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.166.08 Textbook: Calculus by Varberg, Purcell, Rigdon, ninth edition. Friday, April 25 10.7 Calculus Assignment: Page p553: 26-52 Even Only Previous Assignment p551: 2,12,14,22,25 Page 551 Problem 2 Graph the given equation and find the area of the region bounded by it. r = 2 a Cos[t], a > 0. t=Pi 2 INT 1/2 r dt t=0 t=Pi 2 2 INT 1/2 4 a Cos [t] dt t=0 t=Pi 2 INT 1/2 4 a (1+Cos[2t])/2 dt t=0 2 t=Pi a INT 1+Cos[2t] dt t=0 _ _ 2 | | t=Pi a | t+ 1/2 Sin[2t] | |_ _| t=0 2 AREA = Pi a which is not surprising since it is a circle of radius a. Page 551 Problem 12 Sketch the limacon r = 2 - 4 Cos[t] and find the area of the region inside its small loop. 0 = 2 - 4 Cos[t] 4 Cos[t] = 2 Cos[t] = 1/2 o o t = 60, 300 t= Pi/3 2 INT 1/2 r dt t= -Pi/3 t= Pi/3 2 INT 1/2 (2-4 Cos[t]) dt t= -Pi/3 t= Pi/3 2 INT 1/2 ( 4 - 16 Cos[t] + 16 Cos [t] ) dt t= -Pi/3 t= Pi/3 2 INT 2 - 8 Cos[t] + 8 Cos [t] dt t= -Pi/3 t= Pi/3 2 2 INT 1-4 Cos[t] + 4 Cos [t]) dt t= -Pi/3 t= Pi/3 2 INT 1-4 Cos[t] + 2(1+Cos[2t]) dt t= -Pi/3 _ _ | | t= Pi/3 2 | t-4 Sin[t] + 2 t + Sin[2t]) | |_ _| t= -Pi/3 _ _ | | t= Pi/3 2 | 3t-4 Sin[t] + Sin[2t]) | |_ _| t= -Pi/3 _ _ | | 2 | 2 Pi -4 Sqrt[3] + Sqrt[3]) | |_ _| Area = 4 Pi -6 Sqrt[3] = 2.17407 --------------------------------------------------------------------- Integrate[ 1/2 (2-4 Cos[t])^2, {t,-Pi/3,+Pi/3}]; Get["font.math"]; r[t_] := 2 - 4 Cos[t]; a = ParametricPlot[ r[t] { Cos[t], Sin[t]}, {t,0,2 Pi}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; b = Show[a,PlotLabel->"r = 2 - 4 Cos[t]",AspectRatio->Automatic,PlotRange->All]; Display["p12x.ps",b]; aa = 1; bb = (4 Pi -6 Sqrt[3])/Pi; c = ParametricPlot[ { aa Cos[t], bb Sin[t]}+{-1,0},{t,-Pi,Pi}, PlotStyle->{RGBColor[0,1,0],Thickness[0.01]}, AspectRatio->Automatic,PlotRange->All]; d = Show[a,c,PlotLabel->"red: r = 2- Cos[t]; Green: Equal Area Ellipse", AspectRatio->Automatic]; Display["p12y.ps",d]; ------------------------------------------------------------------- Page 551 Problem 14 Sketch one leaf of the four-leaved rose r = 3 Cos[2t] and find the area of the region enclosed by it. t = Pi/4 2 INT 1/2 (3 Cos[2t]) dt t =-Pi/4 t = Pi/4 2 9/2 INT Cos [2t] dt t =-Pi/4 ^ t = Pi/4 9/2 INT (1/2)(1+ Cos[4t]) dt t =-Pi/4 t = Pi/4 9/4 INT 1+ Cos[4t] dt t =-Pi/4 _ _ | | t = Pi/4 9/4 | t+ 1/4 Sin[4t] | |_ _| t =-Pi/4 9/4 ( Pi/2 ) = 9/8 Pi = 3.53429 --------------------------------------------------------- Integrate[ 1/2 (3 Cos[2t])^2,{t,-Pi/4,Pi/4}]; Get["font.math"]; r[t_] := 3 Cos[2t] ; a = ParametricPlot[ r[t] {Cos[t], Sin[t]}, {t,-Pi/4,Pi/4}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; b = Show[a,PlotLabel->"r = 3 Cos[2t]", AspectRatio->Automatic,PlotRange->All]; Display["p14x.ps",b]; c = ParametricPlot[ {3/2 Cos[t], 3/4 Sin[t]} + {3/2,0},{t,-Pi,Pi}, PlotStyle->{RGBColor[0,1,0],Thickness[0.01]}]; d = Show[a,c,AspectRatio->Automatic, PlotLabel->"Red: