Problem 28 - The Calculus with Analytic Geometry

Question

In Exercises 27 and 28 , a function \(f\) and a function \(g\) are defined. Find \(h(x, y)\) if \(h=f \circ g\), and also find the domain of \(h\). \(f(t)=\tan ^{-1} t ; g(x, y)=\sqrt{x^{2}-y^{2}}\)

Step-by-Step Solution

Answer

\(h(x, y) = \tan^{-1}(\sqrt{x^2 - y^2})\) with domain \(|x| \geq |y|\).

Step 1: Understand the functions given

Here, two functions are provided: function \(f(t) = \tan^{-1}(t)\) and function \(g(x, y) = \sqrt{x^2 - y^2}\). The task is to find the composite function \(h(x, y) = (f \circ g)(x, y)\), which means \(h(x, y) = f(g(x, y))\).

Step 2: Compute \(g(x, y)\)

Evaluate the inside function \(g(x, y) = \sqrt{x^2 - y^2}\). This function requires taking the square root of the difference between the squares of \(x\) and \(y\). Ensure the expression under the square root is non-negative for the function to be real.

Step 3: Determine the composite function \(h(x, y)\)

Substitute \(g(x, y)\) into \(f(t)\). Thus, \(h(x, y) = f(\sqrt{x^2 - y^2}) = \tan^{-1}(\sqrt{x^2 - y^2})\).

Step 4: Find the domain of \(g(x, y)\)

The inside of the square root, \(x^2 - y^2\), must be non-negative. Therefore, \(x^2 - y^2 \geq 0\).

Step 5: Form the domain inequality

Rewrite the inequality: \(x^2 \geq y^2\). This can be expressed as \(|x| \geq |y|\), meaning \(x\) can be greater than or equal to \(y\) or \(x\) is less than or equal to the negative of \(y\).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are functions that reverse the roles of the input and output of the regular trigonometric functions. In this exercise, we deal with the inverse tangent function, denoted as \( \tan^{-1}(x) \) or \( \arctan(x) \). This function gives us the angle whose tangent is the number \( x \).
For instance, if \( \tan(\theta) = x \), then \( \theta = \arctan(x) \). The range of \( \tan^{-1} \) is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), meaning it can return angles only within this interval.
The function \( f(t) = \tan^{-1}(t) \) in this exercise is used to determine the composite function \( h(x, y) = f(g(x, y)) \). Therefore, understanding how \( \tan^{-1} \) works is crucial.

Domain of Functions

The domain of a function is the set of all possible inputs for which the function is defined. Determining the domain is crucial, especially when dealing with composite functions. In this exercise, we need to find the domain of the composite function \( h(x, y) \).
First, we understand the domain of the inside function \( g(x, y) = \sqrt{x^2 - y^2} \). The expression inside the square root, \( x^2 - y^2 \), must be non-negative so that the square root is real. This leads to the inequality \( x^2 - y^2 \geq 0 \), which can be read as \( x^2 \geq y^2 \) or \( |x| \geq |y| \).
This means that the valid pairs \( (x, y) \) are those where the absolute value of \( x \) is greater than or equal to the absolute value of \( y \). By finding the domain of \( g(x, y) \), we indirectly find the domain for the composite function \( h(x, y) \), ensuring the inside of the square root is always non-negative.

Square Root Properties

Square root functions take a non-negative number and return its principal (non-negative) square root. One important property is that the expression inside the square root must be non-negative for the function to produce a real number.
For the exercise given, where \( g(x, y) = \sqrt{x^2 - y^2} \), the term \( x^2 - y^2 \) must be non-negative. This is because the square root of a negative number is not a real number, and we are dealing with real functions in this context.
When computing \( h(x, y) = \tan^{-1}(\sqrt{x^2 - y^2}) \), the function \( g(x, y) \) ensures that the input to the inverse tangent function is a real number by confirming that \( x^2 - y^2 \geq 0 \). This property simplifies our analysis and helps to accurately determine the domain for \( h(x, y) \).