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# View of /trunk/NationalProblemLibrary/Rochester/setIntegrals23Work/S06.04.Work.PTP03.pg

Fri Jan 14 22:16:56 2011 UTC (2 years, 5 months ago) by jj
File size: 3246 byte(s)
Fixed bug 1785, typo in library tags.


    1 #DESCRIPTION
2 #  Integration
3 #  Application-based
4 #  Physics--Work.
5 #ENDDESCRIPTION
6
7 #KEYWORDS('Integration', 'Physics', 'Applications')
8 ## kshort tagged and PAID on 2-20-2004
9 ## DBsubject('Calculus')
10 ## DBchapter('Applications of Integration')
11 ## DBsection('Work')
12 ## Date('6/3/2002')
13 ## Author('Arnie Pizer')
14 ## Institution('rochester')
15 ## TitleText1('Calculus: Early Transcendentals')
16 ## EditionText1('4')
17 ## AuthorText1('Stewart')
18 ## Section1('6.4')
19 ## Problem1('20')
20
21 ## Original: /Library/maCalcDB/setIntegrals23Work/ns6_5_12.pg
22
23
24 DOCUMENT();
25
27 "PGbasicmacros.pl",
28 "PGchoicemacros.pl",
30 "PGauxiliaryFunctions.pl"
31 );
32
33 TEXT(beginproblem());
34 $showPartialCorrectAnswers = 1; 35 36 #Here we ensure that the height of the pool is always greater than the depth of 37 #the water 38$c1 = random(5,12,.5);
39 $c2 = random(1,11,.5); 40 @cs =($c1,$c2); 41 @sortedcs = num_sort(@cs); 42$d = $sortedcs[0]; 43$h = $sortedcs[1]; 44$r = random(8,20,.5) ;
45 $w = random(63,66,.1); 46$pi = 4*arctan(1);
47
48 BEGIN_TEXT
49
50 You are visiting your friend Fabio's house. You find that, as a joke, he filled
51 his swimming pool with Kool-Aid, which dissolved perfectly into the water.
52 However, now that you want to swim, you must remove all of the Kool-Aid
53 contaminated water.  The swimming pool is round, with a $r foot radius. It is$h
54 feet tall and has $d feet of water in it.$BR
55 How much work is required to remove all of the water by pumping it over the
56 side?
57
58 Use the physical definition of work, and the fact that the weight of
59 the Kool-Aid contaminated water is $$\sigma = w lbs/ft^3$$ Don't
60 forget to enter the correct \{ helpLink('units') \}.  (You may enter
61 ${BITALIC}lbf${EITALIC} or ${BITALIC}lb*ft${EITALIC} for
62 ${BITALIC}ft-lb${EITALIC}.)
63
64 $BR \{ans_rule(45)\} 65 66 END_TEXT 67 68 &HINT(EV3(<<'EOT')); 69$HINT $BR 70 The formula for work is:$BR
71
72 $\int_{a}^{b} Force * distance$ $BR 73 74 Where distance is the distance over which the force is exerted. 75 EOT 76 77 &SOLUTION(EV3(<<'EOF')); 78$SOL $BR 79 Consider a horizontal cross-section of the pool, with thickness $$dx$$ if we 80 consider the x-axis to be vertical, in the center of the pool. This is 81 just a very short cylinder, so its volume is:$BR
82
83 $$dV = \pi r^2 dx$$. $BR 84 85 We know r, the radius of the pool, is a constant, $$r= r$$. Now that we have 86 the volume of an arbitrary cross-section of the water, we need to find the 87 force which is exerted on the volume. That force is nothing more than the 88 weight. The constant $$\sigma$$ gives us weight-per-volume of the liquid. 89 Therefore, by multipling the volume of the slice by $$\sigma$$, we find:$BR
90
91 $$dF = \sigma \pi r^2 dx$$ $BR 92 93 Since Work (W) is given by:$BR
94
95 $$W = Fx = F\int_{a}^{b}xdx$$ $BR 96 97 in the case of a constant force F, all that remains is to find an expression 98 for $$D$$, the distance each slice of water is lifted. If we consider the top 99 of the pool as x=0,$h-$d is the distance to the surface of the water, since the 100 height of the pool is$h, and the depth is $d. So we have the distance 101 x from x=$h-$d until x=$h. This results in the integral: $BR 102 103 $W = \int_{h-d}^{h} \sigma \pi r^2 xdx$$BR
104
105 which is simple to evaluate.
106 EOF
107
108 $answer = (.5*$w*$pi*$r**2)*(2*$d*$h - $d**2) ; 109 ANS(num_cmp("$answer", units => 'lbf'));
110 ENDDOCUMENT();
111
112