r=3 Cos[2t]; Green: Equal Area Ellipse"]; Display["p14y.ps",d]; ----------------------------------------------------------------- Page 551 Problem 22 Sketch the region in the second quadrant that is inside the cardioid r = 2 + 2 Sin[t] and outside the cardioid r = 2 + 2 Cos[t] and find its area. t=Pi 2 2 1/2 INT (2+2 Sin[t] ) - (2 + 2 Cos[t]) dt t=Pi/2 t=Pi 2 2 1/2 INT 4 + 8 Sin[t] + 4 Sin[t] - 4 - 8 Cos[t] - 4 Cos[t] dt t=Pi/2 t=Pi 2 2 1/2 INT 8 Sin[t] + 4 Sin[t] - 8 Cos[t] - 4 Cos[t] dt t=Pi/2 t=Pi 2 2 2 INT 2 Sin[t] + Sin[t] - 2 Cos[t] - Cos[t] dt t=Pi/2 t=Pi 2 INT 2 Sin[t] -2 Cos[t] - Cos[2t] dt t=Pi/2 _ _ | | t=Pi 2 | - 2 Cos[t] -2 Sin[t] - 1/2 Sin[2t] | |_ _| t=Pi/2 _ _ | | 2 | 2 - ( -2) | = 8 <=== Area |_ _| -------------------------------------------------------------------- t= Pi 2 INT 1/2 (2+2 Sin[t]) dt = 4 + 3 Pi/2 = 8.71239 Total area t=Pi/2 t= Pi 2 INT 1/2 (2+2 Cos[t]) dt = -4 + 3 Pi/2 = 0.712389 Part removed t=Pi/2 ------------- subtract 8 Part remaining residue = answer = 8 ---------------------------------------------------------------- Get["font.math"]; f[t_] := 2 + 2 Sin[t]; g[t_] := 2 + 2 Cos[t]; Integrate[ 1/2 f[t]^2 ,{t,Pi/2,Pi}]; Integrate[ 1/2 g[t]^2 ,{t,Pi/2,Pi}]; a = ParametricPlot[ f[x] {Cos[x], Sin[x]},{x,-Pi,Pi}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; b = ParametricPlot[ g[x] {Cos[x], Sin[x]},{x,-Pi,Pi}, PlotStyle->{RGBColor[0,1,0],Thickness[0.01]}]; c = Plot[0,{x,-2,0},PlotStyle->{Thickness[0.01]}]; d = ParametricPlot[{0,y},{y,0,4},PlotStyle->{Thickness[0.01]}]; e = Show[a,b,c,d, PlotLabel->"P554 p22; r=2+2 Sin[t];r=2+2 Cos[t],inside red; outside green ", AspectRatio->Automatic,PlotRange->All]; Display["p22x.ps",e]; P1 = ParametricPlot[ f[x] {Cos[x], Sin[x]},{x,Pi/2,Pi}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P2 = ParametricPlot[ g[x] {Cos[x], Sin[x]},{x,Pi/2,Pi}, PlotStyle->{RGBColor[0,1,0],Thickness[0.01]}]; P3 = Plot[0,{x,-2,0},PlotStyle->{Thickness[0.01]}]; P4 = ParametricPlot[{0,y},{y,0,4},PlotStyle->{Thickness[0.01]}]; P5 = Plot[{0,1/2,2/2,3/2,4/2,5/2,6/2,7/2,8/2},{x,-5/2,0}]; P6 = Table[ ParametricPlot[ {n/2,y},{y,0,4}],{n,-5,0}]; P7 = Show[P1,P2,P3,P4,P5,P6, PlotLabel->"P554 p22; r=2+2 Sin[t]; r=2+2 Cos[t],inside red;outside green ", AspectRatio->Automatic,PlotRange->All]; Display["p22y.ps",P7]; ---------------------------------------------------------------- Page 551 Problem 25 Find all points on the limacon r = 1 - 2 Sin[t] where the tangent line is horizontal. 