// Numbas version: exam_results_page_options {"name": "Dynamical system 4", "extensions": ["jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"rulesets": {"std": ["all", "!collectNumbers", "!noLeadingMinus"]}, "metadata": {"notes": "

The definition of the trace is tau=random(+repeat(2*random(0..9)*sign(random(-1,1)),5)), i.e. essentially random(0,n1,n2,n3,n4,n5), where ni is calculated as above.  This is so that there is roughly a one in six chance of getting a \"centre\" fixed point.  Without this manipulation, the chance of a centre is quite small, since it can only occur when $\\tau=0$ (and $\\delta>0$).

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05/11/2013

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Fix definition of c, which should be multiplied by the sign of b, so that the determinant is calculated correctly.

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Fix double shuffling of natures list (once in definition, once in choice list).

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Fix no marks given for correct choice of nature.

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Remove degenerate node and star node from the list of fixed points.  They were never a possibility.

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14/11/2013:

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Added Jsxgraph graphs of the solutions and attempt at a picture of the behaviour at the critical point x=0, y=0. Needs more work on finding the bounding box dimensions dynamically to find the best picture.

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05/03/2014:

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Changed variables so that only complex eigenvalues can occur, hence only centres or spirals.

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04/08/2014

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Changed showNavigation to false  in both functions as this does not work at present - need to upload new version of jsxgraph?

", "description": "

Nature of fixed points of a 2D dynamical system.

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These examples are either centres or spirals.

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a)

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The elements of the matrix $\\mathsf{A}$ are the coefficients of $x$ and $y$ in the system of equations, hence

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\$\\mathsf{A}=\\pmatrix{\\var{a} & \\var{b}\\\\\\var{c} & \\var{d}}.\$

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b)

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The trace $\\tau$ and determinant $\\delta$ of a matrix

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\$\\mathsf{A}=\\pmatrix{a & b\\\\c & d}\$

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are given by $\\tau=a+d$ and $\\delta=ad-bc$.

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A straight forward substitution for $a=\\var{a}$, $b=\\var{b}$, $c=\\var{c}$, and $d=\\var{d}$ reveals that $\\tau=\\var{tau}$ and $\\delta=\\var{delta}$.

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c)

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The eigenvalues $\\lambda$ of a matrix $\\mathsf{A}$ can be calculated from the expression

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\$\\mathsf{A}\\boldsymbol{x}=\\lambda\\boldsymbol{x},\$

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which is equivalent to

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\$\\left(\\mathsf{A}-\\lambda\\mathsf{I}\\right)\\boldsymbol{x}=\\boldsymbol{0},\$

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where $\\mathsf{I}$ is the identity matrix.

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This linear system has a solution only when

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\$\\det\\left(\\mathsf{A}-\\lambda\\mathsf{I}\\right)=0,\$

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i.e. when $(a-\\lambda)(d-\\lambda)-bc=0$.  This leads to $\\lambda^2-(a+d)\\lambda+(ad-bc)=0$, which is equivalent to

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\$\\lambda^2-\\tau\\lambda+\\delta=0.\$

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The eigenvalues $\\lambda_1$ and $\\lambda_2$ of the matrix $\\mathsf{A}$ are therefore given by

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\$\\lambda_{1,2}=\\frac{-\\tau\\pm\\sqrt{\\tau^2-4\\delta}}{2}.\$

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Following through this calculation for the specific values of $a$, $b$, $c$, and $d$ in this question leads to $\\lambda_1=\\var{lam1}$ and $\\lambda_2=\\var{lam2}$.

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d)

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The eigenvectors corresponding to the eigenvalues $\\lambda_1$ and $\\lambda_2$ can be found by solving the linear system of equations given by

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\$\\mathsf{A}\\boldsymbol{x}=\\lambda\\boldsymbol{x},\$

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or

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\\\begin{align}ax+by&=\\lambda x,\\\\cx+dy&=\\lambda y,\\end{align}\

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where $\\lambda$ is either $\\lambda_1$ or $\\lambda_2$.

