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Tue Nov 8 15:17:41 2011 UTC (2 years, 1 month 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('Limits and Derivatives')
3 ## DBsection('Definition of the Derivative')
4 ## KEYWORDS('calculus', 'derivatives', 'slope')
5 ## TitleText1('Calculus: Early Transcendentals')
6 ## EditionText1('2')
7 ## AuthorText1('Rogawski')
8 ## Section1('10.2')
9 ## Problem1('30')
10 ## Author('Keith Thompson')
11 ## Institution('W.H.Freeman')
12
13 DOCUMENT();
19
20 #$showPartialCorrectAnswers=1; 21 22$n0 = random(2,9,1);
23 $base=random(3,7,2); 24$mult=random(2,6,2);
25 $netb=$base-$n0*$mult;
26 $p=$netb+$mult; 27 28$ans1=exp($netb)*exp($mult)/(exp($mult)-1); 29 Context()->texStrings; 30 BEGIN_TEXT 31 \{ beginproblem() \} 32 \{ textbook_ref_exact("Rogawski ET 2e", "10.2","30") \} 33$PAR
34
35 Use the formula for the sum of a geometric series to find the sum or state that the series diverges (enter DIV for a divergent series).
36
37 $\sum_{n=n0}^\infty e^{base-mult n}$
38
39 $$S=$$ \{ans_rule()\}
40 END_TEXT
41
42 Context()->normalStrings;
43
44 ANS(std_num_str_cmp($ans1,['DIV'])). 45 46 Context()->texStrings; 47 SOLUTION(EV3(<<'END_SOLUTION')); 48$PAR
49 \$SOL
50 Re-write the series as
51 $\sum_{n=n0}^\infty e^{base}e^{-mult n}=\sum_{n=n0}^\infty e^{base}\left(\frac{1}{e^{mult}}\right)^n$
52
53 to recognize it as a geometric series with $$c=e^{base}\left(\frac{1}{e^{mult}}\right)^{n0}=e^{netb}$$ and $$0<r=\frac{1}{e^{mult}}<1$$. Thus
54
55 $\sum_{n=n0}^\infty e^{base-mult n}=e^{netb}\frac{1}{1-\frac{1}{e^{mult}}}=e^{netb}\frac{e^{mult}}{e^{mult}-1}=\frac{e^{p}}{e^{mult}-1}.$
56 END_SOLUTION
57 ENDDOCUMENT();