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. Practice Test 4 The real test is Friday, April 27. 1. Find the area inside r = 3 + Sin[t] t= 2Pi 2 INT 1/2 (3+Sin[t]) dt = 19 Pi/2 = 29.8451 t=0 t= 2Pi 2 INT 1/2( 9 + 6 Sin[t] + Sin [t] ) dt t=0 t= 2Pi 1-Cos[2t] INT 1/2( 9 + 6 Sin[t] + ---------- ) dt t=0 2 t= 2Pi Cos[2t] INT 19/4 + 3 Sin[t] - -------- dt t=0 2 | | t=2 Pi | 19/4 t - 3 Cos[t] - 1/4 Sin[2t] | | | t=0 19 Pi/2 -3 - 0 - (0 -3 -0) 19 Pi/2 = 29.8451 --------------------------------------------------- Integrate[ 1/2 (3+Sin[t])^2,{t,0,2 Pi}]; Get["font.math"]; r[t_] := 3 + Sin[t]; P1 = ParametricPlot[ r[t] {Cos[t], Sin[t]},{t,0,2 Pi}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P2 = Table[Plot[n/2,{x,-3.5,3.5}],{n,-4,8}]; P3 = Table[ ParametricPlot[ {n/2,y},{y,-2, 4} ], {n,-7,7} ]; P4 = Show[P1,P2,P3, PlotLabel->"p1: r = 3 + Sin[t]", AspectRatio->Automatic, PlotRange->All]; Display["p1.ps",P4]; ------------------------------------------------------------ 2. Find the area in the first quadrant which is common to both of two curves r = 1 + Sin[t] and r = 1 + Cos[t] 1 + Sin[t] = 1 + Cos[t]; Sin[t] = Cos[t] t = Pi/4 or 5 Pi/4 t=Pi/2 2 A1 = INT 1/2 (1+Cos[t]) dt = 7/8 - 1/Sqrt[2] + 3 Pi/16 t=Pi/4 t=Pi/2 2 = 1/2 INT 1 + 2 Cos[t] + Cos [t] dt t=Pi/4 t=Pi/2 1 + Cos[2t] = 1/2 INT 1 + 2 Cos[t] + ------------ dt t=Pi/4 2 t=Pi/2 3 Cos[2t] = 1/2 INT ---+ 2 Cos[t] + ------------ dt t=Pi/4 2 2 | | t=Pi/2 = 1/2 | 3/2 t + 2 Sin[t] + 1/4 Sin[2t] | | | t=Pi/4 = 1/2 ( 3 Pi/4 + 2 + 0 - (3 Pi/8 + Sqrt[2] +1/4 ) = 1/2 ( 3 Pi/8 - Sqrt[2] + 7/4) = 3 Pi/16 - Sqrt[2]/2 + 7/8 t= Pi/4 2 A2 = INT 1/2 (1+Sin[t]) dt = 7/8 - 1/Sqrt[2] + 3 Pi/16 t=0 t=Pi/4 2 = 1/2 INT 1 + 2 Sin[t] + Sin [t] dt t=0 t=Pi/4 1 - Cos[2t] = 1/2 INT 1 + 2 Sin[t] + ------------ dt t=0 2 t=Pi/4 3 Cos[2t] = 1/2 INT ---+ 2 Sin[t] - ------------ dt t=0 2 2 | | t= Pi/4 = 1/2 | 3/2 t - 2 Cos[t] - 1/4 Sin[2t] | | | t= 0 = 1/2 ( 3 Pi/8 - Sqrt[2] - 1/4 - (0 -2 - 0 ) = 1/2 ( 3 Pi/8 - Sqrt[2] + 7/4 ) = 3 Pi/16 - Sqrt[2]/2 + 7/8 A1 + A2 = 7/4 - Sqrt[2] + 3 Pi/8 = 1.51388 ---------------------------------------------------- Get["font.math"]; A1 = Integrate[ 1/2 (1+Cos[t])^2,{t,Pi/4, Pi/2}]; A2 = Integrate[ 1/2 (1+Sin[t])^2,{t,0,Pi/4}]; f[t_] := 1 + Sin[t]; g[t_] := 1 + Cos[t]; P1 = ParametricPlot[ f[t] { Cos[t],Sin[t] },{t,0,2 Pi}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P2 = ParametricPlot[ g[t] { Cos[t], Sin[t] },{t,0,2 Pi}, PlotStyle->{RGBColor[0,1,0],Thickness[0.01]}]; P3 = Show[P1,P2,PlotLabel->"p2: red: r=1+Sin[t] and green: r=1+Cos[t]", PlotRange->All,AspectRatio->Automatic]; Display["p2.ps",P3]; --------------------------------------------------------- 3. Name the curves. (a) r = Cos[t] Circle -------------------------------------------------------------------- (b) r = 1 + Cos[t] Cardioid -------------------------------------------------------------------- (c) r = 2 + Cos[t] Limacon with no center loop ------------------------------------------------------------------- (d) r = t Spiral of Archimedes -------------------------------------------------------------------- (e) r t = 1 Hyperbolic Spiral -------------------------------------------------------------------- (f) r = Cos[2 t] Four Leaf Rose -------------------------------------------------------------------- 2 (g) r = Cos[2 t] Lemniscate ----------------------------------------------------------- Get["font.math"]; r[t_] := Cos[t]; pa = ParametricPlot[{r[t] Cos[t],r[t] Sin[t]},{t,0,2 Pi}, AspectRatio->Automatic,PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; r[t_] := 1 + Cos[t]; pb = ParametricPlot[{r[t] Cos[t],r[t] Sin[t]},{t,0,2 Pi}, AspectRatio->Automatic,PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; r[t_] := 2 + Cos[t]; pc = ParametricPlot[{r[t] Cos[t],r[t] Sin[t]},{t,0,2 Pi}, AspectRatio->Automatic,PlotStyle->{RGBColor[0,1,0],Thickness[0.01]}]; r[t_] := t; pd = ParametricPlot[{r[t] Cos[t],r[t] Sin[t]},{t,0,2 Pi}, AspectRatio->Automatic,PlotStyle->{RGBColor[0,1,0],Thickness[0.01]}]; r[t_] := 1/t; pe = ParametricPlot[{r[t] Cos[t],r[t] Sin[t]},{t,0, Pi}, AspectRatio->Automatic,PlotStyle->{RGBColor[0,0,1],Thickness[0.01]}]; r[t_] := Cos[2 t]; pf = ParametricPlot[{r[t] Cos[t],r[t] Sin[t]},{t,0,2 Pi}, AspectRatio->Automatic,PlotStyle->{RGBColor[0,0,1],Thickness[0.01]}]; r[t_] := Sqrt[Cos[t]] pg1 = ParametricPlot[{r[t] Cos[t],r[t] Sin[t]},{t,0,2 Pi},PlotPoints->100, AspectRatio->Automatic,PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; pg2 = ParametricPlot[{-r[t] Cos[t],r[t] Sin[t]},{t,0,2 Pi},PlotPoints->100, AspectRatio->Automatic,PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; pg = Show[pg1,pg2] ph = Show[pa,pb,pc,pd,pe,pf,pg]; ans = Show[GraphicsArray[Partition[{pa,pb,pc,pd,pe,pf,pg,ph},4]]] Display["p3.ps",ans]; --------------------------------------------------------------------- 4. Find the area of one leaf of r = 4 Cos[2 t] t = Pi/4 2 INT 1/2 (4 Cos[2 t]) dt = 2 Pi t= -Pi/4 t = Pi/4 2 INT 8 Cos [2t] t = -Pi/4 t = Pi/4 INT 4 ( 1 + Cos[4t] ) dt t = -Pi/4 | | t = Pi/4 | 4 t + Sin[4t] | | | t = -Pi/4 Pi + 0 - (-Pi + 0) 2 Pi ------------------------------------------------------------ Get["font.math"]; A = Integrate[ 1/2 (4 Cos[2 t])^2,{t,-Pi/4, Pi/4}]; r[t_] := 4 Cos[2 t]; p1 = ParametricPlot[ r[t] { Cos[t], Sin[t] },{t,0,2 Pi}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}, PlotRange->All,AspectRatio->Automatic]; p2 = ParametricPlot[{2 Cos[t],Sin[t]}+{2,0},{t,0,2 Pi}, PlotStyle->{RGBColor[0,1,0],Thickness[0.005]}, PlotStyle->{RGBColor[1,0,0]}]; p3 =Show[p1,p2,AspectRatio->Automatic,PlotLabel->"p4 r = 4 Cos[2 t]"]; Display["p4.ps",p3]; -------------------------------------------------------- 5. Find the slope of r = 3 + 2 