Consider a discrete-time dynamical system given by \\[P_{n+1}=-\\sqrt{P_{n}+2}\\] with $P_0=a\\geq-2$.
", "advice": "Let the sequence $(P_n)_{n\\geq0}$ be constant, i.e.
\\[
P_n=P_0=a
\\]
for all $n\\geq0$. Then we get the equation
\\[
a=-\\sqrt{a+2}.
\\]
Note that then $a\\geq-2$ (to ensure that $a+2\\geq0$), but also $a\\leq 0$ been equal to 'minus square root of...'. Then we get the equation \\[
a^2=a+2
\\]
that has two roots $a=2$ and $a=-1$, but the first one does not satisfy the condition $a\\leq0$.
Find $a$ such that the sequence $(P_n)_{n\\geq0}$ is constant.
", "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": "1", "showCellAnswerState": true, "choices": ["$a=-1$", "$a=2$", "$a=2$ and $a=-1$", "Such $a$ does not exist"], "matrix": ["1", 0, 0, 0], "distractors": ["Indeed, the equation $a=-\\sqrt{a+2}$ has only this solution (in real numbers).", "No. If $P_0=2=a$ then $P_1=-2$ and hence the sequence is not constant.", "No. If $P_0=2=a$ then $P_1=-2$ and hence the sequence is not constant.", "It does exist and this is $a=-1$."]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "Malthus equation", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Dmitri Finkelshtein", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2756/"}], "tags": [], "metadata": {"description": "", "licence": "None specified"}, "statement": "Consider the Malthus equation
\\[
x'(t)=r x(t), \\qquad x(t_0)=x_0. \\tag{M}
\\]
b) Note that the answer does not depend on $t_0$, since 'in three years' means that $t-t_0=3$ that appears at the formula.
\nc) The solution to the given equation has the form
\\[
u(t)=u(t_0)e^{\\var{rate}t},
\\]
where $t_0$ is the initial moment of time, i.e. $t_0=\\var{year1}$. Since $\\var{rate}<0$, $u(t)$ is decreasing, i.e. $u(t)<u(t_0)$ for all $t>t_0$. Since $\\var{population2}>\\var{population1}$, the population would not become larger than it was, and will not start to grow ever.
What should be instead of $A$ and $B$ in the expression for the function
\\[
x(t)=A e^{rt-rB}
\\]
to ensure that this function solves the Malthus equation (M)?
The number $x=x(t)$ of fishes in a lake at a moment of time $t$ satisfies the equation \\[ x'(t)=1.1 x(t).\\] Initially, at the moment of time $t_0=\\var{year3}$ there were $\\var{fish}$ fishes. How many fishes does one expect to have in $3$ years?
", "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": "1", "showCellAnswerState": true, "choices": ["$\\var{fish} e^{3.3}$", "$\\var{fishes} e^{1.1}$", "$1.1 e^{\\var{fishes}}$", "$3.3 e^{\\var{fish}}$"], "matrix": ["1", 0, 0, 0], "distractors": ["Correct, it's just $\\var{fish} e^{3\\times 1.1}$.", "No, it should be $\\var{fish} e^{3\\times 1.1}$, according to the formula $x(t)=x(t_0) e^{r(t-t_0)}$ with $t-t_0=3$, $r=1.1$, $x(t_0)=\\var{fish}$.", "No, it should be $\\var{fish} e^{3\\times 1.1}$, according to the formula $x(t)=x(t_0) e^{r(t-t_0)}$ with $t-t_0=3$, $r=1.1$, $x(t_0)=\\var{fish}$.", "No, it should be $\\var{fish} e^{3\\times 1.1}$, according to the formula $x(t)=x(t_0) e^{r(t-t_0)}$ with $t-t_0=3$, $r=1.1$, $x(t_0)=\\var{fish}$."]}, {"type": "1_n_2", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Let population $x=x(t)$ of a city evolve in time according to the equation \\[ x'(t)= \\var{rate} x(t) .\\] Suppose that in {year1} the population was {population1} people. When has the population reached the level of {population2} people?
