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Unformatted text preview: MA‘241 Test 2 Form A Spring 2007 Begin each new problem on the top of a new page, back or front.
NO calculators, No PDA, No cell phones , No notes, no etc!!! 1. (10 points each) For each of the following, show every’ step required to set up the
deﬁnite integrals that will solve the problems but do not integrate. x% a. Find the length of the curve y a from x==0 to x=5. b. A spring has a natural length of 1 m. If a 24 N force is required to keep it stretched 3 meters beyond its natural length, how inuch work is required to
stretch it from3mto4m? c. Find the volume of the solid generated by revolving around the yaxis the
‘ 2 3
region bounded by the xaxis and y = 3x + x from x=0 to x=1. d. The tank shown is full of water: Given that water weighs 62.5 lb per cubic foot,
ﬁnd the work required to pump the water out of the tank. ‘ herriisphcrc e. For the lamina of density P formed by the region bounded by y “W
the xaxis from x=0 to x=8, ﬁnd the xcoordinate of the centroid. 2. (12 points) Use Euler’s method with step size ‘/2 to estimate y(l) where y(x) is the ﬂ=2y+x ,y(0)=l
solution of the initialvalue problem dx
Continue on next page Form A page 2 3. (10 points) .For the following problem, identify all variables.
Then set up the differential equation that will solve the problem, but do not solve. A tank contains 1000 gallons of brine with 15 pounds of dissolved salt. Pure water enters the tank at a rate of 10 gallons per minute. The solution is kept
thoroughly mixed and drains ﬁom the tank at the same rate. How much salt is in
the tank aﬁer t minutes? 4. (14 each) Solve the differential equations ' 512:3 , y(1)==4 implicit solution
a 4" Y
dy E=5y , y(1) = 3 explicit solution I did not give nor receive aid on this test. MA 241 Spring 2007 Test 2 solution key
White version Problem 1 a. Find the average value of the functino on the interval :3 = 1 to x = 10 3 ﬁx) = (1 +£I3)2 Solution. The length of the interval is 10 — 1 = 9. The formula for average value over an
interval is the integral of the function divided by the integral length. Hence, we have 1103 9 , (1+a:)2dm Problem 1 b. A uniform cable hanging over the edge of a tall building is 40ft long and
weighs 50 pounds. How much work is required to pull the cable to the top? Solution. The cable weighs 50 pounds, so it weighs 50/40 = 5/4 pounds per foot. So if
we have 40 — y feet of rope, where y is the amount of rope we have pulled up to the top,
hanging off the top of the building, then it weighs a total of (40 — y) * 5/4. We wish to ﬁnd
the work required, but the weight is changing, so we use an integral [0 (40 — y) * 5/4dy Problem 1 c. A vertical dam has a semicircular gate as shown in the ﬁgure on the test
and on WebAssign. Find the hydrostatic force against the gate.
Solution. The formula for hydrostatic force is top of object
/ Density * Depth * (Tiny amount of area)
bottom of object For the sake of simplicity, we will consider the bottommiddle of the gate to be the origin,
that is, the point (0,0). So, we integrate from 0 to 2, since the object is 2 meters tall, and
we are given the density of water to be 9800. The water is 10 meters deep at the bottom of
the gate, so as we go up the gate, the depth is 10 — 3; meters. We now wish to approximate
the area of the gate with very thin horizontal rectangles. Since the gate is centered at the origin, we have the formula 032 + y2 = 4 for the outside of the gate. The width of the gate at
any given point is 2:2: (twice the width from the center to one side), so we solve for a: and get
a: = (/4 — 1/2, so 2515 = 2V4 — y’. So now we are ready to integrate 2
/ 9800(10 — y)2\/4 — 92
0 Problem 1 d. Find the volume of the solid generated by revolving about the yaxis the
region bounded by the x—axis and y = 32: + $3 from a: = 0 to a: = 5. Solution. Since we are revolving this around the yaxis and we have a function giving y in terms of :t, it would be easiest to make our elements of area parallel to the axis of revolution.
The formula is 6
f 271' * radius :0: height * width, Our radius is the distance from the axis of revolution, or just at, our height is simply our
function 333 +303 and the width is the tiny dz. Our limits of integration are given, so we have 5
/ 21x61: + x3)da:
0 Problem 1 e. For the lamina of density p formed by the region bounded by y = 231/3
and the x—axis from a: = 0 to a: = 0, ﬁnd the ycoordinate of the centroid. Solution. The bottom function here is the x—axis, that is, y = 0. We are given our limits
of integration, so we simply substitute our values into the formula 3 atyawn/a
f0 2 dx foe 11/3dx y: _ £5 w
 p "
: 9X ‘97 : 07" x
d); % = _
Z t 2 + _._
——_——2—‘_—"_——'—“—_‘—_—“8
"nd C 0+) = I 2+ ~
. _ 1 3 _____—____"_ _
_______j=_c_______________.;~
2 . 2 4 WW”_#7 1 m— 2 '1] ...
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This note was uploaded on 03/20/2008 for the course MA 141 and 24 taught by Professor Dempster/mccolum during the Spring '08 term at N.C. State.
 Spring '08
 dempster/mccolum

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