// Numbas version: finer_feedback_settings {"name": "Morten's copy of Functions of two variables: Stationary points 3", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["a", "c", "b", "d", "s3", "s2", "s1", "s5", "s4", "r"], "name": "Morten's copy of Functions of two variables: Stationary points 3", "tags": ["Calculus", "Differentiation", "Simultaneous equations", "calculus", "differentiate", "differentiation", "functions of two variables", "partial derivative", "partial differentiation", "simultaneous equations", "stationary points"], "preamble": {"css": "", "js": ""}, "advice": "\n

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 $y$

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\\[\\begin{eqnarray*} \\partial f \\over \\partial x &=&0\\\\ \\\\ \\partial f \\over \\partial y &=&0 \\end{eqnarray*} \\]

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In this case you get two equations to solve for $x$ and $y$

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\\[\\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   $\\simplify[std]{e^(-(x-{c})^2-(y-{d})^2)} \\neq 0,\\;\\forall x,\\;y$.  

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On solving these we get \\[ x = \\var{c},\\;\\;\\;y=\\var{d}\\]

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So the stationary point is $(\\var{c},\\var{d}) \\in D$.

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On substituting these values into $f(x,y)$ we get:

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\\[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]]

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$y$–coordinate, $b=$ [[1]]

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Input the value of $f(x,y)$ at $(a,b)$:

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$f(a,b)=$ [[2]]

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If you want some help, click on Show steps. You will not lose any marks if you do so.

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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 $y$

\n \n \n \n

\\[\\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 \\]

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In this case you get two equations to solve for $x$ and $y$

\n \n \n ", "scripts": {}, "type": "information", "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 $f: D \\rightarrow \\mathbb{R}$:

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

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where \\[D = \\{(x,y): \\simplify[std]{(x-{c})^2+(y-{d})^2}\\} \\le \\var{r}\\]

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That is, $D$ is a disk of radius $\\simplify[std]{sqrt({r})}$ and centre $(\\var{c},\\var{d})$.

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Input both cooordinates as fractions or integers and not decimals.

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10/07/2012:

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Added tags.

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Question appears to be working correctly.

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\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 $(c,d)$.

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