Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 27-32, describe geometrically the domain of each of the indicated functions of three variables. $$ f(x, y, z)=\sqrt{144-16 x^{2}-9 y^{2}-144 z^{2}} $$

Step-by-Step Solution

Answer

The domain is an ellipsoid centered at the origin, defined by \(\frac{x^2}{9} + \frac{y^2}{16} + z^2 \leq 1\).

Step 1: Understand the Function

The function given is \( f(x, y, z) = \sqrt{144 - 16x^2 - 9y^2 - 144z^2} \). This implies the expression inside the square root must be non-negative for the function to be defined in the real numbers.

Step 2: Set the Inequality

Since the expression inside the square root must be non-negative, we set up the inequality:\[ 144 - 16x^2 - 9y^2 - 144z^2 \geq 0. \]

Step 3: Rearrange the Inequality

Rearrange the inequality to express it in a standard form:\[ 16x^2 + 9y^2 + 144z^2 \leq 144. \]

Step 4: Divide to Simplify

Divide every term of the inequality by 144 to simplify it:\[ \frac{16x^2}{144} + \frac{9y^2}{144} + \frac{144z^2}{144} \leq 1. \] This further simplifies to: \[ \frac{x^2}{9} + \frac{y^2}{16} + z^2 \leq 1. \]

Step 5: Identify the Geometric Shape

The inequality \( \frac{x^2}{9} + \frac{y^2}{16} + z^2 \leq 1 \) represents an ellipsoid in 3D space, centered at the origin, with semi-axis lengths of 3, 4, and 1 along the x, y, and z axes respectively.

Key Concepts

Domain of a Function

The **domain** of a function is the collection of all possible input values that allow the function to be expressed with real numbers.
For the function provided in the exercise, \( f(x, y, z) = \sqrt{144 - 16x^2 - 9y^2 - 144z^2} \), the key is to ensure that the quantity inside the square root is non-negative. This is because the square root of a negative number is not real.
To find the domain, we work through some mathematics:

• We set up the inequality: \( 144 - 16x^2 - 9y^2 - 144z^2 \geq 0 \).
• By re-arranging terms, it becomes \( 16x^2 + 9y^2 + 144z^2 \leq 144 \).
• Simplifying further gives the standard form \( \frac{x^2}{9} + \frac{y^2}{16} + z^2 \leq 1 \).

Thus, the domain consists of all points \((x, y, z)\) that satisfy this inequality, essentially forming an ellipsoid, where any point inside this shape can serve as potential values for \((x, y, z)\).

Geometric Interpretation

Understanding the *geometric interpretation* of the domain helps bridge mathematics with real-world visuals.
In this example, the inequality \( \frac{x^2}{9} + \frac{y^2}{16} + z^2 \leq 1 \) is not just a set of numbers; it outlines an **ellipsoid** in a three-dimensional space:

• The ellipsoid is centered at the origin \((0,0,0)\).
• The semi-axis lengths are 3, 4, and 1 respectively along the x, y, and z axes, derived from the denominators and constant value in the inequality.

Think of this shape as a 'stretched' sphere. Along the x-axis, it stretches out to 3, along the y-axis to 4, and along the z-axis it is more squeezed, extending only to 1.
The spatial understanding of the function's domain aids in visualizing not just where the function is defined but also how changes in variables affect its scope.

Functions of Three Variables

A *function of three variables* like \( f(x, y, z) = \sqrt{144 - 16x^2 - 9y^2 - 144z^2} \) takes in three pieces of information or inputs.
This broadens the capability of explorations and calculations beyond functions of a single or two variables.
In a practical sense, each variable could represent a dimension:

• \(x\), \(y\), and \(z\) might correspond to physical quantities such as length, width, and height, or even abstract properties like time, speed, and temperature.
• The expression provides a way to investigate how these three factors interact within the boundaries defined by the domain.

Because the given function involves a square root, consideration of its domain becomes crucial, ensuring meaningful interactions under real conditions.
Overall, studying such functions opens the door to an enhanced comprehension of multi-variable systems, pivotal in fields like physics, engineering, and beyond.