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Tue Nov 8 15:17:41 2011 UTC (2 years, 5 months ago) by aubreyja
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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('19')
10 ## Author('Christopher Sira')
11 ## Institution('W.H.Freeman')
12
13 DOCUMENT();
19 $context = Context(); 20 21$a = random(2, 9);
22 $asq =$a ** 2;
23 $sixa = 6 *$a;
24 $a64 =$sixa - 4;
25 $term2 = 2 *$a64;
26 $term2on4 =$term2 / 4;
27 $f_rad_num =$term2on4**2 - $a64 *$term2on4 + $asq * 9; 28$f_rad = $f_rad_num / 9; 29 30$ans = Compute("sqrt($f_rad_num / 9)"); 31 32 Context()->texStrings; 33 34 BEGIN_TEXT 35 \{ beginproblem() \} 36 \{ textbook_ref_exact("Rogawski ET 2e", "11.2","19") \} 37$PAR
38 Find the minimum speed of a particle with trajectory
39 $$c(t) = (t^3 - a t, \, t^2 + 1 )$$ for $$t \ge 0$$.
40 Hint: it is easier to find the minimum of the square of the speed.
41 $PAR 42 \{ ans_rule() \} 43$PAR
44 END_TEXT
45 Context()->normalStrings;
46
47 ANS($ans->cmp); 48 49 Context()->texStrings; 50 SOLUTION(EV3(<<'END_SOLUTION')); 51$PAR
52 \$SOL
53 We first find the speed of the particle.  We have $$x(t) = t^3 - a t, \, y(t) = t^2 + 1$$ hence $$x' = 3 t^2 - a$$ and $$y'(t) = 2t$$.  The speed is thus
54 $\frac{ds}{dt} = \sqrt{ \left( 3 t^2 - a \right)^2 + \left(2t \right)^2 } = \sqrt{ 9t^4 - sixa t^2 + asq + 4 t^2 } = \sqrt{ 9t^4 - a64 t^2 + asq }$
55 The square root function is an increasing function, hence the minimum speed occurs at the value of $$t$$ where the function $$f(t) = 9 t^4 - a64 t^2 + asq$$ has minimum value.  Since $$\lim_{t \to \infty} f(t) = \infty$$, $$f$$ has a minimum value on the interval $$0 \le t < \infty$$, and it occurs at a critical point or at the endpoint $$t = 0$$.  We find the critical point of $$f$$ on $$t \ge 0$$:
56 $f'(t) = 36t^3 - term2 t = 4t \left( 9 t^2 - term2on4 \right) \Rightarrow t = 0, \, t = \sqrt{ \frac{term2on4}{9} }$
57 We compute the values of $$f$$ at these points:
58 $f(0) = 9 \cdot 0^4 - a64 \cdot 0^2 + asq = asq$
59 $f \left( \sqrt{ \frac{term2on4}{9} } \right) = 9 \left( \sqrt{ \frac{term2on4}{9} } \right) ^4 - a64 \left( \sqrt{ \frac{term2on4}{9} } \right) ^2 + asq = \frac{f_rad_num}{9}$
60 We conclude that the minimum value of $$f$$ on $$t \ge 0$$ is $$\frac{f_rad_num}{9}$$.  The minimum speed is therefore
61 $\left( \frac{ds}{dt} \right) _{min} \approx \sqrt{\frac{f_rad_num}{9}} \approx ans$
62 END_SOLUTION
63
64 ENDDOCUMENT();