Problem 7 - Calculus for Biology and Medicine

Question

Consider $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{1}+x_{2} \\ \frac{d x_{2}}{d t}=3 x_{1}-x_{2} \end{array} $$ Determine the direction vectors associated with the following points in the \(x_{1}-x_{2}\) plane, and graph the direction vectors in the \(x_{1}-x_{2}\) plane: \((1,0),(0,1),(-1,0),(0,-1),(-1,-1),(0,0)\), and \((1,2) .\)

Step-by-Step Solution

Answer

Direction vectors are: (1,3), (1,-1), (-1,-3), (-1,1), (-2,-2), (0,0), and (3,1).

Step 1: Understanding the Differential Equations

The given differential equations describe the rate of change of variables \(x_1\) and \(x_2\) with respect to time \(t\). They are: \[ \frac{d x_{1}}{d t} = x_{1} + x_{2} \] \[ \frac{d x_{2}}{d t} = 3x_{1} - x_{2} \] Our goal is to find the direction vectors at specific points in the \(x_{1}-x_{2}\) plane by evaluating these equations.

Step 2: Calculate Direction Vectors

For each point \((x_1, x_2)\), substitute the values into both differential equations to find \(\left( \frac{d x_{1}}{d t}, \frac{d x_{2}}{d t} \right)\). This vector represents the direction vector at that point.

Step 3: Calculation for Point (1,0)

Substitute \(x_1 = 1\) and \(x_2 = 0\):\[ \frac{d x_{1}}{d t} = 1 + 0 = 1 \]\[ \frac{d x_{2}}{d t} = 3(1) - 0 = 3 \]Direction vector: \((1, 3)\).

Step 4: Calculation for Point (0,1)

Substitute \(x_1 = 0\) and \(x_2 = 1\):\[ \frac{d x_{1}}{d t} = 0 + 1 = 1 \]\[ \frac{d x_{2}}{d t} = 3(0) - 1 = -1 \]Direction vector: \((1, -1)\).

Step 5: Calculation for Point (-1,0)

Substitute \(x_1 = -1\) and \(x_2 = 0\):\[ \frac{d x_{1}}{d t} = -1 + 0 = -1 \]\[ \frac{d x_{2}}{d t} = 3(-1) - 0 = -3 \]Direction vector: \((-1, -3)\).

Step 6: Calculation for Point (0,-1)

Substitute \(x_1 = 0\) and \(x_2 = -1\):\[ \frac{d x_{1}}{d t} = 0 + (-1) = -1 \]\[ \frac{d x_{2}}{d t} = 3(0) - (-1) = 1 \]Direction vector: \((-1, 1)\).

Step 7: Calculation for Point (-1,-1)

Substitute \(x_1 = -1\) and \(x_2 = -1\):\[ \frac{d x_{1}}{d t} = -1 + (-1) = -2 \]\[ \frac{d x_{2}}{d t} = 3(-1) - (-1) = -2 \]Direction vector: \((-2, -2)\).

Step 8: Calculation for Point (0,0)

Substitute \(x_1 = 0\) and \(x_2 = 0\):\[ \frac{d x_{1}}{d t} = 0 + 0 = 0 \]\[ \frac{d x_{2}}{d t} = 3(0) - 0 = 0 \]Direction vector: \((0, 0)\).

Step 9: Calculation for Point (1,2)

Substitute \(x_1 = 1\) and \(x_2 = 2\):\[ \frac{d x_{1}}{d t} = 1 + 2 = 3 \]\[ \frac{d x_{2}}{d t} = 3(1) - 2 = 1 \]Direction vector: \((3, 1)\).

Step 10: Graph the Direction Vectors

Plot each point on the \(x_1-x_2\) plane. From each point, draw an arrow representing the direction vector calculated in the previous steps. These arrows show the direction of change of \(x_1\) and \(x_2\) at those points.

Key Concepts

Direction Vectors

Direction vectors are critical in understanding how systems described by differential equations evolve over time. In the context of the given problem, direction vectors can be seen as small arrows that indicate the direction in which a point moves as time progresses. This gives us insight into how the values of \(x_1\) and \(x_2\) change at any given point.

• To find a direction vector at a specific point, substitute the \(x_1\) and \(x_2\) values into the differential equations.


• The resulting values of \(\frac{d x_{1}}{d t}\) and \(\frac{d x_{2}}{d t}\) form a vector that points in the direction of change.

In this exercise, direction vectors were calculated for several points in the \(x_1-x_2\) plane, helping us visualize the system's behavior. These vectors not only show how each point behaves but also reveal patterns or trends as a group of vectors.

Rate of Change

The rate of change is a fundamental concept in calculus and differential equations. It provides us with the derivative of a function, which represents the speed at which variables change. In our case, the rate of change is expressed through the derivatives: \(\frac{d x_{1}}{d t}\) and \(\frac{d x_{2}}{d t}\).
Understanding the rate of change helps to:

• Determine how rapidly "\(x_1\)" and "\(x_2\)" are changing at a specific point.
• Analyze how interactions affect the movement of points in a phase plane.

Each substitution into the differential equations provides a snapshot of motion at particular coordinates. These snapshots, when calculated for various points, allow us to piece together an understanding of the entire dynamic system.

x1-x2 Plane

The \(x_1-x_2\) plane is a two-dimensional graph where each coordinate pair \((x_1, x_2)\) specifies a point in this system. Visualizing concepts in this plane is immensely helpful for understanding how differential equations govern movements and interactions of variables.

In the context of this problem, each point in the \(x_1-x_2\) plane:

• Provides an initial condition for the system of differential equations.
• When paired with direction vectors, illustrates the path the system will take as time changes.

By plotting the points and their corresponding direction vectors, one can immediately see not only the effect of time on \(x_1\) and \(x_2\) but also how different points interact or align with each other. This graphical interpretation is a powerful tool for intuition in dynamic system analysis.