Problem 16

Question

From $-x\left(4-y^{2}\right)=0$ we obtain $x=0, y=2,$ or $y=-2 .$ If $x=0,$ then substituting into $4 y\left(1-x^{2}\right)$ yields $y=0 .$ Likewise $y=2$ gives $8\left(1-x^{2}\right)=0$ or $x=1,-1 .$ Finally $y=-2$ yields $-8\left(1-x^{2}\right)=0$ or $x=1,-1$ The critical points of the system are therefore $(0,0),(1,2),(-1,2),(1,-2),$ and (-1,-2)

Step-by-Step Solution

Verified

Answer

The critical points are (0,0), (1,2), (-1,2), (1,-2), and (-1,-2).

1Step 1: Analyze the Initial Equation

We start with the given equation \[-x\left(4-y^{2}\right)=0\]For a product to be zero, at least one of the factors must be zero. This means:1. $-x = 0$, which implies $x = 0$.2. $4 - y^{2} = 0$, which implies solving $y^2 = 4$. Therefore, $y = 2$ or $y = -2$.

2Step 2: Evaluate Possibility When x=0

Substitute $x = 0$ into the second expression given,\[4y(1-x^2)\]This simplifies to:\[4y(1-0^2) = 4y\]Given that this should equal zero, solve:\[4y = 0\]So, $y = 0$. Therefore, one critical point is $(0,0)$.

3Step 3: Evaluate Possibility When y=2

Substituting $y=2$ into the expression,\[8(1-x^2) = 0\]Solve for $x$:- Divide by 8, resulting in $1-x^2 = 0$.- This gives us $x^2 = 1$, so $x = 1$ or $x = -1$.This provides critical points $(1, 2)$ and $(-1, 2)$.

4Step 4: Evaluate Possibility When y=-2

Similarly, substitute $y=-2$ into the expression,\[-8(1-x^2) = 0\]Solve for $x$:- Divide by -8, resulting in $1-x^2 = 0$.- This gives us $x^2 = 1$, so $x = 1$ or $x = -1$.This provides critical points $(1, -2)$ and $(-1, -2)$.

5Step 5: List the Critical Points

The final list of critical points where either $x$ or $y$ satisfy the equations given the initial conditions are \[(0, 0), (1, 2), (-1, 2), (1, -2), (-1, -2)\].

Key Concepts

Systems of EquationsFactorizationSolving Quadratic EquationsTwo-variable Systems

Systems of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. In the exercise given, the system is comprised of the original equation \[-x(4-y^2)=0\] and the constraints provided by substituting possible values of $x$ and $y$. Solving such a system involves finding the values of the variables that satisfy all equations in the system simultaneously.

This typically involves techniques such as substitution or elimination, depending on the complexity and nature of the equations.
For the equation provided, once one variable is known or assumed (e.g., $x=0$), it becomes easier to solve for the other variable $(y)$.
The goal is to find the 'critical points' — the variable pairs ($x, y$) that satisfy both equations.

Understanding how to manipulate and solve systems of equations is crucial for solving real-world problems in engineering, physics, and economics.

Factorization

Factorization is the process of breaking down an expression into a product of its simpler counterparts or 'factors'. It is a key tool in solving equations, as demonstrated in the exercise. For an equation like \[-x(4-y^2)=0\], it can be viewed as a product of two factors:

$-x$
$(4-y^2)$

This implies that for the product to be zero, at least one of these factors must be zero.

We set each factor equal to zero independently: $-x=0$ leading to $x=0$ and $4-y^2=0$ leading to $y=2$ or $y=-2$.

The solving becomes significantly easier after this breakdown, highlighting the power of factorization in simplifying and tackling equations.

Solving Quadratic Equations

Quadratic equations are polynomials of degree two of the form $ax^2 + bx + c=0$. In our exercise, the expression $4-y^2=0$ simplifies into a quadratic form:

$y^2 = 4$

Solving this involves taking the square root of both sides, which introduces two potential solutions because of the fundamental nature of squaring:

$y = 2$
$y = -2$

This approach of solving quadratic equations is fundamental, as it surfaces in diverse contexts in mathematics where finding roots of polynomials is essential. Spotting a quadratic form and knowing how to solve it—including handling cases of positive and negative roots—is crucial in higher-level math problem-solving.

Two-variable Systems

Two-variable systems involve equations with two unknowns, customarily $x$ and $y$. The exercise provides an excellent example, where each equation has two variables that must be solved together.

Our starting point was the equation $-x(4-y^2)=0$, making way for analyzing separate conditions where either $x=0$ or solving for $y$.
For $y=2$ or $y=-2$, after substitution into an additional given equation, additional critical points are determined, like $ (1, 2), (-1, 2)$, illustrating variable interdependence.

The process often requires iterative solving, testing conditions for each variable, and results in exact pairs of $x, y$ that fulfill both equations. This is a foundational part of algebra that underpins many engineering and economic models, where multiple factors influence outcomes simultaneously.

Problem 16

Problem 17

Other exercises in this chapter

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Problem 16

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Problem 17

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Problem 19

We have $d y / d x=y^{\prime} / x^{\prime}=-f(x) / y$ and so, using separation of variables, $$\frac{y^{2}}{2}=-\int_{0}^{x} f(\mu) d \mu+c \quad \text { or }

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