Problem 27 - 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 \(h\). \(f(t)=\sin ^{-1} t ; g(x, y)=\sqrt{1-x^{2}-y^{2}}\)

Step-by-Step Solution

Answer

The composite function is \(h(x, y) = \sin^{-1}(\sqrt{1 - x^2 - y^2})\). The domain of \(h\) is \(x^2 + y^2 \leq 1\).

Step 1: Define the composition function

First, understand that the composition function is defined as \[h = f \circ g\]. This means that \(h(x, y) = f(g(x, y))\). Substitute the given functions: \(f(t) = \sin^{-1} t\) and \(g(x, y) = \sqrt{1 - x^2 - y^2}\). Thus, we have \(h(x, y) = \sin^{-1} (\sqrt{1 - x^2 - y^2})\).

Step 2: Simplify the composition function

Substitute the expression for \(g(x, y)\) into \(f(t)\): \ \(h(x, y) = \sin^{-1}(\sqrt{1 - x^2 - y^2})\). This is a simplification of the composite function \(h(x, y)\).

Step 3: Determine domain constraints for \(g(x,y)\)

For \(g(x, y) = \sqrt{1 - x^2 - y^2}\) to be defined, the expression under the square root must be non-negative. Therefore, \ \[1 - x^2 - y^2 \geq 0\]. This implies that \(x^2 + y^2 \leq 1\). Thus, \(x\) and \(y\) must lie within or on the boundary of the unit circle.

Step 4: Analyze range constraints for the inverse sine function

The range of the inverse sine function, \(\sin^{-1} t\), is \([-\frac{\pi}{2}, \frac{\pi}{2}]\). Since \(\sqrt{1 - x^2 - y^2}\) produces values between 0 and 1, this fits within the acceptable range for \(\sin^{-1} t\).

Step 5: Combine domain constraints

The final domain of \(h(x, y)\) is the set of all \((x, y)\) such that \(x^2 + y^2 \leq 1\). Therefore, \ the domain of \(h\) is the unit disk in the \(xy\) plane.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions help us find angles when we know the trigonometric values. For the sine function, its inverse is written as \(\sin^{-1} t\). This is called arcsine and it tells us what angle corresponds to a given sine value. The range of \(\sin^{-1} t\) is \[ -\frac{\pi}{2}, \frac{\pi}{2} \], meaning it only returns values within these limits.

Function Domain

The domain of a function is the set of input values (x, y) that are allowed. For the function \( g(x, y) = \sqrt{1 - x^2 - y^2} \), the domain requires the expression inside the square root to be non-negative. This means \[ 1 - x^2 - y^2 \geq 0 \], or \[ x^2 + y^2 \leq 1 \]. So, \( g(x, y) \) is defined only within or on the boundary of the unit circle. In our example, we need to consider this constraint when finding the domain of the composite function \( h(x, y) = \sin^{-1}(\frac{1 - x^2 - y^2}) \). Both the domains of \( f(t) \) and \( g(x, y) \) are necessary to determine the final domain of the composite function.

Unit Circle

The unit circle is a circle with a radius of 1 centered at the origin of a coordinate plane. Formally, it is represented by the equation \( x^2 + y^2 = 1 \). When we say \( x \) and \( y \) must lie within or on the boundary of the unit circle, we mean they must satisfy \[ x^2 + y^2 \leq 1 \]. This ensures \( g(x, y) \) is valid. Places within this unit circle provide the inputs for the composite function to work correctly.