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

Sat Jun 3 13:50:11 2006 UTC (7 years ago) by gage
File size: 2009 byte(s)
Cleaned up code using the convert-functions.pl script


    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
23            "PGbasicmacros.pl",
24            "PGchoicemacros.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.