Problem 2 - Thomas Calculus

Question

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 \(\partial(x, y) / \partial(u, v)\) \begin{equation} \begin{array}{l}{\text { b. Find the image under the transformation } u=x+2 y} \\\ {v=x-y \text { of the triangular region in the } x y \text { -plane bounded }} \\ {\text { by the lines } y=0, y=x, \text { and } x+2 y=2 . \text { Sketch the trans- }} \\ {\text { formed region in the } u v \text { -plane. }}\end{array} \end{equation}

Step-by-Step Solution

Answer

\(x = \frac{u + v}{2}\), \(y = \frac{u - v}{3}\), Jacobian = \(-\frac{1}{3}\).

Step 1: Solve for x in terms of u and v

We start with the equations \( u = x + 2y \) and \( v = x - y \). To solve for \( x \), add the equations:\[(u + v) = (x + 2y) + (x - y)\]Simplified, this gives:\[(u + v) = 2x + y\]or:\[ x = \frac{u + v}{2} \]

Step 2: Solve for y in terms of u and v

To find \( y \), subtract the second equation \( v = x - y \) from the first equation \( u = x + 2y \):\[(u - v) = (x + 2y) - (x - y)\]Simplifying, we get:\[(u - v) = 3y\]leading to:\[ y = \frac{u - v}{3} \]

Step 3: Jacobian Calculation

To find the Jacobian \( \frac{\partial(x, y)}{\partial(u, v)} \), compute partial derivatives of \( x \) and \( y \) with respect to \( u \) and \( v \).The Jacobian \( J \) is:\[J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}\]Compute the partial derivatives:\( \frac{\partial x}{\partial u} = \frac{1}{2} \), \( \frac{\partial x}{\partial v} = \frac{1}{2} \)\( \frac{\partial y}{\partial u} = \frac{1}{3} \), \( \frac{\partial y}{\partial v} = -\frac{1}{3} \)Thus, the Jacobian determinant is:\[J = \begin{vmatrix} \frac{1}{2} & \frac{1}{2} \ \frac{1}{3} & -\frac{1}{3} \end{vmatrix} = \left(\frac{1}{2}\right)\left(-\frac{1}{3}\right) - \left(\frac{1}{2}\right)\left(\frac{1}{3}\right)\]\[J = -\frac{1}{6} - \frac{1}{6} = -\frac{1}{3}\]

Step 4: Find image of triangular region

The triangular region in the \(xy\)-plane is bounded by the lines \(y = 0\), \(y = x\), and \(x + 2y = 2\).**Transform each line using the equations for \(u\) and \(v\):**- For \( y = 0 \), substituting in \( y \) gives \( u = x \) and \( v = x \), so \( u = v \).- For \( y = x \), substituting in \( y \) gives \( u = 3x \) and \( v = 0 \).- For \( x + 2y = 2 \), substituting in the expression gives \( u = 2 \).Thus, the transformed region in the \(uv\)-plane is bounded by \(u = v\), \( v = 0 \), and \( u = 2 \).

Step 5: Sketch Transformed Region

In the \(uv\)-plane sketch:- A line \( u = v \) from \( v = 0 \) to \( u = 2 \).- A horizontal line \( v = 0 \) from the origin up to \( u = 2 \).- A vertical line \( u = 2 \) going from \( v = 0 \) to the intersection with \( u = v \).This forms a triangular area in the \( uv \)-plane defined by \( 0 \leq v \leq u \leq 2 \).

Key Concepts

Linear Transformations

A linear transformation is a crucial mathematical concept that deals with mapping between two vector spaces. It enables us to transform vectors in one space into vectors in another space without altering the vector structure. This exercise involves such a transformation where the system of equations \( u = x + 2y \) and \( v = x - y \) maps coordinates from the \(xy\)-plane to the \(uv\)-plane.

Understanding linear transformations helps in simplifying complex geometrical problems and allows us to express complex systems in simpler terms. They are instrumental in various fields such as graphics, physics, and engineering.

• Vector operations such as addition and scalar multiplication remain consistent after transformation.
• Linear transformations can be represented with matrices, providing a powerful tool for computation.

System of Equations

A system of equations involves multiple equations that share common variables and are solved simultaneously. In our context, the system given by \( u = x + 2y \) and \( v = x - y \) requires us to find \(x\) and \(y\) in terms of \(u\) and \(v\). This requires algebraic manipulation to isolate the variables of interest.

Solving these equations typically involves techniques such as addition or subtraction and substitution. By manipulating these equations, we reformulate them, finding expressions for \(x\) and \(y\). This essentially "unlocks" the structure of the transformation, detailing how coordinates from one system relate to another. Here, we added the two equations to simplify and solve for \(x\), and then subtracted to find \(y\).

• Solving systems of equations is foundational in fields that require optimization and modeling.
• Representation of solutions often involves expressions or matrices, aiding in further analyses like finding Jacobians or derivatives.

Partial Derivatives

Partial derivatives are vital in multi-variable calculus. They measure how a function changes as each variable changes, holding others constant. Here, partial derivatives are used to compute the Jacobian matrix, which describes how transformations affect differential areas.

We found the partial derivatives of \(x\) and \(y\) with respect to \(u\) and \(v\). Each partial derivative reveals how slight changes in \(u\) or \(v\) affect \(x\) or \(y\). For instance, if \(u\) increases slightly, the partial \( \frac{\partial x}{\partial u} \) tells us how much \(x\) will change as a result, while keeping \(v\) constant.

• Partial derivatives are crucial in the differentiation of functions with several variables.
• They provide insights into the function's behavior near a point, useful in optimization and error analysis.

Coordinate Transformation

Coordinate transformation refers to expressing the position of a point differently based on its transformation from one coordinate system to another. In exercises like this, these transformations often simplify the geometry of a problem, revealing hidden structures.

In the problem, the coordinate transformation from \(xy\)-plane to \(uv\)-plane is achieved using the given equations. By transforming this into the \(uv\)-space, the bounding triangle also transforms, simplifying the representation of the area into a new set of coordinates.

• Transformations provide clear pathways to solve integrals, analyze curves, and determine images of regions.
• They are highly utilized in physics to switch between Cartesian and polar coordinates when solving problems.