View of /trunk/NationalProblemLibrary/WHFreeman/Rogawski_Calculus_Early_Transcendentals_Second_Edition/14_Differentiation_in_Several_Variables/14.5_The_Gradient_and_Directional_Derivatives/14.5.36.pg

    1 ## DBsubject('Calculus')
    2 ## DBchapter('Differentiation in Several Variables')
    3 ## DBsection('The Gradient and Directional Derivatives')
    4 ## KEYWORDS('calculus')
    5 ## TitleText1('Calculus: Early Transcendentals')
    6 ## EditionText1('2')
    7 ## AuthorText1('Rogawski')
    8 ## Section1('14.5')
    9 ## Problem1('36')
   10 ## Author('JustAsk - Kobi Fonarov')
   11 ## Institution('W.H.Freeman')
   12 
   13 DOCUMENT();
   14 
   15 loadMacros("PG.pl","PGbasicmacros.pl","PGanswermacros.pl");
   16 loadMacros("Parser.pl");
   17 loadMacros("freemanMacros.pl");
   18 loadMacros("PGauxiliaryFunctions.pl");
   19 loadMacros("PGgraphmacros.pl");
   20 loadMacros("PGchoicemacros.pl");
   21 
   22 TEXT(beginproblem());
   23 
   24 Context()->texStrings;
   25 
   26 $a=random(1,9);
   27 $b=$a**2;
   28 $nablafp=(1+2*$a**2)**2+$a**2;
   29 
   30 $answerab=Real(sqrt($nablafp));
   31 $answerc=Real(sqrt($nablafp/2));
   32 
   33 BEGIN_TEXT
   34 \{ textbook_ref_exact("Rogawski ET 2e", "14.5","36") \}
   35 $PAR
   36 Let \(f(x,y) = xe^{x^2-y}\) and \(P=($a,$b)\). $BR
   37 $BBOLD (a) $EBOLD Calculate \(\|\nabla f_P\|\). $BR
   38 $BBOLD (b) $EBOLD Find the rate of change of \(f\) in the direction \(\nabla f_P\). $BR
   39 $BBOLD (c) $EBOLD Find the rate of change of \(f\) in the direction of a vector making an angle of \(45^\circ\) with \(\nabla f_P\).
   40 $PAR
   41 Answers : $BR
   42 $BBOLD (a) $EBOLD \{ans_rule()\} $BR
   43 $BBOLD (b) $EBOLD \{ans_rule()\} $BR
   44 $BBOLD (c) $EBOLD \{ans_rule()\}
   45 $PAR
   46 
   47 END_TEXT
   48 
   49 Context()->normalStrings;
   50 
   51 ANS($answerab->cmp);
   52 ANS($answerab->cmp);
   53 ANS($answerc->cmp);
   54 
   55 Context()->texStrings;
   56 
   57 SOLUTION(EV3(<<'END_SOLUTION'));
   58 $PAR
   59 $SOL  $PAR
   60 $BBOLD (a) $EBOLD We compute the gradient of \(f( x,y )=x{e}^{{x}^2-y}\). The partial derivatives are
   61 \[\frac{\partial f}{\partial x}=1\cdot {e}^{{x}^2-y}+x{e}^{{x}^2-y}\cdot 2x={e}^{{x}^2-y}\left(1+2{x}^2\right)\]
   62 \[\frac{\partial f}{\partial y}=-x{e}^{{x}^2-y}\]
   63 The gradient vector is thus
   64 \[\nabla f = \left< \frac{\partial f}{\partial x},\frac{\partial f}{\partial y} \right>
   65 =
   66 \]\[ \left< {e}^{{x}^2-y}\left(1+2{x}^2\right),-x{e}^{{x}^2-y} \right>
   67 ={e}^{{x}^2-y}\left< 1+2{x}^2,-x \right>\]
   68 At the point \(P=( $a,$b )\) we have
   69 \[\nabla {f}_{P}={e}^0\left< 1+\{2*$a**2\},-$a \right> = \left< \{1+2*$a**2\},-$a \right>
   70 \]\[\Downarrow\]\[ \| \nabla {f}_{P}\| =\sqrt{{\{1+2*$a**2\}}^2+{( -$a )}^2}=\sqrt{$nablafp}
   71 \]
   72 $BBOLD (b) $EBOLD The rate of change of \(f\) in the direction of the gradient vector is the length of the gradient, that is,$BR \(\| \nabla {f}_{P}\| =\sqrt{$nablafp}\). $PAR
   73 $BBOLD (c) $EBOLD Let \({\mathbf{e}}_{\mathbf{v}}\) be the unit vector making an angle of \({45}^{\circ }\) with \(\nabla {f}_{P}\). $BR The rate of change of \(f\) in the direction of \({\mathbf{e}}_{\mathbf{v}}\) is the directional derivative of \(f\) in the direction \({\mathbf{e}}_{\mathbf{v}}\), $BR which is the following dot product:
   74 \[
   75 {D}_{\mathbf{{e}_{v}}} f(P) =\nabla {f}_{P}\cdot \mathbf{{e}_{v}}=\| \nabla {f}_{P}\| \| \mathbf{{e}_{v}}\| \cos {45}^{\circ }=
   76 \]\[\sqrt{$nablafp}\cdot 1\cdot \frac{1}{\sqrt{2}}=\sqrt{\{$nablafp/2\}}\approx $answerc
   77 \]
   78 END_SOLUTION
   79 
   80 ENDDOCUMENT();