Problem 63 - Calculus for Biology and Medicine

Question

Draw the vector field plot of the differential equation. Then, using the given initial conditions, sketch the solutions (i.e., draw a graph showing the dependent variable as a function of the independent variable). \(\frac{d N}{d t}=N(N-1)(5-N)\) (a) \(N(0)=1\), (b) \(N(0)=1 / 2\), (c) \(N(0)=3 / 2\), (d) \(N(0)=7\).

Step-by-Step Solution

Answer

Plot vector field; solutions: (a) stays at 1, (b) decreases to 0, (c) increases to 5, (d) decreases to 5.

Step 1: Understand the Differential Equation

The differential equation given is \( \frac{dN}{dt} = N(N-1)(5-N) \). It represents the rate of change of \( N \) with respect to \( t \). The form suggests there are critical points (equilibria) at \( N = 0 \), \( N = 1 \), and \( N = 5 \). By analyzing these critical values, we can determine the behavior of \( N \) over time based on its initial conditions.

Step 2: Determine Equilibrium Points

Set the equation \( N(N-1)(5-N) = 0 \) to find equilibrium points: \( N = 0 \), \( N = 1 \), and \( N = 5 \). At these points, \( \frac{dN}{dt} = 0 \), so the population does not change. These are important when sketching the vector field and solutions as they represent stable and unstable states.

Step 3: Plot the Vector Field

The vector field of a differential equation shows the direction and rate of change of \( N \) at different values. Calculate \( \frac{dN}{dt} \) for values of \( N \) around the critical points to understand how \( N \) behaves: - For \( N < 0 \), \( \frac{dN}{dt} < 0 \) (decreases).- Between \( N = 0 \) and \( N = 1 \), \( \frac{dN}{dt} < 0 \) (decreases).- Between \( N = 1 \) and \( N = 5 \), \( \frac{dN}{dt} > 0 \) (increases).- For \( N > 5 \), \( \frac{dN}{dt} < 0 \) (decreases). Draw arrows pointing accordingly in each region to represent the vector field.

Step 4: Graph the Solution Curves

Using the initial conditions, sketch solution curves on the vector field:- (a) \( N(0) = 1 \): Since \( N = 1 \) is an equilibrium point, the solution remains constant.- (b) \( N(0) = \frac{1}{2} \): Curve moves towards \( N = 0 \) over time since \( \frac{dN}{dt} < 0 \).- (c) \( N(0) = \frac{3}{2} \): Curve moves upwards towards \( N = 5 \) over time as \( \frac{dN}{dt} > 0 \).- (d) \( N(0) = 7 \): Curve moves down towards \( N = 5 \) since \( \frac{dN}{dt} < 0 \) for \( N > 5 \). Plot these to visualize how solutions evolve according to their initial conditions.

Key Concepts

Vector Field

A vector field in the context of differential equations offers a visual representation of the rate and direction of change of a function. For the given equation \( \frac{dN}{dt} = N(N-1)(5-N) \), the vector field essentially tells us how the value of \( N \) changes over time at different points. Imagine a field of arrows, where each arrow shows the direction (up or down) and the speed of change (length of the arrow) for \( N \).

The direction of these arrows is determined by the sign of \( \frac{dN}{dt} \):

• If \( \frac{dN}{dt} > 0 \), arrows point upward, showing that \( N \) is increasing.
• If \( \frac{dN}{dt} < 0 \), arrows point downward, indicating that \( N \) is decreasing.

This helps predict how solutions to the differential equation behave without solving the equation explicitly. It's an intuitive way to see equilibrium points, even for complex behaviors.

Equilibrium Points

Equilibrium points occur where the rate of change \( \frac{dN}{dt} = 0 \). At these points, the function \( N \) does not change over time. For our equation, setting \( N(N-1)(5-N) = 0 \) yields equilibrium points at \( N = 0 \), \( N = 1 \), and \( N = 5 \). These are particularly important because they represent long-term behaviors of the system.

Understanding these points tells us:

• Stable Equilibrium: If small perturbations in \( N \) return to the equilibrium point, it is stable. For a point inside the domain to be stable, the surrounding vector field should direct towards it.
• Unstable Equilibrium: If small changes in \( N \) move it away from the equilibrium point, it is unstable. Typically, the vector field arrows near an unstable equilibrium point push away from it.

In this example:
- \( N = 1 \) is stable for initial conditions between \( 0 \) and \( 1 \) since any perturbation returns the function back to this state.
- \( N = 0 \) and \( N = 5 \) are unstable, as small changes move \( N \) away from these values.

Initial Conditions

Initial conditions specify the state of the system at the beginning of the observation, in this case at \( t = 0 \). They are immensely crucial as they determine the trajectory or path that the solution will follow. In our exercise, the initial conditions are:

• \( N(0) = 1 \)
• \( N(0) = \frac{1}{2} \)
• \( N(0) = \frac{3}{2} \)
• \( N(0) = 7 \)

Each of these beginning values determines how \( N \) will evolve over time under the differential equation's rule. For example, starting at \( N(0) = 1 \), since this is an equilibrium point, \( N \) will not change, staying constant. Similarly, other initial conditions lead \( N \) towards or away from equilibrium points, shaping distinct solution curves.

Solution Curves

Solution curves are graphical representations of how the dependent variable, in this case \( N \), changes over time based on the given initial conditions and the differential equation. They provide a visual storyline for how \( N \) behaves.

By starting from an initial condition:

• When \( N(0) = 1 \), the curve remains flat since it's an equilibrium point.
• For \( N(0) = \frac{1}{2} \), the curve decreases towards zero, indicating that \( N \) will gradually approach \( N = 0 \), an attractor under this condition.
• For \( N(0) = \frac{3}{2} \), the curve increases towards 5, demonstrating stable behavior towards that value.
• Lastly, \( N(0) = 7 \) leads to \( N \) decreasing down to \( 5 \), emphasizing the influence of the vector field directions.

These paths are traced according to the vector field and the differential equation, showing not just where \( N \) will be, but how it gets there, showcasing the interplay between the rate of change and the initial state.