2 + y^2] Sqrt[x^2 + y^2] x^2 + y^2 = ax + by x^2 - ax + y^2 - by = 0 x^2 - ax +(a/2)^2 + y^2 - by + (b^2/^2) = (a/2)^2 + (b/2)^2 ( x-a/2)^2 + (y-b/2)^2 = 1/2 Sqrt[a^2+b^2]. It is a circle with center (a/2,b/2) and radius 1/2 Sqrt(a^2 + b^2). ---------------------------------------------------- a=4; b=5; f[t_,a_,b_] := a Sin[t] + b Cos[t]; p1 = ParametricPlot[f[t,a,b] {Cos[t],Sin[t]},{t,0,2 Pi}]; p2 = Show[p1,PlotLabel->"P 544 P38 r = a Cos[t] + b Sin[t] ",AspectRatio->Automatic]; Display["38.ps",p2]; -------------------------------------------------- ================================================================ New Material Somewhat: The basic curve is the circle r = Cos[theta] and r = Sin[theta]. Remember that it traces the curve twice, once with positive r and once with negative r. a. Graph: r = Cos[theta] b. Graph: r = Cos[2 theta] c. Graph: r = Cos[3 theta] 2 d. Graph: r = Cos[theta] e. Graph: r = 1 + Cos[theta] f. Graph: r = 1/2 + Cos[theta] g. Graph: r = 2 + Cos[theta] --------------------------------------------