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", "contributors": [{"name": "Dmitri Finkelshtein", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2756/"}]}]}], "contributors": [{"name": "Dmitri Finkelshtein", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2756/"}]}