[npl] / trunk / NationalProblemLibrary / WHFreeman / Rogawski_Calculus_Early_Transcendentals_Second_Edition / 14_Differentiation_in_Several_Variables / 14.5_The_Gradient_and_Directional_Derivatives / 14.5.1.pg Repository: Repository Listing bbplugincoursesdistsnplrochestersystemwww

Tue Nov 8 15:17:41 2011 UTC (18 months, 1 week ago) by aubreyja
File size: 2748 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('1')
11 ## Institution('W.H.Freeman')
12
13 DOCUMENT();
14
21
22 TEXT(beginproblem());
23
24 Context()->texStrings;
25
26 $xpow=random(1,3,1); 27$ypow=random(1,3,1);
28 $a=random(1,2,1); 29$b=random(1,2,1);
30 $t=non_zero_random(-1,1); 31 32$context = Context();
33 $context->variables->add(y=>'Real'); 34$context->variables->add(t=>'Real');
35
36 $f=Formula("x^($xpow)*y^($ypow)")->reduce(); 37$cx=Formula("$a*t^2")->reduce(); 38$cy=Formula("$b*t^3")->reduce(); 39$fx=Formula("$xpow*x^($xpow-1)*y^($ypow)")->reduce(); 40$fy=Formula("$ypow*x^($xpow)*y^($ypow-1)")->reduce(); 41$cxx=Formula("2*$a*t")->reduce(); 42$cyy=Formula("3*$b*t^2")->reduce(); 43$answer=Formula("(2*$a*$xpow)*x^($xpow-1)*y^($ypow)*t+(3*$b*$ypow)*x^($xpow)*y^($ypow-1)*t^2")->reduce();
44
45 $fc=Formula("($a^($xpow))*t^(2*$xpow)*($b^($ypow))*t^(3*$ypow)")->reduce(); 46$fc1=Formula("$a^($xpow)*$b^($ypow)*t^(2*$xpow+3*$ypow)")->reduce();
47 $fct=Formula("(2*$xpow+3*$ypow)*$a^($xpow)*$b^($ypow)*t^(2*$xpow+3*$ypow-1)")->reduce(); 48$answer2=$fct->eval(t=>$t);
49
50 BEGIN_TEXT
51 \{ textbook_ref_exact("Rogawski ET 2e", "14.5","1") \}
52 $PAR 53 Let $$f(x,y)=f$$ and $$c(t)=\left(cx,cy\right)$$ 54$PAR
55 (a) Calculate:
56 $BR 57 $$\nabla f \cdot c'(t)=$$ \{ans_rule()\} 58$PAR
59 (b) Use the Chain Rule for Paths to evaluate $$\frac{d}{dt}f(c(t))$$ at $$t=t$$.
60 $BR 61 $$\frac{d}{dt}f(c(t))=$$\{ans_rule()\} 62$BR
63 END_TEXT
64
65 Context()->normalStrings;
66 ANS($answer->cmp); 67 ANS($answer2->cmp);
68 Context()->texStrings;
69
70 SOLUTION(EV3(<<'END_SOLUTION'));
71 $PAR 72$SOL
73 $BR 74 (a) We compute the partial derivatives of $$f(x,y)=f$$ 75 $\frac{\partial{f}}{\partial{x}}=fx, \quad \frac{\partial{f}}{\partial{y}}=fy$ 76 The gradient vector is thus 77 $\nabla f=\left<fx,fy\right>$ 78 Also, 79 $c'(t)=\left< \left(cx\right)', \left(cy\right)' \right>=\left<cxx, cyy\right>$ 80 $\nabla f \cdot c'(t)=\left<fx,fy\right>\cdot\left<cxx, cyy\right>=answer$ 81$PAR
82 (b) Using the Chain Rule and substituting $$x=cx, y=cy$$ gives
83 $\frac{d}{dt}f(c(t))=\frac{d}{dt}\left(fc\right)=\frac{d}{dt}\left(fc1\right)=fct$
84 At the point $$t=t$$, we obtain
85 $\left.\frac{d}{dt}f(c(t))\right|_{t=t}=\left.fct\right|_{t=t}=answer2$
86 \$BR
87 END_SOLUTION
88
89 ENDDOCUMENT();