Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the image of the rectangle with the given corners and find the Jacobian of the transformation. $$ x=u^{2}+v^{2}, y=v ;(0,0),(1,0),(1,1),(0,1) $$

Step-by-Step Solution

Answer

The image corners in the (x,y) plane are (0,0), (1,0), (2,1), and (1,1). The Jacobian determinant is 2u.

Step 1: Identify the Transformation Functions

We begin by observing the transformation functions given in the problem: \( x = u^2 + v^2 \) and \( y = v \). These functions will be used to transform the points of the rectangle in the \((u, v)\) coordinate system.

Step 2: Find Image of Each Corner

Transform each corner of the rectangle:- For \((u,v) = (0,0)\), \(x = 0^2 + 0^2 = 0\), \(y = 0\). Image: \((0,0)\).- For \((u,v) = (1,0)\), \(x = 1^2 + 0^2 = 1\), \(y = 0\). Image: \((1,0)\).- For \((u,v) = (1,1)\), \(x = 1^2 + 1^2 = 2\), \(y = 1\). Image: \((2,1)\).- For \((u,v) = (0,1)\), \(x = 0^2 + 1^2 = 1\), \(y = 1\). Image: \((1,1)\).The images of the corners are \((0,0), (1,0), (2,1), (1,1)\).

Step 3: Set Up the Jacobian Matrix

The Jacobian matrix of the transformation is calculated as follows:\[J = \begin{bmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix}\].Compute the partial derivatives:- \(\frac{\partial x}{\partial u} = 2u\).- \(\frac{\partial x}{\partial v} = 2v\).- \(\frac{\partial y}{\partial u} = 0\).- \(\frac{\partial y}{\partial v} = 1\).Thus, \[J = \begin{bmatrix} 2u & 2v \ 0 & 1 \end{bmatrix}\].

Step 4: Compute the Determinant of the Jacobian

The determinant of the Jacobian matrix is given by\[ \text{det}(J) = \left(2u \cdot 1\right) - \left(2v \cdot 0\right) = 2u \].Thus, the determinant of the Jacobian is \(2u\).

Key Concepts

Calculus

Calculus is a vast area of mathematics that deals with change and motion. It encompasses two main branches: differential calculus and integral calculus.
While differential calculus focuses on derivatives and rates of change, integral calculus involves the summation of infinitely small factors to determine a whole.
The exercise you're working with involves using calculus to transform a coordinate system, showing how mathematical functions can map one space to another.

• Calculus is not just about equations; it's a tool to model real-world phenomena.
• Its concepts are used in various fields such as physics, engineering, economics, and more.

By applying calculus, especially through transformations and derivatives, we can solve complex problems and understand systems more clearly.

Partial Derivatives

Partial derivatives are an extension of the concept of a derivative to functions of several variables.
Where a regular derivative would measure the rate of change of a function with respect to a single variable, a partial derivative looks at the rate of change in relation to one of multiple variables, holding the others constant.
In the given exercise, the transformation involves solving for partial derivatives within the Jacobian matrix.

• For a function \( f(u, v) \), the partial derivative with respect to \( u \) is written as \( \frac{\partial f}{\partial u} \).
• Similarly, the partial derivative with respect to \( v \) is \( \frac{\partial f}{\partial v} \).

The calculation of these partial derivatives helps in understanding how each input variable contributes to the change in the transformed coordinates.

Determinant of Jacobian

The Jacobian matrix is crucial when transforming coordinates, especially from one dimension to another.
It requires partial derivatives of the transformation equations. The determinant of the Jacobian provides a scalar value that offers insights into the transformation's local behavior.
- In this exercise, the Jacobian matrix is: \[ J = \begin{bmatrix} 2u & 2v \ 0 & 1 \end{bmatrix} \]- The determinant of this matrix, \( \text{det}(J) = 2u \), indicates scaling effects and orientation changes in the plane.

• A non-zero determinant suggests that the transformation is invertible.
• If the determinant is zero, it implies potential issues like collapse of dimensions or non-reversible transformations.

Understanding the determinant enables us to predict how space is manipulated via transformation functions.

Coordinate Transformation

Coordinate transformations involve changing the system by which a point in space is described, such as moving from one set of variables to another.
They are key tools in physics and engineering to simplify problems.
The exercise demonstrates a transformation from the \((u, v)\) to the \((x, y)\) space using given functions \( x = u^2 + v^2 \) and \( y = v \).

• Each original point on the rectangle is substituted into the transformation functions to find its new location.
• The resulting new points form a shape that may be analyzed or utilized for further applications.

By understanding coordinate transformations, we can effectively remap systems and gain insights into their structures without changing inherent properties.