Practice TEST 3 The actual Test 3 is Friday, November 13, 2009 During Class 7:30 to 9:00 AM Practice test. Each problem has two parts. First write the integral as INT f(u) du and then integrate it. 1. INT Sqrt[3x+2] dx 1/2 (a) 1/3 INT (3x+2) 3 dx 3/2 (3x+2) (b) 1/3 --------- + C 3/2 2 2. INT x Sqrt[x + 4] dx 2 1/2 (a) 1/2 INT (x + 4) 2 x dx 2 3/2 (x + 4) (b) 1/2 ------------- + C 3/2 2 3 3. INT x Sin[ 6 x - 7 ] dx 3 2 (a) 1/18 INT Sin[ 6 x - 7 ] 18 x dx 3 (b) 1/18 (-Cos[ 6x - 7]) + C 3 4 4. INT x Sin[ x + Pi ] dx 4 3 (a) 1/4 INT Sin[ x + Pi ] 4 x dx 4 (b) 1/4 (-Cos[x + Pi] ) + C 5. INT Sin[x] + Cos[x] dx (a) INT Sin[x] + Cos[x] dx (b) - Cos[x] + Sin[x] + C 3 6. INT Sin[x] Cos[x] dx 3 (a) INT Sin[x] Cos[x] dx 4 Sin[x] (b) ---------- + C 4 7. INT Sec[x] Tan[x] dx (a) INT Sec[x] Tan[x] dx (b) Sec[x] + C 3 8. INT (3x+1) dx 3 (a) 1/3 INT (3x+1) 3 dx 4 (3x+1) (b) 1/3 ----------- dx 4 x 9. INT -------------- dx 2 2 ( x + 9 ) 2 -2 (a) 1/2 INT ( x + 9) 2 x dx 2 -1 ( x + 9 ) ) (b) 1/2 ---------------- + C -1 1/3 2/3 10. INT x( x + 3 x + 7) dx 4/3 5/3 (a) INT x + 3 x + 7 x dx 7/3 8/3 2 x x x (b) ---------- + 3 -------- + 7 ---- + C 7/3 8/3 2 ##################################################################### 11. The strength of a beam is proportional to its width and the square of its depth. Find the dimensions of the strongest beam that can be cut from a cylindrical log 4 feet in diameter. .._'_'_-. . ` | /| ` . . | / | . . d| / | . . | \ | . . | \ | . ` . |__w__\| .' ` .... ' 2 2 2 (d/2) + (w/2) = 2 2 Maximize S = k w d 2 2 d + w = 16 2 S = k w(16-w ) 3 S = k 16 w - k w / 2 S = k 16 - k 3 w 2 dS/dw = 0 when w = 16/3 w = 4/Sqrt[3] <========= width 2 d = 16 - 16/3 = 32/3 d = 4 Sqrt[2/3] <======== depth ===================================== Get["font.math"]; p1 = ParametricPlot[ 2 {Cos[t],Sin[t]},{t,0,2 Pi}]; p2 = Plot[(1/10) 2 Abs[x] (2 Sqrt[4-x^2])^2,{x,-2,2}, PlotStyle->{RGBColor[1,0,0]}]; p3 = ParametricPlot[ {2/Sqrt[3],y},{y,-2 Sqrt[2/3],2 Sqrt[2/3]}, PlotStyle->{RGBColor[0,0,1],Thickness[0.01]}]; p4 = ParametricPlot[ {-2/Sqrt[3],y},{y,-2 Sqrt[2/3],2 Sqrt[2/3]}, PlotStyle->{RGBColor[0,0,1],Thickness[0.01]}]; p5 = Plot[2 Sqrt[2/3],{x,-2/Sqrt[3],2/ Sqrt[3]}, PlotStyle->{RGBColor[0,0,1],Thickness[0.01]}]; p6 = Plot[-2 Sqrt[2/3],{x,-2/Sqrt[3],2/ Sqrt[3]}, PlotStyle->{RGBColor[0,0,1],Thickness[0.01]}]; p7 = ListPlot[{ { 2/Sqrt[3], (1/10) 4/Sqrt[3] (4 Sqrt[2/3])^2}, {-2/Sqrt[3], (1/10) 4/Sqrt[3] (4 Sqrt[2/3])^2}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; p8 = Show[p1,p2,p3,p4,p5,p6,p7,PlotLabel->"P 11; Red = 1/10 w d^2 ", PlotRange->All,AspectRatio->Automatic]; Export["/math/www/hentzel/class.165.09/nov11.p11.pdf",p8]; ============================================================== 12. Use Newton's method to approximate a root of y = f(x) / given that f(2) = 3 and f (2) = 1/4. / X = x - f[x]/f [x] / X = 2 - f[2]/f [2] X = 2 - 3/(1/4) = -10 <======= Answer ======================================== Get["font.math"]; t = 1; p1 = Graphics[ Arrow[ { {-10,0}, {2+t, 3+t/4}} ] ]; p2 = ListPlot[{{-10,0},{2,3}},PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; p3 = Show[p1,p2,PlotLabel->"P 12; f(2) = 3 and f'(2) = 1/4", PlotRange->All,AspectRatio->Automatic,Axes->True]; Export["/math/www/hentzel/class.165.09/nov11.p12.pdf",p3]; ====================================== 13. Use Newton's method to approximate a root to five decimal places. x Sin[x] + Cos[x] = 2 f[x_] := x Sin[x] + Cos[x] - 2; fp[x_] := Sin[x] + x Cos[x] - Sin[x]; n[x_] := x - f[x]/fp[x]; x = 1.0; Do[ (Print[N[i,3]," ",N[x,10]," ",N[f[x],10]]; x = n[x]),{i,1,8}]; x f(x) 1. 