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Chapter 5: Problem 385

In the following exercises, find the Jacobian \(J\) of the transformation. $$ x=v \cosh \left(\frac{1}{u}\right), y=v \sinh \left(\frac{1}{u}\right), z=u+w^{2} $$

Short Answer

Expert verified
The Jacobian matrix \(J\) is \(\begin{bmatrix} -v \frac{\sinh(\frac{1}{u})}{u^2} & \cosh(\frac{1}{u}) & 0 \\ v \frac{\cosh(\frac{1}{u})}{u^2} & \sinh(\frac{1}{u}) & 0 \\ 1 & 0 & 2w \end{bmatrix}\).

Step by step solution

01

Identify the given transformation

In the problem statement, we are given a set of transformations for the variables \(x\), \(y\), and \(z\) in terms of \(u\), \(v\), and \(w\):\[ x = v \cosh\left(\frac{1}{u}\right) \]\[ y = v \sinh\left(\frac{1}{u}\right) \]\[ z = u + w^2 \]
02

Setup the Jacobian matrix

The Jacobian matrix \(J\) for a function \((x, y, z)\) with respect to \((u, v, w)\) is given as:\[J = \begin{bmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \\end{bmatrix}\]
03

Compute partial derivatives of \(x\)

For \(x = v \cosh\left(\frac{1}{u}\right)\):- \(\frac{\partial x}{\partial u} = v \cdot \left(-\frac{\sinh\left(\frac{1}{u}\right)}{u^2}\right), \) using the chain rule.- \(\frac{\partial x}{\partial v} = \cosh\left(\frac{1}{u}\right)\) directly, since \(x\) is linear in \(v\).- \(\frac{\partial x}{\partial w} = 0\) because \(x\) is independent of \(w\).
04

Compute partial derivatives of \(y\)

For \(y = v \sinh\left(\frac{1}{u}\right)\):- \(\frac{\partial y}{\partial u} = v \cdot \left(\frac{\cosh\left(\frac{1}{u}\right)}{u^2}\right), \) using the chain rule.- \(\frac{\partial y}{\partial v} = \sinh\left(\frac{1}{u}\right)\) directly, since \(y\) is linear in \(v\).- \(\frac{\partial y}{\partial w} = 0\) because \(y\) is independent of \(w\).
05

Compute partial derivatives of \(z\)

For \(z = u + w^2\):- \(\frac{\partial z}{\partial u} = 1\) directly, since \(z\) is linear in \(u\).- \(\frac{\partial z}{\partial v} = 0\) because \(z\) is independent of \(v\).- \(\frac{\partial z}{\partial w} = 2w\) because of the \(w^2\) term.
06

Construct the Jacobian matrix \(J\)

Using the partial derivatives computed in Steps 3, 4, and 5, the Jacobian matrix \(J\) is:\[J = \begin{bmatrix}-v \frac{\sinh\left(\frac{1}{u}\right)}{u^2} & \cosh\left(\frac{1}{u}\right) & 0 \v \frac{\cosh\left(\frac{1}{u}\right)}{u^2} & \sinh\left(\frac{1}{u}\right) & 0 \1 & 0 & 2w \\end{bmatrix}\]

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Key Concepts

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

Partial Derivatives
Partial derivatives are a fundamental concept when working with multivariable functions. When we have a function of several variables, a partial derivative represents the rate at which the function changes as only one of the variables is varied, keeping all other variables constant.

In the context of this exercise, we have functions like:
  • \(x = v \cosh\left(\frac{1}{u}\right)\)
  • \(y = v \sinh\left(\frac{1}{u}\right)\)
  • \(z = u + w^2\)
To find the partial derivatives, we differentiate each function with respect to the variables \(u\), \(v\), and \(w\). For instance, when finding \(\frac{\partial x}{\partial u}\), we differentiate the function \(x\) with respect to \(u\) assuming \(v\) and \(w\) remain constant. This gives us insight into how \(x\) changes specifically as \(u\) changes.

