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# View of /trunk/NationalProblemLibrary/Rochester/setDerivatives8RelatedRates/s2_8_21_mo.pg

Wed May 26 16:26:09 2010 UTC (2 years, 11 months ago) by gage
File size: 2487 byte(s)
Adding MathObjects versions of many WeBWorK questions
For use in model_Calculus_1 in particular


    1 #DESCRIPTION
2 #KEYWORDS('derivatives', 'related rates')
3 #TYPE('word problem')
4 # Related rates -- gravel dumped in a pile in a shape of a right circular
5 #  cone with diameter = height.  Find the rate at which the height is growing
6 #  given the volume, height and rate of change of the volume
7 #ENDDESCRIPTION
8
9 ##KEYWORDS('Derivatives')
10 ##Tagged by ynw2d
11
12 ## Modified ('06/20/2008')
13 ## ModifiedBy('nbennett')
14
15 DOCUMENT();        # This should be the first executable line in the problem.
16
18 "PG.pl",
19 "PGbasicmacros.pl",
21 "PGauxiliaryFunctions.pl",
22 "MathObjects.pl",
23 "PGcourse.pl"         # Customization file for the course
24 );
25
26 TEXT(beginproblem());
27 ########################################
28 # Setup
29
30 Context("Numeric");
31 $showPartialCorrectAnswers = 1; 32 33 Context()->variables->add(y=>'Real'); 34 35 Context()->flags->set(reduceConstants=>0); 36 Context()->flags->set(reduceConstantFunctions=>0); 37 38 39$a = random(10,50,10);
40 $h = random(10,25,1); 41 42$rate = Compute("$a*12/(3*($h^2))");
43
44 ########################################
45 # Main Text
46
47 Context()->texStrings;
48 BEGIN_TEXT
49 Gravel is being dumped from a conveyor belt at a rate of $$a$$ cubic feet per minute.
50 It forms a pile in the shape of a right circular cone whose base diameter and height
51 are always the same.  How fast is the height of the pile increasing when the pile is
52 $$h$$ feet high? Recall that the  volume of a right circular cone with height $$h$$
53 and radius of the base $$r$$ is given by $$V= \frac{1}{3}\pi r^2h$$.
54 $BR 55 \{ans_rule(20) \} 56 END_TEXT 57 Context()->normalStrings; 58 59 ######################################## 60 # Answers 61 62 ANS($rate->cmp);
63
64 ########################################
65 # Solution
66
67 Context()->texStrings;
68 SOLUTION(EV3(<<'END_SOLUTION'));
69 We are given that the diameter and height are always the same, so $$2r=h$$ or
70 $$r=\frac{h}{2}$$.
71 $BR$BR The equation $$V= \frac{1}{3}\pi r^2h$$ can be rewritten as
72 $$V= \frac{1}{3}\pi \left( \frac{h}{2} \right)^2h$$ or $$V= \frac{1}{12}\pi h^3$$.
73 $BR$BR Differentiate both sides of the last equation with respect to time $$(t)$$:
74 $BR$BR
75 $$V' = \frac{1}{12}\pi 3 h^2 h'$$
76 $BR$BR
77 $$V' = \frac{1}{12} \pi 3 h^2 h'$$
78 $BR$BR
79 If $$V'=a$$ and $$h=h$$, then
80 $BR$BR
81 $$a = \frac{1}{12} \pi (3) h^2 h'$$
82 $BR$BR
83 $$\displaystyle h' = \frac{a(12)}{(3)h^2\pi}$$
84 END_SOLUTION
85 Context()->normalStrings;
86
87 ########################################
88
89 ENDDOCUMENT();        # This should be the last executable line in the problem.