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Calculate the Laplacian \(\nabla^{2}\) of each of the following scalar fields. $$x y z\left(x^{2}-2 y^{2}+z^{2}\right)$$

Short Answer

Expert verified
\[ abla^2 f = 12 x y z - 6 x z \]

Step by step solution

01

- Understand the Laplacian

The Laplacian of a scalar field \(\f\) is defined as the divergence of the gradient of \(\f\). In three dimensions, it is given by \[ abla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2}. \] Here, \(\f = x y z (x^2 - 2y^2 + z^2)\).
02

- Calculate the second partial derivative with respect to x

First, find the partial derivative \(\frac{\partial f}{\partial x}\). Then find the second partial derivative \(\frac{\partial^2 f}{\partial x^2}\). \[ \frac{\partial f}{\partial x} = y z (3x^2 - 2y^2 + z^2) \] \[ \frac{\partial^2 f}{\partial x^2} = y z (6x) = 6x y z \]
03

- Calculate the second partial derivative with respect to y

First, find the partial derivative \(\frac{\partial f}{\partial y}\). Then find the second partial derivative \(\frac{\partial^2 f}{\partial y^2}\). \[ \frac{\partial f}{\partial y} = x z (x^2 - 6y + z^2) \] \[ \frac{\partial^2 f}{\partial y^2} = x z (-6) = -6 x z \]
04

- Calculate the second partial derivative with respect to z

First, find the partial derivative \(\frac{\partial f}{\partial z}\). Then find the second partial derivative \(\frac{\partial^2 f}{\partial z^2}\). \[ \frac{\partial f}{\partial z} = x y (x^2 - 2y^2 + 3z^2) \] \[ \frac{\partial^2 f}{\partial z^2} = x y (6z) = 6 x y z \]
05

- Sum the second partial derivatives

Add the second partial derivatives obtained: \[ abla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} \] \[ abla^2 f = 6 x y z - 6 x z + 6 x y z = 12 x y z - 6 x z \]

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Key Concepts

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

Second Partial Derivatives
Partial derivatives measure how a function changes as its variables change.
In our case, we need second partial derivatives, which involve taking the partial derivative twice.
Here's how it works for the given function:
First, find the partial derivative, then take it again with respect to the same variable.
For example, if we start with the function \(\f = x y z (x^2 - 2y^2 + z^2)\):

- \(\frac{\frac{\f}{x}}{x}\):
First, \(\frac{\f}{x} = y z (3x^2 - 2y^2 + z^2)\).
Second, \(\frac{\frac{f}{x}}{x} = y z (6x) = 6 x y z\).
We do the same for \(\frac{\f}{y}\) and \(\frac{\f}{z}\). This gives us the second partial derivatives for our function in each variable.
Scalar Fields
A scalar field is a mathematical function that assigns a single scalar value to every point in space.
The concept is particularly important in fields like physics and engineering.
In simpler terms, imagine a field where each point has a specific temperature; this temperature would be the scalar value.
For our function \(\f = x y z (x^2 - 2y^2 + z^2)\), each point in the \(\f\)-space is assigned a value based on its coordinates (x, y, z).
Scalar fields are often visualized using contour plots, where lines connect points of equal scalar value.
Understanding scalar fields helps in grasping more complex topics such as gradients and divergence.
Divergence
Divergence measures the magnitude of a field's source or sink at a given point.
It's a concept often used in vector calculus and has applications in fluid dynamics, electromagnetism, and more.
In simple terms, divergence can tell us if there's something 'coming out' or 'going in' at a point in a field.
For scalar fields, divergence is an intermediate step in calculating the Laplacian.
More formally, if we have a vector field \(\f\), its divergence is given by:
\(abla \bullet \f\).
In our context, we are using this concept to understand how \(abla^2 f\) extends from the gradient of \(\f\).
This helps in calculating how values change and accumulate.
Gradient
The gradient of a scalar field is a vector field that points in the direction of the greatest rate of increase of the scalar field.
Its magnitude represents how fast the scalar field is changing in that direction.
Mathematically, the gradient for a function \(\f = f(x, y, z)\) is:
\(abla f = \frac{\f}{x}\frac{x} + \frac{\f}{y}\frac{y} + \frac{\f}{z}\frac{z}\).
For our function, the gradient vectors at each point give us critical information on how the values are changing.
Understanding gradients is key to later understanding the Laplacian, as the Laplacian is essentially the divergence of the gradient.
Three Dimensions
In this exercise, we are working in three-dimensional space, meaning our function depends on three variables: x, y, and z.
Calculations in three dimensions can be more complex than in two dimensions.
However, the principles remain the same: each variable is a dimension and contributes to the overall shape and behavior of the function.
Understanding three-dimensional analysis is vital in fields like physics, engineering, and computer graphics.
Within this three-dimensional context, we calculate the Laplacian by summing the second partial derivatives of our function with respect to each of the three dimensions.
This allows us to understand how the function behaves in a full \(\f (x, y, z)\)-equation, giving us a comprehensive view.

