Problem 66 - Calculus Early Transcendentals

Question

Find an equation for the family of level surfaces pe corresponding to \(f .\) Describe the level surfaces.$$f(x, y, z)=\frac{1}{x^{2}+y^{2}+z^{2}}.$$

Step-by-Step Solution

Answer

The equation for the family of level surfaces corresponding to the given function is $$c(x^2+y^2+z^2) - 1 = 0$$ or $$x^2+y^2+z^2 = \frac{1}{c}.$$ These level surfaces are spheres centered at the origin with radius equal to $$\sqrt{\frac{1}{c}}.$$

Step 1: Find the equation for the level surfaces

We need to find the function $$g(x, y, z, c) = 0$$ such that $$f(x, y, z) = c \Rightarrow g(x, y, z, c) = 0.$$ The function we start with is $$f(x, y, z) = \frac{1}{x^2+y^2+z^2}.$$ Let's set this equal to \(c\): $$ f(x, y, z) = c = \frac{1}{x^2+y^2+z^2}. $$

Step 2: Rewrite the equation

We want the level surface equation to be in the form $$g(x, y, z, c) = 0,$$ so we need to manipulate the equation from the previous step. We start by multiplying both sides by the denominator $$x^2+y^2+z^2,$$ and we get: $$ c(x^2+y^2+z^2) = 1. $$ Now, we subtract 1 from both sides to get: $$ g(x, y, z, c) = c(x^2+y^2+z^2) - 1 = 0. $$

Step 3: Describe the level surfaces

The level surfaces have the equation: $$ c(x^2+y^2+z^2) - 1 = 0. $$ We can rewrite this equation as: $$ x^2+y^2+z^2 = \frac{1}{c}. $$ This equation is the equation for spheres centered at the origin with radius equal to the square root of the reciprocal of \(c\). So, the level surfaces corresponding to the given function are spheres centered at the origin with radius $$\sqrt{\frac{1}{c}}.$$

Key Concepts

Multivariable Calculus

Multivariable calculus is an extension of calculus that involves multiple variables. Instead of dealing with functions of a single variable, like the familiar ones in basic calculus, it considers functions of two or more variables. For example, in our exercise, the function \(f(x, y, z)\) depends on three variables—\(x\), \(y\), and \(z\).

This branch of calculus allows us to examine how these variables change together and interact. We analyze using concepts like partial derivatives, gradients, and level surfaces. Level surfaces represent visual ways to understand how variables interplay in space. These are useful in fields such as physics and engineering, where systems are often described with multiple dimensions.

In our exercise, discovering a set of level surfaces is key. These surfaces help in visualizing how a function like \(f(x, y, z) = \frac{1}{x^2 + y^2 + z^2}\) behaves when plotted in three-dimensional space. By setting this function equal to a constant \(c\), we reveal a series of related geometrical shapes.

Equation of Spheres

The equation of a sphere in three-dimensional Cartesian coordinates is typically represented as \(x^2 + y^2 + z^2 = r^2\). Here, \(r\) stands for the radius of the sphere.

In our particular exercise, the level surfaces of \(f(x, y, z) = \frac{1}{x^2 + y^2 + z^2}\) form spheres. By manipulating the equation \(f(x, y, z) = c\), where \(c\) is a constant, we reach the form \(x^2+y^2+z^2=\frac{1}{c}\), which fits the general sphere equation with radius \(r = \sqrt{\frac{1}{c}}\).

These spheres are centered at the origin \((0, 0, 0)\), as there are no linear terms in \(x\), \(y\), or \(z\) to imply a shift from the center. The radius of these spheres changes inversely with \(c\); as \(c\) increases, the spheres get smaller, and vice versa. This relationship highlights the flexibility of spheres in describing spatial relationships in multivariable calculus.

Scalar Field

In mathematics, a scalar field assigns a single scalar value to every point in space. Unlike vectors, which have both magnitude and direction, scalars are single numbers. They are particularly helpful in describing temperature, pressure, or gravitational potential at every point in space.

In this exercise, the function \(f(x, y, z) = \frac{1}{x^2 + y^2 + z^2}\) acts as a scalar field over three-dimensional space. Here, \(f\) assigns a single value to each point \((x, y, z)\), and this value decreases as you move further from the origin, representing 'potential' or 'density' diminishing with distance.

Analyzing level surfaces within a scalar field, like in our exercise, offers a clear method to visualize areas of constant value. These levels show where in space the function remains unchanged, providing insight into how the scalar field behaves over different areas of the coordinate space. Scalar fields are crucial in many scientific disciplines for visualizing various physical phenomena without needing vectors.