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The purpose of this lab is to use Maple to study applications of exponential and logarithmic functions. These are used to model many types of growth and decay, as well as in many scales, such as the Richter and decibel scales.
Separating the variables and integrating (see section 4.4 of the text), we have
(If y is a positive quantity, we may drop the absolute value signs around y.) Solving for y
y = ekt + Cwhich we may write in the form y = Aekt, where A is an arbitrary positive constant.
where k>0 is a constant. This is the same equation as in exponential growth, except that -k replaces k. The solution is
A(t) = A0 e-ktwhere A0 is a positive constant. Physically, A0 is the amount of material present at t=0.
Radioactivity is often expressed in terms of an element's half-life.
For example, the half-life of carbon-14 is 5730 years. This statement means
that for any given sample of , after 5730 years, half of it
will have undergone decay.
So, if the half-life is of an element Z is c years, it must be
that , so that and .
where k is the constant of proportionality and is the temperature of the environment. Using a technique called separation of variables it isn't hard to derive the solution
where T0 is the temperature of the object at t=0.
C(t) = C0 e-ktA problem facing physicians is the fact that for most drugs, there is a concentration, m, below which the drug is ineffective and a concentration, M, above which the drug is dangerous. Thus the physician would like the have the concentration C(t) satisfy
m < C(t) < MThis means that the initial dose must not produce a concentration larger than M and that another dose will have to be administered before the concentration reaches m.
> f := x -> exp(-2*x);
Sometimes you need to use experimental data to determine the value of constants in models. For example, suppose that for a particular drug, the following data were obtained. Just after the drug is injected, the concentration is 1.5 mg/ml (milligrams per milliliter). After four hours the concentration has dropped to 0.25 mg/ml. From this data we can determine values of C0 and k as follows. The value of C0 is the initial concentration, so we have
C0 = 1.5To find the value of k we need to solve the equation
0.25 = 1.5 e-4kwhich we get by plugging in t=4 and using the data C(4)=0.25. Maple commands for solving for k and defining and plotting the function C(t) are shown below.
> k1 := solve(0.25=1.5*exp(-4*k),k);
> C1 := t -> 1.5*exp(-k1*t);
f(x) = bxand
Jane E Bouchard