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DM

# 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).(a) Find the average rate of 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) $[2.0, 3.0]$ In each case, include the units.(b) Estimate the instantaneous rate of change at $t = 2$ and interpret your result. What are the units?

## a) (i) $-0.15 \mathrm{mg} / \mathrm{mL} \cdot \mathrm{hr}$ $\\$(ii) $-0.12 \mathrm{mg} / \mathrm{mL} \cdot \mathrm{hr},$$\\(iii) -0.12 \mathrm{mg} / \mathrm{mL} \cdot \mathrm{hr}$$\\$(iv) $-0.11 \mathrm{mg} / \mathrm{mL} \cdot \mathrm{hr}$$\\$b) -0.12

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Raisa B.

September 17, 2019

how did u determined the table

##### Top Calculus 1 / AB Educators   ##### Samuel H.

University of Nottingham Lectures

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### Video Transcript

So in this problem we are given that researchers measured blood alcohol concentration of eight men Uh starting one hour after consumption of a couple of drinks. And we're giving a table rooty isn't ours. One 1.5, 2.2.5 and 3.0 and the concentration C of T in milligrams per millimeter 0.31 0.24, 0.18, 0.12 And 0.08. First of all, we're asked to find the average change of C with respect to T over each time interval. So going from 1.02, we have 0.18 -0.3, 1 Over 2 -1 Is a -0.18. This is milligrams per millimeter and time is in ours. Okay. And from 1.5 two, We have 0.18 -0.24 over 2 -1.5. So that a negative 0.06 over 0.5. So that's a negative 0.12 again, milligrams per mil later our Okay, The next one from 2.0 To 2.5. Well, from our table This will be 0.1, two -0.18. We're just using our average rate of change equation here between these two timeframes, keeping everything coordinated minus two. and so I have a negative 0.06 over 0.5, Which again is a negative 0.12 milligrams per million. Later our Okay. And the last one here, we're going from 2.0 23 oh. And so that's 0.08 minus 0.18 over 3.0 -2. And so that's a negative 0.10 over one. That's a negative 0.10 milligrams per million later. Power. All right. In part b says to estimate the instantaneous rate of change, estimate the instantaneous right of change. Ah T equals two. Okay, so then at T equals two. What we can do is come up here and think about what happens as we approach two from each direction left and right. Well from 1.5 to 2 we did right here A- -12. And from right here we did it from the right And got a negative 0.12. So we have the same rate coming in from both sides. So normally we would average these two. But since they're the same, that means we're going to have the same instantaneous rate as we have a constant rate coming in from both sides left and right.

DM
Oklahoma State University

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##### Top Calculus 1 / AB Educators   ##### Samuel H.

University of Nottingham Lectures

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