Q. 88 - Calculus

Question

Many prescribed drugs must reach a “maintenance level” in the bloodstream to be effective. Say a person takes 300 milligrams of wonder drug Excellente´ per day and that whatever level of Excellente´ is in the bloodstream, p% is eliminated in one 24-hour period. 

Find the approximate level of drugs during the first week.

Step-by-Step Solution

Answer

$$\begin{gathered} L_2=140 \\ {L}_3=356 \\ {L}_5=479.96 \\ {L}_6=490.784 \\ {L}_7=496.31 \\ {L}_4=442.40 \end{gathered}$$

Step 1. Given

The amount of drug in milligrams a person takes D=300 mg.

The percentage p of the drug eliminated from the blood stream after 24 hours = 60

Step 2. Dose on Second day

$$\begin{gathered} \mathrm{The} \mathrm{daily} \mathrm{dose} \mathrm{of} \mathrm{the} \mathrm{medicine} \mathrm{is} \mathrm{D}. \\ \mathrm{Let} {\mathrm{L}}_{1 }\mathrm{is} \mathrm{the} \mathrm{amount} \mathrm{of} \mathrm{drug} \mathrm{in} \mathrm{the} \mathrm{blood} \mathrm{stream} \mathrm{at} \mathrm{the} \mathrm{start}. \\ \mathrm{Hence},{\mathrm{L}}_{ 1}=\mathrm{D} \\ \mathrm{After} 24 \mathrm{hours} 50\mathrm{\%} \mathrm{of} \mathrm{the} \mathrm{drug} \mathrm{has} \mathrm{been} \mathrm{eliminated} \mathrm{from} \mathrm{the} \mathrm{blood} \mathrm{stream} \mathrm{and} \mathrm{the} \mathrm{second} \mathrm{dose} \mathrm{is} \mathrm{taken}. \\ Hence,L2=(1-\frac{p}{100})L_1+D......\left(1\right) \\ \mathrm{In} \mathrm{general},\mathrm{after} k days.L_{k+1}=(1-\frac{p}{100})L_k+D..... \\ \mathrm{Substituting} k=1 in\left(2\right) \\ \mathrm{Therefore},L_2=(1-\frac{60}{100})\left(300\right)+300 \\ Simplify, \\ {L}_2=(1-0.60)\left(300\right)+300 \\ {L}_2=(0.40)\left(300\right)+300 \\ =40+100 \\ {L}_2=140 \end{gathered}$$

Step 3. Dose on Third day

$$\begin{gathered} L_3=(1-\frac{p}{100})L_1+D......\left(1\right) \\ \mathrm{In} \mathrm{general},\mathrm{after} k days.L_{k+1}=(1-\frac{p}{100})L_k+D..... \\ \mathrm{Substituting} k=2 in\left(2\right) \\ \mathrm{Therefore},L_2=140 \\ Simplify, \\ {L}_3=(1-0.60)\left(140\right)+300 \\ {L}_3=(0.40)\left(140\right)+300 \\ {L}_3=356 \end{gathered}$$

Step 4. Dose on fourth day

$$\begin{gathered} L_4=(1-\frac{p}{100})L_3+D......\left(1\right) \\ \mathrm{In} \mathrm{general},\mathrm{after} k days.L_{k+1}=(1-\frac{p}{100})L_k+D..... \\ \mathrm{Substituting} k=3 in\left(2\right) \\ \mathrm{Therefore},L_3=356 \\ Simplify, \\ {L}_4=(1-0.60)\left(356\right)+300 \\ {L}_4=(0.40)\left(356\right)+300 \\ {L}_4=442.40 \end{gathered}$$

Step 5. Dose on fifth day

$$\begin{gathered} L_5=(1-\frac{p}{100})L_4+D......\left(1\right) \\ \mathrm{In} \mathrm{general},\mathrm{after} k days.L_{k+1}=(1-\frac{p}{100})L_k+D..... \\ \mathrm{Substituting} k=4 in\left(2\right) \\ \mathrm{Therefore},L_4=442.40 \\ Simplify, \\ {L}_5=(1-0.60)(442.40)+300 \\ {L}_5=(0.40)(442.40)+300 \\ {L}_5=479.96 \end{gathered}$$

Step 6. Dose on sixth day

$$\begin{gathered} L_6=(1-\frac{p}{100})L_5+D......\left(1\right) \\ \mathrm{In} \mathrm{general},\mathrm{after} k days.L_{k+1}=(1-\frac{p}{100})L_k+D..... \\ \mathrm{Substituting} k=5 in\left(2\right) \\ \mathrm{Therefore},L_5=476.96 \\ \mathrm{Simplify}, \\ {L}_6=(1-0.60)(476.96)+300 \\ {L}_6=(0.40)(476.96)+300 \\ {L}_6=490.784 \end{gathered}$$

Step 7. Dose on seventh day

$$\begin{gathered} L_7=(1-\frac{p}{100})L_6+D......\left(1\right) \\ \mathrm{In} \mathrm{general},\mathrm{after} k days.L_{k+1}=(1-\frac{p}{100})L_k+D..... \\ \mathrm{Substituting} k=5 in\left(2\right) \\ \mathrm{Therefore},L_6=490.78 \\ \mathrm{Simplify}, \\ {L}_7=(1-0.60)(490.78)+300 \\ {L}_7=(0.40)(490.78)+300 \\ {L}_7=496.31 \end{gathered}$$