Problem 25 - Calculus for Biology and Medicine

Question

For make vector field plots of each of the differential equations. Find any equilibria of each differential equation and use your vector field plot to classify whether each equilibrium is stable or unstable. $$ \frac{d x}{d t}=\frac{x}{x+1}, x \neq-1 $$

Step-by-Step Solution

Answer

Equilibrium at \( x = 0 \) is unstable.

Step 1: Understanding the Differential Equation

The differential equation given is \( \frac{dx}{dt} = \frac{x}{x+1} \). This equation describes how the variable \( x \) changes with respect to time \( t \). The denominator indicates that \( x eq -1 \), as division by zero is undefined.

Step 2: Finding Equilibria

Equilibria occur when \( \frac{dx}{dt} = 0 \). So, set \( \frac{x}{x+1} = 0 \). This implies that the numerator \( x = 0 \). Thus, there is an equilibrium point at \( x = 0 \).

Step 3: Analyzing Stability Using Derivative

To determine stability, calculate the derivative \( \frac{d}{dx} \left( \frac{x}{x+1} \right) \). Using the quotient rule, \[ \frac{d}{dx} \left( \frac{x}{x+1} \right) = \frac{(x+1)(1) - x(1)}{(x+1)^2} = \frac{1}{(x+1)^2} \]. Evaluating at \( x = 0 \), \( \frac{1}{1^2} = 1 \), which is positive, indicating instability.

Step 4: Vector Field Plot

Construct the vector field for the differential equation \( \frac{dx}{dt} = \frac{x}{x+1} \). For points where \( x > 0 \), \( \frac{dx}{dt} > 0 \) indicating that the vectors point to the right. For points where \( x < 0 \), \( \frac{dx}{dt} < 0 \) indicating that the vectors point to the left.

Step 5: Classifying Equilibria Using Vector Field

In the vector field, observe that near \( x = 0 \), the vectors point away from the equilibrium, confirming instability. This visual inspection supports the analysis from the derivative, as vectors moving away characterize an unstable equilibrium.

Key Concepts

Vector Field

A vector field can be thought of as a map that assigns a vector to every point in a subset of space. In the context of differential equations, it helps us visualize how a variable changes over time at different points.

For our exercise, the vector field involves plotting the differential equation \( \frac{dx}{dt} = \frac{x}{x+1} \). At each point \( x \), the direction and magnitude of the vector indicates how \( x \) changes.

• Positive values of \( \frac{dx}{dt} \) suggest the vector points to the right.
• Negative values indicate the vector points to the left.

Observing these vectors helps us understand how the system behaves around equilibrium points. It’s like seeing the wind directions on a map, knowing if it's pushing an object right or left, up or down.

Equilibrium Points

Equilibrium points are fundamental in understanding differential equations. They are points where the system doesn't change, meaning the rate of change, \( \frac{dx}{dt} \), equals zero.

For the equation \( \frac{dx}{dt} = \frac{x}{x+1} \), to find equilibria, we set \( \frac{x}{x+1} = 0 \). This equation implies that the numerator \( x \) must be zero, giving us an equilibrium point at \( x = 0 \). Identifying equilibrium points is crucial as these points often represent states of rest or constant movement within dynamic systems. Understanding them allows us to predict long-term outcomes of the system, especially in determining where it might settle naturally.

Stability Analysis

Stability analysis involves determining whether an equilibrium point is stable or unstable. A stable point means that the system tends to return to this point if slightly disturbed. Conversely, an unstable point will repel any nearby trajectories.

In our case, we analyze the stability at \( x = 0 \) by computing the derivative of \( \frac{dx}{dt} \) with respect to \( x \). Using the quotient rule: \[\frac{d}{dx} \left( \frac{x}{x+1} \right) = \frac{(x+1) \cdot 1 - x \cdot 1}{(x+1)^2} = \frac{1}{(x+1)^2}\]Evaluating this at \( x = 0 \): \[\frac{1}{(0+1)^2} = 1\]A positive derivative, like here, indicates that small disturbances will grow, confirming that \( x = 0 \) is indeed an unstable equilibrium point. If disturbances shrink back to equilibrium, it would depict stability, but that’s not the case here.

Quotient Rule

The quotient rule is a method in calculus to find the derivative of a function that is a ratio of two other functions. It is especially handy when dealing with rational expressions, as in our differential equation.

The formula for the quotient rule is: \[\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2}\]Here, \( u(x) = x \) and \( v(x) = x+1 \), giving the derivatives \( u'(x) = 1 \) and \( v'(x) = 1 \). Applying it to our function helps us see how sensitive \( \frac{dx}{dt} \) is to changes in \( x \). It becomes clear how small variations at \( x \) influence the rate of change, an essential aspect for understanding the system’s dynamics and stability. Proper grasp of the quotient rule is key in evaluating and predicting the behavior of differential equations.