Question

Analyzing models The following models were discussed in Section 9.1 and reappear in later sections of this chapter. In each case carry out the indicated analysis using direction fields. Chemical rate equations Consider the chemical rate equations \(y^{\prime}(t)=-k y(t)\) and \(y^{\prime}(t)=-k y^{2}(t),\) where \(y(t)\) is the concentration of the compound for \(t \geq 0,\) and \(k>0\) is a constant that determines the speed of the reaction. Assume the initial concentration of the compound is \(y(0)=y_{0}>0\). a. Let \(k=0.3\) and make a sketch of the direction fields for both equations. What is the equilibrium solution in both cases? b. According to the direction fields, which reaction approaches its equilibrium solution faster?

Short answer

Answer: The reaction with the rate equation \(y^{\prime}(t)=-0.3y^2(t)\) approaches its equilibrium solution faster.

Step-by-step solution

Sketch the direction fields for both equations with \(k=0.3\)

To sketch the direction fields, we need to find the slope of the tangent for different values of \(t\) and \(y\) using the given differential equations. For this purpose, plug in the value of \(k=0.3\): For the first equation: \(y^{\prime}(t)=-0.3y(t)\) For the second equation: \(y^{\prime}(t)=-0.3y^2(t)\) Sketch the direction fields according to these equations by plotting the slope of the tangent at different points on the \(ty\)-plane.

Determine the equilibrium solution in both cases

An equilibrium solution is a constant solution where \(y^{\prime}(t)=0\). We will find \(y(t)\) that satisfies this condition for both equations. For the first equation: \(0=-0.3y(t)\) \(y(t)=0\) The equilibrium solution for the first equation is \(y(t)=0\). For the second equation: \(0=-0.3y^2(t)\) \(y^2(t)=0\) \(y(t)=0\) The equilibrium solution for the second equation is also \(y(t)=0\).

Compare the speed of both reactions according to the direction fields

To determine which reaction approaches its equilibrium solution faster, we need to compare the slopes of the tangents in their respective direction fields. For the first equation, the slope is directly proportional to the concentration \(y(t)\), so the reaction is faster when the concentration is higher. For the second equation, the slope is proportional to the square of the concentration \(y^2(t)\), so the reaction is even faster when the concentration is higher compared to the first equation. According to the direction fields, the reaction with the rate equation \(y^{\prime}(t)=-0.3y^2(t)\) approaches its equilibrium solution faster.

Key concepts

equilibrium solution

In the context of differential equations, an equilibrium solution is a stable state where the rate of change is zero, meaning that the variable of interest does not change over time. To find an equilibrium solution, we set the derivative (rate of change) equal to zero and solve for the function.

For instance, in the exercise regarding chemical rate equations, we have two scenarios. The first equation is linear: \(y'(t) = -0.3y(t)\) and the second is nonlinear: \(y'(t) = -0.3y^2(t)\). In both cases, setting the derivative equal to zero leads to the conclusion that \(y(t)=0\) is the equilibrium solution. This state signifies that the concentration of the compound is no longer changing, which is typically reached as the reaction progresses and the compound is used up or transformed.

This concept is critical in chemistry and biology, where understanding the equilibrium state can indicate stability within a system, such as a chemical reaction reaching completion or a population's size stabilizing.

chemical rate equations

Chemical rate equations are mathematical expressions that describe the rate at which reactants are converted into products in a chemical reaction. They are a form of a differential equation and can vary in complexity. The rate typically depends on the concentration of the reactants, and possibly other factors like temperature and catalysts.

In the provided exercise, the equation \(y'(t)=-ky(t)\) is a first-order rate equation, whereas \(y'(t)=-ky^2(t)\) is a second-order rate equation. These orders refer to the power of the concentration in the rate equation. First-order reactions have a rate directly proportional to the reactant's concentration, while second-order reactions are proportional to the concentration squared, meaning they will speed up as the concentration increases.

This affects how quickly the system will reach equilibrium. Understanding the order of the reaction is crucial for predicting how the concentration of a compound changes over time and for designing processes in industries such as pharmaceuticals and environmental engineering.

direction fields

Direction fields, also known as slope fields, visually represent solutions to first-order differential equations without actually solving them. Each point in a direction field has a small line segment (or arrow) that shows the slope of the solution curve at that point. By observing these fields, one can gain qualitative understanding of the behavior of solutions.

When sketching direction fields for the exercise's chemistry-related differential equations, we look at how the slope of the tangent, depicted by the line segments, varies with different concentrations of the chemical compound. For \(y'(t)=-0.3y(t)\), the field will show a straight line under equilibrium, with slopes becoming flatter as the concentration decreases. In contrast, the field for \(y'(t)=-0.3y^2(t)\) will depict steeper slopes at higher concentrations, illustrating a faster approach to equilibrium.

This visual tool is particularly useful in situations where an analytical solution to the equation is complicated or unknown, allowing for an intuitive grasp of system dynamics over time.

calculus

Calculus, a branch of mathematics, deals with change and motion. It is divided into two key parts: differential calculus and integral calculus. Differential calculus focuses on derivatives, which measure how a function changes as its inputs change. Integral calculus, on the other hand, concerns integrals, which can be interpreted as the area under a curve or the accumulation of quantities.

In our exercise, differential calculus is applied to find the rate at which the concentration of a chemical compound changes over time through differential equations. By finding the derivative of the concentration function concerning time (represented by \(y'(t)\)), we can describe the behavior of chemical reactions, calculate equilibrium solutions, and predict how long it will take for a system to stabilize.

Calculus provides the fundamental tools for modeling and analyzing physical phenomena in many fields, including engineering, physics, economics, and biology. It allows us to predict the future state of dynamic systems and understand complex interactions between changing quantities.