// Numbas version: finer_feedback_settings {"name": "Population growth with plot", "extensions": ["jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Population growth with plot", "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "", "advice": "
We have \\(P_0=\\var{P_0}\\) and \\(P=\\var{pop}\\) at \\(t = \\var{t}\\) years, hence:
\n\\begin{align}\\var{pop}&=\\var{P_0}\\left(1-e^{-\\var{t}k}\\right)\\\\\\frac{\\var{pop}}{\\var{P_0}}&=\\left(1-e^{-\\var{t}k}\\right)\\\\e^{-\\var{t}k}&=1-\\frac{\\var{pop}}{\\var{P_0}}\\quad\\text{taking natural logs of both sides we have}\\\\-\\var{t}k&=\\ln\\left( 1-\\frac{\\var{pop}}{\\var{P_0}}\\right)\\\\k&=-\\frac{\\ln\\left({1-\\frac{\\var{pop}}{\\var{P_0}}}\\right)}{\\var{t}}\\\\k&\\approx\\var{k} \\end{align}
\nHence the formula for population as a function of time is \\(P(t)=\\var{P_0}\\left(1-e^{-\\var{k}t}\\right)\\)
\nThe graph of the population vs time function will be:
\n{sol_graph}
", "rulesets": {}, "extensions": ["jsxgraph"], "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"pop": {"name": "pop", "group": "Ungrouped variables", "definition": "random(0.50*P_0..0.95*P_0 #2000)", "description": "Population density at time t years
", "templateType": "anything", "can_override": false}, "k": {"name": "k", "group": "Ungrouped variables", "definition": "precround(-ln(1-pop/P_0)/t,3)", "description": "", "templateType": "anything", "can_override": false}, "P_0": {"name": "P_0", "group": "Ungrouped variables", "definition": "random(20000..35000 #5000)", "description": "maximum population density
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", "templateType": "anything", "can_override": false}, "sol_graph": {"name": "sol_graph", "group": "Ungrouped variables", "definition": "jessiecode(800,500,[-t/10,1.1*P_0,1.1*t,-0.1*P_0],\"\"\"\n line([0,{P_0}],[1,{P_0}])<The population \\(P\\) of a new suburb can be shown to grow according to an equation given by \\[P=P_0\\left(1-e^{-kt}\\right) \\] where \\(P_0\\) is the maximum population for the suburb and \\(t\\) is measured in years. If the maximum population for a suburb is \\(\\var{P_0}\\) and at time \\(\\var{t}\\) years the population is \\(\\var{pop}\\), find the value of \\(k\\) accurate to 3 decimal places. Hence give the formula for the population density as a function of time. Show full working in your handwritten working.
\n\\(k=\\;\\) [[0]] hence
\n\\(P(t)=\\;\\)[[1]] (note: to enter \\(e^{-kt}\\) type e^(-kt).)
Graph of population growth will be shown below.
\n{graph(P_0,t)}
\nUsing the above graph or otherwise, estimate the time taken for the population to reach \\(\\var{P_1}\\). (Note: your approximation should be given to the precision readable from the graph)
\n\\(t\\approx\\)[[2]] years
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