Calculus Chapter Task

Question

I have created a scatter plot and the exponential function I found is y = 30e^-0.004t. I need help solving the rest of the questions. A brief explanation of each part will also suffice.

Codeine phosphate is a drug used as a painkiller. Generally, it is mixed with acetaminophen in tablet form. It is rapidly absorbed into the bloodstream from the gastrointestinal tract and is gradually eliminated from the body via the kidneys. A common brand contains 30 mg of codeine. Since it is physically addictive and has other unwanted side effects, it is important to avoid an overdose while helping to relieve pain symptoms such as those caused by a headache. Samples of blood were taken at regular time intervals from a patient who had taken a pill containing 30 mg of codeine. The amount of codeine in the bloodstream was determined every 30 min for 3 h. The data are shown in the table below.

Time after Consumption: Amount of Codeine in Blood

30 min 27 mg

60 min 23.5 mg

90 min 21.2 mg

120 min 18.7 mg

150 min 16.6 mg

180 min 14.5 mg

a) determine the instantaneous rate of change in the amount of codeine at each time given in the chart. How does it relate to the amount of codeine in the blood?

b) It is recommended that a second pill be taken when 90% of the codeine is eliminated from the body. When would this occur?

c) Assume that the same model applies to the second pill as to the first. Suppose the patient took a second pill 1 h after consuming the first pill.

Create a model for the amount of codeine in the patient’s bloodstream t minutes after taking the first pill.

Determine the maximum amount of codeine in the patient’s bloodstream.

Determine when 90% of the maximum amount would be eliminated from the body.

Answer 1

Question a) seems to be worded kind of funny. I will assume that they want you to use the equation that you developed to find the instantaneous rate of change (although that isn't very clear).

I would use a couple of additional decimals in the equation and then round off at the end. So, I ran the curve-fit and got y = 30.37e^-0.004065t

The rate of change is found by taking the derivative. The derivative of an e^x function is just e^x but the chain rule requires you to multiply by the derivative of the "something else" if its something other that "x" in the exponent. So the derivative is y' = (-0.004065)(30.37e^-0.004065t) or y' = -0.12345e^-0.004065t

Please note that if you have a TI-84 calculator, you can type the equation y = 30.37e^-0.004065t into "Y1" and use the math button and then "8:nDeriv(" to get the value of the derivative at each value of "x"

Either way, you find that:

The instantaneous rate of change at t = 30 is -0.12345e^{-0.004065(30)} = -0.11 mg/min

The instantaneous rate of change at t = 60 is -0.12345e^{-0.004065(60)} = -0.10 mg/min

The instantaneous rate of change at t = 90 is -0.12345e^{-0.004065(90)} = -0.09 mg/min

The instantaneous rate of change at t = 120 is -0.12345e^{-0.004065(120)} = -0.08 mg/min

The instantaneous rate of change at t = 150 is -0.12345e^{-0.004065(150)} = -0.07 mg/min

The instantaneous rate of change at t = 180 is -0.12345e^{-0.004065(180)} = -0.06 mg/min

You can say that the rate of codeine in the blood is negative because codeine is leaving the blood. You can say that the rate that codeine is leaving the blood is dropping off (codeine is leaving the blood at a higher rate at the beginning and then leaving at a smaller rate after time goes by.

90% gone means 10% is left. 10% of 30 is 3 mg. Using your equation you can plug in y = 3 and solve for "t":

3 = 30.37e^-0.004065t

3/30.37 = e^-0.004065t

0.09878 = e^-0.004065t

Now take the natural log of both sides of the equation:

ln(0.09878) = ln(e^-0.004065t)

-2.3148 = (-0.004065t)(ln(e))

-2.3148 = -0.004065t

-2.3148/-0.004065 = t

t = 569.5 minutes or 9.5 hours so give the second pill 9.5 hours after the first.

For question c), after 1 hour the *from the data table) the amount of codeine in the body (from the first pill only) is 23.5 mg. Taking a 2nd pill at this time would add 30 mg to that making a total of 53.5 mg. If you use you equation to model this, you would first put in t = 60 and get:

(first pill only) 30.37e^{-0.004065(60)} = 23.8 mg

(plus from the second pill t = 0) 30.37e^-0.004065(0) = 30.37 mg

23.8 + 30.37 = 54.2 mg

You can see that this method would over exaggerate the amount of codeine in the body but if you round to 2 digits, they both would be 54 so maybe it's OK.

You could use a piecewise function to model this situation:

The max amount is the 54 mg

90% eliminated means 10% is left which is 5.4 mg:

5.4 = 54e^-0.004065t

5.4/54 = e^-0.004065t

0.10 = e^-0.004065t

ln(0.1) = ln(e^-0.004065t)

-2.3026 = (-0.004065t)(ln(e))

-2.3026 = -0.004065t

t = -2.3026/-0.004065

t = 566 minutes = 9.44 hours (this is actually the same as what you got (or should have gotten) from question b) but there was a little error because of the equation so maybe you want to round off your answer. Anyway, you would still need to add the 1 hour between the 1st and 2nd pill to get a total of 10.5 hours (or there abouts)