January 26 Review Practice test. The real test is Monday, February 4, 2008 2 1. Find the volume when the portion of the curve y = x + x - 6 which lies below the x-axis, and to the right of the y-axis is rotated about the y axis. ---------------------------------------------------------- 2 Solution: y = x + x - 6 = (x-2)(x+3) x=2 2 INT 2 Pi x (x + x - 6) dx x=0 x=2 3 2 2 Pi INT x + x - 6 x dx x=0 | 4 3 2 | x=2 2 Pi | x /4 + x /3 -6 x /2 | | | x=0 -32 Pi 2 Pi ( 4 + 8/3 - 12) = ------- 3 Probably better to give the surface area as 32 Pi/3 rather than the negative. --------------------------------------------------------------------- a = Plot[(x-2)(x+3),{x,-3,2}]; b = Show[a,PlotLabel->"P1: y = (x-2)(x+3)",AspectRatio->Automatic]; Display["p1x.ps",b]; Integrate[2 Pi x (x^2 + x -6 ),{x, 0, 2}] a = ParametricPlot3D[{ x Cos[t], x Sin[t], x^2 + x - 6}, {x, 0, 2},{t,-Pi/4, 3 Pi/2 }]; b = Show[a, PlotLabel->"P1,y = x^2 + x - 6 about y axis", AspectRatio->Automatic]; Display["p1y.ps",b]; --------------------------------------------------------------------- 2. Find the volume when the portion of the curve 2 y = x + x - 6 which lies below the x-axis, is rotated about the x axis. ---------------------------------------------- 2 Solution: x + x - 6 = (x+3)(x-2) x= 2 2 2 INT Pi (x + x - 6) dx x=-3 x= 2 4 3 2 Pi INT x + 2x -11 x -12 x + 36 dx x=-3 | 5 4 3 2 |x=2 Pi | x /5 + 2x /4 -11 x /3 -12 x /2 + 36 x | | |x=-3 625 Pi Pi ( 32/5 + 8 - 88/3 - 24 + 72 -(-243/5 + 81/2 + 99 -54 -108)) = ------- 6 Pi ( 32/5 + 8 - 88/3 - 24 + 72 + 243/5 - 81/2 - 99 + 54 + 108 ) Pi ( 8-24+72-99+54+108 + 32/5 + 243/5 - 88/3 -81/2 ) Pi ( 8+72+54+108-24-99 + 275/5 - 176/6 - 243/6 ) Pi ( 242 - 123 + 55 - 419/6 ) Pi ( 119 + 55 - 419/6 ) Pi ( 174 - 419/6 ) Pi ( 1044 - 419)/6 = 625 Pi/6 --------------------------------------------------------------------- Integrate[ Pi (x^2 + x -6)^2,{x,-3,2}]; a = ParametricPlot3D[{ x , (x^2 + x - 6) Cos[t], (x^2+x-6) Sin[t]}, {x,-3,2},{t, 0 ,7 Pi/4}]; b = Show[a,PlotLabel->"P2, y = x^2 + x - 6 about x axis", AspectRatio->Automatic]; Display["p2.ps",b]; ----------------------------------------------------------------------------- 3. Find the y coordinate of the Centroid of the area under y = Tan[x] from x=0 to x = Pi/4. --------------------------------------------------------- Solution: x=Pi/4 | | x=Pi/4 A = INT Tan[x] dx = | - ln(Cos[x]) | x=0 | | x=0 A = - ln(1/Sqrt[2]) = 1/2 ln(2) x=Pi/4 2 x=Pi/4 2 Mx = INT 1/2 Tan[x] dx = 1/2 INT Sec[x] -1 dx x=0 x=0 | | x= Pi/4 = 1/2 | Tan[x] - x | = 1/2 (1-Pi/4) | | x=0 _ 1/2(1-Pi/4) 4 - Pi y = Mx/A = --------------- = ---------- = 0.309605 1/2 ln(2) 4 ln(2) -------------------------------------- Get["font.math"]; Integrate[Tan[x],{x,0,Pi/4}]; Integrate[1/2 Tan[x]^2,{x,0,Pi/4}]; a = Plot[ Tan[x], {x,0,Pi/4}]; ybar = Integrate[ 1/2 Tan[x] Tan[x],{x,0,Pi/4}]/Integrate[Tan[x],{x,0,Pi/4}]; xbar = Integrate[ x Tan[x],{x,0,Pi/4}]/Integrate[Tan[x],{x,0,Pi/4}]; b = ListPlot[{{xbar,ybar}},PlotStyle->{RGBColor[1,0,0],PointSize[0.02]}]; c = ParametricPlot[{Pi/4,y},{y,0,Tan[Pi/4]}]; d = ListPlot[ { {0,0}, 1/2 {Pi/4, Tan[Pi/4]} }, PlotStyle->{RGBColor[0,1,0]},PlotJoined->True]; e = ListPlot[ { {Pi/8,0}, {Pi/4, Tan[Pi/4]} }, PlotStyle->{RGBColor[0,1,0]},PlotJoined->True]; f = ListPlot[ { 1/2{Pi/4, Tan[Pi/4]},{Pi/4,0} }, PlotStyle->{RGBColor[0,1,0]},PlotJoined->True]; h = ListPlot[ { {0,0},{Pi/4,Tan[Pi/4]} },PlotJoined->True, PlotStyle->{RGBColor[0,1,0]}]; i = Show[a,b,c,d,e,f,h,PlotLabel->"P3 y = tan[x]",AspectRatio->Automatic]; Display["p3.ps",i]; ----------------------------------------------------------- 4. Find the length of the curve r(t) = {1-Cos[t], Sin[t]} from t = 0 to t = Pi. --------------------------------------------------------- Solution: t=Pi 2 2 INT Sqrt[ Sin [x] + Cos [x] ] dx t=0 t=Pi INT dx = Pi. t=0 ----------------------------------------------------------------------- Get["font.math"]; a = ParametricPlot[{1-Cos[t],Sin[t]},{t,0,Pi}]; b = Show[a,PlotLabel->"P4, r(t) = (1-Cos[t],Sin[t])",AspectRatio->Automatic]; Display["p4.ps",b]; --------------------------------------------------------------------- 5. A spring requires 8 pounds to stretch it 3 inches. (a) How much will a force of 82.5 pounds stretch it? (b) How much work is required to stretch it 18 inches? ----------------------------------------------------- Solution: k = 8/3 (* k = force/distance *) (a) 8/3 x = 82.5 x= 30.9375 inches x=18 | 2 |x=18 (b) INT 8/3 x dx = 8/3 | x /2 | = 4/3*18*18 = 432 in lbs x=0 | |x=0 -------------------------------------------------------- Get["font.math"]; a = Plot[8/3 x,{x,0,31}]; b = ListPlot[ {{3,8},{30.9375,82.5},{18,18*8/3}}, PlotStyle->{RGBColor[1,0,0],PointSize[0.05]}]; c = ParametricPlot[{18,y},{y,0,18*8/3}]; d = Show[a,b,c,PlotLabel->"P5 y = 8/3 x", AspectRatio->Automatic,PlotRange->All]; Display["p5.ps",d]; Integrate[8/3 x,{x,0,18}] -------------------------------------------------------- 6. Find the Centroid of the following figure. --------------- | 10 | | | 9 | |8 | | 10 | |______________ | 8 | ---------- | | | 7 6| | | 12 | -------------------- ------------------------------------------------------------------- (0,15) (10,15) (20,15) ---------------............... | | . | | . | p1 | . | | . (0,7 |_____________|______________. (20,7) | p2 | (0,6)-----------------------------| (20,6) . | | . | p3 | . | | (0,0) *.........-------------------- (8,0) (20,0) p1 = (5, 11) of size 80 ( 400, 880 ) p2 = (10, 6.5) of size 20 ( 200, 130 ) p3 = (14,3) of size 72 (1008, 216 ) ---- ---------------- 172 (1608,1226 ) (1608,1226) 402 613 ---------- = (-----,---- } = {9.34884, 7.12791} 172 43 86 Answer 9.34884 units to the right from left edge and 7.12791 units up from the bottom edge. ----------------------------------------------------- Get["font.math"]; P1 = ListPlot[{ {8,0},{20,0},{20,7},{10,7},{10,15},{0,15},{0,6},{8,6},{8,0} }, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]},PlotJoined->True]; P2 = ListPlot[{{402/43,613/86}},PlotStyle->{PointSize[0.04]}]; P3 = Show[P1,P2,PlotLabel->"Problem 6",AspectRatio->Automatic]; Display["p6.ps",P3]; -------------------------------------------------------- 7. A car weights 5000 pounds. It is on a bridge 500 feet from end A and 900 feet from end B. (a) How much of the weight of the car supported by end A. (b) How much of the weight of the car supported by end B. A======================================================B <-----------500-----><----------------900--------------> 