1. -0.6182267093 2. 2.144223711 -0.7412667515 3. 1.506998186 -0.4323127969 4. 6.006578326 -2.678368019 5. 6.470103835 0.1849337689 6. 6.441014327 -0.00006543163385 -12 7. 6.441024614 -1.314504061 10 -15 8. 6.441024614 1.776356839 10 x = 6.441024614 ============================================= Get["font.math"]; f[x_] := x Sin[x] + Cos[x] -2; fp[x_] = D[f[x],x]; n[x_] := x - f[x]/fp[x]; p1 = Plot[f[x],{x,-10,10}]; A = Table[{0,0},{i,1,8}]; A[[1]] = {1.0, f[1.0]}; Do[ A[[i]] = { n[ A[[i-1,1]] ],f[n[ A[[i-1,1]] ] ]},{i,2,8}]; p2 = ListPlot[ {A[[1]], A[[2]]*{1,0}, A[[2]], A[[3]]*{1,0}, A[[3]], A[[4]]*{1,0}, A[[4]], A[[5]]*{1,0}, A[[5]], A[[6]]*{1,0}, A[[6]], A[[7]]*{1,0}, A[[7]], A[[8]]*{1,0}, A[[8]]},PlotJoined->True,PlotStyle->{RGBColor[1,0,0]},PlotRange->All]; p3 = ListPlot[A,PlotStyle->{RGBColor[1,0,0],PointSize[0.005]}]; p4 = Show[p1,p2,p3,PlotLabel->"P13; f[x_] = x Sin[x] + Cos[x] - 2", PlotRange->All, AspectRatio->Automatic]; Export["/math/www/hentzel/class.165.09/nov11.p13.pdf",p4]; ==================================================================== 14. A car can accelerate from 45 to 60 miles an hour in 10 seconds. How far will it travel as it accelerates from 0 to 75 miles/hr. 66 ft/sec to 88 ft/sec in 10 seconds means an acceleration of 2.2 ft/sec^2 a = 2.2 v = 2.2 t 2 s = 1.1 t 75 mph is 75*88/60 = 110 ft/sec. Thus t = 50 seconds. s(50) = 1.1 2500 = 2750 feet ========================= Get["font.math"]; a = 2.2; p1 = Plot[2.2 t,{t,0,50}]; p2 = ListPlot[{{66/a,66}, {88/a,88},{110/a,110}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.02]}]; p3 = ParametricPlot[ {66/a,y},{y,0,66},PlotStyle->{RGBColor[1,0,0]}]; p4 = ParametricPlot[ {88/a,y},{y,0,88},PlotStyle->{RGBColor[1,0,0]}]; p5 = ParametricPlot[ {110/a,y},{y,0,110},PlotStyle->{RGBColor[1,0,0]}]; p6 = Plot[ 66,{x,0,30},PlotStyle->{RGBColor[0,0,1],Thickness[0.001]}]; p7 = Plot[ 88,{x,0,40},PlotStyle->{RGBColor[0,0,1],Thickness[0.001]}]; p8 = Plot[110,{x,0,50},PlotStyle->{RGBColor[0,0,1],Thickness[0.001]}]; p9 = Show[p1,p2,p3,p4,p5,p6,p7,p8, PlotLabel->"P 14; a = 2.2, 66 to 88 takes 10 seconds", Axes->True, PlotRange->All, AxesOrigin->{0,0}, AxesLabel->{TIME,VELOCITY}, AxesStyle->Directive[Orange,12] ]; Export["/math/www/hentzel/class.165.09/nov11.p14.pdf",p9]; #################################################################### 15. A car can accelerate from 45 mph to 60 mph in 176 feet. How far will it travel to accelerate from 45 mph to 75 mph? 45 mph 60 mph 75 mph 66 fps 88 fps 110 fps <--- 176'------> v = a t + 66 2 s = 1/2 a t + 66 t Using the distance of 176 feet to compute the acceleration. 