/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 From \(y^{2}-x=0\) we have \(x=y... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 8

From \(y^{2}-x=0\) we have \(x=y^{2} .\) Substituting into \(x^{2}-y=0,\) we obtain \(y^{4}-y=0\) or \(y\left(y^{3}-1\right)=0 .\) It follows that \(y=0,1\) and so (0,0) and (1,1) are the critical points of the system.

Short Answer

Expert verified
The critical points are (0,0) and (1,1).

Step by step solution

01

Identify Original Equations

We start with two equations given in the problem: \( y^2 - x = 0 \) and \( x^2 - y = 0 \).
02

Express x in terms of y

From the equation \( y^2 - x = 0 \), we can solve for \( x \) to get \( x = y^2 \).
03

Substitute x into Second Equation

Substitute \( x = y^2 \) from the first equation into the second equation \( x^2 - y = 0 \). This gives \( (y^2)^2 - y = 0 \), which simplifies to \( y^4 - y = 0 \).
04

Factor the Simplified Equation

Factor the equation \( y^4 - y = 0 \) to get \( y(y^3 - 1) = 0 \). This gives us two possible scenarios: \( y = 0 \) or \( y^3 - 1 = 0 \).
05

Solve for y

For \( y = 0 \), that is one solution. For \( y^3 - 1 = 0 \), solve to find \( y^3 = 1 \), which results in \( y = 1 \). Thus, we have the solutions \( y = 0 \) and \( y = 1 \).
06

Determine Corresponding x Values

Substitute the values for \( y \) back into \( x = y^2 \). For \( y = 0 \), \( x = 0^2 = 0 \). For \( y = 1 \), \( x = 1^2 = 1 \).
07

Identify Critical Points

The critical points of the system are the pairs \((x, y)\) obtained: \((0, 0)\) and \((1, 1)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

System of Equations
A system of equations consists of two or more equations that share common variables. The goal is often to find the values of these variables that satisfy all the equations simultaneously. In the provided exercise, we have two equations: \(y^2 - x = 0\) and \(x^2 - y = 0\). The variables \(x\) and \(y\) must satisfy both equations at the same time.

When dealing with systems of equations, there are several methods to find solutions. You can use graphical methods to visualize where the equations intersect. Alternatively, algebraic methods, such as substitution and elimination, are preferred for precise solutions. In our exercise, substitution is chosen to solve the system.
Polynomial Factoring
Polynomial factoring is a mathematical process used to express a polynomial as a product of simpler polynomials, which can then be used to solve equations. It is akin to breaking down a number into its prime factors, but in this case, for polynomials.

In our exercise, the equation \(y^4 - y = 0\) results from substituting \(x = y^2\) into another equation. Factoring involves identifying common factors or special polynomial forms. Here, \(y\) is a common factor, so we rewrite the equation as \(y(y^3 - 1) = 0\).
  • \(y = 0\) is one factor, resulting directly in \(y = 0\).
  • For \(y^3 - 1 = 0\), this can be further solved to find the remaining solutions, such as \(y = 1\), by recognizing special identities like \(a^3 - b^3\).
This factoring approach simplifies complex polynomials and aids in finding solutions.
Substitution Method
The substitution method is an algebraic technique for solving a system of equations. It involves expressing one variable in terms of another and substituting this expression into the other equation(s). This helps to reduce the number of variables and simplify the solving process.

In the exercise, substitution starts with the equation \(y^2 - x = 0\), from which we easily find \(x = y^2\). This expression of \(x\) is then used in the second equation, \(x^2 - y = 0\), replacing \(x\) with \(y^2\). This transformation leads to a new equation \((y^2)^2 - y = 0\), or \(y^4 - y = 0\).

The substitution method is beneficial because it often reduces the number of steps required to solve complex systems. With fewer variables and simpler forms, finding solutions becomes more manageable.
Nonlinear Equations
Nonlinear equations are those in which the variables are raised to a power other than one, or they involve products of variables. They are distinguished from linear equations, which graph as straight lines. Nonlinear equations present more complexity due to curves and multiple solutions.

In the original exercise, the equations \(y^2 - x = 0\) and \(x^2 - y = 0\) are nonlinear, as both involve squares of variables. Nonlinear systems can result in multiple critical points, as seen in the critical points \((0, 0)\) and \((1, 1)\).
  • Tackling nonlinear equations often requires specific strategies: substitution, factoring, graphing, or numerical methods.
  • Recognizing the form and degree of nonlinearity helps in choosing the right approach for solving.
Understanding nonlinear equations opens the door to solving real-world problems that involve curves and complex interactions between variables.