2 y = r Sin[t] = (1-2 Sin[t]) Sin[t] = Sin[t] - 2 Sin [t] dy/dt = Cos[t] - 4 Sin[t] Cos[t] dy/dt = 0 when Cos[t] = 0 or when Sin[t] = 1/4 t = Pi/2, 3 Pi/2, ArcSin[1/4], Pi-ArcSin[1/4] r=0 r=3 r=1/2 r = 1/2 The x-y Coordinates (0,0) (0,-3) (Sqrt[15/8],1/8) (-Sqrt[15/8],1/8) ------------------------------------------------ Get["font.math"]; P1 = ParametricPlot[ (1-2 Sin[t]){Cos[t],Sin[t]},{t,0,2 Pi}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P2 = ListPlot[{ {0,0},{0,-3},{Sqrt[15/8],1/8},{-Sqrt[15/8],1/8} }, PlotStyle->{RGBColor[0,0,0],PointSize[0.02]}]; P3 = Show[P1,P2,"551 p25, r = 1 - 2 Sin[t]", AspectRatio->Automatic,PlotRange->All]; Display["p25.ps",P3]; ------------------------------------------------------ Beginning of Review Page 553 Problem 27 Eliminate the parameter to obtain the corresponding Cartesian equation. Sketch the given curve. x = 4 Sin[t] - 2 y = 3 Cos[t] + 1 0 <= t <= 2 Pi _ _ 2 _ _ 2 | | | | | x+2 | | y-1 | | --- | + | -----| = 1 | 4 | | 3 | |_ _| |_ _| Ellipse centered at (-2,1) Semi Major axis is 4; Semi Minor axis is 3; Distance from center to focus is Sqrt[7]. ------------------------------------------------ Get["font.math"]; P1 = ParametricPlot[ {4 Sin[t] -2, 3 Cos[t]+1},{t, 0 , 2 Pi}, PlotStyle->{RGBColor[1,1,0],Thickness[0.01]}]; P2= Table[ Graphics[ Text[ FontForm[ t ,{"Courier-Bold",26}],{4 Sin[t]-2,3 Cos[t]+1} ]], {t,0, 7}]; P3 = Show[P1,P2,PlotLabel->"P553 p27 {4 Sin[t] -2, 3 Cos[t]+1}"]; Display["p27x.ps",P3]; a = 4; b = 3; c = Sqrt[7]; o = {-2,1}; l = b^2/a; P4 = ListPlot[{o, o-{a,0},o+{a,0},o-{0,b},o+{0,b},o+{c,0},o-{c,0}, o-{c,0}+{0,l},o-{c,0}-{0,l},o+{c,0}+{0,l},o+{c,0}-{0,l} }, PlotStyle->{RGBColor[1,0,0],PointSize[0.015]}]; P5 = Graphics[{RGBColor[0,0,1],Thickness[0.005] , Line[{o-{c,0}+{0,l},o-{c,0}-{0,l}}], Line[{o+{c,0}+{0,l},o+{c,0}-{0,l}}], Line[{o-{a,0},o+{a,0}}], Line[{o-{0,b},o+{0,b}}] }]; P6 = Show[P1,P4,P5,PlotLabel->"P553 p27 Ellipse with Reference Points"]; Display["p27y.ps",P6]; ------------------------------------------------ Page 553 Problem 29 Find the equations of the tangent line at t=0 3 x = 2 t - 4 t + 7 y = t + ln(t+1) dy/dt = 1 + 1/(t+1); 2 dx/dt = 6 t - 4 1 1 + ------ t+1 dy/dx = ---------------- 2 6 t -4 | dy/dx | = -1/2 |t=0 y-0 = -1/2(x-7) ------------------------------------------------ Get["font.math"]; f[t_] := 2 t^3 - 4 t + 7; g[t_] := t + Log[t+1]; fp[t_] = D[f[t],t]; gp[t_] = D[g[t],t]; P1 = ParametricPlot[ {f[t],g[t]},{t, -0.95 ,1.35}, PlotStyle->{RGBColor[1,0,0],Thickness[0.005]}]; P2 = ListPlot[ {{f[0],g[0]}},PlotStyle->{RGBColor[1,0,0],PointSize[0.02]}]; P3 = ParametricPlot[ t {fp[0],gp[0]}+{f[0],g[0]},{t,-1,1}, PlotStyle->{RGBColor[0,0,1],Thickness[0.001]}]; P4 = Show[P1,P2,P3,PlotLabel->"P553 p29 {2 t^3 - 4 t + 7,t + Log[t+1]} "]; Display["p29.ps",P4]; ------------------------------------------------ Page 553 Problem 31 Find the length of the curve 3/2 x = 1+t 3/2 y = 2 + t From t = 0 to t = 9. 