This is last years test for practice. Use it if you want to. NAME____________________________________ SCHOOL__________________________________ TEST 1 Monday, January 29, 2007 #### Fax the test answers to: Fax 515-294-5454 and also Snail Mail your test answers to: Irvin Roy Hentzel Department of Mathematics 432 Carver Hall Iowa State University Ames, Iowa 50011-2064 1. Find the volume when the portion of the curve y = (x-2)(x-4) which lies below the x-axis, and to the right of the y-axis is rotated about the y axis. ------------------------------------------------------------ Solution: x=4 V = INT 2 Pi x(x-2)(x-4) dx x=2 = 2 Pi INT x^3 - 6 x^2 + 8 x dx | | x=4 = 2 Pi | x^4/4 -6 x^3/3 + 8 x^2/2 | | | x=2 | | = 2 Pi | 64 -128 + 64 - ( 4 - 16 + 16 ) | | | = 2 Pi ( -4 ) = -8 Pi <== Solution is 8 Pi. Integrate[ 2 Pi x(x-2)(x-4),{x,2,4}] = -8 Pi P1 = Plot[(x-2)(x-4),{x,2,4}]; P2 = Show[P1,PlotLabel->"P1: y = (x-2)(x-4) "AspectRatio->Automatic"]; Display["p1.ps",P2]; P3 = ParametricPlot3D[ { x Cos[t],x Sin[t],(x-2)(x-4) },{x,2,4},{t,0,2 Pi}]; P4 = Show[P3,PlotLabel->"P2: y = (x-2)(x-4) ",AspectRatio->Automatic]; Display["p1x.ps",P4]; ---------------------------------------------------------------------------- 2. Find the volume when the portion of the curve y = x(x-1) which lies below the x-axis, is rotated about the x axis. ---------------------------------------------------------------- Solution: x=1 Volume = INT Pi y^2 dx x=0 = Pi INT x^2 (x-1)^2 dx = Pi INT (x^2-x)^2 dx = Pi INT x^4 -2 x^3 + x^2 dx | | x=1 = Pi | x^5/5 -2 x^4/4 + x^3/3 | | | x=0 = Pi ( 1/5 -2/4 + 1/3 ) = Pi (12 - 30 + 20)/60 = 2 Pi/60 = Pi/30 <=== Answer Integrate[Pi x^2 (x-1)^2,{x,0,1}] = Pi/30 P1 = Plot[ x(x-1),{x,0,1}]; P2 = Show[P1,PlotLabel->"P2: y = x(x-1) "AspectRatio->Automatic"]; Display["p2.ps",P2]; P3 = ParametricPlot3D[ { x ,x(x-1) Sin[t], x(x-1) Cos[t] },{x,0,1},{t,0,2 Pi}]; P4 = Show[P3,PlotLabel->"P3: y = x (x-1) ",AspectRatio->Automatic]; Display["p2x.ps",P4]; ------------------------------------------------------------------------- 3. Find the y coordinate of the Centroid of the area below the x axis and above the curve y = (x+1)(x-1). ---------------------------------------------------------------------- Solution: x=1 A = INT (x+1)(x-1) dx x=-1 x=1 = INT x^2 -1 dx x=-1 | | x=1 = | x^3/3 - x | | | x=-1 = 1/3 -1 -(-1/3 + 1) = -4/3 x=1 Mx = INT 1/2 (x+1)(x-1) (x+1)(x-1) dx x=-1 x=1 = 1/2 INT (x^4 - 2 x^2 + 1) x=-1 | | x=1 = 1/2 | x^5/5 - 2 x^3/3 + x | | | x=-1 = 1/2 ( 1/5 - 2/3 + 1 -( -1/5 + 2/3 - 1) ) = 1/2 ( 3 - 10 + 15 +3 - 10 + 15 )/15 = 16/30 = 8/15 ybar = Mx/area = (8/15)/(-4/3) = -24/60 = -2/5 <== Answer area = Integrate[ (x+1)(x-1),{x,-1,1}] = -4/3 Mx = Integrate[ 1/2(x+1)(x-1)(x+1)(x-1),{x,-1,1}] = 8/15 ybar = (8/15)/(-4/3) = -2/5 ---------------------------------------------------------------------- 3 2 4. Find the length of the curve r(t) = ( t /3, t /2 ) from t = 0 to t = 2. ----------------------------------------------------------- Solution: t=2 INT Sqrt[ t^4 + t^2] dt t=0 t=2 INT t Sqrt[t^2 + 1 ] dt t=0 | 2 3/2 | t=2 | ( t + 1 ) | | ----------- | | (3/2) 2 | t=0 3/2 1/3 (5 - 1) = 3.39345 5 Sqrt[5] -1 Integrate[ Sqrt[ t^4 + t^2 ],{t,0,2}] = --------------- = 3.39345 3 P1 = ParametricPlot[ {t^3/3,t^2/2},{t,0,2}]; P2 = Show[P1,PlotLabel->"P4: r(t) = (t^3/3,t^2/2},PlotRange->All, AspectRatio->Automatic]; Display["p4.ps",P2]; ------------------------------------------------------------------------ 5. A spring requires 3.5 pounds to stretch it 2 inches. (a) How much will a force of 42 pounds stretch it? (b) How