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Because this is an eigenvalue problem, we can arbitrarily choose one of the eigenvector components to be $1$, the $x$-component in this case.  The $y$-component can then be found from

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\$y=\\frac{x(\\lambda-a)}{b}\\quad\\text{or}\\quad y=\\frac{cx}{\\lambda-d},\$

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with $x=1$.  Both expressions are equivalent, and will lead to the same value for $\\lambda$.

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Making the necessary substitutions reveals that the $y$-component of the eigenvector corresponding to $\\lambda_1$ is $\\var{vec1}$, and the $y$-component of the eigenvector corresponding to $\\lambda_2$ is $\\var{vec2}$.

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e)

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The nature of the fixed point at the origin can be determined by examining the eigenvalues.  If the eigenvalues are real, then the sign of each eigenvalue determines the nature of the fixed point.  If the eigenvalues are complex, then the sign of the real part of each eigenvalue determines the nature of the fixed point.

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In this case, {naturetext}.

\n", "name": "Dynamical system 4", "parts": [{"prompt": "\n

The system can be written in the form $\\dot{\\boldsymbol{x}}=\\mathsf{A}\\boldsymbol{x}$, where $\\boldsymbol{x}=\\pmatrix{x,y}^\\mathsf{T}$.

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Identify the components of the matrix $\\mathsf{A}$.

\n \n \n \n \n\n \n \n\n \n \n \n \n \n \n
 $\\mathsf{A}=\\Bigg($ [] [] $\\Bigg)$ [] []
\n \n", "type": "gapfill", "scripts": {}, "showCorrectAnswer": true, "gaps": [{"showPrecisionHint": false, "type": "numberentry", "maxValue": "a", "scripts": {}, "showCorrectAnswer": true, "marks": 1, "minValue": "a"}, {"showPrecisionHint": false, "type": "numberentry", "maxValue": "b", "scripts": {}, "showCorrectAnswer": true, "marks": 1, "minValue": "b"}, {"showPrecisionHint": false, "type": "numberentry", "maxValue": "c", "scripts": {}, "showCorrectAnswer": true, "marks": 1, "minValue": "c"}, {"showPrecisionHint": false, "type": "numberentry", "maxValue": "d", "scripts": {}, "showCorrectAnswer": true, "marks": 1, "minValue": "d"}], "marks": 0}, {"prompt": "\n

Calculate the trace $\\tau$, and the determinant $\\delta$ of the matrix $\\mathsf{A}$.

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Trace $\\tau=$ [].

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Determinant $\\delta=$ [].

\n \n", "type": "gapfill", "scripts": {}, "showCorrectAnswer": true, "gaps": [{"showPrecisionHint": false, "type": "numberentry", "maxValue": "tau", "scripts": {}, "showCorrectAnswer": true, "marks": 1, "minValue": "tau"}, {"showPrecisionHint": false, "type": "numberentry", "maxValue": "delta", "scripts": {}, "showCorrectAnswer": true, "marks": 1, "minValue": "delta"}], "marks": 0}, {"prompt": "\n

Calculate the eigenvalues $\\lambda_1$ and $\\lambda_2$ of the matrix $\\mathsf{A}$.

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If the eigenvalues have zero imaginary part, enter the eigenvalue with the largest real part first.  If the eigenvalues have non-zero imaginary part, enter the eigenvalue with the largest imaginary part first.

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$\\lambda_1=$ [].

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$\\lambda_2=$ [].

\n \n", "type": "gapfill", "scripts": {}, "showCorrectAnswer": true, "gaps": [{"showpreview": true, "expectedvariablenames": [], "scripts": {}, "checkingaccuracy": 0.001, "showCorrectAnswer": true, "checkvariablenames": false, "vsetrangepoints": 5, "type": "jme", "vsetrange": [0, 1], "marks": 1, "answer": "{lam1}", "checkingtype": "absdiff"}, {"showpreview": true, "expectedvariablenames": [], "scripts": {}, "checkingaccuracy": 0.001, "showCorrectAnswer": true, "checkvariablenames": false, "vsetrangepoints": 5, "type": "jme", "vsetrange": [0, 1], "marks": 1, "answer": "{lam2}", "checkingtype": "absdiff"}], "marks": 0}, {"prompt": "\n

Assuming that the $x$-component of each eigenvector is $1$, find the $y$-components of the eigenvectors corresponding to the eigenvalues $\\lambda_1$ and $\\lambda_2$.

