View of /branches/UGA/3.4.1.pg

    1 ## DESCRIPTION
    2 ##   Find Gradient and Directional Derivative
    3 ## ENDDESCRIPTION
    4 
    5 ## KEYWORDS('Gradient', 'Directional Derivative')
    6 ## Tagged by nhamblet
    7 
    8 ## DBsubject('Calculus')
    9 ## DBchapter('Partial Derivatives')
   10 ## DBsection('Directional Derivatives and the Gradient Vector')
   11 ## Date('9/18/2009')
   12 ## Author('Ted Shifrin')
   13 ## Institution('University of Georgia')
   14 
   15 
   16 DOCUMENT();        # This should be the first executable line in the problem.
   17 
   18 loadMacros("PG.pl",
   19            "PGbasicmacros.pl",
   20            "PGchoicemacros.pl",
   21            "PGmatrixmacros.pl",
   22            "PGanswermacros.pl",
   23            "PGauxiliaryFunctions.pl",
   24            "Parser.pl",
   25            "weightedGrader.pl"
   26 );
   27 
   28 install_weighted_grader();
   29 
   30 TEXT( beginproblem() );
   31 $showPartialCorrectAnswers = 1;
   32 
   33 Context("Vector")->variables->are(x=>'Real',y=>'Real');
   34 
   35 $a = random(1, 5, 1 );
   36 $b = non_zero_random( -3, 3, 1 );
   37 $c = non_zero_random( -3, 3, 1 );
   38 
   39 $v = non_zero_random(-4,4);
   40 $w = non_zero_random(-4,4);
   41 
   42 $f = Formula("$a x^2+ $b x y^2 + $c y^3")->reduce;
   43 $arg = '\left(\begin{array}{c} x\\y \end{array}\right)';
   44 
   45 $d = non_zero_random( -3, 3, 1 );
   46 do{$e = non_zero_random( -3, 3, 1 );
   47 $fx = $f -> D('x') -> eval(x=>$d, y=>$e);
   48 $fy = $f -> D('y') -> eval(x=>$d, y=>$e);} until $fx*$d+$fy*$e!=0;
   49 $P = ColumnVector($d, $e);
   50 $V = ColumnVector($v, $w);
   51 
   52 $fval = $f -> eval(x=>$d, y=>$e);
   53 $dirder = $fx*$v + $fy*$w;
   54 $RHS = $fx*$d+$fy*$e;
   55 $ans1 = $fx/$RHS;
   56 $ans2 = $fy/$RHS;
   57 
   58 Context() -> texStrings;
   59 
   60 BEGIN_TEXT
   61 Suppose \( f $arg = $f \), \( \mathbf{a} = $P \), and  \(\mathbf{v} = $V\).
   62 
   63 $PAR
   64 Compute the gradient of f at \(\mathbf{a}\). $BR
   65 
   66 \{mbox('\(\nabla f(\mathbf{a}) = \)', answer_matrix(2,1,15) ) \}
   67 
   68 $BR
   69 
   70 $PAR
   71 Compute the directional derivative of \(f\) at \(\mathbf{a}\) in the direction \(\mathbf{v}\). $BR
   72 \{ ans_rule(30) \}
   73 $BR
   74 
   75 $PAR
   76 For what value \( c\) does \(\mathbf{a}\) lie on the level curve \(f $arg = c\) ?
   77 $BR
   78 \{ mbox ('\( c = \)', ans_rule(15) ) \}
   79 
   80 $PAR
   81 For that value of \( c\), give the equation of the the tangent line to the level curve \(f $arg = c\) at the point \(\mathbf{a}\). Note that the right-hand side is already filled in for you.
   82 
   83 $BR
   84 \{mbox( ans_rule(15), '\(x + \)', ans_rule(15), '\( y = 1 \).') \}
   85 $PAR
   86 
   87 END_TEXT
   88 
   89 WEIGHTED_ANS($fx->cmp,15,
   90 $fy->cmp,15,
   91 $dirder->cmp,20,
   92 $fval->cmp,10,
   93 $ans1->cmp,20,
   94 $ans2->cmp,20);
   95 
   96 
   97 
   98 ENDDOCUMENT();        # This should be the last executable line in the problem.
   99 
  100 # display_matrix([[NAMED_ANS_RULE('opt1',15)],
  101 # [NAMED_ANS_RULE('opt2',15)]]) ) \}
  102 # \{ NAMED_ANS_RULE('opt3',30)\} $BR
  103 # \{ mbox ('\( c = \)', NAMED_ANS_RULE('opt4',15) ) \}
  104 # \{mbox(NAMED_ANS_RULE('opt5',15), '\(x + \)', NAMED_ANS_RULE('opt6',15), '\( # y = 1 \)', '.') \}
  105 
  106 NAMED_WEIGHTED_ANS('opt1',$fx->cmp,15);
  107 NAMED_WEIGHTED_ANS('opt2',$fy->cmp,15);
  108 NAMED_WEIGHTED_ANS('opt3',$dirder->cmp,20);
  109 NAMED_WEIGHTED_ANS('opt4',$fval->cmp,10);
  110 NAMED_WEIGHTED_ANS('opt5',$ans1->cmp,20);
  111 NAMED_WEIGHTED_ANS('opt6',$ans2->cmp,20);
  112 
  113 
  114 
  115