Solving a function using periods of time.

Question

I'm just working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem. Here is the problem:

Jose takes medication. After $t$ minutes, the concentration of medication left in his bloodstream is given by $A(t) = 10(0.5) ^{0.014t}$ , where $A$ is in milligrams per litre.

a. Write down $A(0)$
Would this = $0$ because nothing has happened yet?

b. Find the concentration of medication left in his bloodstream after $50$ minutes.
Do I plug $50$ in for $t$?

c. At 13:00, when there is no medication in Jose’s bloodstream, he takes his first dose of medication. He can take his medication again when the concentration of medication reaches $0.395$ milligrams per litre. What time will Jose be able to take his medication again?
Do I plug in $0.395$ for $A$ and then solve for $t$?

Answer 1

On point a, to explain a little, since it seems to be causing you some confusion.

The question is vague about the way Jose "takes" his medication – was it injected, or did he swallow it?

If it was injected, which is easier to consider for this question, you will get a curve like the following.

That is, a rapid increase in blood concentration, to a maximum amount, followed by an exponential decline.

What is equal to 0 in part a, is the time in minutes, and obviously not the concentration, which is it at its maximum amount at 0 minutes. It is all “downhill” after that, so to speak.

The way I look at these sort of exponential decay rates is as follows:

CurrentConcentration = InitialAmount × RateOfChangePerUnitTime × AmountOfTimeUnits

Which for you, converts to:

A(t) = 10 × (0.5)^0.014 × ^t

Since t = 0, this then resolves to:

A(t) = 10 × (0.5)⁰

And I interpret (0.5)⁰ to physically mean “the decay rate has not been applied”, i.e. the drug concentration has not begun to decay, according to its RateOfChangePerUnitTime.

All of which means that:

A(0) = 10

which I presume to be in units of concentration (mg/mL etc.)

The pharmacokinetics are different, and more confusing for these calculations, if he swallowed it – the curve looks like this.

Answer 2

1. calculate $A(0)$ means put $t=0$ in the given eqn
2. calculate $A(50)$ means put $t=50$ in the given eqn
3. plug $0.395$ in place of $A(t)$ and calculate $t$ and add the time $t$ to $13:00$ to get the required time

$$0.395=10(0.5)^{0.014t}$$ now we can write it as $$0.395/10=(0.5)^{0.014t}$$ now I am taking the log of both side $$log(0.0395)=log{(0.5^{0.014t})}$$ $$log(0.0395)=0.014t*log{(0.5)}$$ $$\frac{log(0.0395)}{log(0.5)}=0.014t$$ $$\frac{-1.4034029}{-0.301029996}=0.014t$$ $$4.66200352=0.014t$$ $$\frac{4.66200352}{0.014}=t$$

which gives t=333.0002526 minutes

thus total $\frac{333}{60}$ hours which is 5 hours 33 minutes thus new time is 18:33