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 16 Chapter review p556 Sample Test Problems: (4),(8),(12),(16),(20),(24),(28),(32),(36),(40),(44) Main Idea: Chapter Review Key Words: Kepler, String, Cone, nappes, Goal: Use knowledge of Conics to answer questions ------------------------------------------------------------------ Previous assignment: Page 554 Problem 4 Sketch the graph of the given equation and find the area of the region bounded by it. r = 5 + 4 Cos[t] t=2 Pi INT 1/2 (5 + 4 Cos[t])^2 dt t=0 t=2 Pi 1/2 INT 25 + 40 Cos[t] + 16 Cos[t]^2 dt t=0 t=2 Pi 1/2 INT 25 + 40 Cos[t] + 8 (1+Cos[2t]) dt t=0 t=2 Pi 1/2 INT 33 + 40 Cos[t] + 8 Cos[2t] dt t=0 | | t=2 Pi 1/2 | 33 t + 40 Sin[t] + 4 Sin[2t] | | | t=0 33 Pi. ------------------------------------------------------ Integrate[1/2 (5+4 Cos[t])^2 ,{t,0,2 Pi}] f[t_] := 5 + 4 Cos[t]; p1 = ParametricPlot[f[t]{Cos[t],Sin[t]},{t,0,2 Pi}]; p2 = ParametricPlot[ {9-Sqrt[33],0}+Sqrt[33] {Cos[t],Sin[t]},{t,0,2 Pi}, PlotStyle->{RGBColor[1,0,0]}]; p3 = Show[p1,p2,AspectRatio->Automatic, PlotLabel->"P554 P4 r = 5 + 4 Cos[t] red:circle of same area"]; Display["4.ps",p3]; --------------------------------------------------------------------- Page 554 Problem 10 Sketch the graph of the given equation and find the area of the region bounded by it. 2 r = a Cos[2t] a > 0 We will figure the region to the right of the y-axis. t = Pi/4 INT 1/2 a Cos[2t] dt t = -Pi/4 | | t = Pi/4 | 1/4 a Sin[2t] | | | t = -Pi/4 1/4 a (1-(-1)) = 1/2 a The combined area is then a. ------------------------------------------------------ Integrate[1/2 (a Cos[2 t]) ,{t,-Pi/4,Pi/4}] a = 16 f[t_] := Sqrt[a Cos[2t]]; p1 = ParametricPlot[f[t]{Cos[t],Sin[t]},{t,-Pi/4,Pi/4}]; p2 = ParametricPlot[{2,0}+{2 Cos[t],4/Pi Sin[t]},{t,0,2 Pi},PlotStyle->{RGBColor[1,0,0]}]; p3 = Show[p1,p2,AspectRatio->Automatic, PlotLabel->"P554 P10 r^2 = a Cos[2t] red=ellipse of same area"]; Display["10.ps",p3]; ---------------------------------------------------- Page 554 Problem 16 Sketch the three-leaved rose r = 2 Sin[3t], and find the area of the region bounded by it. Find area of one leaf: t=Pi/3 INT 1/2 (2 Sin[3t])^2 dt t=0 t=Pi/3 INT 2 Sin[3t]^2 dt t=0 t=Pi/3 INT 1 - Cos[6t] dt t=0 | | t=Pi/3 | t - Sin[6t]/6 | | | t=0 Pi/3 Area of one leaf Total area is Pi. ----------------------------------------------------- Integrate[1/2 (2 Sin[3 t])^2 ,{t,0,Pi/3}] f[t_] := 2 Sin[3t]; p1 = ParametricPlot[f[t]{Cos[t],Sin[t]},{t,0,2 Pi}]; p2 = ParametricPlot[ {Cos[Pi/6],Sin[Pi/6]}+ Cos[t] { Cos[Pi/6], Sin[Pi/6]}+ 1/3 Sin[t] {-Sin[Pi/6], Cos[Pi/6]}, {t,0,2 Pi},PlotStyle->{RGBColor[1,0,0]},AspectRatio->Automatic]; p3 = Show[p1,p2,AspectRatio->Automatic, PlotLabel->"P554 P16 r = 2 Sin[3t] red=ellipse of same area"]; Display["16.ps",p3]; -------------------------------------------------------- Page 554 Problem 20 Sketch the limacon r = 3 - 6 Sin[t], and find the area of the region that is inside its large loop but outside its small loop. small loop t= 5 Pi/6 INT 1/2 (3 - 6 Sin[t])^2 dt t= Pi/6 1/2 INT 9 - 36 Sin[t] + 36 Sin[t]^2 dt 1/2 INT 9 - 36 Sin[t] + 18(1 - Cos[2t] dt | | t = 5 Pi/6 1/2 | 9t + 36 Cos[t] + 18 t -9 Sin[2t] | | | t = Pi/6 | | t = 5 Pi/6 1/2 | 27t + 36 Cos[t] -9 Sin[2t] | | | t = Pi/6 1/2 ( 27(2 Pi/3) +36(Cos[5 Pi/6]-Cos[Pi/6]) -9 (Sin[5 Pi/3) - Sin[Pi/3]) ) 1/2 ( 18 Pi +36(-Sqrt[3]/2 - Sqrt[3]/2) -9 (-Sqrt[3]/2 - Sqrt[3]/2 ) 9 Pi - 18 Sqrt[3] +9/2 Sqrt[3] 9 Pi + -27/2 Sqrt[3] -27 Sqrt[3] ----------- + 9 Pi = 4.89165 2 Large loop t= 11 Pi/6 INT 1/2 (3 - 6 Sin[t])^2 dt t= 5Pi/6 | | t = 11 Pi/6 1/2 | 27t + 36 Cos[t] -9 Sin[2t] | | | t = 5 Pi/6 1/2 ( 27 Pi + 36 ( Cos[11 Pi/6] - Cos[ 5 Pi/6]) - 9 (Sin[11 Pi/3] - Sin[10 Pi/3] ) ) 1/2 ( 27 Pi + 36 ( Sqrt[3]/2 + Sqrt[3]/2 ) - 9 (-Sqrt[3]/2 + Sqrt[3]/2) ) 27 Pi/2 + 18 Sqrt[3] 27 Pi 18 Sqrt[3] + ----- = 73.5884 2 63 Sqrt[3] 9 Pi Large loop - Small loop = ---------- + ---- = 68.6968 2 2 ----------------------------------------------------- f[t_] := 3 - 6 Sin[t] p1 = ParametricPlot[f[t]{Cos[t],Sin[t]},{t,0,2 Pi}]; p2 = Show[p1,AspectRatio->Automatic, PlotLabel->"P554 P20 r = 3 - 6 Sin[t] "]; Display["20.ps",p2]; ---------------------------------------------------------------- Page 554 Concepts Review; Page 555 Chapter Review, odd questions: ---