Problem 33 - Calculus Early Transcendentals

Question

\(29-38\) Determine the set of points at which the function is continuous. $$G(x, y)=\ln \left(x^{2}+y^{2}-4\right)$$

Step-by-Step Solution

Answer

The function is continuous for \((x, y)\) such that \(x^2 + y^2 > 4\).

Step 1: Understand the Domain of the Logarithmic Function

The logarithmic function \(\ln(z)\) is defined only when \(z > 0\). For the given function \(G(x, y) = \ln(x^2 + y^2 - 4)\), we need the argument of the logarithm, \(x^2 + y^2 - 4\), to be positive. Thus, we require： \\[ x^2 + y^2 - 4 > 0 \]

Step 2: Solve the Inequality for the Domain

To find where \(x^2 + y^2 - 4 > 0\), rewrite the inequality as:\[ x^2 + y^2 > 4 \]This is the equation of a circle with center at the origin and radius 2, but the inequality denotes the region outside the circle (not including the boundary).

Step 3: Identify the Set for Continuity

The function \(G(x, y)\) is continuous everywhere in its domain. The domain consists of all points \((x, y)\) such that \[ x^2 + y^2 > 4 \] This is the exterior of the circle with radius 2 centered at the origin. Thus, \(G(x, y)\) is continuous at all points \((x, y)\) satisfying this condition.

Key Concepts

Domain of Logarithmic Functions

Understanding the domain of a logarithmic function is essential since it determines where the function is defined and operative. In any logarithmic expression like \( \ln(z) \), the argument \( z \) must be greater than zero. This is because logarithms of non-positive values are undefined in real numbers.
In our problem, the function is given as \( G(x, y) = \ln(x^2 + y^2 - 4) \). Here, the expression inside the logarithm, \( x^2 + y^2 - 4 \), must satisfy \( x^2 + y^2 - 4 > 0 \) for the function \( G(x, y) \) to be valid.
If you're ever unsure about the domain of a logarithmic function, remember these key steps:

• Set the expression inside the logarithm greater than zero.
• Solve this inequality to find the valid range of inputs for the function.

In this exercise, solving \( x^2 + y^2 - 4 > 0 \) helps us identify the domain where the function is continuous.

Inequality Solving

Inequality solving is a fundamental skill in mathematics, essential in determining ranges of variables. Inequalities often describe a set of solutions instead of a specific answer.
In the function \( G(x, y) = \ln(x^2 + y^2 - 4) \), we solve the inequality \( x^2 + y^2 - 4 > 0 \) to find where the function is defined.
Here's how to approach solving such inequalities:

• Rearrange the inequality to isolate terms involving the variables. For our problem, it becomes \( x^2 + y^2 > 4 \).
• Identify what the inequality represents. In this case, it's a circle's equation, \( x^2 + y^2 = 4 \), and the inequality \( x^2 + y^2 > 4 \) represents the region outside the circle.

Understanding how to solve and interpret inequalities helps define the domain and understand the continuity of mathematical functions.

Circle in the Coordinate Plane

A circle's equation in the coordinate plane is centered at point \((h, k)\) with radius \( r \) is generally given by the formula \((x - h)^2 + (y - k)^2 = r^2\).
In this problem, the circle is described by \( x^2 + y^2 = 4 \), which can be rewritten using our formula to identify the center \((h, k)\) as \((0,0)\) and the radius \( r \) as 2.
This circle in the exercise is centered at the origin with a radius of 2. The inequality \( x^2 + y^2 > 4 \) represents the set of points outside this circle.

• Remember, the boundary \( x^2 + y^2 = 4 \) itself is not included in the solution set since the inequality strictly excludes it (\( > \)).
• Points that are at any position on the plane but outside this boundary satisfy the inequality.

Comprehending how circles are represented on the coordinate plane supports your understanding of geometric regions linked to mathematical functions.