much work is required to stretch it 28 inches? ----------------------------------------------------------- Solution: k = 3.5/2 = 7/4 lbs/in 42/(7/4) = 24 inches <========== answer. x=28 W = INT (7/4)x dx x=0 | | x=28 = (7/4) | x^2/2 | | | x=0 = (7/4) 28^2 /2 = 686 inch-pounds or 343/6 ft-pounds = 57.1667 ft-lbs ----------------------------------------------------------------------------- 6. Find the Centroid of the following figure. /\ / \ / \ / \ 13 / \13 / \ / \ / \ /________________\ | 10 | | | 8 | | 8 | | |________________| 10 ---------------------------------------------------------------- Solution: The altitude of the triangle is Sqrt[169-25] = Sqrt[144] = 12 The centroid of the triangle is 4 units above its base. 12*(1/2)(10*12) + 4*10*8 = h( (1/2)*10*12 + 8*10 ) 1040 = 140 h h = 52/7 The centroid is 52/7 above the base on the centerline. ----------------------------------------------------------------- 7. A car weights 7000 pounds. It is on a bridge 200 feet from end A and 500 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. --------------------------------------------------------- 7000 Solution A-------------------------------------------------B 200 500 Moment about B. 7000 500 = a 700 a = 5000 lbs. <=== answer Moment about A. 7000 200 = b 700 b = 2000 lbs <=== answer ------------------------------------------------------------------ 8. A plane passes 5 inches from the center of a sphere of radius 6. What is the surface area of the slice. ------------------------------------------------------------ Solution: ....... _________.______|________.___________________ . | . . | . ._________|___________. . | . 2 2 . | . x + y = 36 .......... x = Sqrt[ 36 - y^2 ] -2y dx/dy = (1/2) --------------- Sqrt[36-y^2] y=6 SA = INT 2 Pi x ds y=5 y=6 y^2 SA = 2 Pi INT Sqrt[36-y^2] Sqrt[ 1 + -----------] dy y=5 36-y^2 y=6 SA = 2 Pi INT Sqrt[ 36-y^2 + y^2 ] dy y=5 | | y=6 SA = 2 Pi 6 | y | = 12 Pi <=== Answer | | y=5 Integrate[ 2 Pi Sqrt[36-y^2] Sqrt[1+y^2/(36-y^2)],{y,5,6}] = 12 Pi ---------------------------------------------------------------------- 9. How much work is done pumping the water from a pyramid out through the top of the pyramid. The base is 50 feet on each side and the height of the pyramid is 60 feet. ___________________________ /\ . /|\ /.|\ . | / | \. . | / . | \ . 60 / | \ . | / . | \ . | / . .| . .\ . . . . | / . |_ \ / \|/ / . |.|.....\ .../.50..........V /. \ / /____________________\/ 50 ---------------------------------------------------- Solution: The length of the side at height s = 50/60 (60-y) 60 Work = INT 62.4 [(50/60)(60-y)]^2 (60-y) dy y=0 60 Work = 62.4 (50/60)^2 INT (60-y)^3 dy y=0 | | y=60 Work = 62.4 (50/60)^2 | -(60-y)^4/4 | | | y=0 Work = 62.4 (50/60)^2 60^4/4 = Work = 62.4 50^2 60^2 /4 = 140,400,000 ft-lbs. Integrate[ 62.4 ( (50/60) (60-y))^2 (60-y),{y,0,60}] = 1.404 10^8 .................................................................... 10. Find the surface area of the surface formed when the portion 3/2 of the curve y = x from x=0 to x= 4 is rotated about the vertical line x = -4/9. ------------------------------------------------------------ Solution: x=4 SA = INT 2 Pi (x+4/9) Sqrt[ 1 + 9/4 x ] dx x=0 x=4 SA = INT 2 Pi (4/9) (9/4x+1) Sqrt[ 1 + 9/4 x ] dx x=0 SA = 8 Pi/9 INT (1+9/4 x)^(3/2) dx 5/2 | (1+9/4 x) | x=4 SA = 8 Pi/9 | ---------------- | | (5/2)(9/4) | x=0 8 Pi/9 2/5 4/9 ( 10^(5/2) - 1) 64 Pi 5/2 ------ (10 - 1) = 156.495 <== Answer 405 -64 Pi 1280 Sqrt[10] Pi Integrate[ 2 Pi (x+4/9) Sqrt[1+9/4 x],{x,0,4}] = ------ + ---------------- = 156.495 405 81