Problem 4 - Advanced Engineering Mathematics

Question

Find the image of the set \(S\) under the given transformation. $$ S:-1 \leq u \leq 4,1 \leq v \leq 5 ; u=x-y, v=x+2 y $$

Step-by-Step Solution

Answer

The image of set \( S \) is \( \left\{ (x, y) \mid \frac{1}{3} \leq x \leq \frac{14}{3}, -1 \leq y \leq \frac{4}{3} \right\} \).

Step 1: Express Transformation Equations

The given transformation is defined by the equations \( u = x - y \) and \( v = x + 2y \). We want to find expressions for \( x \) and \( y \) in terms of \( u \) and \( v \).

Step 2: Solve for x and y

To find \( x \) and \( y \), solve the equations:1. \( u = x - y \)2. \( v = x + 2y \)By adding these equations, we get \( u + v = 2x + y \). By solving the system of equations, we get:\[ x = \frac{u+2v}{3}, \quad y = \frac{-u+v}{3} \]

Step 3: Determine the Range for x and y

Given \(-1 \leq u \leq 4\) and \(1 \leq v \leq 5\), compute the range for \( x \) and \( y \) using the expressions found in Step 2:- For \( x = \frac{u+2v}{3} \), calculate the minimum and maximum by evaluating \(x\) at the bounds of \(u\) and \(v\): - Minimum: \( x_{min} = \frac{-1 + 2(1)}{3} = \frac{1}{3} \) and \( x_{min} = \frac{-1 + 2(5)}{3} = 3 \) - Maximum: \( x_{max} = \frac{4 + 2(5)}{3} = \frac{14}{3} \)- Thus, \( \frac{1}{3} \leq x \leq \frac{14}{3} \).- For \( y = \frac{-u+v}{3} \): - Minimum: \( y_{min} = \frac{-4 + 1}{3} = -1 \) - Maximum: \( y_{max} = \frac{-1 + 5}{3} = \frac{4}{3} \)- Thus, \( -1 \leq y \leq \frac{4}{3} \).

Step 4: Find the Image of S

The image of set \( S \) under the given transformation, using the ranges computed, is:\[ \left\{ (x, y) \mid \frac{1}{3} \leq x \leq \frac{14}{3}, -1 \leq y \leq \frac{4}{3} \right\} \]

Key Concepts

Coordinate Transformation

A coordinate transformation involves changing the context in which data in a specific coordinate system is expressed. It helps us understand how one system relates to another.
In this case, our initial coordinates, defined by variables \(u\) and \(v\), are related to another set involving \(x\) and \(y\). The given transformation is characterized by the equations \(u = x - y\) and \(v = x + 2y\). These can be thought of as a recipe for translating each point from the \(x, y\) plane to the \(u, v\) plane.
The transformation aims to capture the relationship between these sets of coordinates, offering a map from one domain to another.

• In practical terms, this transformation helps visualize and work with data in different orientations or perspectives within mathematics.

By establishing a framework of translation between \(x, y\) and \(u, v\) while maintaining geometric or algebraic properties, we can query or analyze aspects from both coordinate systems efficiently.

Inverse Transformation

An inverse transformation is essential for converting data back from the newly defined coordinate system to its original form.
This process involves reversing the initial transformation equations. Here, our task is to find expressions for \(x\) and \(y\) in terms of \(u\) and \(v\).
Solving the given equations:
1. \(u = x - y\)
2. \(v = x + 2y\)
By solving these two equations simultaneously, we extract
\[ x = \frac{u + 2v}{3} \]
\[ y = \frac{-u + v}{3} \]

• This provides the mechanism to express original coordinates based on transformed values, effectively reversing the transformation process.

The inverse transformation is critical in applications where one needs to understand how changes in one system affect another or revert changes to restore the original data context.

Range Calculation

Range calculation in this context refers to determining where the transformed coordinates lie.
Using the previously acquired expressions of \(x\) and \(y\) in terms of \(u\) and \(v\), we compute the new boundaries of our data set under the transformation.
Given:

• \(-1 \leq u \leq 4\)
• \(1 \leq v \leq 5\)

we substitute these into the expressions for \(x\) and \(y\):

• For \(x = \frac{u + 2v}{3}\),
  • the minimum \(x_{min}\) becomes \( \frac{1}{3}\),
  • the maximum \(x_{max}\) is \( \frac{14}{3}\).
• For \(y = \frac{-u + v}{3}\),
  • the minimum \(y_{min}\) is \(-1\)
  • and the maximum \(y_{max}\) reads \( \frac{4}{3}\).

This computation gives us the boundaries of the image set: \(\left\{ (x, y) \mid \frac{1}{3} \leq x \leq \frac{14}{3}, -1 \leq y \leq \frac{4}{3} \right\}\).
Range calculations are vital for understanding the limits and constraints of transformed sets, which is essential for analysis and problem-solving in fields like physics, graphics, and engineering.