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Most popular questions from this chapter

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)

We have \(d y / d x=y^{\prime} / x^{\prime}=-f(x) / y\) and so, using separation of variables, $$\frac{y^{2}}{2}=-\int_{0}^{x} f(\mu) d \mu+c \quad \text { or } \quad y^{2}+2 F(x)=c$$ . We can conclude that for a given value of \(x\) there are at most two corresponding values of \(y .\) If (0,0) were a stable spiral point there would exist an \(x\) with more than two corresponding values of \(y .\) Note that the condition \(f(0)=0\) is required for (0,0) to be a critical point of the corresponding plane autonomous system \(x^{\prime}=y, y^{\prime}=-f(x)\).

If \(\operatorname{det}(s \mathbf{I}-\mathbf{A})=0,\) then \(s\) is an eigenvalue of \(\mathbf{A} .\) Thus \(s \mathbf{I}-\mathbf{A}\) has an inverse if \(s\) is not an eigenvalue of \(\mathbf{A} .\) For the purposes of the discussion in this section, we take \(s\) to be larger than the largest eigenvalue of A. Under this condition \(s \mathbf{I}-\mathbf{A}\) has an inverse.

(a) If \(f(x)=x^{2} / 2, f^{\prime}(x)=x\) and so $$\frac{d y}{d x}=\frac{y^{\prime}}{x^{\prime}}=-g \frac{x}{1+x^{2}} \frac{1}{y}$$. We can separate variables to show that \(y^{2}=-g \ln \left(1+x^{2}\right)+c .\) But \(x(0)=x_{0}\) and \(y(0)=x^{\prime}(0)=v_{0}\) Therefore \(c=v_{0}^{2}+g \ln \left(1+x_{0}^{2}\right)\) and so $$y^{2}=v_{0}^{2}-g \ln \left(\frac{1+x^{2}}{1+x_{0}^{2}}\right)$$. Now $$v_{0}^{2}-g \ln \left(\frac{1+x^{2}}{1+x_{0}^{2}}\right) \geq 0 \quad \text { if and only if } \quad x^{2} \leq e^{v_{0}^{2} / g}\left(1+x_{0}^{2}\right)-1$$. Therefore, if \(|x| \leq\left[e^{v_{0}^{2} / g}\left(1+x_{0}^{2}\right)-1\right]^{1 / 2},\) there are two values of \(y\) for a given value of \(x\) and so the solution is periodic. (b) since \(z=x^{2} / 2,\) the maximum height occurs at the largest value of \(x\) on the cycle. From (a), \(x_{\max }=\) \(\left[e^{v_{0}^{2} / g}\left(1+x_{0}^{2}\right)-1\right]^{1 / 2}\) and so $$z_{\max }=\frac{x_{\max }^{2}}{2}=\frac{1}{2}\left[e^{v_{0}^{2} / g}\left(1+x_{0}^{2}\right)-1\right]$$.

Switching to polar coordinates, $$\begin{array}{l} \frac{d r}{d t}=\frac{1}{r}\left(x \frac{d x}{d t}+y \frac{d y}{d t}\right)=\frac{1}{r}\left(-x y-x^{2} r^{4}+x y-y^{2} r^{4}\right)=-r^{5} \\ \frac{d \theta}{d t}=\frac{1}{r^{2}}\left(-y \frac{d x}{d t}+x \frac{d y}{d t}\right)=\frac{1}{r^{2}}\left(y^{2}+x y r^{4}+x^{2}-x y r^{4}\right)=1 \end{array}$$ If we use separation of variables on \(\frac{d r}{d t}=-r^{5}\) we obtain $$r=\left(\frac{1}{4 t+c_{1}}\right)^{1 / 4} \quad \text { and } \quad \theta=t+c_{2}$$ since \(\mathbf{X}(0)=(4,0), r=4\) and \(\theta=0\) when \(t=0 .\) It follows that \(c_{2}=0\) and \(c_{1}=\frac{1}{256} .\) The final solution can be written as $$r=\frac{4}{\sqrt[4]{1024 t+1}}, \quad \theta=t$$ and so the solution spirals toward the origin as \(t\) increases.
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