Problem 6 - Calculus for Biology and Medicine

Question

Consider $$ \begin{array}{l} \frac{d x_{1}}{d t}=2 x_{1}-x_{2} \\ \frac{d x_{2}}{d t}=-x_{2} \end{array} $$ 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: \((2,0),(1.5,1),(1,0),(0,-1),(1,1),(0,0)\), and \((-2,-2)\).

Step-by-Step Solution

Answer

Compute the gradient for each point and graph them as direction vectors in the \(x_1-x_2\) plane.

Step 1: Understand the System of Differential Equations

The given system consists of two differential equations: \(\frac{dx_{1}}{dt} = 2x_{1} - x_{2}\) and \(\frac{dx_{2}}{dt} = -x_{2}\). These equations describe how the variables \(x_1\) and \(x_2\) change over time.

Step 2: Compute the Gradient at Each Point

For each point, calculate the gradient, or direction vector, using the given system. Substitute the coordinates \((x_1, x_2)\) into the equations:- For \((2,0)\), \(\frac{dx_{1}}{dt} = 4, \frac{dx_{2}}{dt} = 0\); direction vector is \((4, 0)\).- For \((1.5,1)\), \(\frac{dx_{1}}{dt} = 2, \frac{dx_{2}}{dt} = -1\); direction vector is \((2, -1)\).- For \((1,0)\), \(\frac{dx_{1}}{dt} = 2, \frac{dx_{2}}{dt} = 0\); direction vector is \((2, 0)\).- For \((0, -1)\), \(\frac{dx_{1}}{dt} = 1, \frac{dx_{2}}{dt} = 1\); direction vector is \((1, 1)\).- For \((1,1)\), \(\frac{dx_{1}}{dt} = 1, \frac{dx_{2}}{dt} = -1\); direction vector is \((1, -1)\).- For \((0,0)\), \(\frac{dx_{1}}{dt} = 0, \frac{dx_{2}}{dt} = 0\); direction vector is \((0, 0)\).- For \((-2,-2)\), \(\frac{dx_{1}}{dt} = -4, \frac{dx_{2}}{dt} = 2\); direction vector is \((-4, 2)\).

Step 3: Graph the Direction Vectors

Draw the \(x_1-x_2\) plane and, at each specified point, plot the direction vectors calculated in the previous step. The direction of each vector shows the instantaneous rate of change of \(x_1\) and \(x_2\) at that point.

Key Concepts

Direction Vectors

Direction vectors are essential in understanding the behavior of solutions in a system of differential equations. When you have a system with two variables, like our example with \(x_1\) and \(x_2\), the direction vectors indicate the rate and direction of change at any given point in the \(x_1\)-\(x_2\) plane.

Imagine you are observing a trajectory. At any snapshot, the direction vector shows how quickly and in what direction the trajectory is moving. In simpler terms, if you stand at a point in this plane, the direction vector tells you which way to head next.

• For a point like \((2,0)\), the direction vector is \((4, 0)\), meaning move 4 units along the \(x_1\) axis and none along the \(x_2\) axis.
• For \((-2,-2)\), the direction vector \((-4, 2)\) indicates moving backwards 4 units on \(x_1\) and up 2 units on \(x_2\).

By plotting these vectors, we visualize how the solution behaves around each point. They provide vital geometric insight into the dynamics of the system.

System of Equations

A system of equations is a set of equations with multiple variables. In differential equations, we often deal with how these variables change with respect to another variable, typically time. Consider the system given:\[\begin{align*}\frac{dx_1}{dt} &= 2x_1 - x_2 \\frac{dx_2}{dt} &= -x_2\end{align*}\]Let's break this down:

• \(\frac{dx_1}{dt} = 2x_1 - x_2\) tells us that the rate of change of \(x_1\) is influenced by twice its current value minus \(x_2\).
• \(\frac{dx_2}{dt} = -x_2\) simplifies things as it depends only on \(x_2\).

In tasking with determining direction vectors, we substitute specific \((x_1, x_2)\) values into these equations. This substitution gives us the gradient of the change at that particular point. Understanding these relationships is key; it explains why systems of differential equations are so powerful. They help us model complex phenomena where variables continuously evolve over time.

Gradient Calculation

Gradient calculation in the context of differential equations involves determining the directional change rate at a point. For a system like ours, the gradient at any point is actually the direction vector at that point.
Here’s how it works: By plugging in the coordinates of a point, \((x_1, x_2)\), into the differential equations, we calculate the derivatives \(\frac{dx_1}{dt}\) and \(\frac{dx_2}{dt}\). These values together form the direction vector.

• At point \((1.5, 1)\), compute \(\frac{dx_1}{dt}\) as \(2\) and \(\frac{dx_2}{dt}\) as \(-1\), forming a direction vector of \((2, -1)\).
• For \((0, 0)\), both derivatives are zero, resulting in a stationary vector \((0, 0)\).

The gradient tells us not only how fast a point is changing but also in what direction. This is why it's crucial for drawing phase portraits and understanding solution behavior. Gradient calculations give us a snapshot of dynamic systems in a quantified form, allowing us to predict future states based on current conditions.