// Numbas version: finer_feedback_settings {"name": "Using partial derivatives to find stationary points", "type": "exam", "duration": 0, "metadata": {"notes": "
Finding stationary points of a function of two varaibles
", "description": "\n \t\t3 questions. Finding the stationary points of functions of 2 variables.
\n \t\tPartial differentiation.
\n \t\t", "licence": "Creative Commons Attribution 4.0 International"}, "allQuestions": true, "shuffleQuestions": false, "questions": [], "percentPass": 50, "timing": {"allowPause": true, "timeout": {"action": "none", "message": ""}, "timedwarning": {"action": "none", "message": ""}}, "pickQuestions": 0, "navigation": {"onleave": {"action": "none", "message": ""}, "reverse": true, "allowregen": true, "preventleave": false, "browse": true, "showfrontpage": false, "showresultspage": "never"}, "feedback": {"showtotalmark": true, "advicethreshold": 0, "showanswerstate": true, "showactualmark": true, "allowrevealanswer": true, "enterreviewmodeimmediately": false, "showexpectedanswerswhen": "never", "showpartfeedbackmessageswhen": "always", "showactualmarkwhen": "always", "showtotalmarkwhen": "always", "showanswerstatewhen": "always", "showadvicewhen": "never"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": [{"name": "Functions of two variables: Stationary points 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "ungrouped_variables": ["rawstatval", "statval", "sol", "lmin", "q1", "dvalue", "s3", "s2", "s1", "s5", "s4", "neither", "b", "c1", "a", "c", "p1", "d", "f", "lmax", "y", "x"], "tags": ["Calculus", "Differentiation", "calculus", "derivative", "differentiation", "function of 2 variables", "functions of 2 variables", "functions of two variables", "partial derivatives", "partial differentiation", "stationary points", "stationary points of functions of two variables"], "advice": "The $(x,y)$ coordinates of the stationary point of a function of 2 variables $f(x,y)$ are given by solving
the following 2 equations for $x$ and $y$:
\\[\\begin{eqnarray*} \\partial f \\over \\partial x &=&0\\\\ \\\\ \\partial f \\over \\partial y &=&0 \\end{eqnarray*} \\]
\nIn this case you get two linear equations to solve for $x$ and $y$:
\n\\[\\begin{eqnarray*} \\simplify[std]{{2*a}x+{b}y+{d}}&=&0\\\\ \\\\ \\simplify[std]{{b}x+{2*c1}y+{f}}&=&0 \\end{eqnarray*} \\]
On solving these we get \\[ x = \\simplify[std]{{2*c1*d-b*f}/{b^2-4*a*c1}},\\;\\;\\;y=\\simplify[std]{{2*a*f-b*d}/{b^2-4*a*c1}}\\]
On substituting these values into $f(x,y)$ we get:
\\[f\\left(\\simplify[std]{{2*c1*d-b*f}/{b^2-4*a*c1}},\\simplify[std]{{2*a*f-b*d}/{b^2-4*a*c1}}\\right) = \\var{rawstatval}\\approx\\var{statval}\\]
to 2 decimal places.
Input both cooordinates as fractions or integers and not decimals.
\n$x$–coordinate, $a=$ [[0]].
\n$y$–coordinate, $b=$ [[1]].
\nInput value of $f(x,y)$ at $(a,b)$:
\n$f(a,b)=\\;\\;$[[2]] (Input to 2 decimal places).
\nIf you want some help, click on Show steps. You will not lose any marks if you do so.
