Consider the Malthus equation

\\[

x'(t)=r x(t), \\qquad x(t_0)=x_0. \\tag{M}

\\]

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?

"}, {"shuffleChoices": true, "marks": 0, "type": "1_n_2", "showCorrectAnswer": true, "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. "], "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."], "variableReplacements": [], "matrix": [0, 0, 0, "1"], "minMarks": 0, "extendBaseMarkingAlgorithm": true, "displayType": "radiogroup", "maxMarks": 0, "unitTests": [], "showFeedbackIcon": true, "scripts": {}, "displayColumns": "1", "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "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?

"}], "preamble": {"js": "", "css": ""}, "metadata": {"licence": "None specified", "description": ""}, "ungrouped_variables": ["population1", "population2", "year1", "year2", "rate", "year3", "fish", "fishes"], "rulesets": {}, "variablesTest": {"maxRuns": 100, "condition": ""}, "functions": {}, "tags": [], "variables": {"fishes": {"templateType": "anything", "description": "", "name": "fishes", "definition": "3 fish", "group": "Ungrouped variables"}, "year1": {"templateType": "anything", "description": "", "name": "year1", "definition": "2000+random(1..16)", "group": "Ungrouped variables"}, "rate": {"templateType": "anything", "description": "", "name": "rate", "definition": "random(-10..-2)", "group": "Ungrouped variables"}, "population2": {"templateType": "anything", "description": "", "name": "population2", "definition": "random(population1+5000..population1+50000)", "group": "Ungrouped variables"}, "year3": {"templateType": "anything", "description": "", "name": "year3", "definition": "random(2001..2017)", "group": "Ungrouped variables"}, "fish": {"templateType": "anything", "description": "", "name": "fish", "definition": "random(101..115)", "group": "Ungrouped variables"}, "year2": {"templateType": "anything", "description": "", "name": "year2", "definition": "random(year1+1..2017)", "group": "Ungrouped variables"}, "population1": {"templateType": "anything", "description": "", "name": "population1", "definition": "random(100000..120000)", "group": "Ungrouped variables"}}, "variable_groups": [], "advice": "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.