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

    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 
   26 loadMacros(
   27 "PGbasicmacros.pl",
   28 "PGchoicemacros.pl",
   29 "PGanswermacros.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