DE for the rate at which chemicals clear from blood

[pope4]

For my Differential Calculus uni course, we were assigned a question about the rate of medication filtration in the blood and I’m so confused. The question states:

The rate for the clearance of a drug is a certain constant r per hour. Assume the mass of the drug in the body at t=0 is 0 when they take a 300mg pill. What is the differential equation for the mass of the drug remaining in the body at time t?

I figured out that the change in mass has to be m(t-dt)=-rm(t), but I cannot for the life of me figure out what m(t) would be, much less how to include the mass at any given time in the differential equation. I’d appreciate any advice I can get!

 

[BigBeachBanana]

For my Differential Calculus uni course, we were assigned a question about the rate of medication filtration in the blood and I’m so confused. The question states:

The rate for the clearance of a drug is a certain constant r per hour. Assume the mass of the drug in the body at t=0 is 0 when they take a 300mg pill. What is the differential equation for the mass of the drug remaining in the body at time t?

I figured out that the change in mass has to be m(t-dt)=-rm(t), but I cannot for the life of me figure out what m(t) would be, much less how to include the mass at any given time in the differential equation. I’d appreciate any advice I can get!

Find the equation for [imath]m(t)[/imath], then differentiate.

 

[limiTS]

The question seems to be testing your understanding of a common differential equation, that the rate of change in something is proportional to the amount of that something.

Amount of drug: this can be expressed as [imath]C_t\cdot V[/imath] where [imath]C_t[/imath] is the concentration of drug at time [imath]t[/imath] and [imath]V[/imath] is the drug's volume of distribution. To keep it simple, let's say that [imath]V[/imath] is constant.

Then the rate of change in the amount of drug can be written as [imath]\displaystyle\frac{\text{d}\left(C_t\cdot V\right)}{\text{d}t}=V\cdot\frac{\text{d}C_t}{\text{d}t}[/imath]

Logically, the rate of change in drug amount must be equal to the rate of drug going in minus the rate of drug going out.

Rate In: the drug is given as a one-time 300 mg pill. Let's assume the drug is not being given continuously, so that the Rate In is zero. The 300 mg will be absorbed quickly and be distributed over its [imath]V[/imath]. Let's say that [imath][\text{drug peak}]=C_p[/imath]

Rate Out: by definition, the clearance rate times the current [drug] equals drug excretion rate. Note that the excretion rate decreases as the [drug] dwindles. So, [imath]\text{Rate Out}=r\cdot C_t[/imath]

Putting it all together,
[math]\begin{aligned} V\cdot\frac{\text{d}C_t}{\text{d}t}&=0-r\cdot C_t\\&=-r\cdot C_t \end{aligned}[/math]
This is a separable differential equation that you can solve. At the end, don't forget to include the arbitrary constant of integration. Find its value using initial conditions: at time zero, the [drug] is [imath]C_p[/imath]. (Don't worry about the part that says the mass of drug at t=0 is zero. That's just so there's no leftover drug from before.)

To find the mass of drug at a given time, all you have to do is multiply your solution for [imath]C_t[/imath] by [imath]V[/imath]

I've assumed a lot of things to keep the question simple. If the assumptions are wrong, please post the entire question.