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. January 19 5.5 Work & Force p306: 2, 4,12,16 Main Idea: Work = Force x distance. Key Words: Work, Spring Constant, 62.4, Hook's Law Goal: Learn that Work = INT Force ds -------------------------------------------------------- Previous assignment: p300:(18),(20),(21),(28) January 18 6.4 Length Page 300 Problem 18 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. Derive parametric equations for the cycloid. | | ..... | ' ` | ' ` | . . |. . |. .| . |. .' | . | .' t| | o' |a . | . | . _____________|______ ....._________________________ (P,Q) = (a t ,0) + (0,a) +(-a Sin[t], - a Cos[t]) (P,Q) = a( t-Sin[t],1-Cos[t]) ---------------------------------------------------------------------- Get["font.math"]; a = 1; P1 = ParametricPlot[ {t-Sin[t],1-Cos[t]},{t,0,2 Pi}, PlotStyle->{Thickness[0.01]}]; f[w_] := ParametricPlot[ {w,a} + a{Cos[t],Sin[t]},{t,3 Pi/2-w,3 Pi/2}, PlotStyle->{RGBColor[1,0,0],Thickness[0.01]}]; g[w_] := ParametricPlot[ {w,a} + a{Cos[t],Sin[t]},{t, -Pi/2, 3 Pi/2-w }, PlotStyle->{RGBColor[0,0,1],Thickness[0.01]}]; 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->"P300 P18 Cycloid",PlotRange->All, AspectRatio->Automatic]; Display["p18.ps",P2]; ---------------------------------------------------------------------- Page 300 Problem 20 Suppose that the wheel of Problem 18 turns at a constant rate w = dtheta/dt, where t is time. Then theta = wt. (a) Show that the speed ds/dt of P along the cycloid is | | ds/dt = 2 a w | Sin[w t/2] | | | R = a( wt-Sin[wt],1+Cos[wt]) dR/dt = a(w-Cos[wt] w, - Sin[wt] w ) dR/dt = a w (1-Cos[wt], -Sin[wt]) | dR/dt | = a w Sqrt[ 1 - 2 Cos[wt] +Cos[wt]^2 + Sin[wt]^2 ] | dR/dt | = a w Sqrt[ 2 - 2 Cos[wt] ] 1- Cos[wt] | dR/dt | = a w Sqrt[4(---------------)] 2 | dR/dt | = a w Sqrt[4( Sin[wt/2]^2 ] | dR/dt | = a w 2 |Sin[wt/2]| (b) When is the speed Maximum? When Sin[wt/2] = (+/-)1 or wt/2 = Pi/2 or 3 Pi/2 + 2 n Pi or wt = Pi or 3 Pi + 4 n Pi or the odd multiples of Pi. Minimum? When Sin[wt/2] = 0 or wt/2 = n Pi or wt 2 n Pi or the even multiples of Pi. (c) Explain why a bug on a wheel of a car going 60 m/h is itself sometimes traveling 120 miles per hour. At the top of the wheel it is going 120 mph. At the bottom it is going zero. ------------------------------------------------------------- Get["font.math"]; a = 2; w = 4; P1 = ParametricPlot[a{w t-Sin[w t],1-Cos[w t]},{t,0,4 Pi/w}]; P2 = ParametricPlot[ {a w t,2 a w Abs[Sin[w t/2]]}, {t,0,4 Pi/w},PlotStyle->{RGBColor[1,0,0]}] P3 = Show[P1,P2,PlotLabel->"P300 p20: w=4 Velocity ",AspectRatio->Automatic, PlotRange->All]; P4= Table[ ParametricPlot[ {a w t,2 a w Abs[Sin[w t/2]]}, {t,0,4 