Cos[t] at (4, Pi/3) y = (3+2 Cos[t]) Sin[t] dy/dt = -2 Sin[t] Sin[t] + (3+2 Cos[t])Cos[t] dy/dt = -3/2 + (3 + 1) 1/2 dy/dt(Pi/3) = 1/2 x = (3+2 Cos[t]) Cos[t] dx/dt = -2 Sin[t] Cos[t]-(3+2 Cos[t]) Sin[t] dx/dt = -2 Sqrt[3]/2 1/2 -(3+1) Sqrt[3]/2 dx/dt = -Sqrt[3]/2 -2 Sqrt[3] dx/dt(Pi/3) = -5 Sqrt[3]/2 dy/dx (Pi/3 ) = -1/(5 Sqrt[3]) ---------------------------------------------------------- Get["font.math"]; r[t_] := 3 + 2 Cos[ t]; p1 = ParametricPlot[ r[t] { Cos[t], Sin[t]},{t,0,2 Pi}, PlotStyle->{Thickness[0.005]}, PlotRange->All,AspectRatio->Automatic]; p2 = ParametricPlot[ {r[Pi/3] Cos[Pi/3], r[Pi/3] Sin[Pi/3]}+ {t,-t/(5 Sqrt[3])}, {t,-3,3}, PlotStyle->{RGBColor[1,0,0],Thickness[0.005]}]; p3 = ParametricPlot[ {s Cos[Pi/3],s Sin[Pi/3]},{s,0,4}]; p4 = Show[p1,p2,p3, PlotLabel->"p5: r=3+2 Cos[t]", PlotRange->All,AspectRatio->Automatic]; Display["p5.ps",p4]; -------------------------------------------------------------------------- 6. Change this polar equation to rectangular coordinates. 2 r = ----------------- 1 - Cos[t] r(1-Cos[t] = 2 r - r Cos[t] = 2 2 2 Sqrt[x + y ] - x = 2 2 2 Sqrt[x + y ] = x+2 2 2 2 x + y = x + 4 x + 4 2 y = 4(x+1) ---------------------------------------------------- Get["font.math"]; r[t_] := 2/(1-Cos[t]); P1 = ParametricPlot[ r[t] { Cos[t], Sin[t]},{t,0.1, 2Pi-0.1}, PlotStyle->{Thickness[0.005]}, PlotRange->All,AspectRatio->Automatic]; P2 = ParametricPlot[ { y^2/4-1,y},{y,-10,10}, PlotStyle->{RGBColor[1,1,0],Thickness[0.01]}]; P3 = Show[P2,P1, PlotLabel->"p6: r = 2/(1-Cos[t])" PlotRange->All,AspectRatio->Automatic]; Display["p6.ps",P3]; --------------------------------------------------------- 7. Find the area of one of the loops of 2 r = 8 Cos[ 2 t] t = Pi/4 INT 1/2 8 Cos[2 t] dt t =-Pi/4 t = Pi/4 INT 4 Cos[2 t] dt t=-Pi/4 _ _ t = Pi/4 | | | 2 Sin[2 t] | = 4 |_ _| t = -Pi/4 ---------------------------------------------------------- Get["font.math"]; Integrate[ 1/2 8 Cos[2 t],{t,-Pi/4,Pi/4}]; r[t_] := Sqrt[8 Cos[2 t]] P1 = ParametricPlot[ r[t] { Cos[t], Sin[t]},{t,-Pi/4, Pi/4}, PlotStyle->{Thickness[0.01]},PlotPoints->100, PlotRange->All,AspectRatio->Automatic]; P2 = ParametricPlot[ -r[t] { Cos[t], Sin[t]},{t,-Pi/4,Pi/4}, PlotStyle->{Thickness[0.01]},PlotPoints->100, PlotRange->All,AspectRatio->Automatic]; P3 = Show[P2,P1, PlotLabel->"p7: r^2 = 8 Cos[2 t]" PlotRange->All,AspectRatio->Automatic]; Display["p7x.ps",P3]; P4 = Table[ ParametricPlot[ r{Cos[t],Sin[t]},{t,-Pi/4,Pi/4}, PlotStyle->{RGBColor[0,1,0],Thickness[0.003]}],{r,0,3,0.25}]; P5 = Table[ ParametricPlot[ r{Cos[t],Sin[t]},{r,0,3}, PlotStyle->{RGBColor[0,1,0],Thickness[0.003]}],{t,-0.75, 0.75, 0.25}]; P6 = Show[P1,P4,P5,PlotLabel->"P7 r^2 = 8 Cos[2 t Box 0.25x0.25 radian", AspectRatio->Automatic, PlotRange->All]; Display["p7y.ps",P6]; ---------------------------------------------------------- 8. Find the points of