Problem 22
Question
Find the gradient field corresponding to \(f\) Use a CAS to graph it. $$f(x, y, z)=(x-y)^{2}+z$$
Step-by-Step Solution
Verified Answer
The gradient field of the function \(f(x, y, z)=(x-y)^{2}+z\) is given by \(\nabla f = (df/dx, df/dy, df/dz)\), which can be found by calculating the partial derivatives of \(f\) with respect to \(x\), \(y\), and \(z\) respectively. The graph of the gradient field can then be generated using a CAS.
1Step 1: Calculating the partial derivatives
Compute the partial derivative of the function \(f(x, y, z)=(x-y)^{2}+z\) with respect to \(x\), \(y\), and \(z\). The partial derivative of \(f\) with respect to \(x\) (\(df/dx\)) is obtained by differentiating \(f\) with respect to \(x\) while treating \(y\) and \(z\) as constant. Similarly, find the partial derivative with respect to \(y\) (\(df/dy\)) and \(z\) (\(df/dz\)).
2Step 2: Derive the gradient
The gradient of the function (\(\nabla f)\) can be found by combining the partial derivatives from step 1 into a vector. It will be of the form \(\nabla f = (df/dx, df/dy, df/dz)\).
3Step 3: Graphing the gradient field using a CAS
Graphing the gradient field will give an idea of how the function behaves. This can be done using any computer algebra system (CAS). Using the CAS, the gradient field can be graphed by plotting the partial derivatives obtained in Step 1. This will provide a vector field representation of the gradient.
Key Concepts
Partial DerivativesVector Field RepresentationComputer Algebra System (CAS)
Partial Derivatives
Understanding partial derivatives is crucial when dealing with functions of multiple variables, as they allow us to see how a function changes as each variable is varied independently of the others. For a function such as \(f(x, y, z)\), the partial derivative with respect to \(x\), denoted by \(\frac{\partial f}{\partial x}\), measures the rate at which the function's value changes as the \(x\) variable is altered, while keeping \(y\) and \(z\) constant.
Imagine you're on a hillside given by \(f(x, y, z)\) and you only move eastward (along the \(x\)-axis); the steepness of the incline that you experience is what \(\frac{\partial f}{\partial x}\) represents. Similarly, \(\frac{\partial f}{\partial y}\) and \(\frac{\partial f}{\partial z}\) describe changes while moving northward and upward, respectively. In the given exercise, the first step toward understanding the behavior of the function \((x-y)^2+z\) is finding these partial derivatives.
Imagine you're on a hillside given by \(f(x, y, z)\) and you only move eastward (along the \(x\)-axis); the steepness of the incline that you experience is what \(\frac{\partial f}{\partial x}\) represents. Similarly, \(\frac{\partial f}{\partial y}\) and \(\frac{\partial f}{\partial z}\) describe changes while moving northward and upward, respectively. In the given exercise, the first step toward understanding the behavior of the function \((x-y)^2+z\) is finding these partial derivatives.
Vector Field Representation
Once we have the partial derivatives, we can represent them visually using a vector field. A vector field is essentially a map that shows, at each point in space, a vector representing the direction and magnitude of the maximum rate of increase of a function at that point. It is a very powerful tool in visualizing how a function changes in a three-dimensional space.
In the context of our function \(f(x, y, z)\), the gradient \(abla f\), which is a vector made up of the partial derivatives \(\Big(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\Big)\), represents the slope of the function in each direction at any given point. By plotting these vectors, we create a visual landscape showing how the function increases. This representation can reveal patterns and behaviors of the function that may not be immediately evident from the equation alone.
In the context of our function \(f(x, y, z)\), the gradient \(abla f\), which is a vector made up of the partial derivatives \(\Big(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\Big)\), represents the slope of the function in each direction at any given point. By plotting these vectors, we create a visual landscape showing how the function increases. This representation can reveal patterns and behaviors of the function that may not be immediately evident from the equation alone.
Computer Algebra System (CAS)
A Computer Algebra System (CAS) is an invaluable tool for students and researchers dealing with complex mathematical computations. It allows for symbolic manipulation of equations, meaning it can carry out algebraic operations exactly, rather than approximately, and it can also handle graphical representations of functions and vector fields.
In our exercise, using a CAS to plot the gradient field translates our mathematical findings into a visual form, which can be easier to comprehend than raw equations or numerical data sets. Effectively, the CAS takes the vector field data and creates a graphical representation of it, enabling us to see at a glance how the function behaves. For students learning about gradient fields and partial derivatives, this can be a profound aid in gaining a more intuitive understanding of the mathematical concepts involved. A CAS not only provides accuracy but also saves valuable time that may have been spent on manual calculations.
In our exercise, using a CAS to plot the gradient field translates our mathematical findings into a visual form, which can be easier to comprehend than raw equations or numerical data sets. Effectively, the CAS takes the vector field data and creates a graphical representation of it, enabling us to see at a glance how the function behaves. For students learning about gradient fields and partial derivatives, this can be a profound aid in gaining a more intuitive understanding of the mathematical concepts involved. A CAS not only provides accuracy but also saves valuable time that may have been spent on manual calculations.
Other exercises in this chapter
Problem 22
Find the flux of \(\mathbf{F}\) over \(\partial Q\). \(Q\) is bounded by \(y^{2}+z^{2}=4, x=1\) and \(x=8-y\) \(\mathbf{F}=\left\langle x^{2} z, 2 y-e^{z}, \sin
View solution Problem 22
Suppose that \(u\) is a harmonic function (that is, \(\nabla^{2} u=0\) ). Show that \(\iint_{S} \frac{\partial u}{\partial n} d S=0\).
View solution Problem 22
Use a line integral to compute the area of the given region. The ellipse \(4 x^{2}+y^{2}=4\)
View solution Problem 22
Evaluate the line integral. \(\int_{C} x z d s,\) where \(C\) is the line segment from (2,1,0) to (2,0,2)
View solution