Science:Math Exam Resources/Courses/MATH110/December 2015/Question 07 (b)

MATH110 December 2015

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Question 07 (b)

An exponential decay function $P(t)=Ce^{-kt}$, where $C$ and $k$ are positive constants, is used to model the amount of drug in the blood t hours after an initial dose is administered. Suppose an initial dose of 100 mg is administered to a patient. Assume it takes 36 hr for the body to eliminate half of the initial dose from the blood stream. Answer the questions below. You may leave your answers unsimplified, i.e. in ”calculator-ready” form. (b) At what rate (in mg per hr) is the drug eliminated from the blood stream after 36 hr since the initial dose?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
How is the rate of change of a quantity represented in mathematical language?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The rate of change of any quantity is represented by its derivative, so the question is asking to find $P'(t)$ at $t=36\ hr$ . By part (a) we have $P(t)=100e^{-{\frac {\ln 2}{36}}t}$ , applying the chain rule with the exponential function derivative gives ${\begin{aligned}P'(t)=100\cdot \left(-{\frac {\ln 2}{36}}\right)e^{-{\frac {\ln 2}{36}}t}\Rightarrow P'(36)&=100\cdot \left(-{\frac {\ln 2}{36}}\right)e^{-{\frac {\ln 2}{36}}36}\\&=\color {blue}-{\frac {25\ln 2}{9}}e^{-\ln 2}\\&=\color {blue}-{\frac {25\ln 2}{18}}\ mg/hr\end{aligned}}$