## WeBWorK Problems

### Using the form of " ? {$var}" for plus minus displays results in exponential notation for values greater than 1000? ### Using the form of " ? {$var}" for plus minus displays results in exponential notation for values greater than 1000?

by tim Payer -
Number of replies: 1
Greetings,

I am writing solutions for some problems and am using the form of "? {$var}" for a given value within BEGIN_SOLUTION. I like the form of inserting a question mark "?" before the variable as it anticipates whether the sign of a value is positive or negative, However it has the troubling aspect of displaying values that are larger than 1000 in scientific notation: That is 1250 is rendered as "1.25e + 03" or worse 1320x is rendered as "1.32e + 03x" These type of displays are confusing for the student when reading the step by step solutions. Is there a way to turn off the exponential display for values displayed in BEGIN_SOLUTION? I am assuming I can avoid converting it all to BEGIN_PGML_SOLUTION. See Line 154 in the code below if you would like to see a reference. # DESCRIPTION ## Calculus: Volumes by Cylindrical Shells ## ENDDESCRIPTION ## Tagged by XW ## DBsubject(Calculus - single variable) ## DBchapter(Applications of integration) ## DBsection(Volumes by disks) ## Date(6/5/2005) ## Institution(UVA) ## Author(Jeff Holt) ## Level(4) ## TitleText1('Calculus: Early Transcendentals') ## AuthorText1('Stewart') ## EditionText1('5') ## Section1('6.3') ## Problem1('35') ## TitleText2('Calculus: Early Transcendentals') ## AuthorText2('Stewart') ## EditionText2('6') ## Section2('6.3') ## Problem2('') ## TitleText3('Calculus: Early Transcendentals') ## AuthorText3('Rogawski') ## EditionText3('1') ## Section3('6.4') ## Problem3('9') ## KEYWORDS('calculus', 'integrals', 'volumes', 'cylindrical shells') ## Library/UVA-Stew5e/setUVA-Stew5e-C06S03-VolumesShells/6-3-35 DOCUMENT(); # This should be the first executable line in the problem. loadMacros( "PG.pl", "PGbasicmacros.pl", "PGchoicemacros.pl", "PGanswermacros.pl", "PGauxiliaryFunctions.pl" ); TEXT(beginproblem()); #Context()->{format}{number} = "%.8g";$showPartialCorrectAnswers = 1;

$a=random(-8, 8,1);$b=random($a+2,10,1);$p=$a*$b;
$s=-($a+$b);$p1 =$p*(-1);$s1 = $s*(-1);$sp1 =$s1*$p1;
#$sp1 =sprintf("%0.0f",$sp);
$ss =$s1*$s1;$pp =$p1*$p1;
$sps =$s*2;
$ppss =$ss +$p*2;$sp2 = $sp1*2;$soln = $PI*(-$a**5/30+$a**4 *$b/6 -$a**3 *$b**2/3 + $a**2 *$b**3/3 -$a*$b**4/6+$b**5/30);$int1 = ($a +$b)/2;
$ch =floor($int1);
if ($ch ==$a)
{
$int =$int1;
} else {
$int =$ch;
}
$ii =$int**2;
$b1 =$s*$int;$ef1 = $ii +$b1+$p;$aa5 = $a**5;$aa4 = $a**4;$aa3 = $a**3;$aa2 = $a**2;$bb5 = $b**5;$bb4 = $b**4;$bb3 = $b**3;$bb2 = $b**2;$cc5 = $bb5 -$aa5;
$cc4 =$bb4 -$aa4;$cc3 = $bb3 -$aa3;
$cc2 =$bb2 -$aa2;$cc1 = $b -$a;
$n4 =$cc4*$sps;$n3 = $cc3*$ppss;
$n2 =$cc2*$sp2;$n1 = $cc1*$pp;
$m5 =$cc5*12;
$m4 =$n4*15;
$m3 =$n3*20;
$m2 =$n2*30;
$m1 =$n1*60;
$nm =$m1+$m2+$m3+$m4+$m5;
$gc = gcd($nm,60);
if ($gc == 1) {$nm1 = $nm;$nd1 = 60;
$anj1 =$nm*pi/60;
} else {
$nm1 =$nm/$gc;$nd1 = 60/$gc;$anj1 = $nm*pi/60; }$anj2 = sprintf("%0.3f",$anj1); TEXT(EV2(<<EOT));$BR
Find the volume of the solid obtained by rotating the function bounded by the
given curve about the $$x$$-axis.
$BR $f(x) =x^2 ? {s} x ? {p}$$BR
Volume = \{ans_rule( 25) \}
$BR EOT ANS(num_cmp($soln));

