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Show that each function in Exercises \(83-90\) satisfies a Laplace equation. $$f(x, y, z)=x^{2}+y^{2}-2 z^{2}$$

Short Answer

Expert verified
The function satisfies the Laplace equation since the sum of second partial derivatives is zero.

Step by step solution

01

Recall the Laplace Equation

The Laplace equation is given by \( abla^2 f = 0 \), where \( abla^2 \) is the Laplacian operator. In three-dimensional Cartesian coordinates, the Laplacian is defined as \( abla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \).
02

Compute the Second Partial Derivative with Respect to x

For the function \( f(x, y, z) = x^2 + y^2 - 2z^2 \), compute the second partial derivative with respect to \( x \): \[ \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x} \right) = \frac{\partial}{\partial x}(2x) = 2. \]
03

Compute the Second Partial Derivative with Respect to y

Compute the second partial derivative with respect to \( y \):\[ \frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right) = \frac{\partial}{\partial y}(2y) = 2. \]
04

Compute the Second Partial Derivative with Respect to z

Compute the second partial derivative with respect to \( z \):\[ \frac{\partial^2 f}{\partial z^2} = \frac{\partial}{\partial z} \left( \frac{\partial f}{\partial z} \right) = \frac{\partial}{\partial z}(-4z) = -4. \]
05

Sum the Second Partial Derivatives

Sum the second partial derivatives obtained in the previous steps:\[ abla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} = 2 + 2 - 4 = 0. \]
06

Conclude that the Function Satisfies the Laplace Equation

Since \( abla^2 f = 0 \), the function \( f(x, y, z) = x^2 + y^2 - 2z^2 \) satisfies the Laplace equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Partial derivatives are at the heart of understanding changes in multivariable functions. Imagine having a function with multiple variables, like our function of interest, \( f(x, y, z) = x^2 + y^2 - 2z^2 \). Each variable represents a dimension in space. When we calculate a partial derivative, we examine how the function changes as we alter just one of these variables, while keeping the others constant.

For example, to find the partial derivative \( \frac{\partial f}{\partial x} \), we differentiate \( f \) with respect to \( x \), treating \( y \) and \( z \) as constants. This is akin to slicing a piece of cake horizontally and checking how its texture varies along the cut.
  • Example: For \( f(x, y, z) = x^2 + y^2 - 2z^2 \), the partial derivative with respect to \( x \) is \( 2x \).
  • Simplicity: Each step ignores changes in the other variables, simplifying the process.
Partial derivatives are essential in finding the rate of change and in applying the Laplacian operator, which is the gateway to understanding the Laplace equation. These derivatives provide a powerful tool to dissect complex functions into manageable pieces.
Laplacian Operator
The Laplacian operator is a pivotal tool in multivariable calculus, particularly in physics and engineering. It is denoted by \( abla^2 \) and is a differential operator that takes any given function and computes the sum of its second partial derivatives. This might sound complex, but think of it as a way to measure how a function changes at an infinitesimally small region in space.

In three-dimensional Cartesian coordinates, the Laplacian is formulated as:\[ abla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \] This formula actively engages with all three spatial dimensions by taking their second partial derivatives, analogous to taking a closer look at the surface and character of a function in all directions.
  • The Laplacian measures how a function like potential or temperature alters through space.
  • Key in solving the Laplace equation: \( abla^2 f = 0 \), indicating stability and equilibrium in physical systems.
Understanding the Laplacian can offer insights into how various physical phenomena, such as heat distribution or gravitational potential, behave across different regions.
Three-dimensional Cartesian Coordinates
Understanding three-dimensional Cartesian coordinates is foundational when exploring concepts like the Laplacian operator. In this system, every point in space has three coordinates \( (x, y, z) \), representing its position along three perpendicular axes. This allows us to thoroughly describe relationships and changes in multidimensional spaces, such as physical phenomena that depend on more than one dimension.

