Problem 14 - Calculus Early Transcendentals

Question

Find the domain of the following functions. $$f(x, y)=\frac{1}{\sqrt{x^{2}+y^{2}-25}}.$$

Step-by-Step Solution

Answer

Answer: The domain of the function is the set of all points (x,y) outside the circle with a radius of 5 centered at the origin, denoted as \((x,y):\ x^2+y^2 > 25\).

Step 1: Set the denominator equal to zero and solve for x and y

In this case, we need to ensure the denominator of the function is not equal to zero. Set the denominator equal to zero and solve for x and y: $$\sqrt{x^{2}+y^{2}-25}=0$$

Step 2: Square both sides of the equation

We can square both sides of the equation to get rid of the square root: $$(x^{2}+y^{2}-25)=0^2$$ Which simplifies to: $$x^{2}+y^{2}=25$$

Step 3: Analyze the expression and determine the domain

The expression \(x^{2}+y^{2}=25\) represents a circle with a radius of 5, centered at the origin. Therefore, we need to find the domain such that the denominator is not equal to zero. In other words, we are looking for values of x and y that don't satisfy the equation \(x^{2}+y^{2}=25\). We want the expression inside the square root to be positive, or: $$x^{2}+y^{2}>25$$

Step 4: Write the domain

The domain of the function is the set of all ordered pairs of real numbers (x,y) that satisfy the inequality \(x^{2}+y^{2}>25\). In terms of interval notation, this can be written as: $$(x,y):\ x^2+y^2 > 25$$ So, the domain of the function is the set of all points outside of the circle with a radius of 5 centered at the origin.

Key Concepts

Function Domain

In multivariable calculus, the concept of a function's domain can become more intricate compared to single variable functions. The domain of a function refers to the set of all possible inputs, or ordered pairs

• For a function like \( f(x, y) = \frac{1}{\sqrt{x^2 + y^2 - 25}} \), the domain is determined by the constraints that prevent calculation errors.
• This function requires the denominator \( \sqrt{x^2 + y^2 - 25} \) to be non-zero and real. Squaring the expression leads to \( x^2 + y^2 > 25 \).

This inequality indicates that the domain consists of all \( (x, y) \) pairs where the expression inside the square root is positive, ensuring the function is defined.In simpler terms, we are looking for points that lie in the plane but are outside a circle centered at the origin. It also emphasizes the importance of considering what values will keep a function defined and calculable.

Inequalities

Inequalities are a critical tool in determining the domain of functions. They help define the regions where the function's expression holds true. Here’s how they work for our given function:

• We start with the condition \( x^2 + y^2 > 25 \). This inequality ensures that the expression under the square root remains positive.
• Inequalities like this depict regions in a coordinate plane. For this function, \( x^2 + y^2 > 25 \) describes all the points outside a circle that would otherwise make the denominator zero.
• Working with inequalities can involve solving them, graphically understanding them, or both.

Understanding inequalities involves being aware of constraints on variables, which aid in understanding how to keep the expressions valid when using functions in multivariable calculus.

Circle Equation

The circle equation plays a substantial role in this problem, as the expression \( x^2 + y^2 = 25 \) is an equation that represents a circle:

• This circle has a radius of 5, and it is centered at the origin \((0, 0)\).
• In the context of the function \( f(x, y) \), this equation reveals where the function's denominator becomes zero.

It is important to grasp that this circle's equation serves as a boundary. Typically, in multivariable calculus, such boundaries form edges of what needs to be excluded to determine the domain. Given the equation \( x^2 + y^2 = 25 \), our task is to find the domain by excluding the points on this circle and including points where \( x^2 + y^2 > 25 \), which allows the function to remain properly defined.