5000 ----------------------------------------------------- Solution: Moment about B should be zero. 900 5000 (a) 900 5000 = 1400 Fa Fa = ---------- = 3214.29 1400 Moment about A should be zero 500 5000 (b) 500 5000 = 1400 Fb Fb = ------------ = 1785.71 1400 ------------ 5000.00 -------------------------------------------------------------------- 8. A plane passes 5 inches from the center of a sphere of radius 6. What is the volume of the slice. ....... _________.______|______.___________________ . | . ' | ` ._________|_________. . | . 2 2 . | . x + y = 36 ........ y=6 2 V = INT Pi x dy y=5 y=6 2 V = INT Pi (36-y ) dy y=5 | 3 | y=6 17 Pi V = Pi | 36 y - y /3 | = Pi ( 216 -216/3 - (180 - 125/3) ) = ------ | | y=5 3 V = Pi (216 - 72 - 180 + 125/3 ) V = Pi (-36 + 125/3) V = Pi (-108+125)/3 V = 17 Pi/3 ------------------------------------------------------------------- Get["font.math"]; a = ParametricPlot3D[{Sqrt[36-y^2] Cos[t],y,Sqrt[36-y^2] Sin[t]}, {y,-6,6},{t,0,2 Pi}, PlotPoints->100]; b = ParametricPlot3D[{x,-5,z},{x,-4,4},{z,-4,4}]; c = ParametricPlot3D[ {Sqrt[36-5^2] Cos[t], -5,Sqrt[36-5^2] Sin[t], {RGBColor[1,0,0],Thickness[0.01]}}, {t,0,2 Pi}]; d = Show[a,b,c,PlotLabel->"Problem 8"]; Display["p8.ps",d]; Integrate[Pi (36-y^2),{y,5,6}] ------------------------------------------------------------------- 9. How much work is done pumping the water from a cone out over the top of the cone. <------20-----> \ / /|\ \ / | \_______ / | \ //|\ 30 ft \ / | | \ / 8 ft | \/ \|/ \|/ <------20------> \ | / /|\ \ | x / | \====|====/ | \ y| / 30 ft \ | / | \ | / | \|/ \|/ x/y = 10/30 x = y/3 y=8 2 Work = INT 62.4 (30-y) Pi (y/3) dy y=0 y=8 2 3 Work = 62.4 Pi/9 INT 30 y - y dy y=0 | 3 4 | y=8 Work = 62.4 Pi/9 | 30 y /3 -y /4 | | | y=0 Work = 62.4 Pi/9 (5120-1024) = 62.4 4096 Pi/9 = 89217.9 ft lbs. Integrate[62.4 (30-y) Pi (y/3)^2,{y,0,8}] --------------------------------------------- 10. Find the surface area of the surface of the volume formed when one arch of the curve y = Sin[x] is rotated about the x-axis. x=Pi 2 SA = INT 2 Pi Sin[x] Sqrt[ 1+Cos [x] ] dx x=0 x=Pi 2 SA = 2 Pi INT Sqrt[ 1 + Cos [x] ] Sin[x] dx x=0 u = Cos[x] du = -Sin[x] u=-1 x=Pi 2 SA = 2 Pi INT Sqrt[ 1 + u ] (-du) x=0 u= 1 | | | | /| | / | | 2 / | Sqrt[1+u ] / | | / |u | / | | / | | /t | |/________|___________ 1 u=-1 2 SA = 2 Pi INT Sqrt[ 1 + u ] (-du) u= 1 u=+1 2 SA = 2 Pi INT Sqrt[ 1 + u ] du u=-1 From equation 44 at the back cover of your book | 2 2 | u= 1 SA = 2 Pi | u/2 Sqrt[u +1]] + 1/2 Ln[u + Sqrt[u +1] | | | u=-1 | | SA = 2 Pi |1/2 Sqrt[2]+1/2 Ln(1+Sqrt[2])-(-1/2 Sqrt[2]+1/2 Ln(-1+Sqrt[2]) | | | SA = 2 Pi (Sqrt[2] + 1/2 Ln(1+Sqrt[2] - Ln(-1+Sqrt[2] ) SA = 2 Pi Sqrt[2] + 1/2 Ln( (Sqrt[2]+1)/ (Sqrt[2]-1) ) SA = 2 Pi ( Sqrt[2] + 1/2 Ln( (Sqrt[2]+1)^2 ] ) SA = 2 Pi ( Sqrt[2] + Log[ Sqrt[2]+1 ] Integrate[Sqrt[1+u^2],{u,-1,1}] Integrate[2 Pi Sin[x] Sqrt[1+Cos[x]^2],{x,0,Pi}] = 2 Pi (Sqrt[2] + ArcSinh[1]) -------------------------------------------------------------- Get["font.math"]; a = ParametricPlot3D[{x, Sin[x] Cos[t], Sin[x] Sin[t]},{x,0,Pi},{t,0,2 Pi}]; b = Show[a,PlotLabel->"P10 y = Sin[x]"]; Display["p10.ps",b]; --------------------------------------------------------------