Pi/w},PlotStyle->{RGBColor[1,0,0]}],{w,1,5}]; P5 = Show[P1,P2,P4,PlotLabel->"P300 p20: w=1...5 Velocity", AspectRatio->Automatic, PlotRange->All]; Display["p20.ps",P5]; ---------------------------------------------------------------------- Page 300 Problem 21 (a) Find the length of u=x 3 y = INT Sqrt[u -1 ] du 1 <= x <= 2 u=1 x=2 2 INT Sqrt[ 1 + (dy/dx) ] dx x=1 x=2 3 INT Sqrt[ 1 + x -1 ] dx x=1 x=2 3/2 INT x dx x=1 | 5/2 | x=2 | x | | ----- | = 2/5( 2^(5/2) - 1) = (8 Sqrt[2]-2)/5 = 1.86274 | 5/2 | x=1 ------------------------------------------------------------------------ Get["font.math"]; f[x_] := NIntegrate[Sqrt[u^3-1],{u,1,x}]; P1 = Plot[f[x],{x,1,2},PlotStyle->{Thickness[0.01]}]; P2 = ParametricPlot[{0,y},{y,0,1.5}]; P3 = Plot[0,{x,0,2}]; P4 = ListPlot[{{1,f[1]},{2,f[2]}},PlotStyle->{PointSize[0.02]}]; P5 = Show[P1,P2,P3,P4,PlotLabel->"P300 P21 f[x] = INT(u^3-1),{u,1,x}", AspectRatio->Automatic,PlotRange->All]; Display["p21.ps",P5]; g[x_] := (f[x+0.0001]-f[x])/0.0001; NIntegrate[Sqrt[1+g[x]^2],{x,1,2}] = 1.86283 --------------------------------------------------------------- Page 300 Problem 21 (b) Find the length of the curve x = t - Sin[t] y = 1 - Cos[t] 0 <= t <= 2 Pi dx/dt = 1 - Cos[t] dy/dt = Sin[t] t=4 Pi INT Sqrt[ (1-Cos[t])^2 + Sin[t]^2 ] dt t=0 t=4 Pi INT Sqrt[ 1-2 Cos[t] + Cos[t]^2 + Sin[t]^2 ] dt t=0 t=4 Pi INT Sqrt[ 2-2 Cos[t] ] dt t=0 _ _ t=4 Pi | 1 - Cos[t] | INT 2 Sqrt| ----------- | dt t=0 |_ 2 _| _ _ t=4 Pi | 2 | INT 2 Sqrt| Sin [t/2 ] | dt t=0 |_ _| t=4 Pi || || INT 2 || Sin [t/2 ] || dt t=0 || || t=2 Pi || || t=4 Pi || || INT 2 || Sin [t/2 ] || dt INT 2 || Sin [t/2 ] || dt t=0 || || t=2Pi || || t=2 Pi t=4 Pi INT 2 Sin [t/2 ] dt INT -2 Sin [t/2 ] dt t=0 t=2Pi | - Cos[t/2] | t=2 Pi | - Cos[t/2] | t = 4 Pi 2 | ------------- | -2 | -------------- | | 1/2 | t=0 | 1/2 | t=2 Pi | | | | 4| -Cos[ Pi] -(-Cos[0])| -4|-Cos[ 2 Pi] -(-Cos[Pi]) | | | | | 8 - (-1 -1) = 16. ----------------------------------------------------------- s = 4 (-Cos[t/2] + 1) Cos[t/2] = -s/4+1 t = 2 ArcCos[1-s/4] ...................................................... p[t_] = {t-Sin[t],1-Cos[t]}; Get["font.math"]; P1 = ParametricPlot[ p[t], {t,0,4 Pi}]; P2 = Show[P1,PlotLabel->"P300 p21b {t-Sin[t],1-Cos[t]}", AspectRatio->Automatic,PlotRange->All]; Display["p21bx.ps",P2]; h[s_] := 2 ArcCos[1-s/4 ]; A = Table[ p[h[s]],{s,0,8,1.0}]; B = Table[{2 Pi,0}+p[h[s]],{s,0,8,1.0}]; P3 = ListPlot[A,PlotStyle->{RGBColor[1,0,0]}]; P4 = ListPlot[B,PlotStyle->{RGBColor[1,0,0]}]; P5 = Show[P1,P3,P4, PlotLabel->"P300 p21b {t-Sin[t],1-Cos[t]}", AspectRatio->Automatic,PlotRange->All]; Display["p21by.ps",P5]; Page 300 Problem 28 Find the area of the