Question

The concentration C (in mg/cc) of a drug in a patient's bloodstream is related to the time t (in hours) after injection in such a way that as t changes from 0 to 12, C changes from 0.7 to 0.1. Find the average rate of change of C with respect to t.
I have a calculus question similar to this one and I am unsure of how to begin working on this problem.

Answer 1

Answer: -0.05 mg/cc*h

Step-by-step explanation:

Here we have the relation:

ConcentrationVsTime.

If we want to find the average rate of change, we can find the slope of a line that passes through the extremes of each interval: (tmin, Cmin) and (tmax, Cmax)

The interval for the time is:

0h to 12h.

The interval for the concentration is:

0.7 mg/cc to 0.1mg/cc.

Then we can use the points:

(0h, 0.7mg/cc) and (12h, 0.1mg/cc)

For a line that passes through the points (x1, y1) and (x2, y2), the slope can be written as:

a = (y2 - y1)/(x2 - x1).

Then in this case the slope (or rate of change) is:

a = (0.1mg/cc - 0.7mg/cc)/(12h - 0h) = (-0.6mm/cc)/12h = -0.05 mg/cc*h