[npl] / 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 Repository: Repository Listing bbplugincoursesdistsnplrochestersystemwww

Tue Nov 8 15:17:41 2011 UTC (18 months, 1 week ago) by aubreyja
File size: 3037 byte(s)
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('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
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();