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If \(k = 0.05 \text{ hr}^{-1}\) and the highest safe concentration is \(e\) times the lowest effective concentration, find the length of the time between repeated doses that will ensure safe but effective concentrations.
let:
| \(t\) | be the time (variable) since the last dose was administered |
| \(T\) | be the constant time between doses |
| \(c_n\left( t \right)\) | be the blood concentration, after a period of \(t\), following the \(n\)th dose |
| \(c_0\) | be the initial dose administered, which will also be the constant dosage and initial blood concentration at \(t= 0\) |
| \(H= r\cdot L,\) | be the maximum safe dosage, \(\quad \exists ! r \in \mathbb{R^+}\) |
| \(L\) | be the minimum effective dosage |
| \(C_n\) | be the drug level immediately following administration |
| \(R_n\) | be the drug level remainining immediately preceeding administration. |
Presume that the rate of drug metabolism is proportional to the drug levels \(c\left( t \right)\):
\[\begin{aligned} \frac{\operatorname{d}c }{\operatorname{d} t} &\propto c\left( t \right) \\ \ln{ \left| c\left( t \right) \right| } &= -kt+ \lambda, \quad \exists \lambda \in \mathbb{R}\end{aligned}\]
blood levels will be positive and so the absolute value may be dispenced with:
\[\begin{aligned} \ln{ \left( c\left( t \right) \right) } &= -kt + \kappa \notag\\ \implies c\left( t \right) &= \lambda^*\cdot e^{-kt}, \quad \exists \lambda^* \in \mathbb{R}\end{aligned}\]
Applying the initial condition that \(c\left( 0 \right)= c_0\):
\[\begin{aligned} c\left( 0 \right)= c_0 &= \lambda^*\cdot e^{-k0} \notag \\ \implies \lambda^* &= c_0 \end{aligned}\]
Hence the blood concentration levels, as a function of time will be:
\[\begin{aligned} c\left( t \right)&= c_0\cdot e^{-kt} \end{aligned}\]
Presume when a dose is applied that the level instantaneously reaches the higher level as shown in the diagram at .
The blood concentration will not necessarily reach the overdose threshold \(H\) or the minimum effective threshold \(L\) following the first dose, as shown in the figure at , there will however be a maximum \(\left( t_{\textit{max}}, c_\textit{max} \right)\) and a minimum concentration \(\left( t_{\textit{min}}, c_{\textit{min}} \right)\):
\[\begin{aligned} c&= c_0e^{-kt}\notag \\ e^{-kt} &= \frac{c}{c_0}\notag \\ -kt &= \ln{ \left( \frac{c}{c_0} \right)}\notag \\ t &= \frac{1}{-k} \cdot \ln{ \left( \frac{c}{c_0} \right)}\notag \\ \end{aligned}\]
The time between dosages, if the dosages are constant, must be the difference between the maximum concentration and the minimum concentration, if the minimum concentration is limited by the effective minimum dosage \(L\) and the maximum effective dosage is limited by the safe threshod \(H= r\cdot L\) then we have:
\[\begin{aligned} T &= t\left( c_{\text{max}} \right) - t\left( c_{\text{min}} \right)\notag \\ &= \frac{1}{-k}\left[ \ln{ \left( \frac{c_{\text{max}}}{c_0} \right) - \ln{ \left( \frac{c_{\text{min}}}{c_0} \right) } } \right]\notag \\ &= \frac{1}{-k}\left[ \ln{ \left( \frac{L\cdot r}{c_0} \right) - \ln{ \left( \frac{L}{c_0} \right) } } \right]\notag \\ &= \frac{1}{-k}\cdot \ln{ \left( r \right) } \label{eq:T Defininition}\end{aligned}\]
Thus the time between repeated doses must be less than \(T\), where \(T\) is defined as above at .
It is not possible to determine the size of each dose, we would need to know the upper/lower limits or, in this case, \(r\).