Cartesian coordinates are intuitive because they're based on the familiar grid system we often encounter. They enable us to visualize and analyze complex functions like \( f(x, y, z) = x^2 + y^2 - 2z^2 \). This specific function can be seen as a surface within the three-dimensional space, where each point on the surface corresponds to a value of \( f \).
  • Each coordinate denotes a distinct dimensional change. For instance, \( x \), \( y \), and \( z \) impact the function independently.
  • Facilitates tracking of changes: You can easily calculate partial derivatives along any axis using this system.
By understanding these coordinates, we can systematically compute how functions behave as we perform operations like partial differentiation and apply operators in three dimensions.

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Most popular questions from this chapter

In Exercises \(1-3,\) begin by drawing a diagram that shows the relations among the variables. If \(w=x^{2}+y-z+\sin t\) and \(x+y=t,\) find $$ \begin{array}{lll}{\text { a. }\left(\frac{\partial w}{\partial y}\right)_{x, z}} & {\text { b. }\left(\frac{\partial w}{\partial y}\right)_{z, t}} & {\text { c. }\left(\frac{\partial w}{\partial z}\right)_{x, y}} \\ {\text { d. }\left(\frac{\partial w}{\partial z}\right)_{y, t}} & {\text { e. }\left(\frac{\partial w}{\partial t}\right)_{x, z}} &{\text { f. }\left(\frac{\partial w}{\partial t}\right)_{y, z}}\end{array} $$

Three variables Let \(w=f(x, y, z)\) be a function of three independent variables and write the formal definition of the partial derivative \(\partial f / \partial z\) at \(\left(x_{0}, y_{0}, z_{0}\right) .\) Use this definition to find \(\partial f / \partial z\) at \((1,2,3)\) for \(f(x, y, z)=x^{2} y z^{2}\).

The heat equation An important partial differential equation that describes the distribution of heat in a region at time \(t\) can be represented by the one-dimensional heat equation $$\frac{\partial f}{\partial t}=\frac{\partial^{2} f}{\partial x^{2.}}$$ Show that \(u(x, t)=\sin (\alpha x) \cdot e^{-\beta t}\) satisfies the heat equation for constants \(\alpha\) and \(\beta .\) What is the relationship between \(\alpha\) and \(\beta\) for this function to be a solution?

Least squares and regression lines When we try to fit a line \(y=m x+b\) to a set of numerical data points \(\left(x_{1}, y_{1}\right)\) \(\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right),\) we usually choose the line that minimizes the sum of the squares of the vertical distances from the points to the line. In theory, this means finding the values of \(m\) and \(b\) that minimize the value of the function $$w=\left(m x_{1}+b-y_{1}\right)^{2}+\cdots+\left(m x_{n}+b-y_{n}\right)^{2}$$ (See the accompanying figure.) Show that the values of \(m\) and \(b\) that do this are $$ \begin{array}{l}{m=\frac{\left(\sum x_{k}\right)\left(\sum y_{k}\right)-n \sum x_{k} y_{k}}{\left(\sum x_{k}\right)^{2}-n \sum x_{k}^{2}}} \\\ {b=\frac{1}{n}\left(\sum y_{k}-m \sum x_{k}\right)}\end{array} $$ with all sums running from \(k=1\) to \(k=n .\) Many scientific calculators have these formulas built in, enabling you to find \(m\) and \(b\) with only a few keystrokes after you have entered the data. The line \(y=m x+b\) determined by these values of \(m\) and \(b\) is called the least squares line, regression line, or trend line for the data under study. Finding a least squares line lets you 1\. summarize data with a simple expression, 2\. predict values of \(y\) for other, experimentally untried values of \(x\) 3\. handle data analytically.

The fifth-order partial derivative \(\partial^{5} f / \partial x^{2} \partial y^{3}\) is zero for each of the following functions. To show this as quickly as possible, which variable would you differentiate with respect to first: \(x\) or \(y ?\) Try to answer without writing anything down. \begin{equation}\begin{array}{ll}{\text { a. }} & {f(x, y)=y^{2} x^{4} e^{x}+2} \\ {\text { b. }} & {f(x, y)=y^{2}+y\left(\sin x-x^{4}\right)} \\\ {\text { c. }} & {f(x, y)=x^{2}+5 x y+\sin x+7 e^{x}} \\ {\text { d. }} & {f(x, y)=x e^{y^{2} / 2}}\end{array}\end{equation}

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