Arkansas Tech University MATH 2924: Calculus II Dr. Marcel B. Finan. Solutions to Assignment 7.6. sin. sin

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1 Arkansas Tech University MATH 94: Calculus II Dr. Marcel B. Finan Solutions to Assignment 7.6 Exercise We have [ 5x dx = 5 ] = 4.5 ft lb x Exercise We have ( π cos x dx = [ ( π ] sin π x = J. From x = to x =, F (x > and the work that results from this force is ( ( π cos x dx = [ ( π ] sin π x = J. π So the force is accelerating the motion of the particle and hence increasing its kinetic energy. From x = to x =, F (x < and the work that results from this force is ( ( π cos x dx = [ ( π ] sin π x = J. π So the force opposes the motion of the particle and hence decreasing its kinetic energy. Exercise The work is the area under the graph of F (x from x = to x = 8 which is given by 8 F (xdx = 4 F (xdx+ 8 4 F (xdx = (4(+(4( = 8 J Exercise 5 By Hooke s Law, the force is given by F (x = kx where x is measured in

2 feet. We are told that F ( = = k (. Solving for k we find k = lb/ft. Now, the work done in stretching the spring from its natural length to 6 in =.5 ft beyond its natural length is.5 xdx = [ 5x ].5 =.75 ft lb Exercise 7 (a By Hooke s Law, F (x = kx. We are told that =. [ x kxdx = k ]. =.7k. Solving for k we find k = 5 9. Now, the work needed to stretch the spring from 5 cm to 4 cm is. [ ] 5 5 x. xdx = = J..5 (b We want x such that f(x = 5 9 x =. Solving for x, we find x =.8 cm Exercise 9 (a Partition the rope into n equally spaced portions each of length x. The portion of the rope between x i and x i weighs.5 x lb. The work required to lift this portion to the top of the building is.5x i x. Thus, the total work required to lift the rope to the top of the building is n lim.5x i x = i= 5.5.5xdx = [.5x ] 5 = 65 ft lb. (b The work required in pulling half of the rope to the top of the building is 5.5xdx = [.5x ] 5 = ft lb 5 5 Exercise Partition the interval [, ] into n equal subintervals each of length x. At the location x i, the mass of the rope is.8x i Kg, the mass of the water is x i Kg, and the mass of the bucket is Kg so that the total mass is +.8x i. The force required to lift this mass a distance x

3 is 9.8( +.8x i N and the corresponding work is 9.8( +.8x i x. Hence, the total work is lim 9.8( +.8x i x = i= 9.8( +.8xdx = J Exercise 5 Partition the interval [, ] into n equal subintervals each of length x. The volume of the layer with thickness [x i, x i ] and depth x i is x. The corresponding mass is x. The force needed to lift this layer is 9.8( x and the corresponding work is 9.8( x i ( x. The total work needed to pump half of the water out of the tank is lim 9.8( x i ( x = i=.5 9.8( x(dx = 45 J Exercise (a b a G m m [ ] dr = Gm r m b r a = Gm ( m a b. (b Using (a, we have ( = J Exercise 5 Partition the interval [, 6] into n equal subintervals each of length x. The area of the i th layer (see figure is 6 x. The pressure on this layer is δx i where δ = 6.5 lb/ft is the density of

4 water. The corresponding hydrostatic force is δx i (6 x. The total hydrostatic force is F = lim 6δx i x = i= 6 6(6.5xdx 6 lb Exercise 7 Partition the interval [, ] into n equal subintervals each of length x. The area of the i th layer (see figure (which is approximately a rectangle is w i x where w i = 9 (x i +. The pressure on this layer is ρgx i = 98x i. The corresponding hydrostatic force is 98x i (w i x. The total hydrostatic force is F = lim 9, 6x i 9 (xi + x = i= 9, 6x 9 (x + dx 6698 N Exercise 9 Partition the interval [, ] into n equal subintervals each of length x. The area of the i th layer (see figure (which is approximately a rectangle is w i x. Using similar triangles we have w i = xi = w i = x i. 4

5 The pressure on this layer is ρgx i = 98x i. The corresponding hydrostatic force is 98x i w i x. The total hydrostatic force is F = lim i= 9, 6x i ( x i x = ( 9, 6x x dx = 98 N Exercise Partition the interval [, 4 ] into n equal subintervals each of length x. The area of the i th layer (see figure (which is approximately a rectangle is w i x. Using similar triangles we have w i x i = 8 4 = w i = x i. The pressure on this layer is ρg(4 x i = 84(9.8(4 x i. The corresponding hydrostatic force is 84(9.8(4 x i w i x. The total hydrostatic 5

6 force is F = lim = i= 4 84(9.8(4 ( x i x i x 84(9.8(4 x ( x dx = N Exercise 6 The moment of the system about the origin is M = ( + 5( + (8 = 54. The mass of the system is m = = 47. The center of mass is x = Exercise 7 The moment of the system about the x axis is M x = 4( + ( + 4(5 =. The moment of the system about the y axis is M y = 4( + ( + 4( = 4. The total mass of the system is m = =. The center of mass is x = 4 =.4 and x = = Exercise 9 The region is shown below. 6

7 The coordinate of the centroid are x = x(xdx xdx = y =.5(x dx xdx = Exercise 4 The region is shown below. The coordinate of the centroid are x = x(ex dx ex dx = e y =.5(ex dx = e + ex dx 4 Exercise 4 The region is shown below. 7

8 The coordinate of the centroid are x = x( x x dx ( x x dx = 9 y =.5[( x (x ]dx (.45 x x dx Exercise 45 The region is shown below. The coordinate of the centroid are π/4 π x(cos x sin xdx 4 x = (cos x sin xdx = π/4.5(cos x sin xdx y = π/4 = (cos x sin xdx 4( Exercise 47 The equation of the line (hypotenuse of the triangle is y = 4x. Thus, m = ρa = (6 = 6. Hence, M x =ρ M y =ρ 4 4 x = 6 6 = 8 y = 6 6 =.5( 4 x dx = 6 x( xdx = 6 4 8