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.166.05 Textbook: Calculus by Varberg, Purcell, Rigdon, eight edition. April 13 12.8 Polar Calculus p554: 1,(4),6,(10),(16),15-17,(20),25 Main Idea: Finding areas and tangent lines to polar curves 2 Key Words: 1/2 r dt Goal: Learn how to find areas and tangent lines to polar curves. ------------------------------------------------------------------ Previous assignment: Page 548 Problem 8 4 r = ---------- 1+Sin[t] This is a parabola. It does not cross the line theta = 3 Pi/2. So it extends outwards along the negative y-axis. e = 1; d = 4 so p = 2. The focus is at (0,0), the vertex is at (0,2). The same curve in rectangular coordinates should be x^2 = -8 (y-2); ------------------------------------------- f[t_] := 4/(1+Sin[t]); p1 = ParametricPlot[ f[t]{Cos[t],Sin[t]},{t,-Pi/2+1,3 Pi/2-1},PlotStyle->{Thickness[0.01]}]; p2 = Plot[-x^2/8 + 2,{x,-5,5},PlotStyle->{RGBColor[1,0,0],Thickness[0.02]}]; p3 = Show[p2,p1,PlotLabel->"P548 P8 black:r = 4/(1+Sin[t]);red y^2 = -8(x-2)", AspectRatio->Automatic,PlotRange->All]; Display["8.ps",p3]; --------------------------------------------------- Page 548 Problem 16 r = 5 -3 Cos[t] -------------------------------------------------------------------------------- f[t_] := 5 -3 Cos[t]; p1 = ParametricPlot[ f[t]{Cos[t],Sin[t]},{t,0,2 Pi},PlotStyle->{Thickness[0.01]}]; p2 = ParametricPlot[-3 Cos[t] {Cos[t],Sin[t]},{t,0,2 Pi}]; p3 = ParametricPlot[ 5{Cos[t],Sin[t]},{t,0,2 Pi}]; p4 = Show[p1,p2,p3,PlotLabel->"P548 P16 r = 5 - 3 Cos[t] and r = -3 Cos[t]", AspectRatio->Automatic,PlotRange->All]; Display["16.ps",p4]; -------------------------------------------------------------------------------- Page 548 Problem 22 r = 3 Sin[3 t] -------------------------------------------------------------------------------- f[t_] := 3 Sin[3t]; p1 = ParametricPlot[ f[t]{Cos[t],Sin[t]},{t,0,2 Pi},PlotStyle->{Thickness[0.01]}]; p2 = ParametricPlot[ 3{Cos[t],Sin[t]},{t,0,2 Pi}]; p3 = Show[p1,p2,PlotLabel->"P548 P22 r = 3 Cos[3t] and r = 3 Cos[t]", AspectRatio->Automatic,PlotRange->All]; Display["22.ps",p3]; -------------------------------------------------------------------------------- Page 548 Problem 38 Sketch the given curves and find their points of intersection. 2 r = 4 Cos[2 t] r = 2 Sqrt[2] Sin[t] They intersect at the origin with r = 0 and no matter what t is. 2 4 Cos[2t] = 8 Sin [t] 2 Cos[2t] = 2 Sin[t] 2 2 1 - 2 Sin[t] = 2 Sin[t] 2 1 = 4 Sin[t] 2 1/4 = Sin[t] Sin[t] = +/- 1/2 t = Pi/6, - Pi/6, 5 Pi/6, 7 Pi/6; (r,t) for 4 = 2 Sqrt[2] Sin[t] f[t_] := 2 Sqrt[2] Sin[t]; g[t_] := 2 Sqrt[Cos[2 t]]; f[{Pi/6,-Pi/6,5 Pi/6, 7 Pi/6}] = {Sqrt[2], -Sqrt[2], Sqrt[2], -Sqrt[2]} g[{Pi/6,-Pi/6,5 Pi/6, 7 Pi/6}] = {+/-Sqrt[2], +/-Sqrt[2],+/- Sqrt[2], +/-Sqrt[2]} +only -only +only -only Besides zero, there are two points of intersection. (Pi/6, Sqrt[2]) = (-Pi/6,-Sqrt[2]) (5 Pi/6, Sqrt[2]) = (7Pi/6, -Sqrt[2]) -------------------------------------------------------------- f[t_] := 2 Sqrt[Cos[2 t]]; p1 = ParametricPlot[ 2 Sqrt[ Cos[2t]] {Cos[t],Sin[t]},{t,0,2 Pi}]; p2 = ParametricPlot[-2 Sqrt[ Cos[2t]] {Cos[t],Sin[t]},{t,0,2 Pi}]; p3 = ParametricPlot[2 Sqrt[2] Sin[t]{Cos[t],Sin[t]},{t,0,2 Pi}]; p4 = ListPlot[{ 2 Sqrt[2] Sin[ Pi/6] {Cos[ Pi/6],Sin[ Pi/6]}, 2 Sqrt[2] Sin[5 Pi/6] {Cos[5 Pi/6],Sin[5 Pi/6]}, 2 Sqrt[2] Sin[7 Pi/6] {Cos[7 Pi/6],Sin[7 Pi/6]}, 2 Sqrt[2] Sin[- Pi/6] {Cos[- Pi/6],Sin[- Pi/6]}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.015]}]; p5 = ListPlot[{ f[ Pi/6] {Cos[ Pi/6 ],Sin[ Pi/6]}, f[5 Pi/6] {Cos[5 Pi/6 ],Sin[5 Pi/6]}}, PlotStyle->{RGBColor[0,1,0],PointSize[0.02]}]; p6 = ListPlot[{ -f[7 Pi/6] {Cos[7 Pi/6],Sin[7 Pi/6]}, -f[- Pi/6] {Cos[- Pi/6],Sin[- Pi/6]}}, PlotStyle->{RGBColor[0,0,1],PointSize[0.03]}]; p7 = Show[p1,p2,p3,p6,p5,p4,PlotLabel->"P548 P38 r^2 = 4 Cos[2 t], r = 2 Sqrt[2] Sin[t]", AspectRatio->Automatic,PlotRange->All]; Display["38.ps",p7]; ------------------------------------------------------------------- New Material The area of a sector of a circle of radius r and angle theta is 2 1/2 r theta. . ''`` /. ' r/ ` rt .' / `. . /t . . /__________. . r . . . . . .... The same formula as for a triangle holds. Area = 1/2 BasexHeight 2 or Area = 1/2 rt * r = 1/2 r theta -------------------------------------------------------------------- Example 1: Find the area of the region inside the limacon r = 2 + Cos[t] ' ` too ' ` big . . ` | . ' . .` .| r ' . . . | ' too . ' . . | ' small ' . | 'dt ' r . |' ' ---------.-------+--------------------.----- | . | . . | . . | . | . . | . . When we draw the approximating sector of the circle, the biggest radius makes the area too big. The smallest radius makes the area too small. Since the limit of the approximations has to be both too big and at the same time too small, then the limit can only be the exact area. t=2Pi 2 INT 1/2 r dt t=0 t=2Pi 2 INT 1/2 (2+Cos[t]) dt t=0 t=2Pi 2 1/2 INT 4 + 2 Cos[t] + Cos[t] dt t=0 t=2Pi 1-Cos[2t] 1/2 INT 4 + 2 Cos[t] + --------- dt t=0 2 | | t = 2 Pi | Sin[2t] | 1/2 | 4 t + 2 Sin[t] + 1/2 t - -------- | | 4 | | | t = 0 1/2 ( 8 Pi + Pi) = 9 Pi/2 ------------------------------------------------------------ r[t_] := 2 + Cos[t] a = ParametricPlot[{ r[t] Cos[t], r[t] Sin[t]}, {t,-Pi, Pi},PlotStyle->{RGBColor[1,0,0]}]; b = Show[a,PlotLabel->"Page 551 Example 1 r = 2 + Cos[t]",AspectRatio->Automatic, PlotRange->All]; Display["ex1.ps",b]; c = ParametricPlot[ 3/Sqrt[2] {Cos[t], Sin[t]}+{1,0}, {t,-Pi, Pi}, PlotStyle->{RGBColor[0,1,0]}]; d = Show[a,c,PlotLabel->"Page 551 Example 1 r = 2 + Cos[t] red:r=3/Sqrt[2]", AspectRatio->Automatic]; Display["ex1x.ps",d]; ----------------------------------------------------------- Page 551 Find the area of one leaf of the four leaved rose r = 4 Sin[2 t]; --------------------------------------------------------- r[t_] := 4 Sin[2 t] a = ParametricPlot[{ r[t] Cos[t], r[t] Sin[t]}, {t,-Pi, Pi},PlotStyle->{RGBColor[1,0,0]}]; b = Show[a,PlotLabel->"Page 551 Example 2 r = 4 Sin[2 t]", AspectRatio->Automatic, PlotRange->All]; Display["ex2.ps",b]; f[t_] := Cos[t] {1/Sqrt[2],-1/Sqrt[2]}+ 2 Sin[t] {1/Sqrt[2],+1/Sqrt[2]} + {3/2,3/2}; c = ParametricPlot[ f[t], {t,-Pi, Pi}]; d = Show[a,c,PlotLabel->"Page 551 Example 2 r = 4 Sin[2 