Problem 2 - University Calculus: Early Transcendentals

Question

Jacobians and Transformed Regions in the Plane. a. Solve the system $$ u=x+2 y, \quad v=x-y $$ for \(x\) and \(y\) in terms of \(u\) and \(v .\) Then find the value of the Jacobian \(\partial(x, y) / \partial(u, v)\) b. Find the image under the transformation \(u=x+2 y\) \(v=x-y\) of the triangular region in the \(x y\) -plane bounded by the lines \(y=0, y=x,\) and \(x+2 y=2 .\) Sketch the transformed region in the \(u v\) -plane.

Step-by-Step Solution

Answer

The Jacobian is \\(-\frac{1}{3}\\), and the transformed region is a triangle with vertices at \\(0,0\\), \\(\frac{8}{3},0\\), \\(2,2\\)."

Step 1: Express x and y in Terms of u and v

We start with the given system of equations: \[ u = x + 2y \quad \text{and} \quad v = x - y. \]From the second equation, express \(x\) in terms of \(v\) and \(y\): \[ x = v + y. \]Substitute \(x = v + y\) into the first equation: \[ u = (v + y) + 2y = v + 3y. \]Thus, solve for \(y\): \[ y = \frac{u - v}{3}. \]Substitute \(y = \frac{u - v}{3}\) back into \(x = v + y\): \[ x = v + \frac{u - v}{3} = \frac{3v + u - v}{3} = \frac{u + 2v}{3}. \]

Step 2: Find Jacobian \(\frac{\partial(x, y)}{\partial(u, v)}\)

The Jacobian \(\frac{\partial(x, y)}{\partial(u, v)}\) is given by the determinant of the matrix of partial derivatives:\[\begin{vmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{vmatrix}.\]From \( x = \frac{u + 2v}{3} \) and \( y = \frac{u - v}{3} \), compute the partial derivatives:\[ \frac{\partial x}{\partial u} = \frac{1}{3}, \quad \frac{\partial x}{\partial v} = \frac{2}{3}, \quad \frac{\partial y}{\partial u} = \frac{1}{3}, \quad \frac{\partial y}{\partial v} = -\frac{1}{3}. \]Calculate the Jacobian:\[\begin{vmatrix}\frac{1}{3} & \frac{2}{3} \\frac{1}{3} & -\frac{1}{3}\end{vmatrix} = \frac{1}{3} \left(-\frac{1}{3}\right) - \frac{2}{3} \left(\frac{1}{3}\right) = \frac{-1 - 2}{9} = -\frac{1}{3}.\]

Step 3: Determine the Image of the Triangular Region

The triangular region in the \(xy\)-plane is defined by the lines \(y = 0\) (or \(x\)-axis), \(y = x\), and \(x + 2y = 2\). Let's find the vertices:1. Intersection of \(y = 0\) and \(x = 2\) gives \( (2, 0) \).2. Intersection of \(y = x\) and \(x + 2y = 2\): Solve \(x + 2x = 2\) giving \((\frac{2}{3}, \frac{2}{3})\).3. Intersection of \(x = 2\) and \(y = 0\) gives \((2, 0)\).Transform the vertices using \(u = x + 2y\) and \(v = x - y\):- \((0, 0) \rightarrow (0, 0) \)- \((2, 0) \rightarrow (2, 2) \)- \((\frac{2}{3}, \frac{2}{3}) \rightarrow (\frac{8}{3}, 0) \).Thus, the transformed image is a triangle in the \(uv\)-plane with vertices \((0, 0)\), \((2, 2)\), \((\frac{8}{3}, 0)\).

Step 4: Sketch the Transformed Region

Sketch in the \(uv\)-plane:- Place a point at \((0, 0)\).- Place a point at \((2, 2)\), which will be along the line \(u = v\).- Place a point at \((\frac{8}{3}, 0)\), which is along the \(u\)-axis.Connect these points to form a triangle. The base runs from \((0, 0)\) to \((\frac{8}{3}, 0)\) on the \(u\)-axis, and the opposite vertex is \((2, 2)\).

Key Concepts

Partial Derivatives

In math, partial derivatives play an essential role when dealing with functions of several variables. They help us see how a function changes when one of its input variables is varied, while the others remain fixed. It's like examining the effect of an ingredient in a recipe while keeping the quantities of other ingredients constant.

Partial derivatives are particularly vital in finding the Jacobian, which measures how a function alters space when a change of coordinates occurs. For instance, in the Jacobian calculation for the problem,

1. For \( x = \frac{u + 2v}{3} \)\, finding the partial derivative with respect to \( u \) gives \( \frac{1}{3} \). With respect to \( v \) you get \( \frac{2}{3} \).
2. Similarly, for \( y = \frac{u - v}{3} \), the partial derivatives are \( \frac{1}{3} \) for \( u \) and \(-\frac{1}{3}\) for \( v \).

These derivatives are used in determining the Jacobian, showcasing the change from one coordinate system to another.

Coordinate Transformation

Coordinate transformations involve changing from one set of coordinate axes to another. This is a fundamental concept in calculus and geometry because it allows us to solve problems more easily by choosing the best frame of reference.

In the context of the given exercise, the original coordinates, \( (x, y) \), are transformed to new coordinates, \( (u, v) \), through a set of equations:

• \( u = x + 2y \)
• \( v = x - y \)

Utilizing these transformations simplifies the triangular region's analysis from the \(xy\)-plane into the \(uv\)-plane. The significant advantage here is that it can often simplify the boundaries and shape of a region, making integration or other analyses much more manageable.

The Jacobian matrix—found using partial derivatives—provides the necessary information to understand how space is stretched or compressed during this transformation. It's like using a lens to view a map; different lenses or perspectives can give new insights.

Triangular Region Transformation

Triangular transformations involve understanding how shapes change under different coordinate systems. Transforming a triangular region allows for the simplification of complex boundary problems.

In our exercise, the original triangle in the \(xy\)-plane is bounded by the lines \( y = 0 \), \( y = x \), and \( x + 2y = 2 \). The vertices of this triangle are specifically:

• \( (0, 0) \)
• \( (2, 0) \)
• \( \left( \frac{2}{3}, \frac{2}{3} \right) \)

Transforming these to the \(uv\)-plane using the defined equations, \( u = x + 2y \) and \( v = x - y \), results in a new set of vertices:

• \( (0, 0) \) transforms to \( (0, 0) \)
• \( (2, 0) \) transforms to \( (2, 2) \)
• \( \left( \frac{2}{3}, \frac{2}{3} \right) \) transforms to \( \left( \frac{8}{3}, 0 \right) \)

The transformed region in the \(uv\)-plane now forms a different triangular shape, which is often easier to work with in calculations and visual representations. This process highlights how mathematical transformations can simplify geometry by altering perspectives and focus.