", "marks": 0, "gaps": [{"notallowed": {"message": "Input answer as a fraction or an integer, not a decimal
", "showStrings": false, "strings": ["."], "partialCredit": 0}, "expectedvariablenames": [], "checkingaccuracy": 0.001, "type": "jme", "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "marks": 2, "answer": "{2*c1*d-b*f}/{b^2-4*a*c1}", "checkingtype": "absdiff", "checkvariablenames": false, "vsetrange": [0, 1]}, {"notallowed": {"message": "Input answer as a fraction or an integer, not a decimal
", "showStrings": false, "strings": ["."], "partialCredit": 0}, "expectedvariablenames": [], "checkingaccuracy": 0.001, "type": "jme", "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "marks": 2, "answer": "{2*a*f-b*d}/{b^2-4*a*c1}", "checkingtype": "absdiff", "checkvariablenames": false, "vsetrange": [0, 1]}, {"allowFractions": false, "scripts": {}, "maxValue": "statval+0.01", "minValue": "statval-0.01", "correctAnswerFraction": false, "showCorrectAnswer": true, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "showCorrectAnswer": true, "scripts": {}, "steps": [{"prompt": "\n \n \nThe $(x,y)$ coordinates of the stationary point of a function of 2 variables $f(x,y)$ are given by solving
the following 2 equations for $x$ and $y$
\\[\\begin{eqnarray*}\n \n \\partial f \\over \\partial x &=&0\\\\\n \n \\\\\n \n \\partial f \\over \\partial y &=&0\n \n \\end{eqnarray*}\n \n \\]
\n \n \n \nIn this case you get two linear equations to solve for $x$ and $y$
\n \n \n ", "type": "information", "marks": 0, "showCorrectAnswer": true, "scripts": {}}], "type": "gapfill"}], "statement": "In the following question find the $(x,y)$ coordinates of the single stationary point $(a,b)$ of the function
\n\\[f(x,y)=\\simplify[std]{{a}*x^2+{b}*x*y+{c1}*y^2+{d}*x+{f}*y}\\]
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\n \t\tAdded tags.
\n \t\tQuestion appears to be working correctly.
\n \t\t", "description": "Find the stationary point $(p,q)$ of the function: $f(x,y)=ax^2+bxy+cy^2+dx+gy$. Calculate $f(p,q)$.
", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Functions of two variables: Stationary points 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "ungrouped_variables": ["a", "c", "ch", "d", "m", "s5", "a1", "other", "b1", "c2", "c1", "b", "check", "d1"], "tags": ["Calculus", "Differentiation", "calculus", "derivative", "differentiation", "functions of two variables", "partial derivative", "partial differentiation", "stationary points", "stationary points of functions of two variables"], "preamble": {"css": "", "js": ""}, "advice": "\\[\\begin{eqnarray*} {\\partial f \\over \\partial x} &=&\\simplify[std]{(({a} * (x ^ 2)) + ({b} * x * y) + ({c} * (y ^ 2)))}\\\\ \\\\ \\partial f \\over \\partial y &=&\\simplify[std]{((({b} / 2) * (x ^ 2)) + ({(2 * c)} * x * y) + {d})} \\end{eqnarray*}\\]
\n$(a,b)$ is a stationary point for the function $f(x,y)$ if $f_x=0,\\;\\;f_y=0$,where the partial derivatives are evaluated at $x=a,\\;\\;y=b$.
So you have to make sure that both these partial derivatives are $0$ at the stationary point.
For this example we have from the above equations that:
\\[\\begin{eqnarray*} \\simplify[std]{(({a} * (x ^ 2)) + ({b} * x * y) + ({c} * (y ^ 2)))}&=&0,\\qquad &\\mathbf{(1)}&\\\\ \\\\ \\simplify[std]{((({b} / 2) * (x ^ 2)) + ({(2 * c)} * x * y) + {d})}&=&0, \\qquad &\\mathbf{(2)}& \\end{eqnarray*}\\]
The left hand side of equation (1) can be factorised as:
\n\\[\\simplify[std]{({a1}x+{b1}y)*({c1}x+{d1}y)=0}\\]
\nand so we have:
\\[y=\\simplify[std]{{-a1}/{b1}*x},\\mbox{ or } y= \\simplify[std]{{-c1}/{d1}*x}\\]
Substituting this into equation (2) gives:
\\[\\simplify[std]{{b}/2*x^2-{2c*a1}/{b1}*x^2+{d}}=0 \\Rightarrow \\simplify[std]{{-b*b1+4*c*a1}/{2*b1}*x^2={d}}\\]
Hence $x=\\var{m}\\mbox{ or } x = \\var{-m}$ and the stationary points which are on the list and which you had to choose are:
\\[\\left(\\var{m},\\simplify[std]{-{a1*m}/{b1}}\\right)\\mbox{ and }\\left(\\var{-m},\\simplify[std]{{a1*m}/{b1}}\\right)\\]
{check}
\nSubstituting this into equation (2) gives:
\\[\\simplify[std]{{b}/2*x^2-{2c*c1}/{d1}*x^2+{d}}=0 \\Rightarrow \\simplify[std]{{-b*d1+4*c*c1}/{2*d1}*x^2={d}}\\]
{other}
", "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "parts": [{"prompt": "Enter the partial derivatives here. Note if you want to enter a product of unknowns, such as $xy$ then you input the expression in the form x*y
.