1/2 dy/dt = 3/2 t 1/2 dx/dt = 3/2 t t=9 S = INT Sqrt[ 9/4 t + 9/4 t] dt t=0 t=9 S = INT Sqrt[ 9/2 t ] dt t=0 t=9 1/2 S = INT 3/Sqrt[2] t dt t=0 3/2 | t=9 t | 3/Sqrt[2] -------- | 3/2 | t=0 27 Sqrt[2] = 38.1838 ------------------------------------------------ Get["font.math"]; f[t_] := 1 + t^(3/2); g[t_] := 2 + t^(3/2); fp[t_] = D[f[t],t]; gp[t_] = D[g[t],t]; P1 = ParametricPlot[ {f[t],g[t]},{t, 0 , 9}, PlotStyle->{RGBColor[1,0,0],Thickness[0.0005]}]; P2 = ListPlot[ { {f[0],g[0]}, {f[6],g[6]}, {f[9],g[9]}}, PlotStyle->{RGBColor[0,0,1],PointSize[0.005]}]; P3= Table[ Graphics[ Text[ FontForm[ t ,{"Courier-Bold",26}],{f[t],g[t]} ]],{t,0,9 }]; P4 = Show[P1,P2,P3,PlotLabel->"P553 p31 {1 + t^(3/2), 2 + t^(3/2)"]; Display["p31.ps",P4]; ----------------------------------------------------------- Integrate[Sqrt[fp[t]^2 + gp[t]^2],{t,0,9}]; Sqrt[ (f[9]-f[0])^2 + (g[9]-g[0])^2]; Actual distance is: 27 Sqrt[2] = 38.1838 Straight Line distance is: 27 Sqrt[2] Same distance. It must be a straight line. Can you prove it? ------------------------------------------------ Page 553 Problem 33 Analyze the given polar equation and sketch its graph. r = 6 Cos[t] ------------------------------------------------------- Get["font.math"]; p[r_,t_] := r{Cos[t],Sin[t]}; P1 = ParametricPlot[ p[6 Cos[t],t],{t,0,2 Pi}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P2 = Show[P1,PlotLabel->"P553 p33 r = 6 Cos[t]",PlotRange->All, AspectRatio->Automatic]; Display["p33.ps",P2]; ---------------------------------------------------------- Page 553 Problem 35 Analyze the given polar equation and sketch its graph. r = Cos[2 t] ------------------------------------------------------- Get["font.math"]; p[r_,t_] := r{Cos[t],Sin[t]}; P1 = ParametricPlot[ p[Cos[ 2 t],t],{t,0,2 Pi}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P2 = Show[P1,PlotLabel->"P553 p35 r = Cos[2 t] ",PlotRange->All, AspectRatio->Automatic]; Display["p35.ps",P2]; ---------------------------------------------------------- Page 553 Problem 37 Analyze the given polar equation and sketch its graph. r = 4 ------------------------------------------------------- Get["font.math"]; p[r_,t_] := r{Cos[t],Sin[t]}; P1 = ParametricPlot[ p[4,t],{t,0,2 Pi}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P2 = Show[P1,PlotLabel->"P553 p37 r = 4 ",PlotRange->All, AspectRatio->Automatic]; Display["p37.ps",P2]; ---------------------------------------------------------- Page 553 Problem 39 Analyze the given polar equation and sketch its graph. r = 4 - 3 Cos[t] ------------------------------------------------------- Get["font.math"]; p[r_,t_] := r{Cos[t],Sin[t]}; P1 = ParametricPlot[ p[4 - 3 Cos[t],t],{t,0,2 Pi}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P2 = Show[P1,PlotLabel->"P553 p39 r = 4 - 3 Cos[t] ",PlotRange->All, AspectRatio->Automatic]; Display["p39.ps",P2]; ---------------------------------------------------------- Page 553 Problem 41 Analyze the given polar equation and sketch its graph. theta = 2/3 Pi ------------------------------------------------------- Get["font.math"]; p[r_,t_] := r{Cos[t],Sin[t]}; P1 = ParametricPlot[ p[r, 2 Pi/3],{r,-3,3}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P2 = Show[P1,PlotLabel->"P553 p41 theta = 2 Pi/3 ",PlotRange->All, AspectRatio->Automatic]; Display["p41.ps",P2]; ------------------------------------------------------- Page 553 Problem 43 Analyze the given polar equation and sketch its graph. 