// Numbas version: exam_results_page_options {"name": "Further Integration", "metadata": {"description": "", "licence": "None specified"}, "duration": 0, "percentPass": "0", "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", ""], "variable_overrides": [[], []], "questions": [{"name": "Integration: Solving Separable Differential Equations - Radioactive Decay", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Evi Papadaki", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18113/"}], "tags": [], "metadata": {"description": "

Solving a separable differential equation that describes the rate of decay of radioactive isotopes over time with a known initial condition to calculate the mass of the isotope after a given time and the time taken for the mass to reach $m$ grams. 

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Decay Constant - Radioactivity - Nuclear Power (nuclear-power.com)

", "licence": "None specified"}, "statement": "

The rate of decay of the radioactive isotope $\\var{isotopes_names[j]}$ is given by:

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\\[\\frac{\\mathrm{d}M}{\\mathrm{d}t}=-\\var{lambda}M\\]

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where $M$ is the mass of the isotope in grams and $t$ is time measured in $\\var{timescale[j]}$.
A sample initially contains $\\var{m0}$ grams of the isotope.

", "advice": "

a) To calculate the mass of the isotope at a specific time, we need an expression for how the mass, $M$, changes in terms of time, $t$. This is given by the differential equation,

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\\[\\frac{\\mathrm{d}M}{\\mathrm{d}t}=-\\var{lambda}M\\]

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which we can solve using separation of variables.

\n

First we need to separate the variables between the two sides:

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\\[\\frac{1}{M}dM=-\\var{lambda}dt\\]

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Now we can integrate both sides: 

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\\[ \\begin{split} \\int\\frac{1}{M}dM&=\\int -\\var{lambda}dt \\\\ \\ln{M}&= -\\var{lambda}t + c \\end{split} \\]

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Taking the exponential of both sides:

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\\[ \\begin{split} M&= e^{ -\\var{lambda}t + c} \\\\ M&=e^c\\times e^{-\\var{lambda}t }\\end{split} \\]

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Now, $e^c$ is a constant which we can calculate using the initial contition that $m=\\var{m0}$ when $t=0$. By substituting in the values in the equation we get:

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\\[ \\begin{split}  \\var{m0}&=e^c\\times e^{-\\var{lambda}\\times 0 } \\\\ \\var{m0}&=e^c\\times e^0 \\\\ \\var{m0}&=e^c\\times 1 \\\\ \\var{m0}&=e^c \\end{split} \\]

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So, $c=\\var{m0}$. Therefore,

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\\[ \\begin{split} M&=\\var{m0} e^{-\\var{lambda}t }\\end{split} \\]

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We can now substitute $t=\\var{t1}$ to calculate the mass at that time: 

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\\[ \\begin{split} M&=\\var{m0} e^{-\\var{lambda} \\times \\var{t1} } \\\\ M&= \\var{m0}\\times e^{\\simplify{{-decayconstant[j]*t1}}}\\\\ M&= \\var{m0} \\times \\simplify{{e^(-decayconstant[j]*t1)}} = \\var{ansa} ~~\\text{[3 s.f.]} \\end{split} \\]

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b) To calculate how long it takes for the mass of the isotope in the sample to fall to $\\var{m1}$ gram, we will use the expression 

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\\[ \\begin{split} M&=\\var{m0} e^{-\\var{lambda}t }\\end{split} \\]

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found in part (a). We can substitute $M=\\var{m1}$ and solve the equation for t.

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\\[ \\begin{split} \\var{m1}&=\\var{m0} e^{-\\var{lambda}t } \\\\ \\frac{\\var{m1}}{\\var{m0}}&=e^{-\\var{lambda}t }\\end{split} \\]

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Taking $\\ln()$ of both sides:

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\\[ \\begin{split} \\ln \\left(\\frac{\\var{m1}}{\\var{m0}}\\right)&=-\\var{lambda}t \\\\ -\\frac{\\ln \\left(\\frac{\\var{m1}}{\\var{m0}}\\right)}{\\var{lambda}}&=t \\end{split} \\]

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Thus, $m=\\var{m1}$ when $t=\\var{ansb}$ {timescale[j]}.