surface generated by revolving the given curve about the x-axis. 2 x = 1-t y = 2 t 0 <= t <= 1 t=1 t=1 2 INT 2 Pi y ds = INT 2 Pi 2t Sqrt[ 4t + 4] dt t=0 t=0 t=1 2 = 8 Pi INT t Sqrt[t + 1] dt t=0 | 2 3/2 | t=1 | (t + 1) | 3/2 = 8 Pi | ---------------| = 8 Pi/3 (2 -1 ) = 15.3178 | 3/2 2 | t=0 Surface area of 1/2 of a sphere of radius 2 is 1/2( 4 Pi r^2 ) = 8 Pi Surface area of 1/2 of a sphere of radius 1 is 1/2( 4 Pi r^2 ) = 2 Pi ---------------------------------------------------------------- Get["font.math"]; Integrate[ 2 Pi 2 t Sqrt[4 t^2 + 4],{t,0,1}] = 15.3178 P1 = ParametricPlot3D[ {(1-t^2), 2 t Cos[theta], 2t Sin[theta]},{t,0,1}, {theta,-3 Pi/3.5,3 Pi/3.5}]; theta = -3 Pi/3.5; P2 = ParametricPlot3D[ {(1-t^2), 2 t Cos[theta], 2t Sin[theta], {RGBColor[1,0,0],Thickness[0.01]} },{t,0,1}]; theta = 3 Pi/3.5; P3 = ParametricPlot3D[ {(1-t^2), 2 t Cos[theta], 2t Sin[theta], {RGBColor[1,0,0],Thickness[0.01]} },{t,0,1}]; P4 = Show[P1,P2,P3, PlotLabel->"P300 P28 R = {1-t^2,2 t}, 0 <= t <= 1", PlotRange->All ]; Display["p28x.ps",P2]; P5 = ParametricPlot[{1-t^2,2t},{t,0,1},PlotStyle->{Thickness[0.01]}]; P6 = Show[P5,PlotLabel->"Page 300 Problem 28 R = {1-t^2,2t}", AspectRatio->Automatic]; Display["p28y.ps",P6]; P7 = ParametricPlot3D[{ 1-t^2, 2t Cos[u], 2t Sin[u]},{t,0,1},{u,-Pi/2, Pi/2}]; r = 1; P8 = ParametricPlot3D[{ r Cos[t] Sin[w], r Sin[t] Sin[w], r Cos[w]}, {t,0, Pi/2 },{w,0,Pi }]; r = 2; P9 = ParametricPlot3D[{ r Cos[t] Sin[w], r Sin[t] Sin[w], r Cos[w]}, {t,0, Pi/2 },{w,0,Pi }]; P10 = Show[P7,P8,P9,PlotLabel->"Page 300 Problem 28 R = {1-t^2, 2t}" ]; Display["p28z.ps",P10]; --------------------------------------------------------------------------- New Material: Page 302 Example 1: If the natural length of a spring is 0.2 meter and if it takes a force of 12 newtons to keep it extended 0.04 meter, find the work done in stretching the spring from its natural length to a length of 0.3 meter. Force in newtons (10,30) | /| | / | | / | | / | | / | | / | | / | | / | | / | | / | | / | | /|(4,12) | | / | | ./\ /\ /\_|/__|__________|____________ Extension \/ \/ 0 4 10 in centimeters <-----20----> The spring constant is 12/4 = 3 newtons/cm. work done to stretch 10 cm is x=10 2 |x=10 INT 3x dx = 3 x /2 | = 150 newton cm x=0 |x=0 or 1.5 newton meters which is called 1.5 joules. ---------------------------------------------------------------- Page 303 Example 2 A tank in the shape of a right circular cone is full of water. If the height of the tank is 10 feet and the radius of its top is 4 feet, find the work done in (a) pumping the water over the top edge of the tank. (b) pumping th water to a height 10 feet above the top of the