[Stack-cvs] stack-dev/sample_questions/oer CR_Int_16_trigonometric.xml, NONE, 1.3.6.2 CR_Diff_03_h

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Update of /cvsroot/stack/stack-dev/sample_questions/oer
In directory sfp-cvsdas-3.v30.ch3.sourceforge.com:/tmp/cvs-serv29766/sample_questions/oer

Added Files:
      Tag: question_reporting
	CR_Int_16_trigonometric.xml CR_Diff_03_higher.xml 
	CR_Diff_01_basic.xml licence.txt CR_Diff_05_quotient.xml 
	CR_Int_15_partialfrac.xml CR_Int_11_linearity.xml CR_all.xml 
	CR_Int_12_definite.xml CR_Int_10_basic.xml 
	CR_Diff_06_chain.xml CR_Diff_02_linearity.xml 
	CR_Int_17_misc.xml AlgMap.xml CR_Int_13_parts.xml Barnard.tex 
	CR_Diff_09_misc.xml CR_Diff_04_product.xml 
	CR_Diff_07_implicit.xml CR_Int_14_substitution.xml 
	CR_Diff_08_parametrical.xml readme.txt 
Log Message:
Merging from the current HEAD into question_reporting. Apologies in advance if this all goes horribly wrong.

--- NEW FILE: licence.txt ---

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--- NEW FILE: readme.txt ---

This folder contains STACK questions which are released as Open
Educational Resources.

==Licence==

These materials are released under the Creative Commons Attribution-Share Alike 3.0 Unported
http://creativecommons.org/licenses/by-sa/3.0/
http://creativecommons.org/licenses/by-sa/3.0/legalcode

Chris Sangwin, 5 November 2009
University of Birmingham

==Background==

Each of these questions should have
* randomly generated versions
* full worked solution

==AlgMap: Algebra questions==

These are approximately 140 questions in algebra, taken from A
Pocket Map of Algebraic Manipulation, by Dr Tony Barnard,
Department of Mathematics. King's College London. They are
designed to demonstrate many of the major problems in
elementary algebra.

The LaTeX source file of this "Map" is also included as
Barnard.tex

==CR: Calculus refresher.==

These are approximately 215 questions in calculus.  They are inspired by, with permission, the questions in the Calculus Refresher
http://www.mathcentre.ac.uk/resources.php/1

CR_all  is a single file with all questions, but this might be difficult to upload.

The following files have questions grouped by topic.

CR_Diff_01_basic 		Derivatives of basic functions
CR_Diff_02_linearity 	Linearity in differentiation
CR_Diff_03_higher  		Higher derivatives
CR_Diff_04_product		The product rule for differentiation
CR_Diff_06_chain  		The chain rule for differentiation
CR_Diff_07_implicit  	Differentiation of functions defined implicitly
CR_Diff_08_parametrical Differentiation of functions defined perimetrically
CR_Diff_09_misc  		Miscellaneous differentiation exercises
CR_Int_10_basic 		Integrals of basic functions
CR_Int_11_linearity 	Linearity in integration
CR_Int_12_definite 		Evaluating definite integrals
CR_Int_13_parts 		Integration by parts
CR_Int_14_substitution 	Integration by substitution
CR_Int_15_partialfrac 	Integration using partial fractions
CR_Int_16_trigonometric Integration using trigonometrical identities
CR_Int_17_misc 			Miscellaneous integration exercises

===Acknowledgements===

The work on the calculus refresher has been made possible, in part, by support from the FETLAR Project.
http://www.fetlar.bham.ac.uk/
The FETLAR project is the MSOR community's bid through the Subject Centre strand for the Open Educational Resources (OER) call from HEFCE/JISC.
http://www.jisc.ac.uk/fundingopportunities/funding_calls/2008/12/grant1408.aspx

== AlgMap ==

These questions are based on Tony Barnard's "Map of Algebraic
Manipulation".

