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Tue Nov 8 15:17:41 2011 UTC (18 months, 1 week ago) by aubreyja
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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('Parametric Equations, Polar Coordinates, and Conic Sections')
3 ## DBsection('Arc Length and Speed')
4 ## KEYWORDS('calculus', 'parametric', 'polar', 'conic')
5 ## TitleText1('Calculus: Early Transcendentals')
6 ## EditionText1('2')
7 ## AuthorText1('Rogawski')
8 ## Section1('11.2')
9 ## Problem1('5')
10 ## Author('Christopher Sira')
11 ## Institution('W.H.Freeman')
12
13 DOCUMENT();
19 $context = Context(); 20 21 22 ($a, $b,$s) = @{ list_random(
23             [3, 4, 4],
24             [5, 10, 9],
25 ) };
26
27 $sup = random(3,5); 28$sub = random(1,2);
29
30 $s2 = 2 *$s;
31 $a2 = 2 *$a;
32 $b3 = 3 *$b;
33 $a2sq =$a2 ** 2;
34 $b3sq =$b3 ** 2;
35 $coeff = Formula("2 *$a");
36 $up = 1 +$s * $sup**2; 37$low = 1 + $s *$sub**2;
38
39 #$ans = Formula("$a2/$s2 * 2/3 * ($up**(3/2) - $low**(3/2))"); 40$ans = Compute("(2*$a2)/(3*$s2)*($up**(3/2) -$low**(3/2))");
41
42
43 Context()->texStrings;
44 BEGIN_TEXT
45 \{ beginproblem() \}
46 \{ textbook_ref_exact("Rogawski ET 2e", "11.2","5") \}
47 $PAR 48 Use equation 4 to calculate the length of the path over the given interval. 49 $c(t) = (a t^2, b t^3), \, sub \le t \le sup$ 50$PAR
51 \{ ans_rule() \}
52 $PAR 53 END_TEXT 54 Context()->normalStrings; 55 56 ANS($ans->cmp);
57
58 Context()->texStrings;
59 SOLUTION(EV3(<<'END_SOLUTION'));
60 $PAR 61$SOL
62 We have $$x = a t^2$$ and $$y = b t^3$$.  Hence $$x' = a2 t$$ and $$y' = b3 t^2$$.  By the formula for the arc length we get
63 $S = \int ^{sup} _{sub} \sqrt{x'(t)^2 + y'(t)^2} \, dt = \int ^{sup} _{sub} \sqrt{a2sq t^2 + b3sq t^4} \, dt = coeff \int ^{sup} _{sub} \sqrt{1 + s t^2} \, t \, dt$
64 Using the substitution $$u = 1 + s t^2, \, du = s2 t \, dt$$ we obtain
65 $S = \frac{a2}{s2} \int ^{up} _{low} \sqrt{u} \, du = \frac{a2}{s2} \cdot \frac{2}{3} u^{3/2} \mid ^{up} _{low} = \frac{\{2*a2\}}{\{3*s2\}} ((up)^{3/2} - (low)^{3/2}) = ans$
66 END_SOLUTION
67
68 ENDDOCUMENT();


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