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Find the domain of the following functions. $$f(x, y)=\frac{12}{y^{2}-x^{2}}$$

Short Answer

Expert verified
Question: Determine the domain of the function f(x, y) = 1 / (y^2 - x^2). Answer: The domain of the function f(x, y) is all (x, y) pairs for which y^2 - x^2 ≠ 0. In other words, the domain includes all real number pairs (x, y) except those where y = x or y = -x. Mathematically, the domain can be described as: $$\{(x, y) \in \mathbb{R}^2 | y \neq x, y \neq -x\}$$

Step by step solution

01

Find when the denominator is equal to zero

Let's set the denominator of the function equal to zero and solve for x and y: $$y^{2}-x^{2}=0$$
02

Factor the equation

Factoring the equation, we get: $$(y-x)(y+x)=0$$
03

Solve for x and y

From the factored equation, we have two cases: 1. $$y-x=0 \implies y=x$$ 2. $$y+x=0 \implies y=-x$$
04

Identify the domain of the function

The function is not defined when the denominator is zero, which occurs if any of the two cases are true. Therefore, the domain of the function includes all (x, y) pairs except those where y=x or y=-x. So, the domain of the function f(x, y) is: $$\{(x, y) \in \mathbb{R}^2 | y \neq x, y \neq -x\}$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Multivariable Calculus
Multivariable calculus extends the concepts of single-variable calculus to functions with multiple inputs. This means we are dealing with functions that have more than one independent variable. For example, the function \( f(x, y) \), from the exercise, depends on both \( x \) and \( y \) variables. In multivariable calculus, one of the fundamental aspects is determining the domain, which is the set of all possible input values (in this case, pairs of numbers) that the function can take without causing any issues, like division by zero.

When finding the domain of a multivariable function, it's important to consider the constraints on each variable and how they interact. The interactions between variables can create regions or lines in the two-dimensional plane (or hyperplanes in higher dimensions) where the function cannot be evaluated. This detail is illustrated in the provided exercise, where the domain is restricted because of the denominator \( y^2 - x^2 \).
Factoring Equations in Multivariable Functions
Factoring equations plays a critical role when dealing with the domains of multivariable functions. Factoring is a method of writing an expression as a product of its factors, often simplifying the process of solving equations. It can reveal hidden relationships between variables that aren't immediately apparent. In the exercise, the equation \( y^2 - x^2 = 0 \) was factored into \( (y - x)(y + x) = 0 \), demonstrating that there are two distinct situations to consider. This factoring helps us understand the relationship between \( x \) and \( y \) that leads to the function being undefined.

Understanding how to factor and solve these types of equations is pivotal in multivariable calculus. It allows us to break down complex problems into simpler parts, making it easier to analyze the function and its domain. Factoring is not just a mechanical process but a critical thinking tool that aids in deciphering the intricacies of multivariable relationships.
Solving for Variables and Determining Their Relationships
Solving for variables in a multivariable context requires considering how changes in one variable affect another. In our exercise's Step 3, we see two cases derived from the factored equation: \( y - x = 0 \) and \( y + x = 0 \). These represent the specific instances where the function \( f(x, y) \) fails to be defined. Solving these equations gives us the relationships \( y = x \) and \( y = -x \), respectively, which are essentially the representations of lines where the function cannot be evaluated.

It is through solving for variables that we identify these critical lines and thus are able to characterize the domain accurately. A clear understanding of this process enables us to visualize and describe the behavior of multivariable functions within their domains. Such conceptual clarity is vital for success in multivariable calculus, as it forms the basis for more advanced topics such as gradients, contour plots, and optimization problems. Recognizing how variables interact and influence each other is a skill that extends far beyond the scope of mathematics and into real-world problem-solving.

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Most popular questions from this chapter

Use Lagrange multipliers in the following problems. When the constraint curve is unbounded, explain why you have found an absolute maximum or minimum value. Extreme distances to an ellipse Find the minimum and maximum distances between the ellipse \(x^{2}+x y+2 y^{2}=1\) and the origin.

Potential functions Potential functions arise frequently in physics and engineering. A potential function has the property that a field of interest (for example, an electric field, a gravitational field, or a velocity field) is the gradient of the potential (or sometimes the negative of the gradient of the potential). (Potential functions are considered in depth in Chapter \(17 .)\) The electric field due to a point charge of strength \(Q\) at the origin has a potential function \(\varphi=k Q / r,\) where \(r^{2}=x^{2}+y^{2}+z^{2}\) is the square of the distance between a variable point \(P(x, y, z)\) and the charge, and \(k>0\) is a physical constant. The electric field is given by \(\mathbf{E}=-\nabla \varphi,\) where \(\nabla \varphi\) is the gradient in three dimensions. a. Show that the three-dimensional electric field due to a point charge is given by $$\mathbf{E}(x, y, z)=k Q\left\langle\frac{x}{r^{3}}, \frac{y}{r^{3}}, \frac{z}{r^{3}}\right\rangle$$ b. Show that the electric field at a point has a magnitude \(|\mathbf{E}|=\frac{k Q}{r^{2}} .\) Explain why this relationship is called an inverse square law.

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Graph several level curves of the following functions using the given window. Label at least two level curves with their z-values. $$z=x^{2}+y^{2} ;[-4,4] \times[-4,4]$$

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