88 = a t + 66 2 176 = 1/2 a t + 66 t a t = 22 176 = 1/2 22 t + 66 t 176 = 77 t t = 176/77 = 16/7 a = 22/(16/7) = 77/8 So the formulas are: a = 77/8; v = 77/8 t + 66 2 s = 1/2 77/8 t + 66 t v = 110 fps when 110 = 77/8 t + 66 44 = 77/8 t 44 8/77 = t 32/7 = t 2 s(32/7) = 1/2 77/8 (32/7) + 66 32/7 = 2816/7 feet = 402.286 feet / \/ v = 77/8 t + 66 2 s = 1/2 77/8 t + 66 t When v = 88 88 = 77/8 t + 66 22 = 77/8 t 22 8/77 = t 16/7 = t 2 s(16/7) = 1/2 77/8 (16/7) + 66 (16/7) = 176 feet ============================================ Get["font.math"]; v[t_] := 77/8 t + 66; p1 = Plot[v[t],{t,0,32/7}]; p2 = ListPlot[ {{16/7,v[16/7]},{32/7,v[32/7]}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.01]}]; p3 = ParametricPlot[ {16/7,y},{y,0,v[16/7]},PlotStyle->{RGBColor[1,0,0]}]; p4 = ParametricPlot[ {32/7,y},{y,0,v[32/7]},PlotStyle->{RGBColor[1,0,0]}]; p5 = Show[p1,p2,p3,p4, PlotLabel->"P 18; 66 to 88 in 176 feet Find Distance till 110", PlotRange->All, Axes->True, AxesOrigin->{0,0}, AxesStyle->Directive[Orange,15], TicksStyle->Directive[Red,15], AxesLabel->{x,y} ]; Export["/math/www/hentzel/class.165.09/nov11.p15.pdf",p5]; ============================== _20_ \ 16. Evaluate /__ (i+1)(i-2) i=1 i=20 2 SUM i - i - 2 i=1 i=20 2 i=20 i=20 SUM i - SUM i - 2 SUM 1 i=1 i=1 i=1 20 21 41 20 21 -------- - ------- - 2 (20) 6 2 10 7 41 - 210 - 40 = 2870-250 = 2620 Sum[(i+1)(i-2),{i,1,20}] ..................................................... 17. What is the first fundamental theorem of calculus, how does it involve the area function, and why does the area under y=f(x) from x=a to x=b equal F[b]-F[a] where F is any function whose derivative is f(x). (a) The first fundamental theorem of calculus says that the derivative of the area function is the cross section. (b) If we have a function F(x) whose derivative is f(x), then F(x) and A(x) differ by a constant. A(x) is the area under the curve from a to x. (c) Therefore F(b)-F(a) = A(b)-A(a). Since the right hand side is the area. then the left hand side must also be the area. ............................................................... 18. The cross sections of a pond taken 10 feet apart are: 2 3 7 12 18 18 14 7 6 5 3 What is the surface area of the pond? Work the problem with Simpson's method and also with the Trapezoid method. trapezoid Simpson 2 1 2 1 2 3 2 6 4 12 7 2 14 2 14 12 2 24 4 48 18 2 36 2 36 18 2 36 4 72 14 2 28 2 28 7 2 14 4 28 6 2 12 2 12 5 2 10 4 20 3 1 3 1 3 ------- ------- (10/2) 185 = 925 sq ft 275*(10/3) = 916.667 sq ft ============================================= (* 2 3 7 12 18 18 14 7 6 5 3 *) Get["font.math"]; p1 = ListPlot[{ { 0, 2/2}, { 10, 3/2}, { 20, 7/2}, { 30, 12/2}, { 40, 18/2}, { 50, 18/2}, { 60, 14/2}, { 70, 7/2}, { 80, 6/2}, { 90, 5/2}, {100, 3/2}, {100, -3/2}, { 90, -5/2}, { 80, -6/2}, { 70, -7/2}, { 60, -14/2}, { 50, -18/2}, { 40, -18/2}, { 30, -12/2}, { 20, -7/2}, { 10, -3/2}, { 0, -2/2}, { 0, 2/2}}, PlotJoined->True, PlotStyle->{ RGBColor[0,0,1], Thickness[0.001]} ]; p2 = ListPlot[{ { 0, 2/2}, { 10, 3/2}, { 20, 7/2}, { 30, 12/2}, { 40, 18/2}, { 50, 18/2}, { 60, 14/2}, { 70, 7/2}, { 80, 6/2}, { 90, 5/2}, {100, 3/2}, {100, -3/2}, { 90, -5/2}, { 80, -6/2}, { 70, -7/2}, { 60, -14/2}, { 50, -18/2}, { 40, -18/2}, { 30, -12/2}, { 20, -7/2}, { 10, -3/2}, { 0, -2/2}, { 0, 2/2}}, PlotStyle->{RGBColor[0,0,1],PointSize[0.005]}]; p3 = Show[p1,p2, PlotLabel->"P 18; Cross section 2 3 7 12 18 18 14 7 6 5 3 h=10 ", PlotRange->All,AspectRatio->Automatic]; Export["/math/www/hentzel/class.165.09/nov11.p18.pdf",p3]; ==================================================================