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Revision 2584 - (download) (annotate)
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('12')
   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, 6);
   23 $asq = $a ** 2;
   24 
   25 $ans = Compute("$a*pi/2 * (1 + $asq * pi**2)**.5 + 1/2 * ln($a*pi + (1+$asq*pi**2)**.5)");
   26 
   27 
   28 Context()->texStrings;
   29 BEGIN_TEXT
   30 \{ beginproblem() \}
   31 \{ textbook_ref_exact("Rogawski ET 2e", "11.2","12") \}
   32 $PAR
   33 Find the length of the spiral \( c(t) = (t \cos t, t \sin t) \) for \( 0 \le t \le $a \pi \) to three decimal places.  Hint: use the formula
   34 \[ \int \sqrt{1 + t^2} \, dt
   35 = \frac{1}{2} t \sqrt{1 + t^2} + \frac{1}{2} \ln \left( t + \sqrt{1 + t^2} \right) + C\]
   36 $PAR
   37 \{ ans_rule() \}
   38 $PAR
   39 END_TEXT
   40 Context()->normalStrings;
   41 
   42 ANS($ans->cmp);
   43 
   44 Context()->texStrings;
   45 SOLUTION(EV3(<<'END_SOLUTION'));
   46 $PAR
   47 $SOL
   48 We use the formula for the arc length:
   49 \[ S = \int ^{$a \pi} _0 \sqrt{x'(t)^2 + y'(t)^2} \, dt \]
   50 Differentiating \( x = t \cos t \) and \( y = t \sin t \) yields
   51 \[ x'(t) = \frac{d}{dt} (t \cos t) = \cos t - t \sin t \]
   52 \[ y'(t) = \frac{d}{dt} (t \sin t) = \sin t + t \cos t \]
   53 Thus,
   54 \[ \sqrt{x'(t)^2 + y'(t)^2} = \sqrt{(\cos t - t \sin t)^2 + (\sin t + t \cos t)^2} \]
   55 \[ = \sqrt{\cos^2 t - 2 t \cos t \sin t + t^2 \sin^2 t + \sin^2 t + 2 t \sin t \cos t + t^2 \cos^2 t} \]
   56 \[ = \sqrt{(\cos^2 t + \sin^2 t)(1 + t^2)} = \sqrt{1 + t^2} \]
   57 We substitute into (1) and use the integral given in the hint to obtain the following arc length:
   58 \[ S = \int ^{$a \pi} _0 \sqrt{1 + t^2} \, dt = \frac{1}{2} t \sqrt{1 + t^2} + \frac{1}{2} \ln \left( t + \sqrt{1 + t^2} \right) \mid ^{$a \pi} _0 \]
   59 \[ = \left[\frac{1}{2} \cdot $a \pi \sqrt{1 + ($a \pi)^2} + \frac{1}{2} \ln \left($a \pi + \sqrt{1 + ($a \pi)^2} \right)\right] - \left[ 0 + \frac{1}{2} \ln 1 \right]\]
   60 \[ = \{$a/2\} \pi \sqrt{1 + $asq \pi^2} + \frac{1}{2} \ln \left( $a \pi + \sqrt{1 + $asq \pi^2} \right) \approx {$ans:%0.3f}\]
   61 END_SOLUTION
   62 
   63 ENDDOCUMENT();

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