2 r = 16 Sin[2 t] ------------------------------------------------------- Get["font.math"]; p[r_,t_] := r{Cos[t],Sin[t]}; P1 = ParametricPlot[ p[Sqrt[16 Sin[2 t]],t],{t,0, Pi/2}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P2 = ParametricPlot[ p[-Sqrt[16 Sin[2 t]],t],{t,0, Pi/2}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P3 = Show[P1,P2,PlotLabel->"P553 p43 r^2 = 16 Sin[2 t] ",PlotRange->All, AspectRatio->Automatic]; Display["p43.ps",P3]; ------------------------------------------------------- Page 553 Problem 45 Analyze the given polar equation and sketch its graph. Find a Cartesian equation of the graph of 2 r - 6 r (Cos[t] + Sin[t]) + 9 = 0 and then sketch the graph. ------------------------------------------------------- Get["font.math"]; p[r_,t_] := r{Cos[t],Sin[t]}; f[t_] := 1/2 (6 (Cos[t]+Sin[t]) + Sqrt[ 36 (Cos[t]+Sin[t])^2 - 36]); g[t_] := 1/2 (6 (Cos[t]+Sin[t]) - Sqrt[ 36 (Cos[t]+Sin[t])^2 - 36]); P1 = ParametricPlot[ p[f[t],t],{t,0, Pi/2}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P2 = ParametricPlot[ p[g[t],t],{t,0, Pi/2}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P3 = Show[P1,P2,PlotLabel->"P553 p45 r^2 - 6 r (Cos[t] + Sin[t]) + 9 = 0 ", PlotRange->All, AspectRatio->Automatic]; Display["p45.ps",P3]; ------------------------------------------------------- Page 553 Problem 47 Find the slope of the tangent line to the graph of r = 3 + 3 Cos[t] at the point on the graph where t = Pi/6 y = r Sin[t] = (3+3 Cos[t]) Sin[t] = 3 Sin[t] + 3 Cos[t] Sin[t] 2 2 dy/dt = 3 Cos[t] - 3 Sin [t] + 3 Cos [t] | dy/dt | = 3 Sqrt[3]/2 -3/4 + 9/4 = 3/2 (1+ Sqrt[3]) | t=Pi/6 2 x = r Cos[t] = (3+3 Cos[t]) Cos[t] = 3 Cos[t] + 3 Cos [t] dx/dt = -3 Sin[t] - 6 Cos[t] Sin[t] | dx/dt | = -3/2 -6 Sqrt[3]/2 (1/2) = -3/2(1+Sqrt[3]) | t=Pi/6 | 3/2 (1+ Sqrt[3]) dy/dx | = ------------------- = -1 | t=Pi/6 -3/2 (1+ Sqrt[3]) ------------------------------------------------------- Get["font.math"]; f[t_] := (3+3 Cos[t]) Cos[t]; g[t_] := (3+3 Cos[t]) Sin[t]; fp[t_] = D[f[t],t]; gp[t_] = D[g[t],t]; p[r_,t_] := r{Cos[t],Sin[t]}; P1 = ParametricPlot[ p[3+3 Cos[t],t],{t,0,2 Pi}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P2 = ListPlot[{{f[Pi/6],g[Pi/6]}},PlotStyle->{RGBColor[1,0,0],PointSize[0.02]}]; P3 = ParametricPlot[ {f[Pi/6],g[Pi/6]} + t{fp[Pi/6],gp[Pi/6]},{t,-1,1}, PlotStyle->{RGBColor[0,1,0],Thickness[0.01]}]; P4 = Show[P1,P2,P3,PlotLabel->"P553 p47 r = 3 + 3 Cos[t] ", PlotLabel->"P553 p47 r = 3 + 3 Cos[t] at Pi/6",AspectRatio->Automatic, PlotRange->All]; Display["p47.ps",P4]; ------------------------------------------------------- Page 553 Problem 49 Find the area of the region bounded by the graph of r = 5 - 5 Cos[t] t=2 Pi 2 Area = INT (1/2)(5-5 Cos[t]) dt t=0 t=2 Pi 2 Area = INT (25/2)(1- Cos[t]) dt t=0 t=2 Pi 2 Area = (25/2) INT 1 - 2 Cos[t] + Cos [t] dt t=0 t=2 Pi 1+Cos[2t] Area = (25/2) INT 1 - 2 Cos[t] + ---------- dt t=0 2 t=2 Pi Area = (25/2) INT 3/2 - 2 Cos[t] + (1/2) Cos[2t] dt t=0 _ _ | | t=2 Pi Area = (25/2) | 3/2 t - 2 Sin[t] + 1/4 Sin[2t] | |_ _| t=0 Area = (25/2) ( 3 Pi ) = 75 Pi/2 = 117.81 ------------------------------------------------------- Get["font.math"]; p[r_,t_] := r{Cos[t],Sin[t]}; P1 = ParametricPlot[ p[5-5 Sin[t],t],{t,0,2 Pi}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P2 = Show[P1,PlotLabel->"P553 p49 r = 5 - 5 Cos[t] ", PlotRange->All, AspectRatio->Automatic]; Display["p49x.ps",P2]; Integrate[ 1/2 (5-5 Sin[t])^2,{t,0,2 Pi}]; Answer = 75 Pi/2 = 117.81 P3 = ParametricPlot[ Sqrt[75/2] {Cos[t], Sin[t]} + {0,-4.0},{t,0,2 Pi}, PlotStyle->{RGBColor[0,1,0],Thickness[0.01]}]; P4 = Show[P1,P3,PlotLabel->"P553 p49 r = 5 - 5 Cos[t] ", PlotRange->All, AspectRatio->Automatic]; Display["p49y.ps",P4]; ------------------------------------------------------- Page 553 Problem 51 A racing car driving on the elliptical race track 2 2 x y ---- + ----- = 1 400 100 went out of control at the point (16,6) and thereafter continued on the tangent line until it hit a tree at (14,k) Determine k. 2 x/400 + 2y dy/dx/100 = 0 at (16,6) 32/400 + 12/100 dy/dx = 0 dy/dx = -(32/400)/(12/100) = -2/3 (16,6)+(1,-2/3) t = (14,k) 16 + t = 14 t = -2 6 -2/3 t = k k = 22/3 How far off of the track was the tree? 14^2/400 + y^2/100 = 1 y^2 = 100(1-14^2/400) = 100 - 14^2/4 = 100 - 196/4 = 51 y = Sqrt[51] = 7.14143 22/3 = 7.33333 So the tree was 0.191905 units off of the track. The arc Length of an ellipse is approximately 2 Pi Sqrt[(a^2+b^2)/2] If it is a 1 mile track, then 5280 = 2 Pi Sqrt[ (400t^2 + 100 t^2)/2 ] Solve[ 5280 == 2 Pi Sqrt[ (400t^2 + 100 t^2)/2 ],t] t -> 53.1476 feet and the distance off is about 10.1993 feet off of the track going upwards. ------------------------------------------------------------- Get["font.math"]; f[t_] := 20 Cos[t]; g[t_] := 10 Sin[t]; fp[t_] = D[f[t],t]; gp[t_] = D[g[t],t]; p[r_,t_] := r{Cos[t],Sin[t]}; P1 = ParametricPlot[ {f[t],g[t]},{t,0,2 Pi}, PlotStyle->{RGBColor[1,0,0],Thickness[0.003]}]; P2 = ListPlot[ {{16,6}},PlotStyle->{RGBColor[0,0,1],PointSize[0.008]}]; P3 = ListPlot[ {{14,22/3}},PlotStyle->{RGBColor[0,0,1],PointSize[0.008]}]; theta = ArcCos[16/20]; P4 = ParametricPlot[ t {14,22/3} + (1-t) {f[theta],g[theta]}, {t,-1,2}, PlotStyle->{RGBColor[0,1,0],Thickness[0.003]}]; P5 = Show[P1,P2,P3,P4,PlotLabel->"P553 p41, (16,6) to (14,k}", AspectRatio->Automatic,PlotRange->All]; Display["P51.ps",P5]; ------------------------------------------------------------------ Solve[t{-3,2} + {16,6} == {14,k}] 2 22 Out[24]= {{t -> -, k -> --}} 3 3 -------------------------------------------------------------------------- Page 553 Problem 53 Match each polar equation with its graph. (a) r = 4 Cos[2 t] (b) r = 3 Cos[3 t] (c) r = 5 Cos[5 t] (d) r = 3 Sin[2 t]