View of /trunk/NationalProblemLibrary/WHFreeman/Rogawski_Calculus_Early_Transcendentals_Second_Edition/11_Parametric_Equations_Polar_Coordinates_and_Conic_Sections/11.2_Arc_Length_and_Speed/11.2.18.pg

    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('18')
   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 $t = random(1, 15);
   22 $den = ($t**2 + 1)**2;
   23 $con = $t**2;
   24 
   25 $exp = $t * (4/$den + 9*$con)**.5;
   26 
   27 $ans = Formula("$exp");
   28 
   29 
   30 Context()->texStrings;
   31 BEGIN_TEXT
   32 \{ beginproblem() \}
   33 \{ textbook_ref_exact("Rogawski ET 2e", "11.2","18") \}
   34 $PAR
   35 Determine the speed \( s(t) \) of a particle with a given trajectory at a time \( t_0 \) (in units of meters and seconds).
   36 \[c(t) = (\ln(t^2 + 1), t^3), \, t = $t \]
   37 $PAR
   38 \{ ans_rule() \}
   39 $PAR
   40 END_TEXT
   41 Context()->normalStrings;
   42 
   43 ANS($ans->cmp);
   44 
   45 Context()->texStrings;
   46 SOLUTION(EV3(<<'END_SOLUTION'));
   47 $PAR
   48 $SOL
   49 We have \( x = \ln(t^2 + 1), \, y = t^3 \), so \( x' = \frac{2t}{t^2 + 1} \) and \( y' = 3t^2 \).  The speed of the particle at time t is thus
   50 \[ \frac{ds}{dt} = \sqrt{x'(t)^2 + y'(t)^2} = \sqrt{ \frac{4t^2}{ \left( t^2 + 1 \right) ^2 } + 9t^4 } = t \sqrt{\frac{4}{ \left( t^2 + 1 \right) ^2 } + 9t^2} \]
   51 The speed at time \( t = $t \) is
   52 \[ \frac{ds}{dt} \mid _{t = $t} = $t \sqrt{ \frac{4}{$den} + 9 \cdot $con } = $ans \]
   53 END_SOLUTION
   54 
   55 ENDDOCUMENT();