NAME____________________________________ SCHOOL__________________________________ TEST 4 Monday, April 27, 2007 Fax the test answers to: Fax 515-294-5454 and also Snail Mail your test answers to: Irvin Roy Hentzel Department of Mathematics 432 Carver Hall Iowa State University Ames, Iowa 50011-2064 1. Name the curves. (a) r = Cos[theta] Circle _________________________________ (b) r = 1 + Cos[theta] Cardioid _________________________________ (c) r = 2 + Cos[theta] Limacon _________________________________ (d) r = theta Spiral of Archimedes _________________________________ (e) r theta = 1 Hyperbolic Spiral _________________________________ (f) r = Cos[2 theta] Four leaf rose _________________________________ 2 (g) r = Cos[2 theta] Lemniscate _________________________________ 2. Find the length of the parametric curve defined by x = a ( t - Sin[t] ) y = a ( 1 - Cos[t] ) 0 <= t <= 2 Pi. dx/dt = a(1-Cos[t]) dy/dt = a Sin[t] t=2 Pi 2 2 2 2 INT Sqrt[ a (1 - 2 Cos[t] + Cos [t] + a Sin [t] ] t=0 t=2 Pi 2 2 INT a Sqrt[ (1 - 2 Cos[t] + Cos [t] + Sin [t] ] t=0 t=2 Pi INT a Sqrt[ (2 - 2 Cos[t] ] t=0 _ _ t=2 Pi | 1 - Cos[t] | INT 2 a Sqrt| ------------- | t=0 |_ 2 _| t=2 Pi INT 2 a Sin[ t/2 ] t=0 _ _ t=2 Pi | | | -4a Cos[ t/2] | |_ _| t=0 4a -(-4a) = 8a <======== Answer 2 2 3. Find dy/dx and d y/ dx without eliminating the parameter. 2 2 x = ----------- y = ------------- t =/= 0 2 2 1 + t t ( 1 + t ) 2 2 -4t -2( 1+t + 2 t dx/dt = ------------- dy/dt = ---------------------- 2 2 2 2 2 (1+t ) t (1 + t ) 2 -4t -2 -6t dx/dt = ------------- dy/dt = ---------------------- 2 2 2 2 2 (1+t ) t (1 + t ) 2 -2 -6t ----------------------- 2 2 2 t (1+t ) dy/dx = ==================================== -4 t ---------------- 2 2 (1 + t ) 2 2 2 -2 -6t (1 + t ) dy/dx = ---------------- --------------- 2 2 2 -4 t t (1+t ) 2 -2 - 6 t dy/dx = --------------------- 3 -4 t 2 1 + 3 t dy/dx = --------------------- 3 2 t 3 2 2 2 t 6 t - (1+3 t )6 t d (dy/dx)/dt = ------------------------- 6 4 t 4 2 4 12 t - 6 t - 18 t d (dy/dx)/dt = ------------------------- 6 4 t 4 2 -6 t - 6 t d (dy/dx)/dt = ------------------------- 6 4 t 2 -3 t - 3 d (dy/dx)/dt = ------------------------- 4 2 t 2 -3 t - 3 --------------- 2 4 d y 2 t ---------- = ============================ 2 -4 t dx ----------------- 2 2 (1+t ) 2 2 2 2 2 -3 t - 3 (1 + t ) d y/dx = ------------- ----------- 4 -4 t 2 t 2 3 2 2 -3 (1 + t ) d y / dx = ---------------------- 5 -8 t 2 3 2 2 3 (1 + t ) d y / dx = ---------------------- 