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Numerade Educator

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Problem 85 Hard Difficulty

The average blood alcohol concentration (BAC) of eight male subjects was measured after consumption of 15 mL of ethanol (corresponding to one alcoholic drink). The resulting data were modeled by the concentration function
$ C(t) = 0.0225te^{0.0467t} $
where $ t $ is measured in minutes after consumption and C is measured in mg/mL.
(a) How rapidly was the BAC increasing alter 10 minutes?
(b) How rapidly was it decreasing half an hour later?

Answer

a. $7.5 \times 10^{-3}(\mathrm{mg} / \mathrm{mL}) / \mathrm{min}$
b. $3 \times 10^{-3}(\mathrm{mg} / \mathrm{mL}) / \mathrm{min}$

Discussion

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CA

Catherine A.

October 26, 2020

Heather Z., thanks this was super helpful.

DG

David Base G.

October 26, 2020

That was not easy, glad this was able to help

Video Transcript

in this problem we are given the blood alcohol content and units of milligrams similar leader after two minutes of consumption that E two equals 20.0 to 252.46 17 patients information. We want to answer any which requires us to use differentiation via the chain rule that is a derivative of a function F of X with respect to X. Is F prime G n x n G. Products which is a function inside about seven A. We use the general to find the rate of change of BF 10 minutes first, we find the prime th in the chain rule as 100.2 to 5.46 17 plus. By the product rule 170.467 times 0.0 to +25 to 0.46 17. Thus our solution is for plugging into equals tend to be primed for a. Be prime 10 equals 7.5, 10 to the negative 30 mg per milliliters per minute and b we find the rate of change is 30 minutes. For this we simply forget 30 giving be a third time of 30 equals three times and 93rd milligram milligrams per milliliters per minute.