These derivatives are used to construct the Jacobian matrix, which offers a snapshot of how a function behaves in a local region by providing a linear approximation.
Transformation
Transformation refers to the process of changing variables in a function. In mathematics, especially in calculus and geometry, transforming coordinates can help simplify problems or bring them into a form that's easier to work with.

In the context of this exercise, we're looking at transforming from the variables \(u\), \(v\), and \(w\) to the space described by \(x\), \(y\), and \(z\). These transformations can be thought of as a way of reshaping the coordinate system to match new dimensions resulting from functions:
  • \(x = v \cosh\left(\frac{1}{u}\right)\)
  • \(y = v \sinh\left(\frac{1}{u}\right)\)
  • \(z = u + w^2\)
When we transform coordinates, we must understand how changes in one coordinate system (\(u\), \(v\), \(w\)) affect the new coordinate system (\(x\), \(y\), \(z\)). This understanding is encapsulated in the Jacobian matrix, which provides the rates of change necessary to describe this transformation.
Chain Rule
The chain rule is a critical tool in calculus for computing the derivative of a composite function. When dealing with functions of several variables, the chain rule helps us find partial derivatives where one function is nested within another.

In this exercise, we used the chain rule to find derivatives of \(x\) and \(y\) with respect to \(u\). For example, \(x = v \cosh\left(\frac{1}{u}\right)\) involves a composition of functions since \(\cosh\) is applied to another function, \(\frac{1}{u}\).
  • To find \(\frac{\partial x}{\partial u}\), we first differentiate \(\cosh\left(\frac{1}{u}\right)\) with respect to its inner function, \(\frac{1}{u}\), using the derivative \(\sinh\left(\frac{1}{u}\right)\).
  • Then we multiply by the derivative of the inner function itself with respect to \(u\), which is \(-\frac{1}{u^2}\).
This process of multiplying the derivative of the outer function by the derivative of the inner function is what the chain rule facilitates.

Mastery of the chain rule, especially in multivariable calculus, is vital for tackling complex transformations and understanding the dynamics of change across systems.

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Most popular questions from this chapter

Show that \(\iiint_{R} f\left(\sqrt{16 x^{2}+4 y^{2}+z^{2}}\right) d V=\frac{\pi}{2} \int_{0} f(\rho) \rho^{2} d \rho, \quad\) where \(f\) is a continuous function on [0,1] and \(R\) is the region bounded by the ellipsoid \(16 x^{2}+4 y^{2}+z^{2}=1\) .

Let \(Q\) be the solid bounded by the \(x y\) -plane, the cylinder \(x^{2}+y^{2}=a^{2},\) and the plane \(z=1,\) where \(a>1\) is a real number. Find the moment \(M_{x y}\) of the solid about the \(x y\) -plane if its density given in cylindrical coordinates is \(\rho(r, \theta, z)=\frac{d^{2} f}{d r^{2}}(r), \quad\) where \(f\) is a differentiable function with the first and second derivatives continuous and differentiable on \((0, a)\).

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In the following exercises, consider a lamina occupying the region \(R\) and having the density function \(\rho\) given in the first two groups of Exercises. a. Find the moments of inertia \(I_{x}, I_{y},\) and \(I_{0}\) about the \(x\) -axis, \(y\) -axis, and origin, respectively. b. Find the radii of gyration with respect to the \(x\) -axis, \(\quad y\) -axis, and origin, respectively. \(R\) is the trapezoidal region determined by the lines \(y=0, y=1, y=x\) and \(y=-x+3 ; \rho(x, y)=2 x+y\)

In the following exercises, the transformation \(T: S \rightarrow R, T(u, v)=(x, y)\) and the region \(R \subset \mathrm{R}^{2}\) are given. Find the region \(S \subset \mathrm{R}^{2}\). $$ \begin{array}{l} \text x=a u, y=b v, R=\left\\{(x, y) \mid \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leq 1\right\\}, \\ \text { where } a, b>0 \end{array} $$
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