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Tue Nov 8 15:17:41 2011 UTC (2 years, 5 months ago) by aubreyja
File size: 4470 byte(s)
Rogawski problems contributed by publisher WHFreeman. These are a subset of the problems available to instructors who use the Rogawski textbook. The remainder can be obtained from the publisher.

    1 ## DBsubject('Calculus')
2 ## DBchapter('Limits and Derivatives')
3 ## DBsection('Definition of the Derivative')
4 ## KEYWORDS('calculus', 'derivatives', 'slope')
5 ## TitleText1('Calculus: Early Transcendentals')
6 ## EditionText1('2')
7 ## AuthorText1('Rogawski')
8 ## Section1('10.2')
9 ## Problem1('45')
10 ## Author('LA Danielson')
11 ## Institution('The College of Idaho')
12
13 DOCUMENT();
19
20 #$showPartialCorrectAnswers=1; 21$l=Real(random(1,20,1));
22 $twol = 2 *$l;
23 $fourl = 4 *$l;
24 $lhalves =$l / 2;
25 $lfourths =$l / 4;
26
27 ($angle,$line,$ansfactor,$ymax,$shift) = @{ list_random( 28 [ "\frac{\pi}{6}", Compute("(1/sqrt(3))*(x-1)+1"), 3, 3, 0 ], 29 [ "\frac{\pi}{4}", Compute("x"), 2, 4, .05 ], 30 [ "\frac{\pi}{3}", Compute("sqrt(3)*(x-1)+1"), 1, 5, .1 ] 31 )}; 32 33$firstdiag = (($ansfactor == 1) ? "$l" : "\frac{$l}{\sqrt{$ansfactor}}");
34 $firstdiagflat = (($ansfactor == 1) ? "$l" : "$l/\sqrt{$ansfactor}"); 35 36 if($l % 2 == 0){
37   $seconddiag = (($ansfactor == 1) ? $lhalves 38 : "\frac{$lhalves}{\sqrt{$ansfactor}}"); 39 } else { 40$seconddiag = "\frac{$firstdiagflat}{2}"; 41 } 42 43$lengthdiag = (($ansfactor == 1) ?$twol : "\frac{$twol}{\sqrt{$ansfactor}}");
44 $totallength = (($ansfactor == 1) ? $fourl 45 : "$twol\left(1+\frac{1}{\sqrt{$ansfactor}}\right)"); 46 47$answer = Compute("2*$l * (1 + 1/sqrt($ansfactor))");
48
49  # Graph the triangle
50   $gr = init_graph(0,0,4,$ymax,'pixels'=>[300,300]);
51   # plot hypotenuse
52   add_functions($gr, 53 "$line for x in <1,3> using color:blue and weight:1");
54
56   $gr->lb( new Label(2,.9,$l,
57   'black','center','center'));
58
59
61   $gr->lb( new Label(1.45,1.2+$shift,'a',
62   'black','center','center'));
63   $gr->lb( new Label(2.6,1.2+$shift,'a',
64   'black','center','center'));
65
66   # and points
67   $gr->stamps( open_circle(1.3,1.07+$shift,'black') );
68   $gr->stamps( open_circle(2.75,1.07+$shift,'black') );
69
70   # side leg opp
71   $gr->moveTo(3,1); 72$gr->lineTo(3,$line->eval(x=>3),'blue'); 73 74 #path in red 75 76 # base leg adj 77$gr->moveTo(1,1);
78   $gr->lineTo(3,1,'red'); 79 80$linepower = 0;
81   $prevy = 1; 82 while($linepower < 4){
83     $thisx = Compute("3-2^(-$linepower)");
84     $thisy =$line->eval(x=>$thisx); 85 86 # draw diagonal 87$gr->moveTo(3,$prevy); 88$gr->lineTo($thisx,$thisy, 'red');
89
90     # then horizontal
91     $gr->moveTo($thisx, $thisy); 92$gr->lineTo(3, $thisy, 'red'); 93 94$prevy = $thisy; 95$linepower++;
96   }
97
98
99 Context()->texStrings;
100 BEGIN_TEXT
101 \{ beginproblem() \}
102 \{ textbook_ref_exact("Rogawski ET 2e", "10.2","45") \}
103 $PAR 104 \{ image(insertGraph($gr), height =>300, width =>300) \}
105 $BR 106 Find the total length of the infinite zigzag path in the above figure 107$BR(each zag occurs at an angle of $$a=angle$$).
108 $PAR Total length of the path = \{ans_rule()\} 109 END_TEXT 110 111 Context()->normalStrings; 112 113 ANS($answer->cmp);
114 Context()->texStrings;
115 SOLUTION(EV3(<<'END_SOLUTION'));
116 $PAR 117$SOL
118 Let us consider the total length, $$L$$, as the sum of the horizontal lengths, $$L_H$$, and the diagonal lengths, $$L_D$$. We compute $$L_H$$ and $$L_D$$ separately.
119 $PAR 120 Each horizontal length is the base of an isosceles triangle and is exactly 1/2 the length of the previous horizontal length, moving up the path. For example, the first (lowest) horizontal line has length $$l$$, and the next lowest horizontal line has length $$lhalves$$, then $$lfourths$$, etc. Therefore, the total length of the horizontal segments is 121 $122 L_H = \sum_{n=0}^{\infty}l\left(\frac{1}{2}\right)^n=\frac{l}{1-\frac{1}{2}}=twol. 123$ 124$PAR
125 To calculate the lengths of the diagonal segments, notice that the first diagonal (the longest of the diagonals) has length $$firstdiag$$. The next diagonal has length $$seconddiag$$, and in general, each diagonal has half the length of the previous (as was the case for the horizontal segments). Therefore, the total length of the diagonal segments is
126 $127 L_D = \sum_{n=0}^{\infty}firstdiag \left(\frac{1}{2}\right)^n = \frac{firstdiagflat}{1-\frac{1}{2}} = lengthdiag. 128$
129 \$PAR
130 Therefore, the total length of the path is
131 $132 L=L_H+L_D=twol+ lengthdiag = totallength. 133$
134 END_SOLUTION
135 ENDDOCUMENT();