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    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('19')
   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 $a = random(2, 9);
   22 $asq = $a ** 2;
   23 $sixa = 6 * $a;
   24 $a64 = $sixa - 4;
   25 $term2 = 2 * $a64;
   26 $term2on4 = $term2 / 4;
   27 $f_rad_num = $term2on4**2 - $a64 * $term2on4 + $asq * 9;
   28 $f_rad = $f_rad_num / 9;
   29 
   30 $ans = Compute("sqrt($f_rad_num / 9)");
   31 
   32 Context()->texStrings;
   33 
   34 BEGIN_TEXT
   35 \{ beginproblem() \}
   36 \{ textbook_ref_exact("Rogawski ET 2e", "11.2","19") \}
   37 $PAR
   38 Find the minimum speed of a particle with trajectory
   39 \( c(t) = (t^3 - $a t, \, t^2 + 1 ) \) for \( t \ge 0 \).
   40 Hint: it is easier to find the minimum of the square of the speed.
   41 $PAR
   42 \{ ans_rule() \}
   43 $PAR
   44 END_TEXT
   45 Context()->normalStrings;
   46 
   47 ANS($ans->cmp);
   48 
   49 Context()->texStrings;
   50 SOLUTION(EV3(<<'END_SOLUTION'));
   51 $PAR
   52 $SOL
   53 We first find the speed of the particle.  We have \( x(t) = t^3 - $a t, \, y(t) = t^2 + 1 \) hence \( x' = 3 t^2 - $a \) and \( y'(t) = 2t \).  The speed is thus
   54 \[ \frac{ds}{dt} = \sqrt{ \left( 3 t^2 - $a \right)^2 + \left(2t \right)^2 } = \sqrt{ 9t^4 - $sixa t^2 + $asq + 4 t^2 } = \sqrt{ 9t^4 - $a64 t^2 + $asq } \]
   55 The square root function is an increasing function, hence the minimum speed occurs at the value of \( t \) where the function \( f(t) = 9 t^4 - $a64 t^2 + $asq \) has minimum value.  Since \( \lim_{t \to \infty} f(t) = \infty \), \( f \) has a minimum value on the interval \( 0 \le t < \infty \), and it occurs at a critical point or at the endpoint \( t = 0 \).  We find the critical point of \( f \) on \( t \ge 0 \):
   56 \[ f'(t) = 36t^3 - $term2 t = 4t \left( 9 t^2 - $term2on4 \right) \Rightarrow t = 0, \, t = \sqrt{ \frac{$term2on4}{9} } \]
   57 We compute the values of \( f \) at these points:
   58 \[ f(0) = 9 \cdot 0^4 - $a64 \cdot 0^2 + $asq = $asq \]
   59 \[ f \left( \sqrt{ \frac{$term2on4}{9} } \right) = 9 \left( \sqrt{ \frac{$term2on4}{9} } \right) ^4 - $a64 \left( \sqrt{ \frac{$term2on4}{9} } \right) ^2 + $asq = \frac{$f_rad_num}{9}\]
   60 We conclude that the minimum value of \( f \) on \( t \ge 0 \) is \( \frac{$f_rad_num}{9}\).  The minimum speed is therefore
   61 \[ \left( \frac{ds}{dt} \right) _{min} \approx \sqrt{\frac{$f_rad_num}{9}} \approx $ans \]
   62 END_SOLUTION
   63 
   64 ENDDOCUMENT();