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91Ó°ÊÓ

If \(\mathbf{E}=x \mathbf{i}+z^{2} \mathbf{j}-y z \mathbf{k}\) find (a) \(\mathbf{E} \cdot \mathbf{i}\) (b) \(\mathbf{E} \cdot \mathbf{j}\) (c) \(\mathbf{E} \cdot \mathbf{k}\) (d) \(\nabla \cdot \mathbf{E}\)

Short Answer

Expert verified
(a) The dot product \(\mathbf{E} \) dot \(\mathbf{i} = x\). (b) The dot product \(\mathbf{E} \) dot \(\mathbf{j} = z^2\). (c) The dot product \(\mathbf{E} \) dot \(\mathbf{k} = -yz\). (d) The divergence of \(\mathbf{E} = 1 - y\).

Step by step solution

01

Calculate the dot product with vector i

To find the dot product of vector \(\mathbf{E}\) with the unit vector \(\mathbf{i}\), we only consider the coefficient of \(\mathbf{i}\) in \(\mathbf{E}\), since \(\mathbf{i} \) dot \(\mathbf{j} = \mathbf{i} \) dot \(\mathbf{k} = 0\) and \(\mathbf{i} \) dot \(\mathbf{i} = 1\). Thus, \(\mathbf{E} \) dot \(\mathbf{i} = x \).
02

Calculate the dot product with vector j

To find the dot product of vector \(\mathbf{E}\) with the unit vector \(\mathbf{j}\), we only consider the coefficient of \(\mathbf{j}\) in \(\mathbf{E}\), since \(\mathbf{j} \) dot \(\mathbf{i} = \mathbf{j} \) dot \(\mathbf{k} = 0\) and \(\mathbf{j} \) dot \(\mathbf{j} = 1\). Therefore, \(\mathbf{E} \) dot \(\mathbf{j} = z^2 \).
03

Calculate the dot product with vector k

To calculate the dot product of vector \(\mathbf{E}\) with the unit vector \(\mathbf{k}\), we consider the coefficient of \(\mathbf{k}\) in \(\mathbf{E}\), as \(\mathbf{k} \) dot \(\mathbf{i} = \mathbf{k} \) dot \(\mathbf{j} = 0\) and \(\mathbf{k} \) dot \(\mathbf{k} = 1\). So, \(\mathbf{E} \) dot \(\mathbf{k} = -yz \).
04

Compute the divergence of vector E

The divergence of vector \(\mathbf{E}\) is given by the scalar field \(abla \) dot \(\mathbf{E} = \frac{\partial}{\partial x} (x) + \frac{\partial}{\partial y} (z^2) - \frac{\partial}{\partial z} (yz)\). This can be simplified to \(abla \) dot \(\mathbf{E} = 1 + 0 - y = 1 - y\).

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Key Concepts

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

Divergence of a Vector Field
The concept of the 'divergence of a vector field' is an essential mathematical tool in fields like fluid dynamics and electromagnetism. Divergence provides a measure of how much a vector field spreads out from a particular point.

Technically, it's a scalar quantity that represents the volume density of the outward flux of a vector field from an infinitesimal volume around a given point. To calculate the divergence of a vector field, \textbf{E}, we use the del operator (∇), also known as the gradient operator, which is a vector derivative operator used in vector calculus.

Calculating Divergence

For a three-dimensional vector field \textbf{E} with components \textbf{E}\textsubscript{x}, \textbf{E}\textsubscript{y}, and \textbf{E}\textsubscript{z}, the divergence is given by the formula: \[ abla \cdot \mathbf{E} = \frac{\partial \mathbf{E}_x}{\partial x} + \frac{\partial \mathbf{E}_y}{\partial y} + \frac{\partial \mathbf{E}_z}{\partial z}.\]It essentially involves taking the partial derivatives of each component of \textbf{E} with respect to its corresponding coordinate axis and summing them up. In our example problem, the divergence of \textbf{E} simplifies to 1 - y.
  • To gain a deeper understanding of divergence, visualize the flow of a fluid or the spreading of scent in the air - it 'diverges' from the source.
  • Divergence reveals sources (positive divergence) and sinks (negative divergence) within a field, crucial for physical interpretations in science and engineering.
Vector Calculus
Vector calculus is a branch of mathematics concerned with differentiation and integration of vector fields, primarily in 3-dimensional Euclidean space \textbf{R}\textsuperscript{3}. It involves various operations, including scalar multiplication, addition, dot product (which gives a scalar), and cross product (which gives a vector).

The dot product, also known as the scalar product, is an operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation can be performed between any two vectors \textbf{A} and \textbf{B} as follows: \[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta), \] where |\mathbf{A}| and |\mathbf{B}| are the magnitudes of vectors, and \(\theta\) is the angle between them.

In the context of the exercise, the dot product simplifies calculations since orthogonal vectors (those at a right angle) have a dot product of zero: \[ \mathbf{i} \cdot \mathbf{j} = \mathbf{i} \cdot \mathbf{k} = \mathbf{j} \cdot \mathbf{k} = 0 \] and any vector dotted with itself equals the square of its magnitude: \[ \mathbf{i} \cdot \mathbf{i} = \mathbf{j} \cdot \mathbf{j} = \mathbf{k} \cdot \mathbf{k} = 1. \]
  • Understanding how to manipulate these operations and their properties is fundamental in the study of fields such as physics and engineering.
  • Applications range from work and energy calculations to finding the angle between vectors in space.
Partial Derivatives
Partial derivatives play a central role in multivariable calculus, which extends the concept of derivative to functions of several variables. A partial derivative measures how a function changes as one of its variables is varied, holding the other variables constant.

For example, if there is a function \textbf{f}(x, y, z), then the partial derivative of \textbf{f} with respect to x is denoted as \[ \frac{\partial \mathbf{f}}{\partial x} \] and it represents the rate at which \textbf{f}'s value changes as x increases, holding y and z constant.

Importance in Vector Calculus

In the context of the exercise, partial derivatives are used to calculate both the components of a vector field and its divergence. This operation is used to facilitate understanding of complex systems where the change does not occur uniformly across variables, such as temperature distribution in a room or pressure changes in a moving fluid.
  • The ability to compute and interpret partial derivatives is crucial for performing and understanding gradient, curl, and divergence which are critical operations in vector calculus.
  • Partial derivatives are the building blocks of more complex concepts like the Jacobian matrix, which represents all possible partial derivatives of a vector-valued function.
While partial derivatives provide the rate of change in one specific direction, the gradient generalizes this concept by combining all the partial derivatives into a vector, pointing in the direction of the greatest rate of increase of the function.

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