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The mass is [[0]]grams.

\n

Round your answer to 3 significant figures, if needed.

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How long does it take for the mass of the isotope in the sample to fall to $\\var{m1}$ gram?

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It takes $t=$[[0]]$\\var{timescale[j]}$.

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Round your answer to the nearest integer if needed.

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Solving a separable differential equation that describes the population growth over time with a known initial condition to calculate the population after $n$ years. 

", "licence": "None specified"}, "statement": "

A population of $N$ individuals grows over time, $t$ years, according to the equation:
$\\frac{\\mathrm{d}N}{\\mathrm{d}t}=\\left(\\var{a}+ e^{\\simplify{-{b}t}}\\right)N $

\n

At time $t=0$, the population contains $\\var{N}$ individuals.

", "advice": "

To calculate the population $\\var{N}$ years later, we need an expression for how the population, $N$, changes in terms of time, $t$. This is given by the differential equation,

\n

\\[ \\frac{dN}{dt}=\\left(\\var{a}+ e^{\\simplify{-{b}t}}\\right)N \\]

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which we can solve using separation of variables.

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First, we need to separate the variables between the two sides:

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\\[ \\frac{1}{N}dN=\\left( \\var{a}+e^{\\simplify{-{b}t}}\\right)dt \\]

\n

Now, we integrate both sides:

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\\[ \\begin{split} \\int\\frac{1}{N}dN &=\\int\\left( \\var{a}+e^{\\simplify{-{b}t}}\\right)dt \\\\ \\ln{N} &= \\simplify{{a}*t}+\\left(-\\simplify{1/{b}*e}^{\\simplify{-{b}t}}\\right)+c \\\\ \\ln{N} &= \\simplify{{a}*t}-\\simplify{1/{b}*e}^{\\simplify{-{b}t}}+c \\end{split} \\]

\n

To find the value of $c$, we can use the fact that when $t=0$, $N=\\var{N}$. By substituting in the values in the equation we get:

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\\[ \\begin{split} \\ln(\\var{N}) &= \\var{a} \\times 0-\\simplify{1/{b}*e}^{-\\simplify{{b}* 0}}+c \\\\ \\ln(\\var{N}) &= -\\simplify{1/{b}*e}^{0}+c \\\\ \\ln(\\var{N}) &= -\\simplify{1/{b}* 1}+c \\\\ \\ln(\\var{N}) &= -\\simplify{1/{b}}+c \\\\ c&=\\simplify{ ln{{N}}+1/{b}}\\end{split} \\]

\n

Therefore $c=\\simplify{ln{{N}}+1/{b}}=\\var{c}$. So, 

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\\[ \\begin{split} \\ln({N}) &= \\simplify{{a}*t}-\\simplify{1/{b}*e}^{-\\simplify{{b}*t}}+\\simplify{ln{{N}}+1/{b}} \\end{split} \\]

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Taking the exponential of both sides:

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\\[ \\begin{split} N &= e^{\\simplify{{a}*t}-\\simplify{1/{b}e}^{-\\simplify{{b}*t}}+\\simplify{ln{{N}}+1/{b}}} \\end{split} \\]

\n

We can now substitute $t=\\var{t0}$ to calculate the population:

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\\[ \\begin{split} N &= e^{\\var{a} \\times \\var{t0}-\\simplify{1/{b}*e}^{-\\var{b} \\times \\var{t0}}+\\simplify{ln{{N}}+1/{b}}} \\\\ N&=\\var{step3} \\end{split} \\]

\n

Thus, the population $\\var{t0}$ years later will be $\\var{ans}$ individuals.

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How many individuals are in the population $\\var{t0}$ years later?

\n

The population is [[0]] individuals.

\n

If needed, round your answer to the nearest integer. 

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