--- NEW FILE: CR_Diff_09_misc.xml ---
<?xml version="1.0" encoding="UTF-8"?>
<mathQuiz version="2.0" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:dcterms="http://purl.org/dc/terms/" xmlns:lom="http://www.imsglobal.org/xsd/imsmd_v1p2"><assessmentItem version="2.0" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:dcterms="http://purl.org/dc/terms/" xmlns:lom="http://www.imsglobal.org/xsd/imsmd_v1p2"><questionCasValues><questionStem type="CasText"><castext>Find the following derivative:&#13;
\[ \frac{d}{d@v@}\left[ @p@\right]. \]&#13;
&#13;
#ans1# &lt;IEfeedback&gt;ans1&lt;/IEfeedback&gt;&#13;
&lt;PRTfeedback&gt;Result&lt;/PRTfeedback&gt;</castext><forbidFloats>false</forbidFloats><simplify>true</simplify></questionStem><questionVariables type="RawKeyVal"><rawKeyVals>v = x; rdm1 = 2+rand(4); rdm2 = 2+rand(5); rdm3 = 2+rand(5); ppp1 = sin(rdm1*v); ppp2 = cos(rdm2*v); pp1 = ppp1*ppp2; pp2 = tan(rdm3*v); p = pp1/pp2</rawKeyVals><forbidFloats>false</forbidFloats><simplify>true</simplify></questionVariables><workedSolution type="CasText"><castext>To solve this problem you will need to use the product rule &lt;hint&gt;calc_product_rule&lt;/hint&gt; and the quotient rule&lt;hint&gt;calc_quotient_rule&lt;/hint&gt;. &#13;
&#13;
Let $f(@v@)=@pp1@$ and $g(@v@)=@pp2@$. Then,&#13;
&#13;
\[\frac{d}{d@v@}\left[@p@\right] = \frac{\frac{df}{d@v@}\cdot g - f\cdot \frac{dg}{d@v@}}{g^2}.\]&#13;
&#13;
In order to calculate $\frac{df}{d@v@}$ you need to use the product rule, by defining two new functions: $h(@v@)=@ppp1@$ and $i(@v@)=@ppp2@$ so that $f(@v@)=h(@v@)i(@v@)$.&#13;
&#13;
So,&#13;
&#13;
\[\begin{array}{rcl}\displaystyle &#13;
 \frac{d}{d@v@}[@pp1@] &amp;\displaystyle =&amp;\displaystyle  \frac{dh}{dx}\cdot i + h\cdot \frac{di}{dx}\\&#13;
   &amp;\displaystyle =&amp;\displaystyle  [@diff(ppp1,v)@]@ppp2@ + @ppp1@[@diff(ppp2,v)@] \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  @diff(ppp1,v)*ppp2@ + @ppp1*diff(ppp2,v)@ \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  @diff(pp1,v)@&#13;
 \end{array}\] &#13;
&#13;
Hence,&#13;
&#13;
\[\begin{array}{rcl}\displaystyle &#13;
 \frac{d}{dx}\left[@p@\right] &amp;\displaystyle =&amp;\displaystyle  \frac{\frac{df}{d@v@}\cdot g - f\cdot \frac{dg}{d@v@}}{g^2} \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  \frac{[@diff(pp1,v)@]@pp2@ - @pp1@[@diff(pp2,v)@]}{[@pp2@]^2} \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  \frac{@diff(pp1,v)*pp2@ - @pp1*diff(pp2,v)@}{@pp2^2@} \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  @factor(diff(p,v))@&#13;
 \end{array}\] </castext><forbidFloats>false</forbidFloats><simplify>true</simplify></workedSolution><questionNote type="CasText"><castext>\[\frac{d}{d@v@}(@p@) = @diff(p,v)@ \]</castext><forbidFloats>false</forbidFloats><simplify>true</simplify></questionNote></questionCasValues><questionparts><questionpart><name>ans1</name><inputType type="Meta"><selection>Algebraic Input</selection><default>Algebraic Input</default><values/></inputType><boxsize>15</boxsize><teachersAns type="CasString"><casString>factor(diff(p,v))</casString><forbidFloats>false</forbidFloats><simplify>true</simplify></teachersAns><studentAnsKey>ans1</studentAnsKey><syntax