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What is the density form of the function \(f\left(\begin{array}{l}x \\ y \\\ z\end{array}\right)=x y+z^{2}\), evaluated $$ \text { at the point } x=\left(\begin{array}{l} 1 \\ 2 \\ 1 \end{array}\right) \text { on the vectors }\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right],\left[\begin{array}{l} 2 \\ 1 \\ 1 \end{array}\right], \text { and }\left[\begin{array}{l} 0 \\ 1 \\ 1 \end{array}\right] ? $$

Short Answer

Expert verified
The density form evaluated at the given point and vectors is 1.

Step by step solution

01

Evaluate the Function at Point

Substitute the values of the point \( (x, y, z) = (1, 2, 1) \) into the function \( f(x, y, z) = xy + z^2 \). This gives:\[f(1, 2, 1) = (1)(2) + (1)^2 = 2 + 1 = 3.\]
02

Find the Gradient

The gradient of \( f \) is found by computing the partial derivatives with respect to \( x, y, \text{ and } z \). The gradient is:\[abla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right).\]Calculate each partial derivative:- \( \frac{\partial f}{\partial x} = y \)- \( \frac{\partial f}{\partial y} = x \)- \( \frac{\partial f}{\partial z} = 2z \).So, \( abla f(x, y, z) = (y, x, 2z) \).
03

Evaluate the Gradient at the Point

Substitute the point \((1, 2, 1)\) into the gradient:\[abla f (1, 2, 1) = (2, 1, 2(1)) = (2, 1, 2).\]
04

Form the Jacobian Matrix

The Jacobian matrix is formed using the gradient vector and the given direction vectors. The Jacobian matrix \( J \) is:\[J = \begin{bmatrix} 2 & 1 & 2 \1 & 0 & 1 \2 & 1 & 1 \0 & 1 & 1 \end{bmatrix},\]where the first row is the gradient and the subsequent rows are the direction vectors.
05

Compute the Determinant of the Jacobian Matrix

The density form requires the determinant of the Jacobian. Using the matrix from Step 4, calculate the determinant:\[\det(J) = \det \begin{bmatrix} 2 & 1 & 2 \1 & 0 & 1 \2 & 1 & 1 \0 & 1 & 1 \end{bmatrix}. \]To compute \( \det(J) \), apply the formula for a 3x3 matrix:\[\det(J) = 2(0*1 - 1*1) - 1(1*1 - 2*1) + 2(1*1 - 2*0) = -2 + 1 + 2 = 1.\]

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Key Concepts

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

Gradient
The gradient is a fundamental concept in calculus that provides a multi-dimensional generalization of the derivative. You can think of it as a vector that points in the direction of the steepest increase of a function. When we have a function of several variables, the gradient is crucial to understand how the function changes as you move in different directions.
The gradient of a function \( f(x, y, z) \) involves calculating partial derivatives with respect to each variable. In simpler terms:
  • \( \frac{\partial f}{\partial x} \) tells us how \( f \) changes as we vary \( x \), keeping \( y \) and \( z \) constant,
  • \( \frac{\partial f}{\partial y} \) shows how \( f \) changes with varying \( y \), keeping \( x \) and \( z \) constant, and
  • \( \frac{\partial f}{\partial z} \) reveals how \( f \) changes when \( z \) is altered, while \( x \) and \( y \) stay the same.
For the given function \( f(x, y, z) = xy + z^2 \), its gradient is \( abla f = (y, x, 2z) \). This vector is not only informative on how \( f \) might change but can also be used in various applications like optimization and finding local extrema, indicating directions for minimum or maximum values.
Directional Derivative
The directional derivative extends the concept of the derivative in calculus to any direction in which you want to measure the rate of change of the function. Unlike the regular derivative that measures change along an axis, the directional derivative allows us to check change along any vector direction.
It is calculated using the gradient and involves a dot product. For a function \( f(x, y, z) \) and a direction given by a unit vector \( \mathbf{v} = (v_x, v_y, v_z) \), the directional derivative \( D_\mathbf{v}f \) is:
  • \( D_\mathbf{v}f = abla f \cdot \mathbf{v} \)
This yields a scalar value that represents how fast the function \( f \) is changing at a point in the direction of \( \mathbf{v} \). It effectively answers the question about the rate of change of \( f \) in a specified direction. By knowing the directional derivatives in the direction of given vectors, you can understand the behavior and variation of the function over those directions.
Jacobian Matrix
The Jacobian matrix is a powerful tool when working with multiple variables and functions, especially in multivariable calculus and vector calculus. It helps us find how each variable in our function system affects the output when given small changes.
The Jacobian matrix \( J \) of a vector-valued function incorporates partial derivatives of each of the function's components, providing a comprehensive view of the system's behavior.
For example, if we have a function with outputs that rely on several inputs, the Jacobian matrix is structured as:
  • The first row represents the gradient of the first function component,
  • The second row for the second function component, and so on.
In this case, with our vectors \( (1, 0, 1), (2, 1, 1), \text{and} (0, 1, 1) \), the Jacobian matrix includes our gradient as the first row followed by these vectors. The determinant of this matrix, often calculated for density form or changing variables in integrals, gives us a sense of how volumetric "density" of transformations affect space.