Suppose \(k = 0.05 \text{ hr}^{- 1}\) and \(T = 10 \text{ hr}\); what is the smallest \(n\) such that \(R_n>0.5\cdot R\)
\[\begin{aligned} \frac{\operatorname{d}C }{\operatorname{d} t} \propto C\notag \\ \implies \frac{1}{C}\cdot \frac{\operatorname{d}C }{\operatorname{d} t}&= -k, \quad \exists k\in \mathbb{R}^{- } \notag \\ \implies \ln{ \left| C \right| }&= -k\cdot t+ \lambda, \quad \exists \lambda \in \mathbb{R}\notag \\ \implies C\left( t \right)&= \lambda^*\cdot e^{-k\cdot t}, \quad \exists \lambda^* \in \mathbb{R}\end{aligned}\]
Now by using the initial condition that \(C\left( 0 \right)= C_0\): \[\begin{aligned} C\left( 0 \right)&= \lambda^*\cdot e^{0} \notag \\ \implies \lambda^* &= C_0 \notag \\ \implies C\left( t \right)&= C_0\cdot e^{-k\cdot t} \end{aligned}\]
The function \(C_n\left( t \right)\) that describes drug levels, given the simplifying assumption that drug levels immediately rise following administration of the drug, is described by a sequence of seperate functions, \(\left( C_1\left( t \right), C_2\left( t \right), C_3\left( t \right) \dots C_n\left( t \right) \right)\) corresponding to the domain \(\left( \left( n- 1 \right)\cdot T , T \right)\) respectively.
Following the initial dose of \(C_0\), a subsequent dose will need to be administered after a period of time \(T\), which correspoonds to the constant dosing schedule, at this time the blood levels will be:
\[\begin{aligned} R_1&= C\left( T \right) \\ &= C_0\cdot e^{-k\cdot t}\end{aligned}\]
At this time, the simplifying assumption is made that the levels rise immediately to reach \(C_2\), which is given by the initial value pluse the dose \(C_0\) (which is also assumed constant):
\[\begin{aligned} C_1&= C_0+ R_1 \notag \\ &= C_0+ C_0\cdot e^{-k\cdot t}\end{aligned}\]
Following this the levles will again decrease up until the time of the next dose, after a period of \(t= T\), but this time they will fall from an initial value of \(C_1\):
\[\begin{aligned} R_2&= C_1\cdot e^{- kT}\notag \\ &= \left( C_0 + C_0\cdot e^{- kT} \right)\cdot e^{- kT}\notag \\ &= C_0e^{- kT} + C_0\cdot e^{- 2kT}\end{aligned}\]
following the preceeding logic:
\[\begin{aligned} C_2 &= R_2 + C_0 \notag \\ &= C_0 + C_0 \cdot e^{- kT} + C_0 \cdot e^{- 2kT}\end{aligned}\]
now by the geometric series we have \(\sum^{n- 1}_{i= 0} \left[ r^n \right]= \frac{1- r^n}{1- r}\) so:
\[\begin{aligned} R_3 &= C_2 \cdot e^{- kT}\notag \\ &= \left( C_0 po C_0 \cdot e^{- kT} + C_0 \cdot e^{- 2kT} \right) \cdot e^{- kT}\notag \\ &= c)e^{- kT} + C_0 e^{- 2kT} + C_0 e^{- 3kT}\notag \\ \dots \notag \\ R_n &= C_0\cdot \sum^{n}_{i= 1} \left[ \left( e^{- kT} \right)^i \right] \notag\\ &= \frac{C_0\cdot e^{- kT} \cdot \left( 1- e^{- kTn} \right)}{\left( 1- e^{- kT} \right)}\end{aligned}\]
The long term behaviour of the concentration levels will be:
\[\begin{aligned} R= R_{\infty} &= \lim_{n \rightarrow \infty}\left[ R_n \right]\notag \\ &= \lim_{n \rightarrow \infty}\left[ \frac{C_0 - e^{- kTn}}{\left( e^{- kT} - 1\right)} \right]\notag \\ &= \frac{C_0 - \lim_{n \rightarrow \infty}\left[ e^{- kTn} \right]}{\left( e^{- kT} - 1\right)} \notag\\ &= \frac{C_0 - 0}{\left( e^{- kT} - 1\right)} \notag \\ &= \frac{C_0 }{\left( e^{- kT} - 1\right)} \end{aligned}\]