type="Meta"><selection></selection><default></default><values/></syntax><stackoption><name>formalSyntax</name><type>list</type><default>true</default><values><value>true</value><value>false</value></values><casKey></casKey><casType></casType><selected>true</selected></stackoption><stackoption><name>forbidFloats</name><type>list</type><default>true</default><values><value>true</value><value>false</value></values><casKey>OPT_NoFloats</casKey><casType>ex</casType><selected>true</selected></stackoption><stackoption><name>insertStars</name><type>list</type><default>false</default><values><value>true</value><value>false</value></values><casKey></casKey><casType></casType><selected>false</selected></stackoption><stackoption><name>sameType</name><type>list</type><default>true</default><values><value>true</value><value>false</value></values><casKey></casKey><casType></casType><selected>true</selected></stackoption><stackoption><name>lowestTerms</name><type>list</type><default>true</default><values><value>true</value><value>false</value></values><casKey>OPT_LowestTerms</casKey><casType>ex</casType><selected>true</selected></stackoption><inputTypesParameters/></questionpart></questionparts><PotentialResponseTrees><PotentialResponseTree><prtname>PotResTree_Result</prtname><questionValue>1</questionValue><autoSimplify>true</autoSimplify><feedbackVariables>q 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</ansnote><nextPR>-1</nextPR></true><false><rawModMark>=</rawModMark><rawMark></rawMark><feedback></feedback><ansnote>Result-1-F </ansnote><nextPR>2</nextPR></false><teacherNote></teacherNote></PR><PR id="2"><answerTest>AlgEquiv</answerTest><teachersAns>factor(diff(q,v))</teachersAns><studentAns>ans1</studentAns><testoptions></testoptions><quietAnsTest></quietAnsTest><true><rawModMark>=</rawModMark><rawMark></rawMark><feedback>You appear to have differentiated @q@ instead of the other way up!</feedback><ansnote>Result-2-T </ansnote><nextPR>-1</nextPR></true><false><rawModMark>=</rawModMark><rawMark></rawMark><feedback></feedback><ansnote>Result-2-F </ansnote><nextPR>-1</nextPR></false><teacherNote></teacherNote></PR></PotentialResponses></PotentialResponseTree></PotentialResponseTrees><MetaData><dc:Publisher stackname="lastUserEditor" type="Meta"><selection>2</selection><default></default><values/></dc:Publisher><dc:language stackname="language" 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</value></col></test><test><col><key>IE_ans1</key><value>factor(diff(pp2/pp1,v))</value></col><col><key>PRT_PotResTree_Result</key><value>Result-2-T </value></col></test></ItemTests></assessmentItem><assessmentItem version="2.0" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:dcterms="http://purl.org/dc/terms/" xmlns:lom="http://www.imsglobal.org/xsd/imsmd_v1p2"><questionCasValues><questionStem type="CasText"><castext>Find the following derivative:&#13;
\[ \frac{d}{d@v@}\left[\left(@pp@\right)^{@rdm2@}\right]. \]&#13;
&#13;
#ans1# &lt;IEfeedback&gt;ans1&lt;/IEfeedback&gt;&#13;
&lt;PRTfeedback&gt;Result&lt;/PRTfeedback&gt;</castext><forbidFloats>false</forbidFloats><simplify>true</simplify></questionStem><questionVariables type="RawKeyVal"><rawKeyVals>v = x; rdm1 = 2+rand(7); rdm2 = (2+2*rand(5))/(3+2*rand(5)); ppp1 = 1-v^rdm1; ppp2 = 1+v^rdm1; pp = ppp1/ppp2; p = pp^(rdm2)</rawKeyVals><forbidFloats>false</forbidFloats><simplify>true</simplify></questionVariables><workedSolution type="CasText"><castext>To solve this problem you will need to use the chain rule &lt;hint&gt;calc_chain_rule&lt;/hint&gt; and the quotient rule &lt;hint&gt;calc_quotient_rule&lt;/hint&gt;. &#13;