BEGIN_SOLUTION
$BR$BR
To find the volume of the solid that results from a rotation of $$f(x) =x^2 ? {s} x ? {p}$$ about the $$x$$-axis we must find first find the points of intersection of the function to determine the limits of integration. Then integrate for the square of the function times $$\pi$$ over the limits of intersection by applying the general from for the calculation of a solid: $BR$BR
Volume = $$\displaystyle{ \pi \int_{\text{lower bound}}^{\text{ upper bound}} \left( f(x) \right)^2 \, dx}$$ $BR$BR
\begin{aligned}&\\ x^2 ? {s} x ? {p} & = 0 && \text{Set the function equal to zero and solve for }x.\\ (x -a)(x - b) & = 0 && \text{Factor the trinomial. }\\ \end{aligned}
$BR Then the function $$f(x)$$ intersects the $$x$$ axis at $$x = a$$ and $$x = b$$.$BR
Accordingly the volume of the bounded region can be found by integrating over the limits of integration: $BR$BR
$$\displaystyle{ Volume = \pi \int_{a}^{b} \left( f(x) \right)^2 \, dx}$$ $BR$BR
Substitute the functions of $$f(x)$$ into the volume equation: $BR$BR
$$\displaystyle{ Volume = \pi \int_{a}^{b} \left( \left(x^2 ? {s} x ? {p} \right) \right)^2 \, dx}$$ $BR$BR
Square the quantity: $BR$BR
$$\displaystyle{ Volume = \pi \int_{a}^{b} \left( x^2 ? {s} x ? {p} \right) \left( x^2 ? {s} x ? {p} \right)\, dx}$$ $BR$BR
$$\displaystyle{ Volume = \pi \int_{a}^{b} \left( x^4 ? {s} x^3 ? {p} x^2 ? {s} x^3 ? {ss} x^2 ? {sp1} x ? {p} x^2 ? {sp1} x + {pp}\right) \, dx}$$ $BR$BR
$BR Collect like terms:$BR
$BR $$\displaystyle{ Volume = \pi \int_{a}^{b} \left( x^4 ? {sps} x^3 ? {ppss} x^2 ? {sp2}x + {pp}\right) \, dx}$$$BR
$BR$BR

Integrate term by term: $BR$BR
$$\displaystyle{ Volume = \pi \left[ \frac{1}{5} x^5 +\frac{sps}{4} x^4 +\frac{ppss}{3} x^3 + \frac{sp2}{2} x^2 + {pp} x \right] \Big|_{a}^{b} }$$ $BR$BR
Evaluate term by term: $BR$BR
$$\displaystyle{ Volume = \pi \left[ \frac{1}{5} \left(b^5-a^5\right) +\frac{sps}{4} \left(b^4 - a^4\right) +\frac{ppss}{3} \left(b^3 - a^3\right) + \frac{sp2}{2} \left(b^2 - a^2\right) + {pp} (b - a) \right] }$$ $BR$BR
Raise each base to its respective power: $BR$BR
$$\displaystyle{ Volume = \pi \left[ \frac{1}{5} \left(bb5-aa5\right) +\frac{sps}{4} \left(bb4 - aa4\right) +\frac{ppss}{3} \left(bb3 - aa3\right) + \frac{sp2}{2} \left(bb2 - aa2\right) + {pp} (b - a) \right] }$$ $BR$BR
Gather terms for each quantity: $BR$BR
$$\displaystyle{ Volume = \pi \left[ \frac{1}{5} \left(cc5\right) +\frac{sps}{4} \left(cc4\right) +\frac{ppss}{3} \left(cc3\right) + \frac{sp2}{2} \left(cc2\right) + {pp} (cc1) \right] }$$ $BR$BR
Form single fractions: $BR$BR
$$\displaystyle{ Volume = \pi \left[ \frac{cc5}{5} +\frac{n4}{4} +\frac{n3}{3} + \frac{n2}{2} + n1 \right] }$$
$BR$BR
Form common denominators: $BR$BR
$$\displaystyle{ Volume = \pi \left[ \frac{cc5}{5} \cdot \frac{12}{12} +\frac{n4}{4} \cdot \frac{15}{15}+\frac{n3}{3} \cdot \frac{20}{20} + \frac{n2}{2} \cdot \frac{30}{30} + n1 \cdot \frac{60}{60} \right] }$$ $BR$BR
Combine fraction factors: $BR$BR
$$\displaystyle{ Volume = \pi \left[ \frac{m5}{60} +\frac{m4}{60} +\frac{m3}{60} + \frac{m2}{60} +\frac{m1}{60} \right] }$$ $BR$BR
Combine into one fraction: $BR$BR
$$\displaystyle{ Volume = \pi \left[ \frac{m5 +m4 +m3+m2+m1}{60} \right] }$$ $BR$BR
Simplify the numerator: $BR$BR
$$\displaystyle{ Volume = \pi \left[ \frac{nm}{60} \right] }$$ $BR$BR
Reduce the fraction: $BR$BR
$$\displaystyle{ Volume = \frac{nm1 \pi}{nd1} }$$ cubic units $BR$BR
$$\displaystyle{ Volume \approx anj2 }$$ cubic units. $BR$BR

END_SOLUTION
ENDDOCUMENT(); # This should be the last executable line in the problem.