Answer:
a) Average change over each time interval: <u>see below</u>
b) Instantaneous rate of change at t = 2:<u> - 0.12 mg/mL/h</u>
Units are: <u>mg/mL/h or mg/(mL . h)</u>, they are equivalen.
Explanation:
- <u>Table:</u> Average blood alcohol concentration
t (hours) 1.0 1.5 2.0 2.5 3.0
C(t) (mg/mL) 0.33 0.22 0.16 0.1 0.09
a) <u>Average rate of change:</u>
- Formula: rate of change = [Change in C(t)] / [size of interval]
Interval Rate of change (mg/mL / h)
(i) [1.0, 1.5] [0.22 - 0.33] / [1.5 - 1.0] = - 0.22
(ii) [1.5, 2.0] [0.16 - 0.22] / [2.0 - 1.5] = - 0.12
(iii) [2.0, 2.5] [0.1 - 0.16] / [2.5 - 2.0] = - 0.12
(iv) [2.0, 3.0] [0.09 - 0.1] / [3.0 - 2.5] = - 0.02
b) <u>Estimation of the instantaneous rate of change at t = 2.</u>
- Since, the rate of change in the interval [1.5, 2.0] is equal to - 0.12 mg/mL /h and the rate of change in the interval [2.0, 2.5] is also equal to - 0.12 mg/mL/h, the best estimate for the instantaneous rate of change at t = 2 is the same value - 0.12 mg/ml/h.
- If the values were not equal, the best estimation would be the result of the interpolation which is a weighted average, for equal size intervals it is the equivalent to the average.
You obtain the units by dividing the units of the numerator (mg/mL) by the units of the denominator (h). This is how you obtain them:
- [Rate of change] = [Ct] / [t] = [mg/mL] / [h] = mg/mL/h = mg / (mL . h]
- You can write the units as mg/mL/h or mg/(mL . h). They are equivalent.