Problem 14 - APEX Calculus

Question

(a) Find the domain \(D\) of the given function. (b) State whether \(D\) is an open or closed set. (c) State whether \(D\) is bounded or unbounded. $$ f(x, y)=\frac{x^{2}-y^{2}}{x^{2}+y^{2}} $$

Step-by-Step Solution

Answer

(a) Domain: \( \mathbb{R}^2 \setminus \{(0,0)\} \). (b) \( D \) is open. (c) \( D \) is unbounded.

Step 1: Identify the Domain

The function \( f(x, y) = \frac{x^2 - y^2}{x^2 + y^2} \) is defined wherever the denominator is not zero. Thus, we need to find where \( x^2 + y^2 eq 0 \). The expression \( x^2 + y^2 = 0 \) would only hold if both \( x \) and \( y \) are zero simultaneously. Therefore, the function is defined for all \( (x, y) \) except at the point \( (0, 0) \). Hence, the domain \( D \) is \( \mathbb{R}^2 \setminus \{(0,0)\} \).

Step 2: Determine if the Domain is Open or Closed

A set is considered open if it does not include its boundary points, and closed if it includes them. The domain \( D = \mathbb{R}^2 \setminus \{(0,0)\} \) excludes the point \( (0,0) \). Since the point \( (0, 0) \) is not part of \( D \), it does not contain all its boundary points. Therefore, \( D \) is an open set.

Step 3: Determine if the Domain is Bounded or Unbounded

A set is bounded if it can fit within some ball of finite radius. The domain \( D = \mathbb{R}^2 \setminus \{(0,0)\} \) essentially includes all points in \( \mathbb{R}^2 \) except one point. Since \( \mathbb{R}^2 \) is unbounded, \( D \) is also unbounded.

Key Concepts

Open and Closed Sets

The concept of open and closed sets is foundational in understanding domains in mathematics, particularly in topology.
An **open set** is a set that does not include its boundary points. This means, for any point in the set, you can find a small "neighborhood" around it that is still entirely contained within the set. In simpler terms, you can move a little in any direction from the point and still remain within the set. For our function's domain, which is all points except the origin, this means you are never right on the edge of the actual set.
An **closed set**, on the other hand, includes its boundary points. If you can get to a point simply by being in the set or arriving at a point of closure while still on the set, you have a closed set. Since our domain excludes a specific boundary point, it is indeed an open set because the edge point is not part of it.
These ideas are important not only in theoretical math but also in understanding functions and their domains when designing real-world systems or analyzing data.

Bounded and Unbounded Sets

When talking about sets in the realm of mathematics, gatherings can be described as either bounded or unbounded.
A **bounded set** is one that can fit completely within a finite "ball" or "sphere." Imagine drawing a circle around the entire set; if you can do this without the set spilling outside, it is bounded. Boundedness implies there is a limit or a maximum size to the area or volume that the set occupies.

• For example, the set of all points within a circle of radius 5 is bounded.

An **unbounded set**, conversely, does not have this limitation. It stretches out indefinitely, much like the entire plane of \(\mathbb{R}^2\) or in our situation, the domain of the multivariable function except the origin.
This domain essentially fills up the entire plane except one tiny point, meaning you cannot encompass it within any finite radius. Hence, it is unbounded.
Understanding boundedness helps visualize whether we are dealing with a finite or infinite potential set of values, which can be crucial for practical applications and analytical modeling.

Multivariable Functions

Multivariable functions are an extension of single-variable functions, where instead of having just one input, there are multiple inputs.
For our function, \(f(x, y) = \frac{x^2 - y^2}{x^2 + y^2}\), we have two inputs: \(x\) and \(y\). Each pair \((x, y)\) gives a unique outcome, making the function more dynamic and representative of real-world scenarios where factors seldom exist in isolation.
Understanding the domain of such a function includes determining where it is defined, which is why we examine the condition \(x^2 + y^2 ot= 0\) to avoid division by zero.

• Multivariable functions can model complicated systems like temperature over a geographical area or air pressure in a weather system.

Being comfortable with multivariable functions allows you to interpret complex systems by examining how different factors (or variables) interact with each other. This type of analysis is particularly useful in fields like physics, engineering, and economics, where systems often depend on several changing variables.