91Ó°ÊÓ

(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 $$z_{\max }=\frac{x_{\max }^{2}}{2}=\frac{1}{2}\left[e^{v_{0}^{2} / g}\left(1+x_{0}^{2}\right)-1\right]$$.

Step by step solution

01

Analyze Given Functions and Equations

We start by analyzing the function and its derivative given: \( f(x) = \frac{x^2}{2} \) and \( f'(x) = x \). Also, the differential equation is \( \frac{dy}{dx} = -g \frac{x}{1+x^2} \frac{1}{y} \). We need to separate variables to solve this.

02

Separate Variables

Rearrange the given differential equation to separate variables: \( y \cdot dy = -g \frac{x}{1+x^2} dx \). Integration can now occur on both sides with respect to their respective variables.

03

Integrate Both Sides

Integrate the left side with respect to \( y \) and the right side with respect to \( x \): \( \int y \, dy = \int -g \frac{x}{1+x^2} \cdot dx \).The integrals will result in:- Left: \( \frac{y^2}{2} \)- Right: \( -g \ln(1+x^2)/2 + C \) after simplification.

04

Simplify and Solve for C

From the integration, the equation becomes:\( y^2 = -g \ln(1+x^2) + C \).We use the initial conditions \( x(0) = x_0 \) and \( y(0) = v_0 \) to find \( C \): \( v_0^2 = -g \ln(1+x_0^2) + C \) \( \Rightarrow C = v_0^2 + g \ln(1+x_0^2) \).

05

Substitute Expression for C into Y Equation

Replacing \( C \) in the equation for \( y^2 \):\( y^2 = v_0^2 - g \ln(\frac{1+x^2}{1+x_0^2}) \).

06

Analyze Condition for Periodicity

The condition for the inequality \( v_0^2 - g \ln(\frac{1+x^2}{1+x_0^2}) \ge 0 \) must be satisfied, leading to \( x^2 \le e^{v_0^2/g}(1+x_0^2) - 1 \).

07

Find Maximum Value of X and Maximum Height Z

Given \( z = x^2 / 2 \), the maximum value \( x_{\max} = \sqrt{e^{v_0^2/g} (1+x_0^2) - 1} \). Consequently:\( z_{\max} = \frac{x_{\max}^2}{2} = \frac{1}{2}(e^{v_0^2/g}(1+x_0^2) - 1) \).

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Key Concepts

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

Variable Separation

The method of separating variables is a powerful technique for solving differential equations. It involves rearranging the equation such that each variable appears on a different side of the equation. This allows us to integrate both sides independently.

For the problem at hand, our differential equation is given as \( \frac{dy}{dx} = -g \frac{x}{1+x^2} \frac{1}{y} \). The aim is to move all terms involving \( y \) to one side and all terms involving \( x \) to the other. By rearranging, we can write it as \( y \cdot dy = -g \frac{x}{1+x^2} dx \), thus achieving the separation of variables.

Once this separation is achieved, each side of the equation can be integrated with respect to its variable. This step essentially turns the problem into a more manageable algebraic form.

Initial Conditions

Initial conditions are the values of the variables at a starting point, often given to find a particular solution for differential equations. They allow us to determine unknown constants that emerge during integration.

In our exercise, the initial conditions provided are \( x(0) = x_0 \) and \( y(0) = v_0 \). These conditions are crucial for solving for the integration constant \( C \). After integrating, we found the equation \( y^2 = -g \ln(1+x^2) + C \). Substituting the initial conditions \( x(0) = x_0 \) and \( y(0) = v_0 \) into this equation gives \( v_0^2 = -g \ln(1+x_0^2) + C \), which simplifies to \( C = v_0^2 + g \ln(1+x_0^2) \).

This constant \( C \) is crucial because it ensures that the solution reflects the initial state of the system.

Periodicity

Periodicity in this context refers to the repetitive nature of the solutions as we vary the parameters, particularly \( x \), over a cycle.

By examining the inequality \( v_0^2 - g \ln\left(\frac{1+x^2}{1+x_0^2}\right) \geq 0 \), we determine when the solution remains valid. This is crucial for identifying the range over which the function exhibits periodic behavior.

Simplifying this inequality gives us \( x^2 \leq e^{v_0^2 / g}(1+x_0^2) - 1 \). This inequality highlights the conditions necessary for the periodic solutions, essentially indicating that for each value of \( x \) within this range, there are two possible values for \( y \). The system, therefore, exhibits periodicity as \( x \) cycles through these values.

Maximum Value Analysis

To determine the maximum value of a variable, we should first consider the intervals over which the function is defined and any cyclical nature present. This helps us pinpoint where the variable might attain its peak value.

For this exercise, the maximum value of \( x \) occurs at the largest allowable value dictated by our earlier periodic condition. From our calculations, the equation \( x_{\max} = \sqrt{e^{v_0^2 / g}\left(1+x_0^2\right)-1} \) provides this maximum \( x \).

Using this \( x_{\max} \), we can then calculate the maximum value of \( z \), given by \( z = \frac{x^2}{2} \). Thus, the maximum height \( z_{\max} \) can be determined through \( z_{\max} = \frac{1}{2}\left(e^{v_0^2/g}(1+x_0^2) - 1\right) \). This gives us insight into the behavior of the system and allows for practical applications, such as predicting peak height in a periodic cycle.