$\\displaystyle { \\partial f \\over \\partial x}=$ [[0]]
\n$\\displaystyle {\\partial f \\over \\partial y}=$ [[1]]
", "marks": 0, "gaps": [{"expectedvariablenames": ["x", "y"], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "(({a} * (x ^ 2)) + ({b} * x * y) + ({c} * (y ^ 2)))", "marks": 2, "checkvariablenames": true, "checkingtype": "absdiff", "type": "jme"}, {"expectedvariablenames": ["x", "y"], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "std", "scripts": {}, "answer": "((({b} / 2) * (x ^ 2)) + ({(2 * c)} * x * y) + {d})", "marks": 2, "checkvariablenames": true, "checkingtype": "absdiff", "type": "jme"}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"maxAnswers": 0, "displayColumns": 3, "prompt": "\nTick the two choices which give stationary points for $f(x,y)$.
\nNote that the easiest way to do this question is to substitute the values for $x$ and for $y$ into the expressions for $\\displaystyle {\\partial f \\over \\partial x}$ and $\\displaystyle{\\partial f \\over \\partial y}$ and see if you get $0$ for both.
\n ", "matrix": [2, 2, 0, 0, 0, 0], "minAnswers": 0, "maxMarks": 0, "shuffleChoices": true, "warningType": "none", "displayType": "checkbox", "showCorrectAnswer": true, "scripts": {}, "distractors": ["", "", "", "", "", ""], "marks": 0, "choices": ["$x=\\var{m},\\;\\;y=\\simplify[std]{-{a1*m}/{b1}}$
", "$x=\\var{-m},\\;\\;y=\\simplify[std]{{a1*m}/{b1}}$
", "$x=\\var{m+1},\\;\\;y=\\simplify[std]{-{c1*(m+1)}/{d1}}$
", "$x=\\var{-m-1},\\;\\;y=\\simplify[std]{{c1*(m+1)}/{d1}}$
", "$x=\\var{m-1},\\;\\;y=\\simplify[std]{-{a1+2*b1}/{b1}}$
", "$x=\\var{-m+1},\\;\\;y=\\simplify[std]{{a1+2*b1}/{b1}}$
"], "type": "m_n_2", "minMarks": 0}], "statement": "\nAnswer the following questions about the function:
\n\\[f(x,y)=\\simplify[std]{ ({a} / 3) * x ^ 3 + ({b} / 2) * x ^ 2 * y + {c} * y ^ 2 * x + {d} * y}\\]
\n ", "type": "question", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "variables": {"a": {"definition": "a1*c1", "templateType": "anything", "group": "Ungrouped variables", "name": "a", "description": ""}, "c": {"definition": "b1*d1", "templateType": "anything", "group": "Ungrouped variables", "name": "c", "description": ""}, "ch": {"definition": "if(a1*d1=b1*c1,0,1)", "templateType": "anything", "group": "Ungrouped variables", "name": "ch", "description": ""}, "d": {"definition": "-(b1*c1-3*a1*d1)/2*m^2", "templateType": "anything", "group": "Ungrouped variables", "name": "d", "description": ""}, "m": {"definition": "random(1..4)", "templateType": "anything", "group": "Ungrouped variables", "name": "m", "description": ""}, "s5": {"definition": "if(d*(-b*d1+4*c*c1)<=0,-1,1)", "templateType": "anything", "group": "Ungrouped variables", "name": "s5", "description": ""}, "a1": {"definition": "random(1..5)", "templateType": "anything", "group": "Ungrouped variables", "name": "a1", "description": ""}, "other": {"definition": "if(s5=-1,'There