// Numbas version: exam_results_page_options {"name": "Simon's copy of Functions of two variables: Stationary points 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"statement": "\n

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\$f(x,y)=\\simplify[std]{ ({a} / 3) * x ^ 3 + ({b} / 2) * x ^ 2 * y + {c} * y ^ 2 * x + {d} * y}\$

\n ", "variablesTest": {"maxRuns": 100, "condition": ""}, "tags": ["Calculus", "Differentiation", "calculus", "derivative", "differentiation", "functions of two variables", "partial derivative", "partial differentiation", "stationary points", "stationary points of functions of two variables"], "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "variable_groups": [], "functions": {}, "advice": "

#### a)

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\$\\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*}\$

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#### b)

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$(a,b)$ 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.

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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*}\$

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The left hand side of equation (1) can be factorised as:

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\$\\simplify[std]{({a1}x+{b1}y)*({c1}x+{d1}y)=0}\$

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and so we have:
\$y=\\simplify[std]{{-a1}/{b1}*x},\\mbox{ or } y= \\simplify[std]{{-c1}/{d1}*x}\$

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##### First case: $y= \\simplify[std]{{-a1}/{b1}*x}$
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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}}\$

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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)\$

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##### Second case: $y= \\simplify[std]{{-c1}/{d1}*x}$
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{check}

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Substituting 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}}\$

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{other}

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10/07/2012:

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Question appears to be working correctly.

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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.

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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.

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$\\displaystyle { \\partial f \\over \\partial x}=$ [[0]]

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$\\displaystyle {\\partial f \\over \\partial y}=$ [[1]]

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$x=\\var{m},\\;\\;y=\\simplify[std]{-{a1*m}/{b1}}$

", "

$x=\\var{-m},\\;\\;y=\\simplify[std]{{a1*m}/{b1}}$

", "

$x=\\var{m+1},\\;\\;y=\\simplify[std]{-{c1*(m+1)}/{d1}}$

", "

$x=\\var{-m-1},\\;\\;y=\\simplify[std]{{c1*(m+1)}/{d1}}$

", "

$x=\\var{m-1},\\;\\;y=\\simplify[std]{-{a1+2*b1}/{b1}}$

", "

$x=\\var{-m+1},\\;\\;y=\\simplify[std]{{a1+2*b1}/{b1}}$

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#### Finding Stationary Points.

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Tick the two choices which give stationary points for $f(x,y)$.

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Note 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.

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