", "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": "1", "showCellAnswerState": true, "choices": ["In {year2}", "It has not reached yet, but it will in few years", "Just recently, this year", "It has not reached yet and will not reach ever."], "matrix": [0, 0, 0, "1"], "distractors": ["Since the rate $\\var{rate}$ is negative, the population will always decrease, hence it could not become larger than it was. ", "Since the rate $\\var{rate}$ is negative, the population will always decrease, hence it could not become larger than it was. ", "Since the rate $\\var{rate}$ is negative, the population will always decrease, hence it could not become larger than it was.", "Correct. Since the rate $\\var{rate}$ is negative, the population will always decrease, hence it could not become larger than it was. "]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "Phase portrait", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Dmitri Finkelshtein", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2756/"}], "statement": "Consider a dynamical system of the form
\\[
x'=f(x). \\tag{DS}
\\]
Fixed points are solutions to the equation
\\[
f(x)=0
\\]
If the sign of $f$ by passing a fixed point changes from $+$ to $-$ it means that the point is an attractor. If th sign changes from $-$ to $+$, the point is a repeller.
Let in (DS)
\\[
f(x)=\\var{a}x^2-\\var{b}. \\tag{F}
\\]
How many fixed points this dynamical system does have?
Match choices with answers
", "showFeedbackIcon": true, "answers": ["solutions to the equation $f(x)=0$.", "attractor or sink.", "repeller or source."], "unitTests": [], "layout": {"type": "all", "expression": ""}, "customMarkingAlgorithm": "", "type": "m_n_x", "variableReplacements": [], "matrix": [["1", 0, 0], [0, "1", 0], [0, 0, "1"]], "displayType": "radiogroup", "shuffleAnswers": true}, {"showFeedbackIcon": true, "scripts": {}, "minMarks": 0, "shuffleChoices": true, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "maxMarks": 0, "displayColumns": "1", "choices": ["$-\\var{c}$", "$\\var{c}$", "This dynamical system does not have an attractor.", "$\\pm\\var{c}$"], "showCorrectAnswer": true, "unitTests": [], "type": "1_n_2", "extendBaseMarkingAlgorithm": true, "variableReplacements": [], "distractors": ["Yes, since $f(x)$ is positive in a left neighbourhood of $-\\var{c}$ and is negative in a right neighbourhood.", "No, since $f(x)$ is negative in a left neighbourhood of $\\var{c}$ and is positive in a right neighbourhood.", "Since $f(x)$ is positive in a left neighbourhood of $-\\var{c}$ and is negative in a right neighbourhood, $x=-\\var{c}$ is a stable fixed point.", "No, since $f(x)$ is negative in a left neighbourhood of $\\var{c}$ and is positive in a right neighbourhood."], "marks": 0, "matrix": ["1", 0, 0, 0], "displayType": "radiogroup", "prompt": "Consider the dynamical system (DS) with $f(x)$ given by (F). Find the attractor of this dynamical system.
"}, {"showFeedbackIcon": true, "scripts": {}, "minMarks": 0, "shuffleChoices": true, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "maxMarks": 0, "displayColumns": "1", "choices": ["is a sink.", "is a source.", "is not here a fixed point at all.", "is a fixed point, but it is neither sink nor source."], "showCorrectAnswer": true, "unitTests": [], "type": "1_n_2", "extendBaseMarkingAlgorithm": true, "variableReplacements": [], "distractors": ["No, the sign of $f(x)$ near $x=\\pi$ changes from $-$ to $+$.", "Yes, the sign of $f(x)$ near $x=\\pi$ changes from $-$ to $+$.", "No, $f(\\pi)=\\sin 2\\pi=0$.", "No, the sign of $f(x)$ near $x=\\pi$ changes from $-$ to $+$."], "marks": 0, "matrix": [0, "1", 0, 0], "displayType": "radiogroup", "prompt": "Consider the dynamical system (DS) with
\\[
f(x)=\\sin 2x.
\\]
Using the phase portrait of this system check that $x=\\pi$...
This is an unassessed homework. Note that, however, it compulsory to pass this homework to proceed further.
", "licence": "None specified"}, "percentPass": "100", "type": "exam", "contributors": [{"name": "Dmitri Finkelshtein", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2756/"}], "extensions": [], "custom_part_types": [], "resources": []}