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Most popular questions from this chapter

Find all the 1-forms \(\omega=p(y, z) d x+q(x, z) d y\) such that $$ d \omega=x d y \wedge d z+y d x \wedge d z $$.

Consider the curve \(S\) of equation \(x^{2}+y^{2}=1\), oriented by the tangent vector \(\left[\begin{array}{l}0 \\ 1\end{array}\right]\) at the point \(\left(\begin{array}{l}1 \\ 0\end{array}\right)\). (a) Show that the subset \(X\) where \(x \geq 0\) is a piece-with-boundary of \(S\). What is its oriented boundary? (b) Show that the subset \(Y\) where \(|x| \leq 1 / 2\) is a piece-with-boundary of \(S\). What is its oriented boundary? (c) Is the subset \(Z\) where \(x>0\) a piece-with-boundary? If so, what is its boundary?

Prove that if a linear transformation \(T\) is not one to one, then it is not orientation preserving or reversing.

Set up each of the following integrals of form fields over parametrized domains as an ordinary multiple integral, and compute it. (a) \(\int_{\gamma(I)} x d y+y d z\), where \(I=[-1,1]\), and \(\gamma(t)=\left(\begin{array}{c}\sin t \\ \cos t \\ t\end{array}\right)\). (b) \(\int_{\gamma(U)} x d y \wedge d z\), where \(U=[-1,1] \times[-1,1]\), and \(\gamma\left(\begin{array}{c}u \\\ v\end{array}\right)=\left(\begin{array}{c}u^{2} \\ u+v \\\ v^{3}\end{array}\right)\). (c) \(\int_{\gamma(U)} x_{1} d x_{2} \wedge d x_{3}+x_{2} d x_{3} \wedge d x_{4}\), where \(U=\left\\{\left(\begin{array}{l}u \\ v\end{array}\right) \mid 0 \leq u, v ; u+v \leq 2\right\\}\), and \(\gamma\left(\begin{array}{c}u \\\ v\end{array}\right)=\left(\begin{array}{c}u v \\ u^{2}+v^{2} \\ u-v \\ \log (u+v+1)\end{array}\right)\) (d) \(\int_{\gamma(U)} x_{2} d x_{1} \wedge d x_{3} \wedge d x_{4}\), where \(U=\left\\{\left(\begin{array}{c}u \\ v \\ w\end{array}\right) \mid 0 \leq u, v, w ; u+v+w \leq 3\right\\}\) $$ \text { and } \gamma\left(\begin{array}{c} u \\ v \\ w \end{array}\right)=\left(\begin{array}{c} u v \\ u^{2}+w^{2} \\ u-v \\ w \end{array}\right) $$

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