Difference between revisions of "ExtractingCoordinatesFromPoint"
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| + | <p style="background-color:#eeeeee;border:black solid 1px;padding:3px;"> |
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| + | <em>This code snippet shows the essential PG code to evaluate antderivative and general antiderivative formulas. Note that these are <b>insertions</b>, not a complete PG file. This code will have to be incorporated into the problem file on which you are working.</em> |
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| − | + | loadMacros("MathObjects.pl"); |
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| − | + | In the initialization section, we need to include the macros file <code>MathObjects.pl</code>. |
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| − | tagging and documentation section of the problem file. |
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| − | In the initialization section, we need to include the macros file <code>-------.pl</code>. |
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($d1, $d2) = ($point[0] - $point[1])->value; # $d1 = x1 - x2, $d2 = y1 - y2 |
($d1, $d2) = ($point[0] - $point[1])->value; # $d1 = x1 - x2, $d2 = y1 - y2 |
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| − | # Need to put () around $d1 and $d2 because if $d1 = -6, then -6^2 = -36, not 36, as desired. |
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$length = Compute("sqrt( ($d1)^2+($d2)^2 )"); |
$length = Compute("sqrt( ($d1)^2+($d2)^2 )"); |
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$mid = ( $point[1] + $point[0] ) / 2; |
$mid = ( $point[1] + $point[0] ) / 2; |
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| − | + | In the problem setup section of the file, we don't use the Vector context and <code>norm( $point[0] - $point[1] )</code> |
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| − | to determine length because we don't want to accept an answer like <code>|<5,7>-<7,8>|</code> You could use <code>$length=norm( $point[0] - $point[1] );</code> with the Vector context if you didn't care about accepting answers that are valid in the Vector context. |
+ | to determine length because we don't want to accept an answer like <code>|<5,7>-<7,8>|</code> You could use <code>$length=norm( $point[0] - $point[1] );</code> with the Vector context if you didn't care about accepting answers that are valid in the Vector context. (such as the absolute value of a vector.) |
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| − | Notes: on using this and related Contexts. |
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| + | We need to put parentheses around <code>$d1</code> and <code>$d2</code> because if <code>$d1</code> = -6, then -6^2 = -36, not 36, as desired. However, if the code is <code>($d1)^2</code> then that evaluates as (-6)^2 = 36, as desired. |
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Revision as of 20:51, 13 June 2008
Your title here: PG Code Snippet
This code snippet shows the essential PG code to evaluate antderivative and general antiderivative formulas. Note that these are insertions, not a complete PG file. This code will have to be incorporated into the problem file on which you are working.
This wiki page is under construction as of 6/13/08.
| PG problem file | Explanation |
|---|---|
loadMacros("MathObjects.pl");
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In the initialization section, we need to include the macros file |
Context( "Point" );
push( @point, Point( random(1,5,1) , random(-5,-1,1) ) );
push( @point, Point( random(5,10,1) , random(6,11,1) ) );
($d1, $d2) = ($point[0] - $point[1])->value; # $d1 = x1 - x2, $d2 = y1 - y2
$length = Compute("sqrt( ($d1)^2+($d2)^2 )");
$mid = ( $point[1] + $point[0] ) / 2;
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In the problem setup section of the file, we don't use the Vector context and
We need to put parentheses around |
Context()->texStrings;
BEGIN_TEXT
Consider the two points \( $point[0] \) and \( $point[1] \).
The distance between them is:\{ $length->ans_rule() \}
$BR
The midpoint of the line segment
that joins them is:\{ $mid->ans_rule() \}
$BR
END_TEXT
Context()->normalStrings;
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The problem text section of the file is as we'd expect. |
ANS( $length->cmp ); ANS( $mid->cmp ); |
As is the answer. |
