View of /trunk/NationalProblemLibrary/270/setDerivatives8RelatedRates/SRM_c2s8p3.pg

    1 ## DESCRIPTION
    2 ##  Calculus
    3 ## ENDDESCRIPTION
    4 
    5 ## KEYWORDS('Calculus','Derivatives')
    6 ## Tagged by dgt5v
    7 
    8 ## DBsubject('Calculus')
    9 ## DBchapter('Differentiation')
   10 ## DBsection('Related Rates')
   11 ## Date('')
   12 ## Author('')
   13 ## Institution('')
   14 ## TitleText1('')
   15 ## EditionText1('')
   16 ## AuthorText1('')
   17 ## Section1('')
   18 ## Problem1('')
   19 
   20 DOCUMENT();        # This should be the first executable line in the problem.
   21 
   22 loadMacros("PG.pl",
   23            "PGbasicmacros.pl",
   24            "PGchoicemacros.pl",
   25            "PGanswermacros.pl",
   26 "PGauxiliaryFunctions.pl");
   27 
   28 TEXT(beginproblem());
   29 $showPartialCorrectAnswers = 1;
   30 
   31 $flowRate=random(6000.0,15000.0,100.0);
   32 $height=random(6.0,15.0,1.0);
   33 $diameter=random(3.0,7.0,0.5);
   34 $level=random(1.0,5.0,0.5);
   35 $levelRate=random(15.0,30.0,1.0);
   36 $levelcm=100*$level;
   37 
   38 TEXT(EV2(<<EOT));
   39 Water is leaking out of an inverted conical tank at a rate of \(!{$flowRate:%5.1f}
   40 \)
   41 cubic centimeters per min at the same time that water is being pumped
   42  into the tank at
   43 a constant rate.  The tank has height \(!{$height:%5.1f}\) meters and the diameter at
   44 the top is \(!{$diameter:%5.1f}\) meters.  If the water level is rising at a rate of
   45 \( !{$levelRate:%5.1f}\) centimeters per minute when the height of the water is
   46 \(! {$level:%5.1f}\) meters,
   47 find the rate at which water is being pumped into the tank in cubic
   48 centimeters per minute.
   49 \{ &ans_rule(30) \}
   50 $BR
   51 Note: Let "R" be the unknown rate at which water is being pumped in.  Then you know that if \(V\) is volume of water, \(\frac{dV}{dt}=R-!{$flowRate:%5.1f}\).  Use geometry (similar triangles?) to find the relationship between the height of the water and the volume of the water at any given time.  Recall that the volume of a cone with base radius r and height h is given by \(\frac{1}{3}\pi r^2 h\).
   52 EOT
   53 
   54 $ans=3.1415926*$diameter*$diameter*$levelcm*$levelcm*$levelRate/(4*$height*$height)+$flowRate;
   55 
   56 ANS(num_cmp($ans, format=>"%0.4f", relTol=>1));
   57 
   58 ENDDOCUMENT();        # This should be the last executable line in the problem.