Question

The concentration C C (in mg/cc) of a particular drug in a person's bloodstream is related to the time t t (in hours) after injection in such a way that as t t changes from 0 0 to 8 8, C C changes from 0.9 0.9 to 0.1 0.1. Find the average rate of change of C C with respect to t t.

Answer 1

Answer:

The average rate of change of 'C' with 't' is -0.1. Negative sign means there is decrease in concentration with time.

Step-by-step explanation:

Given:

Initial concentration (C₁) = 0.9

Final concentration (C₂) = 0.1

Starting time (t₁) = 0

End time (t₂) = 8

Now, change in concentration is equal to the difference of final and initial concentration. Therefore,

$\Delta C=C_2-C_1\\\\\Delta C=0.1-0.9=-0.8$

Change in time interval is given as:

$\Delta t=t_2-t_1=8-0=8$

Average rate of change of concentration 'C' with time 't' is given as:

$R=\frac{\Delta C}{\Delta t}\\\\R=\frac{-0.8}{8}\\\\R=-0.1$

Therefore, the average rate of change of 'C' with 't' is -0.1. Negative sign means there is decrease in concentration with time.