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Tue Nov 8 15:17:41 2011 UTC (2 years, 5 months 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('16')
   10 ## Author('Christopher Sira')
   11 ## Institution('W.H.Freeman')
   12 
   13 DOCUMENT();
   14 loadMacros("PG.pl","PGbasicmacros.pl","PGanswermacros.pl");
   15 loadMacros("PGchoicemacros.pl");
   16 loadMacros("Parser.pl");
   17 loadMacros("freemanMacros.pl");
   18 loadMacros("PGchoicemacros.pl");
   19 $context = Context();
   20 
   21 
   22 $a = random(2, 5);
   23 $b = random(6, 9);
   24 $c = random(2, 9);
   25 
   26 $ac = $a * $c;
   27 $bc = $b * $c;
   28 $acsq = $ac**2;
   29 $bcsq = $bc**2;
   30 $csq = $c**2;
   31 $q = $bcsq - $acsq;
   32 $acsqonc = $acsq / $csq;
   33 $bcsqonc = $q / $csq;
   34 
   35 $exp = $c * ($acsqonc + $bcsqonc * (sin($c * (pi/4)))**2)**.5;
   36 
   37 $ans = Formula("$exp");
   38 
   39 
   40 Context()->texStrings;
   41 BEGIN_TEXT
   42 \{ beginproblem() \}
   43 \{ textbook_ref_exact("Rogawski ET 2e", "11.2","16") \}
   44 $PAR
   45 Determine the speed \( s(t) \) of a particle with a given trajectory at a time \( t_0 \) (in units of meters and seconds).
   46 \[c(t) = ($a \sin $c t, $b \cos $c t), \, t = \frac{\pi}{4} \]
   47 $PAR
   48 \{ ans_rule() \}
   49 $PAR
   50 END_TEXT
   51 Context()->normalStrings;
   52 
   53 ANS($ans->cmp);
   54 
   55 Context()->texStrings;
   56 SOLUTION(EV3(<<'END_SOLUTION'));
   57 $PAR
   58 $SOL
   59 We have \( x = $a \sin $c t, \, y = $b \cos $c t \), hence \( x' = $ac \cos $c t, \, y' = - $bc \sin $c t \).  Thus, the speed of the particle at time t is
   60 \[ \frac{ds}{dt} = \sqrt{x'(t)^2 + y'(t)^2} = \sqrt{$acsq \cos^2 $c t + $bcsq \sin^2 $c t} \]
   61 \[ = \sqrt{$acsq \left( \cos^2 $c t + \sin^2 $c t \right) + $q \sin^2 $c t} = $c \sqrt{$acsqonc + $bcsqonc \sin^2 $c t} \]
   62 Thus,
   63 \[ \frac{ds}{dt} = $c \sqrt{$acsqonc + $bcsqonc \sin^2 $c t} \]
   64 The speed at time \( t = \frac{\pi}{4} \) is thus
   65 \[ \frac{ds}{dt} \mid _{t = \pi/4} = $c \sqrt{$acsqonc + $bcsqonc \sin^2 \left( $c \frac{\pi}{4} \right)} \cong $ans \]
   66 END_SOLUTION
   67 
   68 ENDDOCUMENT();

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