(Solved): A patient is taking a medicine that has a half-life of 6 hours in the human body, meaning that the ...

A patient is taking a medicine that has a half-life of 6 hours in the human body, meaning that the rate medicine is broken down is proportional to the amount present with constant of proportionality \( k=\frac{\ln 2}{6} \) Starting at \( t=0 \), they are on an IV drip that delivers a constant \( r \mathrm{mg} \) per hour of medicine. Therefore, the differential equation satisfied by \( Q(t) \), the amount (in \( \mathrm{mg} \) ) of medicine in the patient's bloodstream at time \( t \) (in hours) is \[ \frac{d Q}{d t}=r-\frac{\ln 2}{6} Q, \] and the initial value problem is \[ \left\{\begin{aligned} \frac{d Q}{d t} &=r-\frac{\ln 2}{6} Q \\ Q(0) &=0 . \end{aligned}\right. \] (a) Solve the initial value problem. (b) Suppose that the medicine requires \( 10 \mathrm{mg} \) to be in the bloodstream to be effective. What values of \( r \) are appropriate if we want the patient to have at least \( 10 \mathrm{mg} \) in their blood within three hours? (c) Unfortunately the medicine is toxic if there are ever \( 100 \mathrm{mg} \) in the body. Keeping in mind we still want the patient to have effective levels after three hours, which values of \( r \) are also safe if the patient remains on the IV indefinitely?