Researchers measured the average blood alcohol concentration c(t) of eight men starting one hour after consumption of 30 ml of ethanol (corresponding to two alcoholic drinks). (round your answers to two decimal places.) t (hours) 1.0 1.5 2.0 2.5 3.0 c(t) (mg/ml) 0.31 0.22 0.16 0.1 0.06 (a) find the average change of c with respect to t over each time interval. (i) [1.0, 2.0] (ii) [1.5, 2.0] (iii) [2.0, 2.5] (iv) [2.0, 3.0] (b) estimate the instantaneous rate of change at t = 2. what are the units? h ml/mg h mg/ml h h mg/ml mg/ml ml/mg h

a)

i) at [1.0,2.0] = = -0.15 mg/ml¬hr

ii) at [1.5, 2.0]  = - 0.12 mg/ml¬hr

iii) at [2.0, 2.5] = -0.12  mg/ml¬hr

(iv) at [2.0, 3.0] = -0.1 mg/ml¬hr

b) the instantaneous rate of change at t = 2 = 0.12  mg/ml¬hr and the unit will be h mg /ml

Explanation:

1) according to the table of the change in c (mg/ml) and the change in time (hours) So, we are going to use this formula to solve part (a)

m = ΔC / Δt

when m is the average change of c with respect to t over each time interval

i) [1.0,2.0] :

m = ΔC / Δt

   = (0.16-0.31)/(2-1)

   = -0.15 mg/ml¬hr

ii) [1.5, 2.0]:

m = ΔC / Δt

    = (0.16-0.22)/(2-1.5)

    = - 0.12 mg/ml¬hr

iii) [2.0, 2.5]:

m = ΔC / Δt

    = (0.1 - 0.16)/(2.5-2)

    =  -0.12  mg/ml¬hr

(iv) [2.0, 3.0] :

m = ΔC / Δt

   = (0.06-0.16)/(3-2)

   = -0.1 mg/ml¬hr

(b) to estimate the instantaneous rate of change at t = 2 we can see the results at (ii) and (iii) at time rates [1.5, 2.0] and [2.0, 2.5] both are the same answer = -0.12 mg/ml¬hr so, the instantaneous rate of change at t=2 is equal -0.12 and the unit will be h mg/ml because (mg/ml) / h = mg h /ml

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