&#13;
Let \[f(@v@)=@p@\] and \[g(@v@)=@pp@.\] Then $f(@v@)=[g(@v@)]^{@rdm2@}$. So,&#13;
&#13;
\[\begin{array}{rcl}\displaystyle &#13;
 \frac{d}{d@v@}\left[@p@\right] &amp;\displaystyle =&amp;\displaystyle  \frac{df}{dg}\cdot \frac{dg}{dx} \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  \frac{d}{dg}[g^{@rdm2@}]\cdot \frac{d}{d@v@}[g]&#13;
 \end{array}\] &#13;
&#13;
In order to find $\frac{dg}{d@v@}$ you have to define two new functions: $h(@v@)=@ppp1@$ and $i(@v@)=@ppp2@$. So,&#13;
&#13;
\[\begin{array}{rcl}\displaystyle &#13;
 \frac{d}{d@v@}(g) &amp;\displaystyle =&amp;\displaystyle  \frac{d}{d@v@}\left[@pp@\right] \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  \frac{d}{d@v@}\left(\frac{h}{i}\right) \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  \frac{\frac{dh}{d@v@}\cdot i - h\cdot \frac{di}{d@v@}}{i^2} \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  \frac{\left[\frac{d}{d@v@}(@ppp1@)\right]\cdot (@ppp2@) - (@ppp1@)\cdot \left[\frac{d}{d@v@}(@ppp2@)\right]}{(@ppp2@)^2} \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  \frac{[@diff(ppp1,v)@]\cdot (@ppp2@) - (@ppp1@)\cdot [@diff(ppp2,v)@]}{@ppp2^2@} \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  \frac{@factor(diff(ppp1,v)*ppp2-ppp1*diff(ppp2,v))@}{@ppp2^2@}.&#13;
 \end{array}\] &#13;
&#13;
Now, &#13;
\[\begin{array}{rcl}\displaystyle &#13;
 \frac{d}{dg}[g^{@rdm2@}] &amp;\displaystyle =&amp;\displaystyle  @rdm2@g^@rdm2-1@ \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  @rdm2*pp^(rdm2-1)@&#13;
 \end{array}\] &#13;
&#13;
Hence,&#13;
&#13;
\[\begin{array}{rcl}\displaystyle &#13;
 \frac{d}{d@v@}\left[@p@\right] &amp;\displaystyle =&amp;\displaystyle  \frac{d}{dg}[g^{@rdm2@}]\cdot \frac{d}{d@v@}[g] \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  \frac{@factor(diff(ppp1,v)*ppp2-ppp1*diff(ppp2,v))@}{@ppp2^2@}\cdot @rdm2*pp^(rdm2-1)@ \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  @factor(diff(p,v))@&#13;
 \end{array}\] </castext><forbidFloats>false</forbidFloats><simplify>true</simplify></workedSolution><questionNote type="CasText"><castext>\[\frac{d}{d@v@}(@p@) = @diff(p,v)@ \]</castext><forbidFloats>false</forbidFloats><simplify>true</simplify></questionNote></questionCasValues><questionparts><questionpart><name>ans1</name><inputType type="Meta"><selection>Algebraic Input</selection><default>Algebraic Input</default><values/></inputType><boxsize>15</boxsize><teachersAns type="CasString"><casString>factor(diff(p,v))</casString><forbidFloats>false</forbidFloats><simplify>true</simplify></teachersAns><studentAnsKey>ans1</studentAnsKey><syntax 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\[ \frac{d}{d@v@}[@p@]. \]&#13;
&#13;
#ans1# &lt;IEfeedback&gt;ans1&lt;/IEfeedback&gt;&#13;
&lt;PRTfeedback&gt;Result&lt;/PRTfeedback&gt;</castext><forbidFloats>false</forbidFloats><simplify>true</simplify></questionStem><questionVariables type="RawKeyVal"><rawKeyVals>v = x; rdm1 = 2+rand(2); rdm2 = 1/(2+rand(5)); ppp = v-v^rdm1; pp = ppp^rdm2; p = exp(pp)</rawKeyVals><forbidFloats>false</forbidFloats><simplify>true</simplify></questionVariables><workedSolution type="CasText"><castext>To solve this question, you will have to use the chain rule &lt;hint&gt;calc_chain_rule&lt;/hint&gt; more than once. &#13;
