Problem 9 - Calculus

Question

Given the system of differential equations $$\begin{aligned} &\frac{d x}{d t}=5 x-3 x y\\\ &\frac{d y}{d t}=-8 y+x y \end{aligned}$$ determine whether \(x\) and \(y\) are increasing or decreasing at the point (a) \(\quad x=3, y=2\) (b) \(x=5, y=1\)

Step-by-Step Solution

Answer

At \((3,2)\), both \(x\) and \(y\) are decreasing; at \((5,1)\), \(x\) is increasing and \(y\) is decreasing.

Step 1: Evaluate dx/dt for Point (a)

Substitute the given values \(x = 3\) and \(y = 2\) into the equation \(\frac{d x}{d t}=5 x-3 x y\). We have:\[ \frac{d x}{d t} = 5(3) - 3(3)(2) = 15 - 18 = -3 \]Since \(\frac{d x}{d t} = -3\), the value of \(x\) is decreasing at this point.

Step 2: Evaluate dy/dt for Point (a)

Substitute the same values \(x = 3\) and \(y = 2\) into the equation \(\frac{d y}{d t}=-8 y+x y\). We have:\[ \frac{d y}{d t} = -8(2) + 3(2) = -16 + 6 = -10 \]Since \(\frac{d y}{d t} = -10\), the value of \(y\) is also decreasing at this point.

Step 3: Evaluate dx/dt for Point (b)

Substitute the given values \(x = 5\) and \(y = 1\) into the equation \(\frac{d x}{d t}=5 x-3 x y\). We have:\[ \frac{d x}{d t} = 5(5) - 3(5)(1) = 25 - 15 = 10 \]Since \(\frac{d x}{d t} = 10\), the value of \(x\) is increasing at this point.

Step 4: Evaluate dy/dt for Point (b)

Substitute the same values \(x = 5\) and \(y = 1\) into the equation \(\frac{d y}{d t}=-8 y+x y\). We have:\[ \frac{d y}{d t} = -8(1) + 5(1) = -8 + 5 = -3 \]Since \(\frac{d y}{d t} = -3\), the value of \(y\) is decreasing at this point.

Key Concepts

System of Differential Equations

A system of differential equations is a set of two or more differential equations. These equations involve two or more unknown functions and their derivatives. Systems of differential equations often model real-world phenomena where multiple variables interact and change over time.
For example, in our given system:

• \(\frac{d x}{dt} = 5x - 3xy\)
• \(\frac{d y}{dt} = -8y + xy\)

Both equations describe how the variables \(x\) and \(y\) change with respect to time \(t\). This system can represent situations like population models, chemical reactions, or mechanical systems. Our goal is to find how \(x\) and \(y\) evolve together by analyzing these equations.

Calculus

Calculus is a branch of mathematics that studies how things change. It includes concepts like derivatives and integrals. Derivatives measure the rate at which a quantity changes. They are pivotal in understanding systems of differential equations, as they reveal the rates of change of the variables.
In our exercise, we specifically look at:

• The derivative \(\frac{dx}{dt}\), representing how \(x\) changes over time in relation to \(y\).
• The derivative \(\frac{dy}{dt}\), showing how \(y\) evolves with \(x\).

Using calculus, we substitute specific points into these derivatives to investigate the behavior of the system at those moments. This helps us predict if our system is growing or declining at certain points.

Increasing and Decreasing Functions

In mathematics, a function is said to be increasing if, as the input increases, the output also increases. Conversely, a function is decreasing if the output decreases as the input increases.
To determine whether \(x\) and \(y\) are increasing or decreasing in our differential system, we examine the signs of \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) at given points:

• If \(\frac{dx}{dt} > 0\), then \(x\) is increasing.
• If \(\frac{dx}{dt} < 0\), then \(x\) is decreasing.
• If \(\frac{dy}{dt} > 0\), then \(y\) is increasing.
• If \(\frac{dy}{dt} < 0\), then \(y\) is decreasing.

For instance, at the point \((3, 2)\), both derivatives are negative, indicating both \(x\) and \(y\) are decreasing. In contrast, at \((5, 1)\), while \(x\) is increasing, \(y\) continues to decrease. Understanding these concepts helps predict the trajectory of the system over time.