5 8 t ------------------------------------------------------------------------ 2 2 x = ----------- y = ------------- t =/= 0 2 2 1 + t t ( 1 + t ) f[t_] := 2/(1+t^2); fp[t_] := D[f[t],t]; g[t_] := 2/(t (1+t^2)); gp[t_] := D[g[t],t]; D[gp[t]/fp[t],t]/fp[t] ----------------------------------------------------------------- 4. Change this polar equation to rectangular coordinates. 2 r = a Cos[ 2 theta ] 2 2 r = a( 1 - 2 Sin [theta]) 4 2 2 r = a( r - 2 (r Sin[theta]) _ _ 2 | 2 2 | 2 2 2 |x + y | = a (x + y ) -2 a y |_ _| _ _ 2 | 2 2 | 2 2 | x + y | = a (x - y ) |_ _| 5. Find the points of intersection of the curves r = 1 + Cos[2 theta] r = 1 - Sin[2 theta] 1 + Cos[2 theta] = 1 - Sin[2 theta] Cos[2 theta] = - Sin[2 theta] 1 = - Tan[2 theta] Tan[2 theta] = -1 2 theta = 3 Pi/4 or 7 Pi/4 or 11 Pi/4 or 15 Pi/4 + theta = 3 Pi/8 or 7 Pi/8 or 11 Pi/8 or 15 Pi/8 Cos[2 theta] = -1/Sqrt[2] +1/Sqrt[2] -1/Sqrt[2] +1/Sqrt[2] Sin[2 theta] = +1/Sqrt[2] -1/Sqrt[2] +1/Sqrt[2] -1/Sqrt[2] r 1-1/Sqrt[2] 1+1/Sqrt[2] 1-1/Sqrt[2] 1+1/Sqrt[2] (1-1/Sqrt[2], 3 Pi/8) (1+1/Sqrt[2], 7 Pi/8) (1-1/Sqrt[2],11 Pi/8) (1+1/Sqrt[2],13 Pi/8) and (0,0). ---------------------------------------------------------------- p[r_,t_] := {r Cos[t],r Sin[t]}; P1 = ParametricPlot[ p[1+Cos[2t],t],{t,0,2 Pi}]; P2 = ParametricPlot[ p[1-Sin[2t],t],{t,0,2 Pi}]; a = 1+1/Sqrt[2]; b = 1-1/Sqrt[2]; P3 = ListPlot[{p[b,3Pi/8],p[a,7Pi/8],p[b,11Pi/8],p[a,15Pi/8],p[0,0]}, PlotStyle->{RGBColor[0,1,0],PointSize[0.02]}]; P4 = Show[P1,P2,P3,PlotLabel->"P5 r=1+Cos[2 t] r=1-Cos[2t]", PlotRange->All, AspectRatio->Automatic]; Display["p5.ps".P4]; ---------------------------------------------------------------- 6. Find the area of the inner loop of r = 1 + Cos[ 1/2 theta ] t= 3 Pi 2 INT 1/2 ( 1 + Cos [1/2 t]) dt t= Pi t= 3 Pi 2 1/2 INT 1 + 2 Cos[1/2 t] + Cos [1/2 t] dt t= Pi t= 3 Pi 1 + Cos[t] 1/2 INT 1 + 2 Cos[1/2 t] + ---------- dt t= Pi 2 t= 3 Pi Cos[t] 1/2 INT 3/2+ 2 Cos[1/2 t] + ---------- dt t= Pi 2 _ _ | | t= 3 Pi 1/2 | 3/2 t + 4 Sin[1/2 t] + 1/2 Sin[t] | |_ _| t= Pi 1/2 ( 9/2 Pi -4 + 0 - (3/2 Pi + 4 + 0 ) 1/2 ( 3 Pi - 8 ) = 3/2 Pi - 4 <=================== 3/4 Pi - 2 <==========Answer ------------------------------------------------------- p[r_,t_] := r{Cos[t],Sin[t]}; P1 = ParametricPlot[ p[1+Cos[t/2],t],{t, Pi,3 Pi}]; P2 = Show[P1,PlotLabel->"P6 r = 1 + Cos[theta/2]", AspectRatio->Automatic,PlotRange->All]; Display["p6.ps",P2]; ------------------------------------------------------- 7. Find the