t]",AspectRatio->Automatic]; Display["ex2x.ps",d]; --------------------------------------------- t=Pi/2 2 Area = INT 1/2 r dt t=0 t=Pi/2 2 Area = 1/2 INT 16 Sin[2 t] dt t=0 t=Pi/2 Area = INT 4(1-Cos[4t]) dt t=0 | | t = Pi/2 Area = | 4 t - Sin[4t]| | | t = 0 Area = 2 Pi ------------------------------------------------------- Page 552 Example 3. Find the area of the region outside the cardioid r = 1+Cos[t] and inside the circle r = Sqrt[3] Sin[t]; -------------------------------------------------------------- r[t_] := 1+Cos[t]; a = ParametricPlot[{ r[t] Cos[t], r[t] Sin[t]}, {t,-Pi, Pi},PlotStyle->{RGBColor[1,0,0]}]; b = Show[a,PlotLabel->"Page 552 Example 3: r = 1+Cos[t]",AspectRatio->Automatic, PlotRange->All]; Display["ex3.ps",b]; f[t_] := Sqrt[3] Sin[t] {Cos[t],Sin[t]}; c = ParametricPlot[ f[t], {t,-Pi, Pi}]; d = Show[a,c,PlotLabel->"Page 551 Example 3: 1+Cos[t]",AspectRatio->Automatic]; Display["ex3x.ps",d]; -------------------------------------------------------------- Find the limits of integration by finding where the curves cross. 1 + Cos[t] = Sqrt[3] Sin[t] 2 2 1 + 2 Cos[t] + Cos[t] = 3 Sin[t] 2 2 1 + 2 Cos[t] + Cos[t] = 3 - 3 Cos[t] 2 -2 + 2 Cos[t] + 4 Cos[t] = 0 2 -1 + Cos[t] + 2 Cos[t] = 0 (2 Cos[t] - 1)(Cos[t] + 1) = 0 Cos[t] = 1/2 Cos[t] = -1 t = Pi/3 or t = Pi t = Pi 2 2 1/2 INT 3 Sin[t] - (1+Cos[t]) dt t=Pi/3 t = Pi 2 2 1/2 INT 3 Sin[t] - 1 - 2 Cos[t] - Cos[t] dt t=Pi/3 t = Pi 2 2 1/2 INT 3 - 3 Cos[t] - 1 - 2 Cos[t] - Cos[t] dt t=Pi/3 t = Pi 2 1/2 INT 2 - 4 Cos[t] - 2 Cos[t] dt t=Pi/3 t = Pi 1/2 INT 2 - 2(1+Cos[2t])- 2 Cos[t] dt t=Pi/3 t = Pi 1/2 INT -2 Cos[2t])- 2 Cos[t] dt t=Pi/3 | | t = Pi 1/2 | -Sin[2t] - 2 Sin[t] | | | t=Pi/3 1/2 ( + Sin[2 Pi/3] + 2 Sin[Pi/3] ) 1/2 ( Sqrt[3]/2 + 2 Sqrt[3]/2 ) 3 Sqrt[3]/4 ---------------------------------------------------------- Page 553 Example 4 Consider the polar equation r = 4 Sin[3 t] (a) Find the slope of the tangent line at t = Pi/6 and t = Pi/4 (b) Find the tangent lines at the pole. (c) Sketch the graph. (d) Find the area of one leaf. The slope we are used to is a cartesian coordinate concept. So it is not surprising that we simply compute it by going to the coordinate representation of the graph. x = r Cos[t] y = r Sin[t] x = 4 Sin[3 t] Cos[t] y = 4 Sin[3 t] Sin[t] dx/dt = 12 Cos[3 t] Cos[t] - 4 Sin[3t] Sin[t] dy/dt = 12 Cos[3 t] Sin[t] + 4 Sin[3 t] Cos[t] dx/dt | = -2 dx/dt | = -8 Pi/6 Pi/4 dy/dt | = 2 Sqrt[3] dy/dt | = -4 Pi/6 Pi/4 dy/dx | = -Sqrt[3] dy/dx | = 1/2 Pi/6 Pi/4 -------------------------------------------------------------- r[t_] := 4 Sin[3 t]; a = ParametricPlot[{ r[t] Cos[t], r[t] Sin[t]}, {t,-Pi, Pi},PlotStyle->{RGBColor[1,0,0]}]; b = Show[a,PlotLabel->"Page 553 Example 4: r = 4 Sin[3t]",AspectRatio->Automatic, PlotRange->All]; Display["ex4.ps",b]; c = ParametricPlot[ r[Pi/6]{Cos[Pi/6],Sin[Pi/6]} +{t,t (-Sqrt[3])},{t,-1,1}]; d = ParametricPlot[ r[Pi/4]{Cos[Pi/4],Sin[Pi/4]} +{t,t/2},{t,-1,1}]; e = Show[a,c,d,PlotLabel->"Page 553 Example 4: r = 4 Sin[3t] ",AspectRatio->Automatic]; Display["ex4x.ps",d]; ------------------------------------------------------------------------------