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Using given information to complete the equation $c= A \\cos{ \\left( \\frac{2 \\pi}{P} \\left( t-H \\right) \\right) }+V $ that describes the concentration, $c$, of perscribed drug in a patient's drug over time, $t$. Calculating the maximum concentration and the concentration at a specific time.
", "licence": "None specified"}, "statement": "A patient takes a drug every $\\var{P}$ hours each day. The concentration, $c$, of the drug in the patient’s blood $t$ hours after the start of the treatment is modelled by the equation:
$c=\\var{A}\\cos{\\left(k\\left(t-\\var{H}\\right)\\right)}+\\var{V}$
a) The equation
\n\\[ \\begin{split} c=\\var{A}\\cos{\\left(k\\left(t-\\var{H}\\right)\\right)}+\\var{V} \\end{split} \\]
\nis a wave function.
\nThe general form of a wave function can be written as:
\n\\[ \\begin{split} c=A\\cos{\\left(\\frac{2 \\pi}{P}\\left(t-H\\right)\\right)}+V \\end{split} \\]
\nWhere $V$ is the average value, $A$ is the amplitute, $H$ the phase and $P$ the period.
\nBy comparying the equation with the general form we can notice that
\n\\[ k= \\frac{2 \\pi}{P} \\]
\nWe know that the patient takes the drug every $\\var{P}$ hours each day. Therefore, the period is $P=\\var{P}$. Therefore,
\n\\[ \\begin{split} k&= \\frac{2 \\pi}{\\var{P}} \\\\ &=\\simplify{ {2*pi} / {P}} \\end{split} \\]
\nSo, $k=\\simplify{ {2*pi} / {P}}$ and we can now rewrite the equation as:
\n\\[ \\begin{split} c=\\var{A}\\cos{\\left(\\simplify{ {2*pi} / {P}}\\left(t-\\var{H}\\right)\\right)}+\\var{V} \\end{split} \\]
\n\nb) We know that the wave function is a transformation of the trigonometric function cos(x).
\nClink on the link for a visual representation of the wave function.
\nhttps://www.desmos.com/calculator/ssqdx7ys7k
\nWe also know that the cosine function has a maximum value of 1. So, the maximum value of the wave function will occure when $ \\cos{\\left(\\simplify{ {2*pi} / {P}}\\left(t-\\var{H}\\right)\\right)}=1$. Therefore,
\n\\[ \\begin{split} c_{max}&=\\var{A}\\times 1 +\\var{V} \\\\ &= \\var{A} +\\var{V} \\\\ &=\\var{A+V} \\end{split} \\]
\nSo, $c_{max}=\\var{max}$ mg/L.
\n\nc) To calculate the concentration of the drug $\\var{t1}$ hours after taking it for the first time, we need to subtitute $t=\\var{t1}$ in the equation
\n\\[ \\begin{split} c=\\var{A}\\cos{\\left(\\simplify{ {2*pi} / {P}}\\left(t-\\var{H}\\right)\\right)}+\\var{V} \\end{split} \\]
\nTherefore, when $t=\\var{t1}$ the equation become
\n\\[ \\begin{split} c_{\\var{t1}}&=\\var{A}\\cos{\\left(\\simplify{ {2*pi} / {P}}\\left(\\var{t1}-\\var{H}\\right)\\right)}+\\var{V} \\\\ &=\\var{A}\\cos{\\left(\\simplify{ {2*pi} / {P}} \\times \\var{t1-H}\\right)}+\\var{V} \\\\ &=\\var{A}\\cos{\\left(\\simplify{ {2*pi*(t1-H)}/{P}}\\right)}+\\var{V} \\\\ &=\\var{ansC} \\end{split} \\]
\nSo, $ c_{\\var{t1}}=\\var{ansc} $ mg/L.
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\n$k=$[[0]]
\nGive your answer as a fraction in terms of $\\pi$ (you can type: pi).
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\n$c_{max}=$[[0]] mg/L
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\n$c_{\\var{t1}}=$[[0]]
\nGive your answer rounded to 2 decimal places, if needed.
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