A 0.6 ml dose of a drug is injected into a patient steadily for half a second.

Question

At the end of this time, the quantity, Q, of the drug in the body starts to decay exponentially at a continuous rate of 0.2% per second. Using formulas, express Q as a continuous function of time, t in seconds.

Answer 1

At the end of one second .2% of the drug has been lost or .002(.6) has been lost. That means .998(.6) is still in the body. During the next second another .2% is lost. or .998(.998)(.6) remains.

This continues each second:

Q = .6(.998)^t

Or in terms of "e"", since .998^t = (e^ln.998)^t, Q = .6 e^{ln.998 t} = .6 e^-0.002002t

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Answer 2

Drugs are metabolized according to enzyme kinetics. Some drugs follow zero order kinetics, such as alcohol. In zero order, the rate of decay is unrelated to the concentration, so the formula is a straight line, perhaps Q = mx + b where b is the initial quantity, In first order kinetics, which I believe the problem refers to, the rate of decay or metabolism is related to the concentration of the drug. The rate equation is products divided by reactants. In first order kinetics, only the reactant concentration, Q matters. Substituting Q in the equation, and assuming the concentration of the product, or metabolite has no effect upon the enzyme, the equation would become Q =1/x where x is the current amount. This of course is an exponential first order curve.

When you use the term exponential decay, this can also refer to radiodecay of an isotope, which is, of course, also dependent upon concentration. Drugs are either enzymatically metabolized, or excereted according to kinetics.

Hope this helps.