Find the partial derivatives of $f$ with respect to $x$ and $y$.
\nNote that if you want to enter a product of two 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]]
", "scripts": {}}, {"maxAnswers": 0, "choices": ["$x=\\var{m},\\;\\;y=\\simplify[std]{-{a1*m}/{b1}}$
", "$x=\\var{-m},\\;\\;y=\\simplify[std]{{a1*m}/{b1}}$
", "$x=\\var{m+1},\\;\\;y=\\simplify[std]{-{c1*(m+1)}/{d1}}$
", "$x=\\var{-m-1},\\;\\;y=\\simplify[std]{{c1*(m+1)}/{d1}}$
", "$x=\\var{m-1},\\;\\;y=\\simplify[std]{-{a1+2*b1}/{b1}}$
", "$x=\\var{-m+1},\\;\\;y=\\simplify[std]{{a1+2*b1}/{b1}}$
"], "distractors": ["", "", "", "", "", ""], "showFeedbackIcon": true, "warningType": "none", "maxMarks": 0, "displayColumns": 3, "minMarks": 0, "variableReplacements": [], "showCorrectAnswer": true, "marks": 0, "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 ", "extendBaseMarkingAlgorithm": true, "unitTests": [], "customMarkingAlgorithm": "", "matrix": [2, 2, 0, 0, 0, 0], "variableReplacementStrategy": "originalfirst", "shuffleChoices": true, "minAnswers": 0, "displayType": "checkbox", "type": "m_n_2", "scripts": {}}], "name": "Luis's copy of Find partial derivatives of $f(x,y)$ and identify its stationary points", "ungrouped_variables": ["a", "c", "ch", "d", "m", "s5", "a1", "b", "b1", "c2", "c1", "d1"], "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 ", "preamble": {"css": "", "js": ""}, "variables": {"d": {"description": "", "templateType": "anything", "name": "d", "group": "Ungrouped variables", "definition": "-(b1*c1-3*a1*d1)/2*m^2"}, "d1": {"description": "", "templateType": "anything", "name": "d1", "group": "Ungrouped variables", "definition": "random(2,4,6)"}, "c": {"description": "", "templateType": "anything", "name": "c", "group": "Ungrouped variables", "definition": "b1*d1"}, "c2": {"description": "", "templateType": "anything", "name": "c2", "group": "Ungrouped variables", "definition": "random(1..4)"}, "b": {"description": "", "templateType": "anything", "name": "b", "group": "Ungrouped variables", "definition": "b1*c1+a1*d1"}, "s5": {"description": "", "templateType": "anything", "name": "s5", "group": "Ungrouped variables", "definition": "if(d*(-b*d1+4*c*c1)<=0,-1,1)"}, "b1": {"description": "", "templateType": "anything", "name": "b1", "group": "Ungrouped variables", "definition": "random(2,4,6)"}, "a": {"description": "", "templateType": "anything", "name": "a", "group": "Ungrouped variables", "definition": "a1*c1"}, "ch": {"description": "", "templateType": "anything", "name": "ch", "group": "Ungrouped variables", "definition": "if(a1*d1=b1*c1,0,1)"}, "a1": {"description": "", "templateType": "anything", "name": "a1", "group": "Ungrouped variables", "definition": "random(1..5)"}, "c1": {"description": "", "templateType": "anything", "name": "c1", "group": "Ungrouped variables", "definition": "if(b1*c2=3*a1*d1,c2+1,c2)"}, "m": {"description": "", "templateType": "anything", "name": "m", "group": "Ungrouped variables", "definition": "random(1..4)"}}, "metadata": {"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.
\nInputting the values given into the partial derivatives to see if 0 is obtained is tedious! Could ask for the factorisation of equation 1 as the solution uses this. However there is a problem in asking for the input of the stationary points - order of input and also giving that there is two stationary points.
", "licence": "Creative Commons Attribution 4.0 International"}, "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "functions": {}, "variable_groups": [], "advice": "\\begin{align}
\\partial f \\over \\partial x &= \\simplify[std]{(({a} * (x ^ 2)) + ({b} * x * y) + ({c} * (y ^ 2)))} \\\\[1em]
\\partial f \\over \\partial y &= \\simplify[std]{((({b} / 2) * (x ^ 2)) + ({(2 * c)} * x * y) + {d})}
\\end{align}
$(a,b)$ is a stationary point for the function $f(x,y)$ if $f_x=0$ and $f_y=0$, where the partial derivatives are evaluated at $x=a$, $y=b$.
\nSo you have to make sure that both of these partial derivatives are $0$ at the stationary point.
\nFor this example we have from the above equations that:
\n\\begin{align}
\\simplify[std]{(({a} * (x ^ 2)) + ({b} * x * y) + ({c} * (y ^ 2)))} &= 0, & \\mathbf{(1)}\\\\
\\simplify[std]{((({b} / 2) * (x ^ 2)) + ({(2 * c)} * x * y) + {d})} &= 0, & \\mathbf{(2)}
\\end{align}
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:
\n\\[ y=\\simplify[std]{{-a1}/{b1}*x}, \\text{ or } y= \\simplify[std]{{-c1}/{d1}*x} \\]
\nSubstituting this into equation (2) gives:
\n\\[\\simplify[std]{{b}/2*x^2-{2c*a1}/{b1}*x^2+{d}}=0 \\implies \\simplify[std]{{-b*b1+4*c*a1}/{2*b1}*x^2={d}}\\]
\nHence $x=\\var{m}$ or $x = \\var{-m}$. The corresponding stationary points are:
\n\\[ \\left(\\var{m},\\simplify[std]{-{a1*m}/{b1}}\\right) \\text{ and } \\left(\\var{-m},\\simplify[std]{{a1*m}/{b1}}\\right) \\]
\nSubstituting this into equation (2) gives:
\n\\[\\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}}\\]
\nThere can be no more stationary points as this equation has no real solution.
", "extensions": [], "tags": ["calculus", "functions", "multivariable", "partial derivatives", "stationary points"], "type": "question", "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}, {"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Luis Hernandez", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2870/"}]}]}], "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}, {"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Luis Hernandez", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2870/"}]}