&#13;
Let $f(@v@)=@p@$ and $g(@v@)=@pp@$. Then, $f(@v@)=e^{(g(@v@))}$&#13;
&#13;
\[\begin{array}{rcl}\displaystyle &#13;
 \frac{d}{d@v@}[@p@] &amp;\displaystyle =&amp;\displaystyle  \frac{df}{dg}\cdot \frac{dg}{dx} \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  \frac{d}{dg}(e^{g})\cdot \frac{d}{dx}[@pp@]&#13;
 \end{array}\] &#13;
 &#13;
To find \[\frac{d}{dx}[@pp@]\] you have to define a new function: $h(@v@)=@ppp@$, so that $g(@v@)=h^{@rdm2@}$. So,&#13;
&#13;
\[\begin{array}{rcl}\displaystyle &#13;
 \frac{d}{d@v@}[@pp@] &amp;\displaystyle =&amp;\displaystyle  \frac{dg}{dh}\cdot \frac{dh}{d@v@} \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  \frac{d}{dh}(h^{@rdm2@})\cdot \frac{d}{d@v@}(@ppp@) \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  @rdm2@h^{@rdm2-1@}\cdot (@diff(ppp,v)@) \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  @diff(pp,v)@&#13;
 \end{array}\]  &#13;
&#13;
So,&#13;
&#13;
\[\begin{array}{rcl}\displaystyle &#13;
 \frac{d}{d@v@}[@p@] &amp;\displaystyle =&amp;\displaystyle  \frac{d}{dg}(e^{g})\cdot \frac{d}{dx}[@pp@] \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  e^g \cdot @diff(pp,v)@ \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  e^{@pp@}@diff(pp,v)@ \\&#13;
   &amp;\displaystyle =&amp;\displaystyle  @diff(p,v)@&#13;
 \end{array}\] &#13;
</castext><forbidFloats>false</forbidFloats><simplify>true</simplify></workedSolution><questionNote type="CasText"><castext>\[\frac{d}{d@v@}(@p@) = @diff(p,v)@ \]</castext><forbidFloats>false</forbidFloats><simplify>true</simplify></questionNote></questionCasValues><questionparts><questionpart><name>ans1</name><inputType type="Meta"><selection>Algebraic Input</selection><default>Algebraic Input</default><values/></inputType><boxsize>15</boxsize><teachersAns type="CasString"><casString>diff(p,v)</casString><forbidFloats>false</forbidFloats><simplify>true</simplify></teachersAns><studentAnsKey>ans1</studentAnsKey><syntax type="Meta"><selection></selection><default></default><values/></syntax><stackoption><name>formalSyntax</name><type>list</type><default>true</default><values><value>true</value><value>false</value></values><casKey></casKey><casType></casType><selected>true</selected></stackoption><stackoption><name>forbidFloats</name><type>list</type><default>true</default><values><value>true</value><value>false</value></values><casKey>OPT_NoFloats</casKey><casType>ex</casType><selected>true</selected></stackoption><stackoption><name>insertStars</name><type>list</type><default>false</default><values><value>true</value><value>false</value></values><casKey></casKey><casType></casType><selected>false</selected></stackoption><stackoption><name>sameType</name><type>list</type><default>true</default><values><value>true</value><value>false</value></values><casKey></casKey><casType></casType><selected>true</selected></stackoption><stackoption><name>lowestTerms</name><type>list</type><default>true</default><values><value>true</value><value>false</value></values><casKey>OPT_LowestTerms</casKey><casType>ex</casType><selected>true</selected></stackoption><inputTypesParameters/></questionpart></questionparts><PotentialResponseTrees><PotentialResponseTree><prtname>PotResTree_Result</prtname><questionValue>1</questionValue><autoSimplify>true</autoSimplify><feedbackVariables>q3 = v+v^rdm1; q2 = q3^rdm2; q = exp(q2)</feedbackVariables><description type="Meta"><selection></selection><default></default><values/></description><PotentialResponses><PR id="0"><answerTest>Diff</answerTest><teachersAns>diff(p,v)</teachersAns><studentAns>ans1</studentAns><testoptions>v</testoptions><quietAnsTest></quietAnsTest><true><rawModMark>=</rawModMark><rawMark>1</rawMark><feedback></feedback><ansnote>Result-0-T </ansnote><nextPR>-1</nextPR></true><false><rawModMark>=</rawModMark><rawMark></rawMark><feedback></feedback><ansnote>Result-0-F </ansnote><nextPR>1</nextPR></false><teacherNote></teacherNote></PR><PR id="1"><answerTest>AlgEquiv</answerTest><teachersAns>-diff(p,v)</teachersAns><studentAns>ans1</studentAns><testoptions></testoptions><quietAnsTest></quietAnsTest><true><rawModMark>=</rawModMark><rawMark></rawMark><feedback>You appear to have lost the minus sign as your answer is the negative of what was expected.</feedback><ansnote>Result-1-T </ansnote><nextPR>-1</nextPR></true><false><rawModMark>=</rawModMark><rawMark></rawMark><feedback></feedback><ansnote>Result-1-F </ansnote><nextPR>2</nextPR></false><teacherNote></teacherNote></PR><PR id="2"><answerTest>AlgEquiv</answerTest><teachersAns>diff(q,v)</teachersAns><studentAns>ans1</studentAns><testoptions></testoptions><quietAnsTest></quietAnsTest><true><rawModMark>=</rawModMark><rawMark></rawMark><feedback>You appear to have differentiated @ppp@ incorrectly - there is a sign wrong.</feedback><ansnote>Result-2-T </ansnote><nextPR>-1</nextPR></true><false><rawModMark>=</rawModMark><rawMark></rawMark><feedback></feedback><ansnote>Result-2-F </ansnote><nextPR>3</nextPR></false><teacherNote></teacherNote></PR><PR id="3"><answerTest>AlgEquiv</answerTest><teachersAns>diff(p,v)/diff(pp,v)</teachersAns><studentAns>ans1</studentAns><testoptions></testoptions><quietAnsTest></quietAnsTest><true><rawModMark>=</rawModMark><rawMark></rawMark><feedback>You need to use the chain rule &lt;hint&gt;calc_chain_rule&lt;/hint&gt; to differentiate @p@ with respect to @v@.</feedback><ansnote>Result-3-T </ansnote><nextPR>-1</nextPR></true><false><rawModMark>=</rawModMark><rawMark></rawMark><feedback></feedback><ansnote>Result-3-F </ansnote><nextPR>-1</nextPR></false><teacherNote></teacherNote></PR></PotentialResponses></PotentialResponseTree></PotentialResponseTrees><MetaData><dc:Publisher stackname="lastUserEditor" type="Meta"><selection>2</selection><default></default><values/></dc:Publisher><dc:language stackname="language" type="Meta"><selection>en</selection><default>en</default><values><value key="en">en</value><value key="fr">fr</value><value key="nl">nl</value><value key="es">es</value><value key="unspecified">Unspecified</value></values></dc:language><dc:title stackname="questionName" type="Meta"><selection>CR-Diff-09-misc-4.b</selection><default></default><values/></dc:title><dc:description stackname="questionDescription" type="Meta"><selection>Miscellaneous differentiation exercises</selection><default></default><values/></dc:description><dc:publi...

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