I am skeptical of my results (percent rate of change)

Question

Excuse the png equations, still a MathJax newbie.

I am analyzing data I have computed: Alcohol content and Caffeine content retention after a duration of 8 hours for each.

I had gotten the data in the last four tables by using the first two tables shown below:

In the tables above, I attempt to compute the Average Rate of Change, which I believe to be correct. Using the formula;

In these tables above, I attempt to compute the Average Percent Rate of Change. Using the formula;

I need to construct a model that could find the caffeine content after t hours. I figured it best be accomplished using the formula for exponential decay;

and so I decided to make something like this:

Until I realized I perhaps have not gotten the correct value to make r. Which is correct, because whiles my APRC for caffeine is slightly constant, that for alcohol is completely variant.

In fact the computation results for ARC and APRC in either of the four tables do not seem to make sense to me, why Alcohol that decreases at .015 every time must have a varying percentage results. All even being so far from each other in value.

Basically what I am trying to achieve is the correct rate value to use for my exponential decay formula. How do I derive it? Is it even possible, necessary or suggested that I go along with this formula? Professional insight is welcomed and appreciated.

Thank You.

Below are my scatterplot data that supplements this project.

Answer 1

You can use the value at $t=8$ to calculate the value of $r$.

\begin{array} \\ & 7.58=48\cdot e^{-r\cdot 8}\\ & \\ & \textrm{Solving for r} \\ & \\ &\frac{7.58}{48}=e^{-r\cdot 8} \\ &\\ & \ln\left(\frac{7.58}{48} \right)=-r\cdot 8 \\ &\\ & \frac{\ln\left(\frac{7.58}{48} \right)}{8}=-r \\ &\\ & r=-\frac{\ln\left(\frac{7.58}{48} \right)}{8} \\ &\\ & r=\frac{\ln\left(\frac{48}{7.58} \right)}{8}\approx 0.230711 \\ \end{array}

I checked it for several values of t and it shows that the theoretical results are very close to the empirical results.

For instance $n(4)=19.0746$ and $n(7)=9.547$

Answer 2

OK, so I think the point of this problem is to get you to understand some different kinds of behaviour that drug concentrations might have in the body. In fact these are also important forms of behaviour for all sorts of situations involving variables changing in time, such as population growth, nuclear decay, disease spread, and so on.

Let's look at the case of alcohol first. The increment of decrease is the same for every time interval, that is, the total amount of alcohol lost each hour is constant. This behaviour tells you right away that a linear model is appropriate, and calculating percentage change is actually not necessary. A suitable equation to model this would be $n(t)=n_0-kt$, where $n(t)$ is the concentration at a given time, $t$, $n_0$ is the initial concentration, and $k$ is the rate constant (I have used a minus sign to flag the fact that the amount is decreasing in time). In this case, it is the gradient of the straight line, which is a constant for the whole graph. You could find the gradient using the statistical method of least squares, linked to in the comments, or simply by computing the gradient of your graph 'by hand'.

For the case of caffeine, we see that the amount lost each time interval changes, so we don't have linear behaviour. The proportion lost is constant, which tells us that we have exponential behaviour. We can use logarithms to find your $r$ in $n(t)=n_0e^{rt}$:

$\frac{n(t)}{n_0}=e^{rt}$

$\ln\frac{n(t)}{n_0}=rt$

So if you table $\ln\frac{n(t)}{n_0}$ against $t$, r is found as the gradient of the graph.

Of course many reactions, biological processes, etc. follow neither of these patterns! I hope this gets you started, gives you search terms to explore.