Problem 14 - Calculus Early Transcendentals

Question

Find and sketch the domain of the function. $$f(x, y)=\sqrt{y-x} \ln (y+x)$$

Step-by-Step Solution

Answer

The domain is \( \{(x, y) \,|\, y \geq x, \, y + x > 0\} \), above both lines \( y = x \) and \( y = -x \).

Step 1: Identify Restrictions for the Square Root

The square root function \( \sqrt{y-x} \) is only defined for non-negative values. Therefore, we require that the inside of the square root be greater than or equal to zero. This gives us the inequality:\[ y - x \geq 0 \]which simplifies to \( y \geq x \).

Step 2: Identify Restrictions for the Natural Logarithm

The natural logarithm function \( \ln(y+x) \) is defined for positive arguments only. Thus, we need:\[ y + x > 0 \]This inequality ensures that the logarithm is defined and real.

Step 3: Combine the Restrictions

To find the domain of the function \( f(x, y) \), combine the restrictions from Step 1 and Step 2:1. \( y \geq x \)2. \( y + x > 0 \)The domain of the function consists of all \((x, y)\) pairs that satisfy both conditions.

Step 4: Sketch the Domain

To sketch the domain, use the inequalities:1. The line \( y = x \) represents the boundary where \( y = x \), and the region where \( y \geq x \) lies above this line.2. The line \( y = -x \) represents the region where \( y + x = 0 \), and the area where \( y + x > 0 \) lies above this line.The intersection of these areas is the domain: the region above the line \( y = x \) and above the line \( y = -x \) in the upper right quadrant of a coordinate plane.

Key Concepts

Square Root Function

A square root function involves finding a number that, when multiplied by itself, gives the original value. In mathematical expressions, it appears as \( \sqrt{x} \). Within this context, the square root function \( \sqrt{y-x} \) requires the expression inside to be non-negative. This is because square roots of negative numbers are undefined in the realm of real numbers. As such, we derive the inequality \( y - x \geq 0 \) to ensure the expression remains valid.

By solving \( y - x \geq 0 \), we find that \( y \) must be greater than or equal to \( x \). This inequality lays the foundation for determining the domain where the square root function is real and calculable.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \) is an irrational and transcendental constant approximately equal to 2.718. The logarithmic function is only defined for positive numbers. This is because the logarithm of zero or a negative number is undefined in the real number system.

In our function \( \ln(y+x) \), we require that \( y+x > 0 \) in order for the logarithm to exist. This inequality means that the sum of \( y \) and \( x \) must be greater than zero to maintain a real and defined natural log function. This requirement must be coupled with other conditions in the domain search.

Inequalities in Two Variables

Inequalities in two variables are expressions that establish a region or set of points that satisfy given conditions. For instance, considering our problem, the inequalities \( y \geq x \) and \( y+x > 0 \) represent lines on a coordinate plane.

These inequalities identify specific lines:

• \( y = x \) forms a line where all points are equidistant from the x and y-axis.
• \( y + x = 0 \) represents a line through the origin, dividing the plane into regions where the sum of \( x \) and \( y \) is positive or negative.

Understanding how to manipulate these inequalities helps create visual regions illustrating where a function is defined.

Coordinate Plane

A coordinate plane is a two-dimensional surface where we can plot points, lines, and curves, defined by a pair of numerical coordinates: an x-coordinate and a y-coordinate. This plane is divided by a horizontal x-axis and a vertical y-axis.

In our task, we utilize the coordinate plane to sketch the domain of the function \( f(x, y) = \sqrt{y-x} \ln(y+x) \). By plotting the inequalities \( y \geq x \) and \( y+x > 0 \), we visualize where both conditions overlap. The solution lies in the intersection area of these two conditions, occupying the space in the upper right quadrant of the plane. This intersection effectively outlines where both function constraints are satisfied. Thus, the coordinate plane becomes an essential tool for graphically understanding the function's domain.