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Find the Jacobian \(\partial(x, y, z) / \partial(u, v, w)\). $$ u=x+y+z, v=x+y-z, w=x-y+z $$

Short Answer

Expert verified
The Jacobian is -1.

Step by step solution

01

Write down the function relationships

We are given the following equations that define the transformation: \[ \begin{align*} u &= x + y + z, \ v &= x + y - z, \ w &= x - y + z.\end{align*} \] Our task is to find the Jacobian \( \frac{\partial(x, y, z)}{\partial(u, v, w)} \), which requires us to express \(x\), \(y\), and \(z\) in terms of \(u\), \(v\), and \(w\).
02

Express x, y, z in terms of u, v, w

Add the equations for \(u\), \(v\), and \(w\) to express \(x\), \(y\), and \(z\):From \(u+v = 2x + 2y\), solve for \(x + y = \frac{u + v}{2}\).From \(u - v = 2z\), solve for \(z = \frac{u - v}{2}\).From \(w + y = x + z\) and using \(x + y\) and \(z\), solve for \(x = \frac{w + u}{2} - y\) and \(y = \frac{w - z}{2}\).
03

Calculate partial derivatives

Compute partial derivatives:For \(x\):\(\frac{\partial x}{\partial u} = \frac{1}{2}\), \(\frac{\partial x}{\partial v} = \frac{1}{2}\), \(\frac{\partial x}{\partial w} = \frac{1}{2}\).For \(y\):\(\frac{\partial y}{\partial u} = \frac{1}{2}\),\(\frac{\partial y}{\partial v} = \frac{1}{2}\),\(\frac{\partial y}{\partial w} = -\frac{1}{2}\).For \(z\):\(\frac{\partial z}{\partial u} = \frac{1}{2}\),\(\frac{\partial z}{\partial v} = -\frac{1}{2}\),\(\frac{\partial z}{\partial w} = \frac{1}{2}\).
04

Form the Jacobian matrix

The Jacobian matrix \( J \) is\[J = \begin{bmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \\end{bmatrix}= \begin{bmatrix} \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \end{bmatrix}.\]
05

Calculate the determinant of the Jacobian matrix

The determinant of the Jacobian matrix is found by calculating:\[\det(J) = \begin{vmatrix} \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \end{vmatrix}.\]Expanding along the first row, compute:\[ = \frac{1}{2} \left( \frac{1}{2}\times\frac{1}{2} - (-\frac{1}{2})\times(-\frac{1}{2}) \right) - \frac{1}{2} \left( \frac{1}{2}\times\frac{1}{2} - (-\frac{1}{2})\times(\frac{1}{2}) \right)+ \frac{1}{2} \left((-\frac{1}{2})\times(\frac{1}{2}) - \left(\frac{1}{2}\times\frac{1}{2}\right) \right).\]After simplification, the determinant is -1.

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Key Concepts

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

Determinant
The determinant is a critical tool in mathematics, especially in linear algebra. It is a scalar value derived from a square matrix. This number provides essential information about the matrix, such as whether the system of equations it represents has unique solutions. Also, a determinant of zero indicates that the matrix is singular, meaning it does not have an inverse.
When calculating the determinant of a 3x3 matrix, as in our exercise, we use the specific formula. This involves choosing any row or column (commonly the first) to expand along. The determinant depends on all elements on the chosen row or column, accounting for their signed minors.
Understanding how to expand the determinant can reveal many properties of the transformation represented by the matrix, including how it scales volumes in space.
Partial Derivatives
Partial derivatives are a cornerstone concept in calculus, often used in multivariable functions. They focus on a single variable while keeping others constant. This allows us to observe how changes in one variable influence a function.
In the context of the Jacobian, partial derivatives are used to measure the rate at which the transformed variables change with respect to the original ones. For each of the functions defining our new variables, we calculate partial derivatives for every original variable.
  • For instance, \(\frac{\partial x}{\partial u}\), is the rate of change of \(x\) with respect to \(u\).
  • Each entry in the Jacobian matrix is one of these partial derivatives.
Mastering partial derivatives helps in understanding changes in multidimensional spaces and is vital for optimization problems.
Jacobian Matrix
The Jacobian matrix is constructed from first-order partial derivatives. It is used to describe the transformation between coordinate systems. In this context, the Jacobian matrix \( J \) provides a linear approximation of a transformation between different variables.
The matrix offers a way to relate changes between the original variables \((u, v, w)\) and the new ones \((x, y, z)\). Here's how it works:
  • Each row of the Jacobian corresponds to the partial derivatives of a single transformed variable with respect to each original variable.
  • When multiplied by a vector of small changes in \((u, v, w)\), the Jacobian matrix yields the corresponding changes in \((x, y, z)\).
The capability of the Jacobian to map one space onto another is why it's heavily used in engineering and physics for coordinate transformations.
Transformation
Transformation refers to a way we map one set of variables into another. In mathematics, this often means changing from one coordinate system to another to simplify problem-solving.
In our exercise, we transformed from \((u, v, w)\) to \((x, y, z)\) through a system of equations. Each new variable is expressed as a function of the originals. Transformations are helpful in many fields, such as physics and engineering, where they simplify complex real-world problems.
  • They are crucial for converting complex shapes and motions into simpler forms or rotating and translating systems within \(\mathbb{R}^n\).
  • Understanding transformations allows one to see how certain systems behave under different conditions.
Mastering transformations leads to improved insights and problem-solving abilities in higher-dimensional spaces.

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Most popular questions from this chapter

Find the mass and center of gravity of the solid. The solid that has density \(\delta(x, y, z)=x z\) and is enclosed by \(\left.y=9-x^{2} \text { (for } x \geq 0\right), x=0, y=0, z=0,\) and \(z=1\)

Find the mass and center of gravity of the lamina. A lamina with density \(\delta(x, y)=x^{2}+y^{2}\) is bounded by the \(x\) -axis and the upper half of the circle \(x^{2}+y^{2}=1\)

Use a double integral in polar coordinates to find the area of the region described. The region enclosed by the rose \(r=\sin 2 \theta\)

Use cylindrical coordinates. Find the mass of a right circular cylinder of radius \(a\) and height \(h\) if the density is proportional to the distance from the base. (Let \(k\) be the constant of proportionality.)

These exercises reference the Theorem of Pappus: If \(R\) is a bounded plane region and \(L\) is a line that lies in the plane of \(R\) such that \(R\) is entirely on one side of \(L,\) then the volume of the solid formed by revolving \(R\) about \(L\) is given by $$\text {volume}=(\text {area of } R) \cdot\left(\begin{array}{c}{\text {distance} \text {traveled}} \\ {\text {by the centroid}}\end{array}\right)$$ Perform the following steps to prove the Theorem of Pappus: (a) Introduce an xy-coordinate system so that \(L\) is along the \(y\) -axis and the region \(R\) is in the first quadrant. Partition \(R\) into rectangular subregions in the usual way and let \(R_{k}\) be a typical subregion of \(R\) with center \(\left(x_{k}^{*}, y_{k}^{*}\right)\) and area \(\Delta A_{k}=\Delta x_{k} \Delta y_{k} .\) Show that the volume generated by \(R_{k}\) as it revolves about \(L\) is $$2 \pi x_{k}^{*} \Delta x_{k} \Delta y_{k}=2 \pi x_{k}^{*} \Delta A_{k}$$ (b) Show that the volume generated by \(R\) as it revolves about \(L\) is $$ V=\iint_{R} 2 \pi x d A=2 \pi \cdot \bar{x} \cdot[\text { area of } R] $$

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