The concentration level is hence given by:
\[\begin{aligned} C_n &= C_0+R_n \notag \\ &= c_0 + 1 + \frac{1 - e^{- kTn}}{\left( e^{- kT} - 1 \right)} \notag\\ c_\infty &= \lim_{n \rightarrow \infty}\left[ C_n\left( t \right) \right] \notag\\ &= C_0 + R_\infty \notag\\ &= c_0\left( 1 + \frac{1}{\left( e^{- kT} - 1 \right)} \right) \notag\\\end{aligned}\]
From above we have that the dose schedule is:
\[\begin{aligned} T&= \ln{ \left( r \right) }\cdot \frac{1}{k} \notag\end{aligned}\]
hence by substitution:
\[\begin{aligned} 10 = 100\cdot \ln{ \left( r \right) } \\ r= e^{\frac{1}{10}}\end{aligned}\]
Hence we may conclude: \[\begin{aligned} H= e^{\frac{1}{10}}\end{aligned}\]
Now in order to solve \(n\) :
\[\begin{aligned} r_n &> 0.5\cdot R \\ \frac{C_0 \cdot e^{- kT}\left( 1- e^{- kTn} \right)}{1- e^{- kt}} &> \frac{C_0}{2\left( e^{kt} - 1 \right)} \notag \\ \end{aligned}\]
multiply the RHS by \(\frac{e^{- kt}}{e^{- kt}}\)
\[\begin{aligned} \frac{C_0 \cdot e^{- kT}\left( 1- e^{- kTn} \right)}{1- e^{- kt}} &> \frac{C_0\cdot e^{- kt}}{2\left(1- e^{-kt} \right)} \notag \\ \end{aligned}\]
because \(\left( 1- e^{- kt} \right)>1\):
\[\begin{aligned} C_0\cdot e^{- kT}\left( 1- e^{- kTn} \right) &> C_0\cdot e^{- kT} \cdot \frac{1}{2}\notag \\ 1- e^{- kTn}&>\frac{1}{2} \notag \\ \frac{1}{2}&>e^{- kTn} \notag \\ -\ln{ \left( 2 \right) } &> - kTn\notag \\ \ln{ \left( 2 \right) }&<kTn\notag \\ \frac{\ln{ \left( 2 \right) }}{kT} &< n\notag \\ n&>\frac{\ln{ \left( 2 \right) }}{kT}\notag \\ n&>10\times \ln{ \left( 2 \right) } \notag \\ n&>6.9\end{aligned}\]
\(\therefore\) \(n = 7\) is the minimum value of \(n\) that satisfies that condition.
Suppose that \(k= 0.2 \text{ hr}^{- 1}\) and that the smallest concetration is 0.03 mg/ml. A signle dose that produces a concentration of 0.1 mg/ml is administered. Approximately how many hours will the drug remain effective?
From the previous working we have:
\[\begin{aligned} \frac{\operatorname{d}C }{\operatorname{d} t}= C\left( t \right) \implies C\left( t \right)= C_0\cdot e^{k\cdot t}\end{aligned}\]
and the question provides:
\[\begin{aligned} C_0&= 0.1\\ L&= 0.03\end{aligned}\]
Now substitute the values and find \(t\) :
\[\begin{aligned} L&= C\left( t \right)\\ &= C_0\cdot e^{k\cdot t}\notag \\ kt&= \ln{ \left( \frac{C_0}{L} \right) }\\ t&= \frac{1}{k}\cdot \ln{ \left( \frac{C_0}{L} \right) }\notag \\ &= 5\cdot \ln{ \left( \frac{10}{3} \right) }\notag \\ &= 6 \text{ hours, } 1 \text{ min }\end{aligned}\]
\(\therefore\) the dosage will be effective for only six hours
A patient is given a dosage of \(Q\) of a drug at regular intervals of time \(T\). The concentration of the drug in the blood behaves differently in this scenario, and it has been found that the concentration level \(C\) is given by:
\[\begin{aligned} \frac{\operatorname{d}C }{\operatorname{d} t}&= - ke^C, \qquad \exists k \in \mathbb{R^+}\end{aligned}\]