tank. <----4----> \ | /| \ | / \ | x / | \ |-------/ \ | / | \ y| / 10 \ | / | \ | / \ | / | \ | / \|/ | (a) x/y = 4/10 y=10 2 Work = INT Pi x 62.4 (10-y)dy = y=0 2 y=10 | 4 y | = INT Pi | ---- |(10-y) 62.4 dy = y=0 | 10 | y=10 2 = 0.16 Pi 62.4 INT y (10-y) dy y=0 y=10 2 3 = 0.16 Pi 62.4 INT 10 y - y dy y=0 3 4 | 10 y y | y=10 = 0.16 Pi 62.4 |------- - --- | | 3 4 | y=0 = 0.16 Pi 62.4 ( 10000/3 - 10000/4) = 26138.1 ft lbs -------------------------------------------- (b) y=10 2 Work = INT Pi x 62.4 (20-y)dy = y= 0 y=10 2 = 0.16 Pi 62.4 INT y (20-y) dy y=0 3 4 | 20 y y | y=10 = 0.16 Pi 62.4 |------- - ___ | | 3 4 | y=0 = 0.16 Pi 62.4 ( 20000/3 - 10000/4) = 130,690 ft lbs ------------------------------------------------- y=10 2 10 2 Volume = INT Pi x dy = INT Pi (4/10 y) dy y=0 y=0 | 3 | y=10 = .16 Pi | y /3 | = 160/3 Pi | | y=0 Weight = Pi 160/3 62.4 = 10,455.2 lbs Work from lifting 10 feet is 104,552 ft lbs Difference = 130,690-26,138.1 = 104,552 So the extra work to pump the water 10 additional feet up, is the same as lifting the entire tank 10 feet and then pumping the water over the top of the tank. --------------------------------------------- Page 304 Example 3 Find the work done in pumping the water over the rim of a tank that is 50 feet long and has a semicircular end of radius 10 feet if the tank is filled to a depth of 7 feet. <--10--><--10--> ________________ /. / / . /. / / / ` / ' / . /. / ...../ / . / 50 / . / / . / / . / / . / / . / / . / / . / / . / / . / /_______________/ . . . . . . . ` . ' . . . ..... y=-3 Work = INT 62.4 2 x 50 (-y) dy y=-10 y=-3 2 = -6240 INT y Sqrt[100-y ] dy y=-10 | 2 3/2 | y=-3 | (100-y ) | = -6240 | ---------- | | 3/2 (-2) | y=-10 3/2 = 2080 91 = 1,805,620 ft lbs ================================================= A force of 6 pounds is required to keep a spring stretched 1/2 foot beyond its normal length. (a) Find the value of the spring constant. (b) Find the work done in stretching the spring 1/2 foot beyond its natural length. (a) 12 lbs/ft | 2 | x=1/2 x=1/2 | x | (b) INT 12 x dx = 12 | ---- | = 6/4 = 3/2 ft-lbs x=0 | 2 | x=0 ans 1.5 ft lbs --------------------------------------------------- Find the work done in pump the water 5 feet about the top of the tank. The tank is 10 feet long. <---3--><---3--> ________________ /. / __________ / . /. /_________/| / / | || / ` / ' | ____ ||/ . /. |___| | || ...../ | || . / | /|| . / |/ || . / | || . /10 /| || . / / | || . / / | || . / / | || . / / | || . / / | || . / /______|__|/____/ . . . . . . . ` . ' . . . ..... y= 0 INT 62.4 2 x (5-y) dy y=-3 y= 0 INT 62.4 2 Sqrt[9-y^2] (5-y) dy y=-3 Integrate[62.4 2 Sqrt[9-y^2] (5-y),{y,-3,0}] = 5534 ft lbs