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, May 2. 1. Find the area inside r = 3 + Sin[theta] t=2Pi 2 INT 1/2 (3+Sin[t]) dt t=0 2. Find the area in the first quadrant which is common to both of two curves r = 1 + Sin[theta] and r = 1 + Cos[theta] 3. Name the curves. (a) r = Cos[theta] (b) r = 1 + Cos[theta] (c) r = 2 + Cos[theta] (d) r = theta (e) r theta = 1 (f) r = Cos[2 theta] 2 (g) r = Cos[theta] 4. Find the area of one leaf of r = 4 Cos[2 theta] 5. Find the slope of r = 3 + 2 Cos[theta] at (4, Pi/3) -------------------------------------------------------- 1. Find the area inside r = 3 + Sin[theta] t=2Pi INT 1/2 (3+Sin[t])^2 dt = 19 Pi/2 t=0 --------------------------------------------------- Integrate[ 1/2 (3+Sin[t])^2,{t,0,2 Pi}] r[t_] := 3 + Sin[t]; p1 = ParametricPlot[ r[t] {Cos[t], Sin[t]},{t,0,2 Pi}, PlotLabel->"p6: r= 3 + Sin[theta]"]; AspectRatio->Automatic, PlotRange->All, Display["p6.ps",p1]; ---------------------------------------------------------- 2. Find the area in the first quadrant which is common to both of two curves r = 1 + Sin[theta] and r = 1 + Cos[theta] ---------------------------------------------------- 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]}]; p2 = ParametricPlot[ g[t] { Cos[t], Sin[t] },{t,0,2 Pi}]; p3 = Show[p1,p2,PlotLabel->"p7: (red) r=1+Sin[t] and (black) r=1+Cos[t]", PlotRange->All,AspectRatio->Automatic]; Display["p7.ps",p3]; --------------------------------------------------------- 1 + Sin[theta] = 1 + Cos[theta]; Sin[theta] = Cos[theta] Theta = Pi/4 or 5 Pi/4 t=Pi/2 A1 = INT 1/2 (1+Cos[t])^2 dt = 7/8 - 1/Sqrt[2] + 3 Pi/16 t=Pi/4 t= Pi/4 A2 = INT 1/2 (1+Sin[t])^2 dt = 7/8 - 1/Sqrt[2] + 3 Pi/16 t=0 A1 + A2 = 7/4 -Sqrt[2] + 3 Pi/8; <==== Answer -------------------------------------------------------- 3. Name the curves. (a) r = Cos[theta] Circle (b) r = 1 + Cos[theta] Cardioid (c) r = 2 + Cos[theta] Limacon (d) r = theta Archimedian Spiral (e) r theta = 1 Hyperbolic Spiral (f) r = Cos[2 theta] Four Leaved Rose 2 (g) r = Cos[ 2 theta] Lemniscate ----------------------------------------------------------- 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]}]; 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]}]; 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]}]; r[t_] := t; pd = ParametricPlot[{r[t] Cos[t],r[t] Sin[t]},{t,0,2 Pi}, AspectRatio->Automatic,PlotStyle->{RGBColor[0,1,0]}]; r[t_] := 1/t; pe = ParametricPlot[{r[t] Cos[t],r[t] Sin[t]},{t,0, Pi}, AspectRatio->Automatic,PlotStyle->{RGBColor[0,0,1]}]; 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]}]; r[t_] := Sqrt[Cos[t]] pg = ParametricPlot[{r[t] Cos[t],r[t] Sin[t]},{t,0,2 Pi}, AspectRatio->Automatic,PlotStyle->{RGBColor[1,0,0]}]; ph = Show[pa,pb,pc,pd,pe,pf,pg,PlotLabel->"p8", AspectRatio->Automatic]; Display["p8.ps",ph]; ---------------------------------------------------------- 4. Find the area of one leaf of r = 4 Cos[2 theta] t = Pi/4 INT 1/2 (4 Cos[2 t])^2 dt = 2 Pi t= -Pi/4 ------------------------------------------------------------ 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}, PlotRange->All,AspectRatio->Automatic]; p2 = ParametricPlot[{2 Cos[t],Sin[t]}+{2,0},{t,0,2 Pi}, PlotStyle->{RGBColor[1,0,0]}]; p3 =Show[p1,p2,AspectRatio->Automatic,PlotLabel->"p9 r =4 Cos[2 theta]"]; Display["p9.ps",p3]; -------------------------------------------------------- 5. Find the slope of r = 3 + 2 Cos[theta] 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(Pi/3) = 1/2 x = (3+2 Cos[t]) Cos[t] dx/dt = -2 Sin[t] Cos[t]-(3+2Cos[t]) Sin[t] dx/dt(Pi/3) = -5 Sqrt[3]/2 dy/dx (Pi/3 ) = -1/(5 Sqrt[3]) ---------------------------------------------------------- r[t_] := 3 + 2 Cos[ t]; p1 = ParametricPlot[ r[t] { Cos[t], Sin[t]},{t,0,2 Pi}, PlotStyle->{Thickness[0.02]}, PlotRange->All,AspectRatio->Automatic]; p2 = ParametricPlot[ {r[Pi/3] Cos[Pi/3], r[Pi/3] Sin[Pi/3]}+ {t,-t/(5 Sqrt[3])}, {t,-5,5}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; p3 = ParametricPlot[ {s Cos[Pi/3],s Sin[Pi/3]},{s,0,4}]; p4 = Show[p1,p2,p3, PlotLabel->"p10: r=3+2 Cos[t]", PlotRange->All,AspectRatio->Automatic]; Display["p10.ps",p4]; -------------------------------------------------------------------------- 9. Set up the integral for the length of the Cardioid r = 1+Cos[theta] 0 <= theta <= 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] = -Sin[t]^2 + 1 + Cos[t]^2 = 1 + Cos[t]^2 - Sin[t]^2 = 2 Cos[t]^2 t=2 Pi INT Sqrt[ Sin[t]^2 (1+2 Cos[t])^2 + 4 Cos[t]^4 ] dt t=0 t=2 Pi INT Sqrt[ Sin[t]^2 (1+4 Cos[t] + 4 Cos[t]^2 + 4 Cos[t]^4 ] dt t=0 t=2 Pi INT Sqrt[ (1-Cos[t]^2)(1+4 Cos[t] + 4 Cos[t]^2 + 4 Cos[t]^4 ] dt t=0 t=2 Pi INT Sqrt[ 1+4 Cos[t] + 4 Cos[t]^2 + 4 Cos[t]^4 ] dt t=0 -Cos[t]^2 -4 Cos[t]^3 - 4 Cos[t]^4 t=2 Pi INT Sqrt[ 1+4 Cos[t] + 3 cos[t]^2 - 4 Cos[t]^3 ] dt t=0