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Calculate the Laplacian \(\nabla^{2}\) of each of the following scalar fields. $$x^{3}-3 x y^{2}+y^{3}$$

Short Answer

Expert verified
The Laplacian \(abla^2 f = 6y.\)

Step by step solution

01

- Understand the Laplacian Operator

The Laplacian operator in Cartesian coordinates is given by \[abla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}\] where \(f\) is a scalar field.
02

- Given Scalar Field

The given scalar field is \[f(x, y) = x^3 - 3xy^2 + y^3.\]
03

- Compute the Second Partial Derivative with respect to x

First, find the first partial derivative of \(f\) with respect to \(x\): \[\frac{\partial f}{\partial x} = 3x^2 - 3y^2.\]Now, find the second partial derivative with respect to \(x\): \[\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}(3x^2 - 3y^2) = 6x.\]
04

- Compute the Second Partial Derivative with respect to y

First, find the first partial derivative of \(f\) with respect to \(y\): \[\frac{\partial f}{\partial y} = -6xy + 3y^2.\] Now, find the second partial derivative with respect to \(y\): \[\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}(-6xy + 3y^2) = -6x + 6y.\]
05

- Add the Second Partial Derivatives

Add the second partial derivatives obtained in Steps 3 and 4: \[abla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 6x + (-6x + 6y) = 6y.\]

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Key Concepts

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

Laplacian operator
The Laplacian operator, often denoted by abla^2, is a differential operator used extensively in mathematics and physics. At its core, the Laplacian measures how a scalar field deviates from its average value over a small region.
In Cartesian coordinates, it's the sum of the second partial derivatives with respect to each spatial variable. For a function f(x, y), the Laplacian is expressed as:
\[ abla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} \] This means you calculate the second derivatives of the function f with respect to x and y, and then sum them up.

Importance in Various Fields
\ - In physics, particularly in electromagnetism and fluid dynamics, the Laplacian helps describe phenomena like heat distribution and fluid flow.
\ - In engineering, it contributes to solving problems in signal processing and control systems.
\ - In mathematics, it's fundamental in solving partial differential equations.
Partial derivatives
Partial derivatives are the building blocks for understanding changes in multivariable functions. They show how a function changes with respect to one of its variables, while holding other variables constant.
Let's break it down further:
First Partial Derivative:
For a function f(x, y), the first partial derivative with respect to x, written as \(\frac{\partial f}{\partial x}\), measures the rate of change of f as x varies, keeping y constant.

Second Partial Derivative:
The second partial derivative involves differentiating the first partial derivative. For instance, \(\frac{\partial^2 f}{\partial x^2}\) is obtained by differentiating \(\frac{\partial f}{\partial x}\) with respect to x again.
In our example, for the function \(f(x, y) = x^3 - 3xy^2 + y^3\), the first partial derivatives are:
\(\frac{\partial f}{\partial x} = 3x^2 - 3y^2\)
\(\frac{\partial f}{\partial y} = -6xy + 3y^2\)
Then, taking the second partial derivatives, we get:
\(\frac{\partial^2 f}{\partial x^2} = 6x\)
\(\frac{\partial^2 f}{\partial y^2} = -6x + 6y\)
Understanding these derivatives is crucial for calculating the Laplacian.
Scalar field analysis
A scalar field assigns a scalar value to every point in space. For example, temperature distribution in a room can be represented as a scalar field where each point has a temperature value.
Scalar field analysis focuses on how these values change across space. The gradients, partial derivatives, and the Laplacian are tools to study these changes.

Key Applications:
In physics, scalar fields represent physical quantities like temperature, pressure, and electrostatic potentials.
In geography, they help visualize topographic surfaces to understand elevation changes.
In our exercise, the scalar field is given by \(f(x, y) = x^3 - 3xy^2 + y^3\). By analyzing its derivatives, we delve deeper into how this field changes spatially.
The first partial derivatives give slope information, while the second partial derivatives, summed up in the Laplacian, provide insight into the concavity and local extremum of the field.
Cartesian coordinates
Cartesian coordinates are a system that specifies each point uniquely in a plane by a pair of numerical coordinates. These are distances to the point from two fixed perpendicular lines, measured in the same unit of length.
This system is fundamental in mathematics and physics for locating points, analyzing curves, and plotting graphs.
Structure:
- The coordinates are written as (x, y) in 2D space and (x, y, z) in 3D space.
- The x-axis and y-axis are perpendicular, intersecting at the origin (0,0).
- The position along these axes determines the value of the coordinates.
In our context, Cartesian coordinates allow us to express the scalar field \(f(x, y)\) in a readable form. The Laplacian operator, expressed through second partial derivatives, leverages this coordinate system for straightforward computations.
The use of Cartesian coordinates simplifies the handling of partial derivatives, which are crucial when computing the Laplacian.

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Most popular questions from this chapter

Find the value of \(\int \mathbf{F} \cdot d \mathbf{r}\) along the circle \(x^{2}+y^{2}=2\) from (1,1) to (1,-1) if $$\mathbf{F}=(2 x-3 y) \mathbf{i}-(3 x-2 y) \mathbf{j}$$

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.

Given the vector \(\mathbf{A}=\left(x^{2}-y^{2}\right) \mathbf{i}+2 x y \mathbf{j}\) (a) Find \(\nabla \times \mathbf{A}\) (b) Evaluate \(\iint(\nabla \times \mathbf{A}) \cdot d \sigma\) over a rectangle in the \((x, y)\) plane bounded by the lines \(x=0, x=a, y=0, y=b\) (c) Evaluate \(\oint \mathbf{A} \cdot d \mathbf{r}\) around the boundary of the rectangle and thus verify Stokes' theorem for this case.

Verify that each of the following force fields is conservative. Then find, for each, a scalar potential \(\phi\) such that \(\mathbf{F}=-\nabla \phi\). $$\mathbf{F}=\left(3 x^{2} y z-3 y\right) \mathbf{i}+\left(x^{3} z-3 x\right) \mathbf{j}+\left(x^{3} y+2 z\right) \mathbf{k}$$

If \(\mathbf{A}\) and \(\mathbf{B}\) are unit vectors with an angle \(\theta\) between them, and \(\mathbf{C}\) is a unit vector perpendicular to both \(\mathbf{A}\) and \(\mathbf{B},\) evaluate \([(\mathbf{A} \times \mathbf{B}) \times(\mathbf{B} \times \mathbf{C})] \times(\mathbf{C} \times \mathbf{A})\).

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