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The table shows values of the viral load $ V(t) $ in HIV patient 303, measured in RNA copies/mL, $ t $ days after ABT-538 treatment was begun.

(a) Find the average rate of change of $ V $ with respect to $ t $ over each time interval: (i) $ [4, 11] $ (ii) $ [8, 11] $ (iii) $ [11, 15] $ (iv) $ [11, 22] $ What are the units?

(b) Estimate and interpret the value of the derivative $ V'(11) $.

a) see solutionb) $V^{\prime}(11)=-1.96 \mathrm{RNA}$ copies $/ \mathrm{mL}$ , represents the rate of decrease ofthe viral load in the patient on the 11 th day

06:04

Daniel J.

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 7

Derivatives and Rates of Change

Limits

Derivatives

Baylor University

University of Michigan - Ann Arbor

University of Nottingham

Boston College

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So in this problem were given this table of data which is the viral load bot in this patient. T days after a treatment has begun so those tears in days and this viral load is R. N. A. Copies per mr leader. Okay. And first of all we're asked to find the average rate of change of the respect to the T. Over each of these different time intervals. The first one is from 4 to 11. So the average rate of change abbreviated abbreviated arrow C will be the at 11 minus V. At four over 11 miles four. Well from the table this is 9.4 -53/7. Which yes A- 43.6/11. So then we're asked to the interval 8 to 11. So the average rate of change here again the 11 minus the eight Over 11 -8. And so this is 9.4 minus 18 over three. And that is a negative 8.6 over three. Then we were asked to do From 11 through 15. And then average rate of change Will be v. of 15 on SV of 11 Over 15 -11 Which is 5.2 -9.4 over four. So That is a negative 4.2 over four. Okay and then we're asked to do 11- 22. This average rate of change Will be the of 22 Honest v. of 11 over 22 -11. Okay. From our table Of 3.6. Well it's not really neat. Let me clean that up. 3.6 9.4 over 11. And that's a negative 5.8 over 11. And were asked what are the units on these? Okay well the units we took V. RNA copies per milliliter and we divided them by days didn't we? So the units on all these. Yeah par And a copies. Her middle leader days. Okay then we're asked to estimate interpret the value of the derivative the prime at 11. Well why we do this is we take the average rate of change coming from the left and the average really change coming from the right and we average them And we did that right. We did it from 8 to 11 here And we did it from 11 to 15 here. Okay so this is one half of -8.6 over three plus minus 4.2/4. Okay and add fractions, we have to have the same denominator. So this is one half times Well Common denominator would be 12. So we multiply that by four .6 times four will be -34.4 over 12 -4.2 times three will be 12.6 over 12. And so that's one half times a negative 12.6. 34.4. That's 47 or four. So it's a -47/8. So there's our estimate and interpret the value here. We say that or every eight days we have a change and viral load of the change. We could say a decrease here decrease in viral load of 47 are alright a copies or a little later. Hat. 11 days. So there we go. We found all of the average rate to change. We found, estimated the value of the derivative, The prime at 11 and interpreted it.

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