[npl] / trunk / NationalProblemLibrary / WHFreeman / Rogawski_Calculus_Early_Transcendentals_Second_Edition / 11_Parametric_Equations_Polar_Coordinates_and_Conic_Sections / 11.2_Arc_Length_and_Speed / 11.2.12.pg Repository: Repository Listing bbplugincoursesdistsnplrochestersystemwww

# 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/11.2.12.pg

Tue Nov 8 15:17:41 2011 UTC (2 years, 3 months ago) by aubreyja
File size: 2435 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('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('12')
10 ## Author('Christopher Sira')
11 ## Institution('W.H.Freeman')
12
13 DOCUMENT();
19 $context = Context(); 20 21 22$a = random(2, 6);
23 $asq =$a ** 2;
24
25 $ans = Compute("$a*pi/2 * (1 + $asq * pi**2)**.5 + 1/2 * ln($a*pi + (1+$asq*pi**2)**.5)"); 26 27 28 Context()->texStrings; 29 BEGIN_TEXT 30 \{ beginproblem() \} 31 \{ textbook_ref_exact("Rogawski ET 2e", "11.2","12") \} 32$PAR
33 Find the length of the spiral $$c(t) = (t \cos t, t \sin t)$$ for $$0 \le t \le a \pi$$ to three decimal places.  Hint: use the formula
34 $\int \sqrt{1 + t^2} \, dt 35 = \frac{1}{2} t \sqrt{1 + t^2} + \frac{1}{2} \ln \left( t + \sqrt{1 + t^2} \right) + C$
36 $PAR 37 \{ ans_rule() \} 38$PAR
39 END_TEXT
40 Context()->normalStrings;
41
42 ANS($ans->cmp); 43 44 Context()->texStrings; 45 SOLUTION(EV3(<<'END_SOLUTION')); 46$PAR
47 \$SOL
48 We use the formula for the arc length:
49 $S = \int ^{a \pi} _0 \sqrt{x'(t)^2 + y'(t)^2} \, dt$
50 Differentiating $$x = t \cos t$$ and $$y = t \sin t$$ yields
51 $x'(t) = \frac{d}{dt} (t \cos t) = \cos t - t \sin t$
52 $y'(t) = \frac{d}{dt} (t \sin t) = \sin t + t \cos t$
53 Thus,
54 $\sqrt{x'(t)^2 + y'(t)^2} = \sqrt{(\cos t - t \sin t)^2 + (\sin t + t \cos t)^2}$
55 $= \sqrt{\cos^2 t - 2 t \cos t \sin t + t^2 \sin^2 t + \sin^2 t + 2 t \sin t \cos t + t^2 \cos^2 t}$
56 $= \sqrt{(\cos^2 t + \sin^2 t)(1 + t^2)} = \sqrt{1 + t^2}$
57 We substitute into (1) and use the integral given in the hint to obtain the following arc length:
58 $S = \int ^{a \pi} _0 \sqrt{1 + t^2} \, dt = \frac{1}{2} t \sqrt{1 + t^2} + \frac{1}{2} \ln \left( t + \sqrt{1 + t^2} \right) \mid ^{a \pi} _0$
59 $= \left[\frac{1}{2} \cdot a \pi \sqrt{1 + (a \pi)^2} + \frac{1}{2} \ln \left(a \pi + \sqrt{1 + (a \pi)^2} \right)\right] - \left[ 0 + \frac{1}{2} \ln 1 \right]$
60 $= \{a/2\} \pi \sqrt{1 + asq \pi^2} + \frac{1}{2} \ln \left( a \pi + \sqrt{1 + asq \pi^2} \right) \approx {ans:%0.3f}$
61 END_SOLUTION
62
63 ENDDOCUMENT();