Final answer:
The phase portrait of the differential equation suggests that as t approaches infinity, X(t) will tend to approach B (a stable equilibrium) in the general case, and to approach a when a = B depending on whether X(0) is less than or greater than a.
Step-by-step explanation:
The question deals with an autonomous differential equation, dX/dt = k(a-X)(B-X), that models the rate at which a new compound is formed during a chemical reaction.
A phase portrait helps in understanding the behavior of the solution X(t) over time. In general, for k > 0, B > a > 0, the quantity of new compound X approaches B as t approaches infinity.
The phase portrait would show that X(t) has two equilibrium points at X = a and X = B, with X = a being an unstable equilibrium (since X(t) tends to move away from a) and X = B being a stable equilibrium (since X(t) tends to approach B over time).
For the special case when a = B, if X(0) < a, then X(t) will approach a as t approaches infinity because all initial conditions less than the equilibrium value will move towards it, indicating again a stable equilibrium. When X(0) > a, X(t) will decrease towards a but will not cross it, establishing the nature of a as an attractor in the phase portrait.
The primary focus here is on the application of calculus in predicting the behavior of chemical reactions over time, particularly showing how the evolving concentration of products and reactants affects the reaction rate.
Affected by the concentrations of both reactants and products due to the reversible nature of most chemical reactions, this relationship is often depicted in a standard chemistry course involving chemical kinetics or reaction mechanisms.