The first dose is administered at \(t= 0\), after \(T\) hours the residual in the blood is:
\[\begin{aligned} R_1= - \ln{ \left( kT+ e^{- Q} \right) }\end{aligned}\]
In order to show this consider that The residual corresponds to \(t= T\) and solve \(C\left( t \right)\) :
\[\begin{aligned} e^{- C}\cdot \frac{\operatorname{d}C }{\operatorname{d} t}&= - k \notag \\ \int^{}_{} e^{- C} dC \operatorname{d}C &= - kt+ A_1, \qquad \exists A_i \in \mathbb{R}, \forall i in \mathbb{Z^+} \notag \\ - e^{- C}&= - kt + A_1\notag \\ e^{- C}&= kt+ A_2 \notag \\ - C&= \ln{ \left( kt+ A_2 \right) } \notag \\ C&= - \ln{ \left( kt+ A_2 \right) } \label{standardConstForm}\end{aligned}\]
Consider the initial condition:
\[\begin{aligned} C\left( 0 \right)= Q &= - \ln{ \left( kt+ A_2 \right) } \\ - Q &= \ln{ \left( A_2 \right) } \\ A_2 &= e^{- Q}\end{aligned}\]
hence, given this model, we may conclude: \[\begin{aligned} C\left( t \right)&= - \ln{ \left( kt+ C_2 \right) } \\ &= - \ln{ \left( kt+ e^{- Q} \right) } \end{aligned}\]
Now that we have solved the exponential model corresponding to the time following the initial dose and before the subsequent dose, i.e. \(t\in \left[ 0, T \right]\), in order to solve the first residual:
\[\begin{aligned} R_1&= C\left( T \right)\\ &= - \ln{ \left( kT+ e^{- Q} \right) }\end{aligned}\]
Assume an instantanous rise in concentration whenever the drug is administrated, show that after the second dose and another \(T\) hours have elapsed that the residual concetration in the blood will be given by:
\[\begin{aligned} R_2 = - \ln{ \left[ kT\left( 1+ e^{- Q} \right)+ e^{- 2Q} \right] }\end{aligned}\]
The first thing to be mindful of here is that, owing to the subsequent readministration of the drug and the simplifying assumption that the readministration will lead to an instantaneous rise in blood concentration, the blood concentration will be described by a sequence of function \(\left( C_i\left( t \right) \right)\) corresponding to the domain \(\left( t \in \left[ \left( i- 1 \right)\cdot T, i\cdot T \right] \right)\) wherein \(\frac{\operatorname{d} }{\operatorname{d} t}\left( C_i \right)= - k\cdot e^{C}\) and the constant dose is the initial dose \(C_0\).
As described above, \(\left( C_i\left( t \right) \right)\) is a sequence of functoins describing the blood level over the applicable domain, while \(\left( R_i\left( t \right) \right)\) is a sequence of values describing the blood level at remaining in the blood immediately preceeding the upcoming dose.
Following a subsequent dose, the blood levels will rise to \(C_1\) :
\[\begin{aligned} C_1&= R_1+ Q \\ &= Q - \ln{ \left( kT+ e^{- Q} \right) }\end{aligned}\]
\(R_2= C_1\left( T \right)\) will be such that \(\frac{\operatorname{d} }{\operatorname{d} t}\left( C_i\left( t \right) \right) \propto C_i, \enspace \forall i \in \mathbb{Z^+}\) with the initial condition now being that \(C_1\left( 0 \right)= R_1+ Q\), this change in initial condition being how the seqence of functions evolves over iteration.
