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 24 6.6 Mo Cent p310:1,(2),9,(12),(14),15,17-20,(23),24,28,29, Main Idea: How things balance. Key Words: Center of Mass, Centroid, Mx, My Goal: Find the centroid and the center of mass and know the difference. -------------------------------------------------------- Previous assignment: Page 303 Problem 6 For a certain type of nonlinear spring, the force required to keep the spring stretched a distance s is given by the formula 4/3 F = k s If the force required to keep it stretched 8 inches is 2 pounds, how much work is done in stretching this spring 27 inches. 4/3 Find k: 2 = k 8 = 16 k so k = 1/8 7/3 s=27 4/3 | s | s=27 7 W = INT (1/8) s ds = 1/8 | ----- | = 3/56 3 = 6561/56 = 117.161 s=0 | 7/3 | s=0 Approximation f[27]*27/2 = 136.687 Little too big. Why? -------------------------------------------------- Integrate[(1/8) s^(4/3),{s,0,27}] f[s_] := (1/8) s^(4/3); a = Plot[f[s],{s,0,27}]; b = Show[a,PlotLabel->"P303p6 F = ks^(4/3)" ]; Display["6.ps",b]; ================================================= Page 303 Problem 14 Find the work done in pumping all the oil (density delta = 50 pounds per cubic foot) over the edge of a tank . Assume that the tank has circular cross sections of radius 4+x feet at height x feet above the base. Assume that the height is 10 feet and the tank is full of oil. x=10 Work = INT 50 Pi (4+x)^2 (10-x) dx = 445000 Pi/3 = 466003. x=0 ------------------------------------------------------------ Integrate[50 Pi (4+x)^2 (10-x),{x,0,10}]; r[x_] := 4+x; a = ParametricPlot3D[ {r[x] Cos[t],r[x] Sin[t],x},{x,0,10},{t,0,2 Pi}]; b = Show[a,PlotLabel->"P303p14 r[x_] := 4+x"]; Display["14.ps",b]; ======================================================================= Page 303 Problem 18 One cubic foot of air under a pressure of 80 pounds per square inch expands adiabatically to 4 cubic feet according 1.4 to the law p v = c. Find the work done by the gas. x=3 80 Work = INT ------- dx = 71.1212 x=1 1.4 x ------------------------------------ Integrate[80/x^(1.4),{x,1,3}]; f[v_] := 80/v^(1.4); a = Plot[ f[v],{v,1,4}]; b = Show[a,PlotLabel->"P303p18 pv^(1.4) = c"]; Display["18.ps",b]; ============================================= Page 303 Problem 19 A cable weighing 2 pounds per foot is used to haul a 200-pound load to the top of a shaft that is 500 feet deep. How much work is done. x=500 INT 200 + 2(500-x) dx = 350,000 x=0 Integrate[200+2(500-x),{x,0,500}]; The weight itself is 100,000. The cable weighs 500*2 = 1,000 and if it was lifted the whole way it would be 500,000. But on the average it is only lifted half way, or 250,000. -------------------------------------------------------- New Material -------------------------------------------------------- Example 1. Masses of 4 2 6 and 7 kilograms are located at points 0,1,2,and 4 respectively along the x-axis. Find the center of Mass. _ 4*0 + 2*1 + 6*2 + 7*4 42 x = ----------------------- = ---- 19 19 --------------------------------------------------------------- Example 2. The density d(x) of a wire at a point x-centimeters 2 from one end is given by d(x) = 3x grams per centimeter. Find the center of mass of the piece between x=0 and x=10. x=10 3 4 -- INT 3x dx 3x /4 7500 x=0 x = ------------ = ------------ = ----------- = 7.5 cm x=10 2 3 INT 3x dx 3x /3 1000 x=0 --------------------------------------------------------------------- Example 3. Five particles, having masses 1,4,2,3 and 2 units are located at points (6,-1),(2,3),(-4,2),(-7,4) and(2,-2) respectively. Find the center of mass. 1{6,-1}+ 4 {2,3}+2 {-4,2}+ 3{-7,4}+ 2{2,-2} {-11, 23} --------------------------------------------- = ----------- 1+4+2+3+2 12 Example 4. Find the centroid of the region bounded by the 3 curve y = x and y = Sqrt[x]. x=1 3 Area = INT Sqrt[x]-x dx = 5/12 x=0 x=1 My = INT x(Sqrt[x]-x^3 dx = 1/5 x=0 x=1 3 Mx = INT 1/2(Sqrt[x]+x^3)(Sqrt[x]-x ) dx = 5/28 x=0 _ x = My/A = (1/5)/(5/12) = 12/25 _ y = Mx/A = (5/28)/(5/12) = 12/28 ----------------------------------------------------------------------- a = Plot[{Sqrt[x],x^3},{x,0,1}]; b = ListPlot[{{12/25,12/28}},PlotStyle->{RGBColor[1,0,0]}]; c = Show[a,b,PlotLabel->" Page 308 Example 4",AspectRatio->Automatic]; Display["x4.ps",c]; --------------------------------------------------- Page 311 Problem 21. (-3,1)/14 Page 311 Problem 30. Find the centroid of a triangle.