Problem 15 - Advanced Engineering Mathematics

Question

From \(x\left(1-x^{2}-3 y^{2}\right)=0\) we have \(x=0\) or \(x^{2}+3 y^{2}=1 .\) If \(x=0,\) then substituting into \(y\left(3-x^{2}-3 y^{2}\right)\) gives \(y\left(3-3 y^{2}\right)=0 .\) Therefore \(y=0,1,-1 .\) Likewise \(x^{2}=1-3 y^{2}\) yields \(2 y=0\) so that \(y=0\) and \(x^{2}=1-3(0)^{2}=1\) The critical points of the system are therefore \((0,0),(0,1),(0,-1),(1,0),\) and (-1,0)

Step-by-Step Solution

Answer

The critical points are \((0,0), (0,1), (0,-1), (1,0), (-1,0)\).

Step 1: Understanding the Initial Condition

Given the equation \(x(1-x^2-3y^2)=0\), identify the possible values of \(x\) by solving \(x=0\) or \(x^2+3y^2=1\).

Step 2: Solve for Condition When x=0

If \(x=0\), substitute it into the equation \(y(3-x^2-3y^2)=0\), simplifying it to \(y(3-3y^2)=0\). This implies \(y=0\) or \(3-3y^2=0\), hence \(y=0,1,-1\).

Step 3: Solve for Condition When x^2+3y^2=1

Here, rearrange the equation to \(x^2=1-3y^2\) and find the values of \(y\) that satisfy both \(2y=0\) and \(x^2=1-3(0)^2=1\). The equation \(2y=0\) implies \(y=0\). Thus, \(x^2=1\) yields \(x= \pm 1\).

Step 4: Identifying the Critical Points

Combine the results from previous steps to list all possible critical points: \((0,0), (0,1), (0,-1), (1,0), (-1,0)\). These are obtained by pairing the values of \(x\) and \(y\) that satisfy the given equations.

Key Concepts

Nonlinear Equations

Nonlinear equations are mathematical expressions where variables are not arranged in a straight line when graphed. Instead, they often involve squares, cubes, or other exponents beyond one. These equations usually appear as curves when plotted on a graph. Understanding nonlinear equations can be tricky because they don't always follow the simplest path of solutions, unlike linear equations.
Working with nonlinear equations can involve solving positions or values where conditions like zero slopes or intersections occur. In the main exercise here, we deal with two nonlinear equations resulting from the interaction of the terms involving squares. It leads to a more complex set of possible solutions that need careful solving.

• The equation \(x(1-x^2-3y^2)=0\) is nonlinear.
• Solving nonlinear equations often requires identifying all potential solutions for each variable.
• Nonlinear systems like this might have multiple critical points due to the square terms.

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. In this exercise, the system includes equations that share the variables \(x\) and \(y\). The goal is to find a set of values (or points) that satisfy all equations simultaneously. Solving systems of equations is essential in many fields, such as physics, economics, and engineering.
The exercise involves finding conditions where the equations intersect or hold true at certain points, called critical points.

• The system here includes the primary equations: \(x(1-x^2-3y^2)=0\) and \(y(3-x^2-3y^2)=0\).
• Systems can be solved by substitution, elimination, or graphically, depending on their nature.
• The approach involves examining cases separately, such as when \(x=0\) or \(x^2+3y^2=1\).

Algebraic Solutions

Algebraic solutions involve using mathematical operations to solve equations or systems of equations step by step. It's a methodical way to find the values of variables that satisfy given conditions. The use of algebraic manipulation allows us to break down complicated equations into simpler components.
In this exercise, algebraic solutions are employed to determine critical points by solving each equation for possible values of \(x\) and \(y\). By rearranging terms and isolating variables, we can find all possible combinations.

• By setting parts of the equations to zero, such as \(x=0\), solutions can be checked one at a time.
• Rearranging gives simplified forms like \(x^2=1-3y^2\), making it easier to evaluate.
• Combining results from different parts leads to complete solutions, here resulting in critical points \((0,0), (0,1), (0,-1), (1,0), (-1,0)\).