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$y$-component of eigenvector corresponding to $\\lambda_1$: [].

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$y$-component of eigenvector corresponding to $\\lambda_2$: [].

\n \n", "type": "gapfill", "scripts": {}, "showCorrectAnswer": true, "gaps": [{"showpreview": true, "expectedvariablenames": [], "answersimplification": "all", "scripts": {}, "checkingaccuracy": 0.001, "showCorrectAnswer": true, "checkvariablenames": false, "vsetrangepoints": 5, "type": "jme", "vsetrange": [0, 1], "marks": 1, "answer": "{vec1}", "checkingtype": "absdiff"}, {"showpreview": true, "expectedvariablenames": [], "answersimplification": "all", "scripts": {}, "checkingaccuracy": 0.001, "showCorrectAnswer": true, "checkvariablenames": false, "vsetrangepoints": 5, "type": "jme", "vsetrange": [0, 1], "marks": 1, "answer": "{vec2}", "checkingtype": "absdiff"}], "marks": 0}, {"minAnswers": 0, "prompt": "

Determine the nature of the fixed point at the origin.

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{nature}

", "

{natures}

", "

{natures}

", "

{natures}

", "

{natures}

", "

{natures}

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Consider the two-dimensional dynamical system. Complex eigenvalues case , so centres or spirals.

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\\\begin{align}\\dot{x}&=\\simplify[std]{{a}*x+{b}*y},\\\\\\dot{y}&=\\simplify[std]{{c}*x+{d}*y}.\\end{align}\

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{ode1()}

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The above diagram shows the plot of $(x(t),y(t))$.

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At $t=0$ we have initally $x=-5,\\;\\;y=5$. Moving the point gives phase diagrams for the following initial values at $t=0$:

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$x=\\;$       $y=\\;$

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Dotted lines give points at which $\\dot{x}=0,\\;\\dot{y}=0$.

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The plots below are for $x(t),\\;x(0)=-1$ and $y(t),\\;y(0)=1$, in black and red respectively where the horizontal axis is $t$.

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{ode(xr,yr)}

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", "type": "question", "variable_groups": [], "question_groups": [{"pickingStrategy": "all-ordered", "name": "", "questions": [], "pickQuestions": 0}], "functions": {"ode": {"type": "html", "language": "javascript", "parameters": [["xr", "number"], ["yr", "number"]], "definition": "\nvar a = Numbas.jme.unwrapValue(scope.variables.a);\nvar b = Numbas.jme.unwrapValue(scope.variables.b);\nvar c = Numbas.jme.unwrapValue(scope.variables.c);\nvar d = Numbas.jme.unwrapValue(scope.variables.d);\n\nfunction f(x,yy) {\n var y1 = yy;\n var y2 = yy;\n var z1 = a*y1+b*y2;\n var z2 = c*y1+d*y2;\n return [z1,z2];\n}\n\nvar div = Numbas.extensions.jsxgraph.makeBoard('400px','400px',{axis:true,showNavigation:true, boundingbox:[-xr,yr,xr,-yr]});\nvar brd=div.board;\n\nfunction ode() {\n return JXG.Math.Numerics.rungeKutta('heun', [1,-1], [0, 10], 200, f);\n}\nfunction ode1() {\n return JXG.Math.Numerics.rungeKutta('heun', [1,-1], [0, -10], 200, f);\n} \nvar g1 = brd.create('curve', [,], {strokeColor:'red', strokeWidth:2, name:'y_1', withLabel:false});\nvar g2 = brd.create('curve', [,], {strokeColor:'black', strokeWidth:2, name:'y_2', withLabel:false});\nvar data=ode();\nvar data1=ode1();\nvar l=data.length;\nvar l1=data1.length;\nvar h = 10/200;\ng1.updateDataArray = function() {\nthis.dataX = [];\nthis.dataY = [];\nfor(var i=0; i