can be no more stationary points as we cannot solve this.', if(ch=0,' we get the same stationary points',' You solve this getting 2 values of $x$ and then you obtain two more stationary points (which are not on the list you choose from).'))", "templateType": "anything", "group": "Ungrouped variables", "name": "other", "description": ""}, "b1": {"definition": "random(2,4,6)", "templateType": "anything", "group": "Ungrouped variables", "name": "b1", "description": ""}, "c2": {"definition": "random(1..4)", "templateType": "anything", "group": "Ungrouped variables", "name": "c2", "description": ""}, "c1": {"definition": "if(b1*c2=3*a1*d1,c2+1,c2)", "templateType": "anything", "group": "Ungrouped variables", "name": "c1", "description": ""}, "b": {"definition": "b1*c1+a1*d1", "templateType": "anything", "group": "Ungrouped variables", "name": "b", "description": ""}, "check": {"definition": "if(ch=0, 'The two expressions for y in terms of x are the same and we will get exactly the same so ignore the rest!',' ')", "templateType": "anything", "group": "Ungrouped variables", "name": "check", "description": ""}, "d1": {"definition": "random(2,4,6)", "templateType": "anything", "group": "Ungrouped variables", "name": "d1", "description": ""}}, "metadata": {"notes": "\n \t\t10/07/2012:
\n \t\tAdded tags.
Question appears to be working correctly.
\n \t\t\n \t\t", "description": "
Find the stationary points of the function: $f(x,y)=a x ^ 3 + b x ^ 2 y + c y ^ 2 x + dy$ by choosing from a list of points.
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Functions of two variables: Stationary points 3", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "ungrouped_variables": ["a", "c", "b", "d", "s3", "s2", "s1", "s5", "s4", "r"], "tags": ["Calculus", "Differentiation", "Simultaneous equations", "calculus", "differentiate", "differentiation", "functions of two variables", "partial derivative", "partial differentiation", "simultaneous equations", "stationary points"], "preamble": {"css": "", "js": ""}, "advice": "\nThe $(x,y)$ coordinates of the stationary point of a function of 2 variables $f(x,y)$ are given by solving
the following 2 equations for $x$ and $y$
\\[\\begin{eqnarray*} \\partial f \\over \\partial x &=&0\\\\ \\\\ \\partial f \\over \\partial y &=&0 \\end{eqnarray*} \\]
\nIn this case you get two equations to solve for $x$ and $y$
\n\\[\\begin{eqnarray*} \\simplify[std]{{-2*b}*(x-{c})*e^(-(x-{c})^2-(y-{d})^2)}&=&0\\\\ \\\\ \\simplify[std]{{-2*b}*(y-{d})*e^(-(x-{c})^2-(y-{d})^2)}&=&0 \\end{eqnarray*} \\]
We can cancel off the term $\\simplify[std]{e^(-(x-{c})^2-(y-{d})^2)}$ in both equations as $\\simplify[std]{e^(-(x-{c})^2-(y-{d})^2)} \\neq 0,\\;\\forall x,\\;y$.
On solving these we get \\[ x = \\var{c},\\;\\;\\;y=\\var{d}\\]
\nSo the stationary point is $(\\var{c},\\var{d}) \\in D$.
