Instructor Hentzel Office Phone: 515-294-8141 E-mail: hentzel@iastate.edu Math Department Fax: 515-294-5454 http://orion.math.iastate.edu/hentzel/class.166.05 Textbook: Calculus by Varberg, Purcell, Rigdon, eight edition. January 21 6.5 Work p303:1,3,(6),7,9,(14),(18),(19), Main Idea: Work = Force x distance. Key Words: Work, Spring Constant Goal: Learn that Work = INT Force ds -------------------------------------------------------- Previous assignment: Page 299 Problem 8 Find the length of the indicated curve. 2 3/2 y = 2/3 (x + 1) between x=1 and x=2 2 1/2 y' = (2/3)(3/2)(x +1) (2x) 2 y' = 2x Sqrt[(x +1) 2 2 2 2 4 2 Sqrt[ 1 + (y') ] = Sqrt[ 1 + 4x (x +1) ] = Sqrt[ 1 + 4x + 4x ] = 1+2x x=2 2 x=2 2 | 3 |x=2 INT Sqrt[1+(y') ] dx = INT 1 + 2x dx =|x + 2x /3 | = 2+16/3-(1+2/3) = 17/3 x=1 x=1 | |x=1 ------------------------------------------------------------------- f[x_] := 2/3 (x^2+1)^(3/2) s[x_,h_] := Sqrt[ h^2 + (f[x+h]-f[x])^2 ]; g[n_] := Sum[ s[1+i/n,1/n],{i,0,n-1}]; A = Table[g[n],{n,2,30}]; a = ListPlot[A]; b = Show[a,PlotLabel->"P299p8 y = 2/3(x^2+1)^(3/2) lengths"]; Display["8.ps",b]; c = Plot[f[x],{x,0,4}]; d = Plot[f[x],{x,1,2},PlotStyle->{RGBColor[1,0,0],Thickness[0.008]}]; e = Show[c,d,PlotLabel->"P299p8 y = 2/3(x^2+1)^(3/2)"]; Display["8x.ps",e]; =================================================================== Page 299 Problem 14 Sketch the graph of the given parametric equation and find its length. 3 2 x = t /3 y = t /2 0 <= t <= 1 2 dx/dt = t dy/dt = t 2 2 4 2 2 Sqrt[ (dx/dt) + (dy/dt) ] = Sqrt[ t + t ] = t Sqrt[t + 1] 2 3/2 t=1 2 2 t=1 2 | (t +1) | t=1 INT Sqrt[ (dx/dt) + (dy/dt) ] dt = INT t Sqrt[t +1] = | ------------ | t=0 t=0 | (3/2)(2) | t=0 (1/3)(2 Sqrt[2] - 1 ) = 0.609476 ------------------------------------------------------------------- f[t_] := {t^3/3 , t^2/2 }; s[t_,d_] := Sqrt[(f[t+d]-f[t]).(f[t+d]-f[t])]; g[n_] := Sum[ s[i/n,1/n],{i,0,n-1}]; A = Table[g[n],{n,5,100}]; a = ListPlot[A]; b = Plot[(1/3)(2 Sqrt[2] - 1 ),{x,0,100},PlotStyle->{RGBColor[1,0,0]}]; c = Show[a,b,PlotLabel->"P299p14 p = {t^3/3, t^2/2}"]; Display["14.ps",c]; d = ParametricPlot[f[t],{t,0,1}]; e = Show[d,PlotLabel->"P299p14 f[t_] := {t^3/3 , t^2/2 }"]; Display["14x.ps",e]; ================================================================== Page 299 Problem 21 Find the length of each curve. u=x 3 (a) y = INT Sqrt[ u -1] du 1 <= x <= 2 u=1 3 dy/dx = Sqrt[x - 1] x=2 2 x=2 3 x=2 3/2 INT Sqrt[ 1+ (y') ] dx = INT Sqrt[1 + x - 1] dx = INT x dx x=1 x=1 x=1 | 5/2 | x=2 | x /(5/2) | = 2/5 (4 Sqrt[2]-1) = 1.86274 | | x=1 ------------------------------------------------------------------ f[x_] := NIntegrate[Sqrt[u^3 -1],{u,1,x}]; a = Plot[f[x],{x,1,2},PlotStyle->{RGBColor[1,0,0],Thickness[0.008]}]; b = Show[a,PlotLabel->"P299p21 y = Int Sqrt[u^3-1] {u,1,x}",AspectRatio->Automatic]; Display["21a.ps",b]; ------------------------------------------------------------------------- (b) x = t - Sin[t] y = 1 - Cos[t] 0 <= t <= 4 Pi dx/dt = 1 - Cos[t] dy/dt = Sin[t] 2 2 2 2 (dx/dt) + (dy/dt) = 1 - 2 Cos[t] + Cos[t] + Sin[t] = 2 - 2 Cos[t] t=4 Pi 2 2 t=4 Pi INT Sqrt[(dx/dt) + (dy/dt) ] dt = INT Sqrt[2-2 Cos[t] dt t=0 t=0 _ _ t=4 Pi | 1-Cos[t] | t=4 Pi 2 INT Sqrt| ------- | dt = 2 INT | Sin[t/2] | dt t=0 |_ 2 _| t=0 t=Pi | | t=Pi 8 INT Sin[t/2] dt = 8 |- 2 Cos[t/2] | = 16(0-(-1)) = 16. t=0 | | t=0 --------------------------------------------- f[t_] := {t-Sin[t],1-Cos[t]}; a = ParametricPlot[f[t],{t,0,4 Pi}]; b = Show[a,PlotLabel->"P299p21 {t-Sin[t],1-Cos[t]}",AspectRatio->Automatic]; Display["21x.ps",b]; ============================================================================== Page 299 Problem 24 Find the area of the surface generated by revolving the given curve about the x-axis. 2 y = Sqrt[25-x ], -2 <= x <= 3 2 -1/2 -x dy/dx = 1/2 (25-x ) (-2x) = ------------- Sqrt[25-x^2] x= 3 INT 2 Pi y ds x=-2 2 x= 3 2 x INT 2 Pi Sqrt[25 - x ] Sqrt[ 1 + -------- ] x=-2 2 25-x x= 3 2 2 x= 3 2 Pi INT Sqrt[ 25-x + x ] = 2 Pi INT 5 dx = 50 Pi x=-2 x=-2 ------------------------------------------------------------ f[x_] := Sqrt[25-x^2]; a = ParametricPlot3D[ {x, f[x] Cos[t], f[x] Sin[t]},{x,-2,3},{t,0,2 Pi}]; b = Show[a,PlotLabel->"P299p24 y = Sqrt[25-x^2"]; Display["24.ps",b]; --------------------------------------------------------------------------- New Material: --------------------------------------------------------------------------- Page 301 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. (10,30) | / | /. | / . | / . | / . | / . | / . | / . | / . | / . | / . | (4,12) . | / . |/__________________________ 0 4 10 The spring constant is 12/4 = 3 newtons/cm. work done to stretch 10 cm is x=10 |x=10 INT 3x dx = 3 x^2/2 | = 150 newton cm x=0 |x=0 or 1.5 newton meters which is called 1.5 joules. ---------------------------------------------------------------- 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 Work = INT Pi x^2 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 Work = INT Pi x^2 62.4 (20-y)dy = y=0 y=10 = 0.16 Pi 62.4 INT y^2 (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) = 130690 ft lbs --------------------------------------------- Lifting the water 10 extra feet should require 1/3 Pi 4^2 10 62.4 10 = 104552 ft-lbs 26138.1 + 104552 = 130690 so it checks. --------------------------------------------- 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---> . . .~~~~~~~~~~~~~~~. .~~~~~~~~~~~. ~~~~~~~~~ y=-3 Work = INT 62.4 (2 x) 50 (-y) dy y=-10 y=-3 = 6240 INT -y Sqrt[100-y^2] 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. Force = 12 lbs / ft x=1/2 | |x=1/2 Work = INT 12 x dx = | 6x^2 | = 6/4 = 3/2 ft lbs x=0 | |x=0 --------------------------------------------------- Find the work done in pump the water 5 feet above the top of the tank. The tank is 10 feet long. <------6--------> . . . . . . ..... y= 0 INT 2x 10 (5-y) 62.4 dy y=-3 y= 0 1248 INT x (5-y) dy y=-3 y= 0 1248 INT Sqrt[9-y^2] (5-y) dy y=-3 y= 0 2 y= 0 6240 INT Sqrt[9-y ] + 1248 INT Sqrt[9-y^2] y y=-3 y=-3 | 2 3/2 | y= 0 | (9-y ) | 6240 Pi 3^2/4 + 1248 | ------------ | | 3/2 (-2) | y=-3 6240 Pi 3^3/4 - 1248 27/3 = -11232 + 42120 Pi 6240 27/3 + 6240 Pi 3^2 /4 = 56160 + 14040 Pi Integrate[1248 Sqrt[9-y^2] (5-y),{y,-3,0}] = 11232 + 14040 Pi