intersection of the curves r = 1 + Cos[t] r = 1 - Sin[t] 1 + Cos[t = 1 - Sin[t] Cos[t] = -Sin[t] 1 = - Tan[t] Tan[t] = -1; t = 3 Pi/4 or 7 Pi/4; The points are {3 Pi/4, 1-1/Sqrt[2]} {7 Pi/4, 1+1/Sqrt[2]} {0,#} for {0,Pi/2} and (0,0) --------------------------------------------------- Get["font.math"]; r[t_] := 1+Cos[t]; p[r_,t_] := r{Cos[t],Sin[t]}; P1 = ParametricPlot[ r[t] { Cos[t], Sin[t]},{t,0,2 Pi}, PlotStyle->{Thickness[0.005]},PlotPoints->100, PlotRange->All,AspectRatio->Automatic]; r[t_] := 1-Sin[t]; P2 = ParametricPlot[ r[t] { Cos[t], Sin[t]},{t,0,2 Pi}, PlotStyle->{Thickness[0.005]},PlotPoints->100, PlotRange->All,AspectRatio->Automatic]; P3 = ListPlot[{ p[1-1/Sqrt[2],3 Pi/4],p[1+1/Sqrt[2],7 Pi/4] , p[0,0] }, PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; P4 = Show[P1,P2,P3,PlotLabel->"P8 r = 1 + Cos[t] & r = 1 - Sin[t]"]; Display["p8.ps",P4]; ------------------------------------------------------------------ 9. Set up the integral for the length of the Cardioid r = 1+Cos[t] 0 <= t <= 2 Pi. x = (1+Cos[t]) Cos[t] y = (1+Cos[t]) Sin[t] dx/dt = - Sin[t] Cos[t] - (1 + Cos[t]) Sin[t] = - Sin[t] Cos[t] - Sin[t] - Sin[t] Cos[t] = -Sin[t]( 2 Cos[t] + 1) dy/dt = - Sin[t] Sin[t] + (1 + Cos[t]) Cos[t] 2 2 = -Sin[t] + Cos[t] + Cos[t] 2 2 = Cos[t] + Cos[t] - Sin[t] t = 2 Pi 2 2 2 2 2 INT Sqrt[ Sin [t] (1+2 Cos [t]) + (Cos[t]+Cos [t] - Sin [t]) ] dt t = 0 ----------------------------------------------------------------------------- Get["font.math"]; Integrate[Sqrt[Sin[t]^2 (1+2Cos[t])^2+(Cos[t]+Cos[t]^2-Sin[t]^2)^2],{t,0,2 Pi}]; r[t_] := 1 + Cos[ t]; P1 = ParametricPlot[ r[t] { Cos[t], Sin[t]},{t,0,2 Pi}, PlotStyle->{Thickness[0.02]}, PlotRange->All,AspectRatio->Automatic]; a = 8/(2 Pi); P2 = ParametricPlot[ {a/2+0.1,0}+ a {Cos[t],Sin[t]},{t,0,2 Pi}, PlotStyle->{RGBColor[0,1,0],Thickness[0.01]}]; P3 = Show[P1,P2,PlotLabel->"P9, Black:r=1+Cos[t];Green=Same Length Circle", PlotRange->All, AspectRatio->Automatic]; Display["p9.ps",P3]; ------------------------------------------------------------------------ 10. Find all points on the cardioid r = a(1+Cos[t] where (a) the tangent line is horizontal. (b) the tangent line is vertical. x = a(1+Cos[t]) Cos[t] y = a(1+Cos[t]) Sin[t] x = a( Cos[t] + Cos[t]^2 y = a( Sin[t] + Sin[t] Cos[t]) dx/dt = a(-Sin[t] -2 Cos[t] Sin[t] ) dy/dt = a(Cos[t] + Cos[t]^2 - Sin[t]^2 ) 2 dx/dt = -a Sin[t] ( 1 + 2 Cos[t]) dy/dx = a(2 Cos[t] + Cos[t] -1 ) dy/dx = a(2 Cos[t]-1)(Cos[t]+1) a( 2 Cos[t]-1)(Cos[t]+1) dy/dx = ----------------------------- -a Sin[t] (2 Cos[t]+1) Tangent is horizontal when Cos[t] = -1 or Cos[t] = 1/2 t = Pi Pi/3, -Pi/3 Tangent is vertical when Sin[t] = 0 Cos[t] = -1/2 t = 0, Pi 2 Pi/3, 4 Pi/3 ------------------------------------------------------------- Get["font.math"]; a = 5; f[t_] := { a (1+Cos[t]) Cos[t], a ( 1+Cos[t]) Sin[t] }; fp[t_] = D[f[t],t]; P1 = ParametricPlot[f[t],{t,0,2 Pi}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P2 = ListPlot[ { f[Pi], f[Pi/3], f[-Pi/3], f[0], f[2 Pi/3], f[4 Pi/3]}, PlotStyle->{RGBColor[0,1,0],PointSize[0.04]}]; P3 = Show[P1,P2,PlotLabel->"P10, r = a(1+Cos[t])", AspectRatio->Automatic, PlotRange->All]; Display["p10.ps",P3]; -------------------------------------------------------------- 11. (a) Write the rectangular equations for the cycloid generated by a wheel of radius a with parameter t. (b) Find the slope of the cycloid in terms of the parameter t. A point P on the rim of a wheel of radius a is initially at the origin. As the wheel rolls to the right along the x-axis, P traces out a curve called a CYCLOID. | | ..... | . . | . . | . a . | (P,Q) .________ . | . t | . | | | at. |a . | . | . ______|_____________ ....._________________________ at (P,Q) = (a t ,0) + (0,a) +(-a Sin[t], - a Cos[t]) (P,Q) = a( t-Sin[t],1-Cos[t]) / / (P ,Q ) = a( 1-Cos[t], Sin[t]) dy/dx = Sin[t]/(1-Cos[t]) -------------------------------------------------------------- Get["font.math"]; a = 1; P1 = ParametricPlot[ {t-Sin[t],1-Cos[t]},{t,0,2 Pi}, PlotStyle->{Thickness[0.005]}]; f[w_] := ParametricPlot[ {w,a} + a{Cos[t],Sin[t]},{t,3 Pi/2-w,3 Pi/2}, PlotStyle->{RGBColor[1,0,0],Thickness[0.005]}]; g[w_] := ParametricPlot[ {w,a} + a{Cos[t],Sin[t]},{t, -Pi/2, 3 Pi/2-w }, PlotStyle->{RGBColor[0,0,1],Thickness[0.005]}]; h[w_] := ListPlot[{ {w-Sin[w],1-Cos[w]} },PlotStyle->{RGBColor[0,1,0], PointSize[0.01]}]; P2 = Show[P1, f[1.5], g[1.5], h[1.5], f[4.0], g[4.0], h[4.0], PlotLabel->"P11 Cycloid",PlotRange->All, AspectRatio->Automatic]; q[w_] := Sin[w]/(1-Cos[w]); P3 = Table[ ParametricPlot[ {w-Sin[w],1-Cos[w]} + t {1,q[w]}/Sqrt[1+q[w]^2], {t,0,1},PlotStyle->{RGBColor[1,0,0],Thickness[0.002]} ],{w,1,6}]; P4 = Show[P1,P3,PlotLabel->"P11 Cycloid",AspectRatio->Automatic, PlotRange->All ]; Display["p11.ps",P4]; ------------------------------------------------------------------------------ 12. Let x = 5 Cos[t] and y = 4 Sin[t]. 2 2 Find dy/dx and d y/dx dx/dt = -5 Sin[t] dy/dt = 4 Cos[t] dy/dx = 4 Cos[t]/(-5 Sin[t] = -4/5 Ctn[t] 2 d(dy/dx)/dt = 4/5 Csc [t] 2 2 2 3 d y/dx = 4/5 Csc [t]/(-5 Sin[t] = -4/25 Csc [t] 13. Find the length of the parametric curve defined by 3/2 x = t and y = t 0 <= t <= 3. t=3 S = INT Sqrt[ 1 + 9/4 t ] dt t=0 u = 1+9/4 t du = 9/4 dt t=3 1/2 S = INT u 4/9 du t=0 _ 3/2 _ u=31/4 | u | t=3 S = 4/9 | ---- | |_ 3/2 _| t=0 u=1 3/2 S = 8/27 Sqrt[ (31/4) - 1 ] = 6.09632 ---------------------------------------------------- Integrate[ Sqrt[1+9/4 t],{t,0,3}] P1 = ParametricPlot[{t,t^(3/2)},{t,0,3}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; P2 = Show[P1,PlotLabel->"P13 {t,t^(3/2) }", PlotRange->All,AspectRatio->Automatic]; Display["p13.ps",P2]; #######################################################################33