\[\begin{aligned} C_1\left( 0 \right)= C_1&= R_1+ Q\\ &= Q- \ln{ \left( kT+ e^{- Q} \right) }\\\end{aligned}\]
Now we have from earlier in that:
\[\begin{aligned} {2} & \frac{\operatorname{d} }{\operatorname{d} t}\left( C_i\left( t \right) \right)& \implies C_i(t)= -\ln{ \left( kt + A_2 \right) } \quad \\ % \implies C&= - \ln{ \left( kt&+ A_2 \right) } \quad \\ && \qquad \exists A_i \in \mathbb{R}, \forall \enspace i \in \mathbb{Z^+} \notag\end{aligned}\]
so by applying this new initial condition:
\[\begin{aligned} C_1\left( 0 \right)= C_1&= R_1+Q \notag\\ & \implies - \ln{ \left( k \times 0 + A_2 \right) }= R_1+ Q \notag \\ & \implies A_2 = -e^{R_1}\cdot e^{Q} \label{eq:firstconcconst}\end{aligned}\]
Now by substituting the value from above into the solution from we have:
\[\begin{aligned} C_1\left( t \right)= - \ln{ \left( kt+ e^{R_1}\cdot e^{Q} \right) }\end{aligned}\]
The residual \(R_2\) will correspond to \(C_1\left( T \right)\):
I may have made a mistake with negatives here
\[\begin{aligned} R_2= C_1\left( T \right)&= - \ln{ \left( kTe^{R_1}\cdot e^Q \right) }\notag \\ &= - \ln{ \left( kt + e^{- Q}\left( kT+ e^{- Q} \right) \right) }\notag \\ &= - \ln{ \left( kt\left( 1+ e^{- Q} \right)+ e^{- 2Q} \right) }\end{aligned}\]
Given that the residual in the bloodstreamafter \(nT\) hours is given by
\[R_{n}=-\ln \left[k T\left(1+e^{-Q}+e^{-2 Q}+\cdots+e^{-(n-1) Q}\right)+e^{-n Q}\right]\]
Show that the limiting value R of the residual concentration for doses of \(Q\) mg/ml repeated at intervals \(T\) hours is given by the formula
\[R=-\ln \left(\frac{k T}{1-e^{-Q}}\right) \mathrm{mg} / \mathrm{ml}\]
by continuing this pattern we would find that:
\[\begin{aligned} R_n &= - \ln{ \left( kT\cdot \sum^{n- 1}_{i= 0} \left[ \left( e^{- Q} \right)^n \right]+ e^{- nQ} \right) } \notag \\ &= - \ln{ \left( kT\cdot \frac{1- e^{- Qn}}{1- e^{- Q}} + e^{- nQ} \right) }\notag \\ \lim_{n \rightarrow \infty}\left[ R_n \right]&= \lim_{n \rightarrow \infty}\left[ - \ln{ \left( kT \cdot \frac{1- e^{- Qn}}{1- e^{- Q}}+ e^{- nQ} \right) } \right]\notag \\ \end{aligned}\]
by the definition of continuity
\[\begin{aligned} \lim_{n \rightarrow \infty}\left[ R_n \right]&= - \ln{ \left( kT \cdot \lim_{n \rightarrow \infty}\left[ \frac{1- e^{- Qn}}{1- e^{- Q}} + e^{- nQ} \right] \right) }\notag \\ &= - \ln{ \left( kT\cdot \frac{1- 0}{1- e^{- Q}} + 0 \right) }\notag \\ &= - \ln{ \left( kT\cdot \frac{1}{1- e^{- Q}} \right) } \end{aligned}\]
Assuming the drugis ineffective below a concentration of L mg/ml and can be harmful above concentrations of H mg/ml. Show that the dose schedule T for a safe and effective concentration of the drug in the bloodsatisfies the formula
\[\begin{aligned} T=\frac{1}{k}\left(e^{-L}-e^{-H}\right)\end{aligned}\]
If the drug is effective only above L and inneffective below ht we would want
\[\begin{aligned} T&= C_n- R_n \quad \exists n \in \mathbb{Z^+}: \qquad C_n \leq H \enspace \wedge \enspace R_n \geq L \end{aligned}\] Setting \(C_0\) as \(H\) would not be appropriate, instead: \[C_0 = H - L\] That if the dose is taken when it is not effective at L way the blood levels will i’mediately reach the threshold avoiding an overdose and maximising the dosage schedule
\[\begin{aligned} C_0= H- L &\implies C_n: \notag \\ \ \notag \\ c_n &= - \ln{ \left( kt + e^{L- H} \right) }\notag \\ \end{aligned}\]
and we would hence have \(R_1 = L\) :
\[\begin{aligned} R_{\infty} &=-\ln \left(k T \times \frac{1}{1-e^{-Q} }\right) \\ R_{\infty} &=-\ln \left(k T \times \frac{1}{1-e^{L-H}}\right) \\ -L &=\ln \left(k T \times \frac{1}{1-e^{L-H}}\right) \\ e^{-L} &=e^{-L}\left(1-e^{L-H}\right) \\ &=e^{-L}-e^{-H} \\ T &=1 / k\left(e^{-L}-e^{-H}\right) \end{aligned}\]
\(\therefore\) the dosage schedule must be exactly \(T\).