area of one leaf of r = 16 Cos[8 theta] t=Pi/16 2 2 INT 1/2 256 Cos (8 t) dt t=0 t=Pi/16 1 + Cos[16 t] 256 INT --------------- dt t=0 2 _ _ | t Sin[16 t] | t=Pi/16 256 | - + ---------- | |_ 2 32 _| t=0 256( Pi/32 ) = 8 Pi Integrate[ 2 1/2 256 Cos[8t]^2,{t,0,Pi/16}] = 8 Pi 8. Set up the integral for the arclength of the limacon r = 1 + 2 Cos[theta] 0 <= theta <= 2 Pi. x = (1 + 2 Cos[t]) Cos[t] y = (1+2 Cos[t] Sin[t] 2 x = Cos[t] + 2 Cos [t] y = Sin[t] + 2 Sin[t] Cos[t] 2 2 dx/dt = - Sin[t] - 4 Cos[t] Sin[t] dy/dt = Cos[t] + 2 Cos [t] - 2 Sin [t] _ _ _ 2 _ _ 2 _ t= 2 Pi | | | | 2 2 | | S= INT Sqrt| |-Sin[t] - 4 Cos[t] Sin[t] |+|Cos[t] + 2 Cos [t] -2 Sin [t] | | t=0 |_|_ _| |_ _| _| 9. Find the area inside r = 7 + Sin[theta] t= 2 Pi 2 INT 1/2(49 + 14 Sin[t] + Sin [t] ) dt t=0 t= 2 Pi 1-Cos[2t] 1/2 INT (49 + 14 Sin[t] + ---------- ) dt t=0 2 t= 2 Pi 99 -Cos[2t] 1/2 INT (--- + 14 Sin[t] + ---------- ) dt t=0 2 2 _ _ t= 2 Pi | | 1/2 | 99/2 t - 14 Cos[t] -1/4 Sin[2t] | |_ _| t=0 1/2 ( 99 Pi - 14 +14 ) 99 Pi/2 <=========== Answer 2 10. Find the slope of r = Cos[2 theta] when theta = Pi/6. x = Sqrt[ Cos[2 t] ] Cos[t] y = Sqrt[Cos[2 t] ] Sin[t] -1/2 dx/dt = 1/2 Cos [2t] (-Sin[2t] 2) Cos[t] + Sqrt[ Cos[2t] ] (-Sin[t] ) -Sin[2t] Cos[t] dx/dt = ---------------------- - Sqrt[ Cos[2t] } Sin[t] Sqrt[ Cos[2t] ] | -Sqrt[3]/2 Sqrt[3]/2 dx/dt | --------------------- - Sqrt[1/2] 1/2 |t=Pi/6 1/Sqrt[2] | dx/dt | -3/4 Sqrt[2] -1/(2 Sqrt[2]) = -Sqrt[2] |t=Pi/6 -1/2 dy/dt = 1/2 Cos [2t] (-Sin[2t] 2 ) Sin[t] + Sqrt[ Cos[2t] ] Cos[t] - Sin [2t] Sin[t] dy/dt = -------------------------------- + Sqrt[ Cos[2t] ] Cos[t] Sqrt[ Cos[2t] ] | -Sqrt[3]/2 1/2 dy/dt | ----------------- + Sqrt[1/2] Sqrt[3]/2 |t=Pi/6 Sqrt[1/2] | - Sqrt[3] Sqrt[2] Sqrt[3] dy/dt | -------------- + -------- |t=Pi/6 4 2 Sqrt[2] | - Sqrt[3] Sqrt[2] Sqrt[3] Sqrt[2] dy/dt | -------------- + ---------------- |t=Pi/6 4 4 | dy/dt | | t = Pi/6 = 0 0 dy/dx = ================ = 0 -Sqrt[2] -------------------------------------------------------------- p[r_,t_] := r{Cos[t],Sin[t]}; P1 = ParametricPlot[ p[ Sqrt[ Cos[2 t] ],t], {t,-Pi/2,Pi/2 }]; P2 = ListPlot[ { p[ Sqrt[ Cos[Pi/3] ],Pi/6]}, PlotStyle->{RGBColor[1,0,0],PointSize[0.02]}]; P3 = Show[P1,P2,PlotLabel->"P10",PlotRange->All, AspectRatio->Automatic]; Display["p10.ps",P3]; -------------------------------------------------------------- f[t_] := Sqrt[Cos[2t]] Cos[t]; fp[t_] = D[f[t],t]; g[t_] := Sqrt[Cos[2t]] Sin[t]; gp[t_] = D[g[t],t]; gp[Pi/6]/fp[Pi/6]