# DESCRIPTION

# arc length for (x+a)^3 /12 + (x+a)^(-1) on [L,R] (with a >= 0)

# ENDDESCRIPTION

## DBsubject('Calculus')

## DBchapter('Applications of Integration')

## DBsection('Arc Length')

## KEYWORDS('calculus', 'integrals', 'integration', 'arc length')

## TitleText1('Calculus: Early Transcendentals')

## EditionText1('2')

## AuthorText1('Rogawski')

## Section1('8.1')

## Problem1('3')

## Institution('W.H.Freeman')

## Author('Dick Lane')

## Date('05/19/2011')

DOCUMENT();

loadMacros( "PGstandard.pl" , "MathObjects.pl" ,

"contextFraction.pl" ,

"freemanMacros.pl" ) ;

Context( "Fraction" ) ;

# Context() -> flags -> set(reduceConstants => 0,reduceConstantFunctions => 0);

$a = random(1,7,1) ;

$L = random(0,4,1) ; #### allowing L=0 is ok since a > 0

$R = $L + random(1,3,1) ; #### interval [ L , R ]

$W = Formula( "x + $a" ) ;

$f = Formula( "$W^3 / 12 + $W^(-1)" ) -> reduce ;

$fp = $f -> D -> reduce ;

$integrand = Formula( "$W^2 / 4 + $W^(-2)" ) -> reduce ;

$antideriv = Formula( "$W^3 / 12 - $W^(-1)" ) -> reduce ;

TEXT(beginproblem());

$showPartialCorrectAnswers = 0 ;

Context() -> texStrings ;

BEGIN_TEXT

\{ textbook_ref_exact("Rogawski ET 2e", "8.1", "3") \}$BR

The graph of \(\displaystyle f(x) = $f \)

on the interval \( [$L,$R] \) has

arc length $SPACE \{ ans_rule() \}

END_TEXT

$showHint = 3 ; #### default value is 1

BEGIN_HINT

$HINT

\( 1 + \left( y' \right)^2 \) is a perfect square.

END_HINT

Context() -> normalStrings;

$right = $antideriv -> substitute( x => $R ) -> reduce ;

$left = $antideriv -> substitute( x => $L ) -> reduce ;

$arc = Compute( "$right - $left" ) ;

ANS( $arc -> cmp() );

$Fp = "\frac{$W^2}{4} - \frac{1}{$W^2}" ;

Context() -> texStrings;

BEGIN_SOLUTION

$SOLUTION

\( y = $f \) implies \( y^\prime = $Fp \) and

\[\begin{aligned}

1 + (y^\prime)^2

& = 1 +

\left( \frac{($W)^4}{16} - \frac{1}{2} + \frac{1}{($W)^4} \right)\\

& = \frac{($W)^4}{16} + \frac{1}{2} + \frac{1}{($W)^4}

= \left( \frac{($W)^2}{4} + \frac{1}{($W)^2} \right)^2

\end{aligned} \]

Since \(\displaystyle $integrand > 0 \) on \( [$L,$R] \),

the arc length is

\[ \begin{aligned}

s & = \int_{$L}^{$R} \sqrt{ 1 + (y^\prime)^2 } \; dx \\

& = \int_{$L}^{$R }\left( \frac{($W)^2}{4} + \frac{1}{($W)^2} \right) dx \\

& = \left( $antideriv \right) \Bigg|_{$L}^{$R} \\

& = $right - $left

= $arc

\end{aligned} \]

END_SOLUTION

Context() -> normalStrings;

ENDDOCUMENT();