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Most popular questions from this chapter

For the force field \(\mathbf{F}=(y+z) \mathbf{i}-(x+z) \mathbf{j}+(x+y) \mathbf{k},\) find the work done in moving a particle around each of the following closed curves: (a) the circle \(x^{2}+y^{2}=1\) in the \((x, y)\) plane, taken counterclockwise; (b) the circle \(x^{2}+z^{2}=1\) in the \((z, x)\) plane, taken counterclockwise; (c) the curve starting from the origin and going successively along the \(x\) axis to (1,0,0), parallel to the \(z\) axis to ( 1,0,1 ), parallel to the \((y, z)\) plane to (1,1,1) and back to the origin along \(x=y=z\) (d) from the origin to \((0,0,2 \pi)\) on the curve \(x=1-\cos t, y=\sin t, z=t,\) and back to the origin along the \(z\) axis.

Evaluate each integral in the simplest way possible. $$\oint \mathbf{F} \cdot d \mathbf{r} \text { around the circle } x^{2}+y^{2}+2 x=0, \text { where } \mathbf{F}=y \mathbf{i}-x \mathbf{j}$$

Evaluate each of the integrals as either a volume integral or a surface integral, whichever is easier. \(\iint \mathbf{r} \cdot \mathbf{n} d \sigma\) over the entire surface of the cone with base \(x^{2}+y^{2} \leq 16, z=0,\) and vertex at \((0,0,3),\) where \(\mathbf{r}=\mathbf{i} x+\mathbf{j} y+\mathbf{k} z\).

The force on a charge \(q\) moving with velocity \(\mathbf{v}=d \mathbf{r} / d t\) in a magnetic field \(\mathbf{B}\) is \(\mathbf{F}=q(\mathbf{v} \times \mathbf{B}) .\) We can write \(\mathbf{B}\) as \(\mathbf{B}=\boldsymbol{\nabla} \times \mathbf{A}\) where \(\mathbf{A}\) (called the vector potential) is a vector function of \(x, y, z, t .\) If the position vector \(\mathbf{r}=\mathbf{i} x+\mathbf{j} y+\mathbf{k} z\) of the charge \(q\) is a function of time \(t,\) show that $$\frac{d \mathbf{A}}{d t}=\frac{\partial \mathbf{A}}{\partial t}+\mathbf{v} \cdot \nabla \mathbf{A}$$ Thus show that $$\mathbf{F}=q \mathbf{v} \times(\mathbf{\nabla} \times \mathbf{A})=q\left[\nabla(\mathbf{v} \cdot \mathbf{A})-\frac{d \mathbf{A}}{d t}+\frac{\partial \mathbf{A}}{\partial t}\right]$$

\(\iint \mathbf{V} \cdot \mathbf{n} d \sigma\) over the entire surface of a cube in the first octant with edges of length 2 along the coordinate axes, where $$\mathbf{V}=\left(x^{2}-y^{2}\right) \mathbf{i}+3 y \mathbf{j}-2 x z \mathbf{k}$$

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