\nOn substituting these values into $f(x,y)$ we get:
\n\\[f(\\var{c},\\var{d})=\\simplify[std]{{a}+{b}*e^0={a+b}}\\]
\n ", "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "parts": [{"stepsPenalty": 0, "prompt": "$x$–coordinate, $a=$ [[0]]
\n$y$–coordinate, $b=$ [[1]]
\nInput the value of $f(x,y)$ at $(a,b)$:
\n$f(a,b)=$ [[2]]
\nIf you want some help, click on Show steps. You will not lose any marks if you do so.
", "marks": 0, "gaps": [{"allowFractions": false, "marks": 2, "maxValue": "{c}", "minValue": "{c}", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "marks": 2, "maxValue": "{d}", "minValue": "{d}", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "marks": 1, "maxValue": "{a+b}", "minValue": "{a+b}", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}], "showCorrectAnswer": true, "scripts": {}, "steps": [{"type": "information", "showCorrectAnswer": true, "scripts": {}, "prompt": "\n \n \nThe $(x,y)$ coordinates of the stationary point of a function of 2 variables $f(x,y)$ are given by solving
the following 2 equations for $x$ and $y$
\\[\\begin{eqnarray*}\n \n \\partial f \\over \\partial x &=&0\\\\\n \n \\\\\n \n \\partial f \\over \\partial y &=&0\n \n \\end{eqnarray*}\n \n \\]
\n \n \n \nIn this case you get two equations to solve for $x$ and $y$
\n \n \n ", "marks": 0}], "type": "gapfill"}], "statement": "In the following question find the $(x,y)$ coordinates of the single stationary point $(a,b) \\in D$ of the continuous function $f: D \\rightarrow \\mathbb{R}$:
\n\\[f(x,y) = \\simplify[std]{{a} + {b}*e^(-(x-{c})^2-(y-{d})^2)}\\]
\nwhere \\[D = \\{(x,y): \\simplify[std]{(x-{c})^2+(y-{d})^2}\\} \\le \\var{r}\\]
\nThat is, $D$ is a disk of radius $\\simplify[std]{sqrt({r})}$ and centre $(\\var{c},\\var{d})$.
\nInput both cooordinates as fractions or integers and not decimals.
", "type": "question", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "variables": {"a": {"definition": "s1*random(1..9)", "templateType": "anything", "group": "Ungrouped variables", "name": "a", "description": ""}, "c": {"definition": "s3*random(1..9)", "templateType": "anything", "group": "Ungrouped variables", "name": "c", "description": ""}, "b": {"definition": "s2*random(1..9)", "templateType": "anything", "group": "Ungrouped variables", "name": "b", "description": ""}, "d": {"definition": "s4*random(1..9)", "templateType": "anything", "group": "Ungrouped variables", "name": "d", "description": ""}, "s3": {"definition": "random(1,-1)", "templateType": "anything", "group": "Ungrouped variables", "name": "s3", "description": ""}, "s2": {"definition": "random(1,-1)", "templateType": "anything", "group": "Ungrouped variables", "name": "s2", "description": ""}, "s1": {"definition": "random(1,-1)", "templateType": "anything", "group": "Ungrouped variables", "name": "s1", "description": ""}, "s5": {"definition": "random(1,-1)", "templateType": "anything", "group": "Ungrouped variables", "name": "s5", "description": ""}, "s4": {"definition": "random(1,-1)", "templateType": "anything", "group": "Ungrouped variables", "name": "s4", "description": ""}, "r": {"definition": "random(2,3,5,6,7)", "templateType": "anything", "group": "Ungrouped variables", "name": "r", "description": ""}}, "metadata": {"notes": "\n \t\t10/07/2012:
\n \t\tAdded tags.
Question appears to be working correctly.
\n \t\t\n \t\t", "description": "
Find the coordinates of the stationary point for $f: D \\rightarrow \\mathbb{R}$: $f(x,y) = a + be^{-(x-c)^2-(y-d)^2}$, $D$ is a disk centre $(c,d)$.
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}]}], "contributors": [{"name": "Nick McCullen", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/953/"}], "extensions": [], "custom_part_types": [], "resources": []}