91Ó°ÊÓ

The rate of change is a fundamental concept in calculus. It measures how a quantity changes over time or another variable. When considering motion, as in the movement of a particle, rate of change could refer to speed or velocity. In this exercise, we are examining how the position of a particle, expressed by the variable \( x \), changes over time.
In mathematical terms, the rate of change of \( x \) with respect to time is expressed as \( \frac{dx}{dt} \). This represents the velocity of \( x \) regarding time. Similarly, the rate of change of \( y \), which is also a function of \( x \), relative to time is given as \( \frac{dy}{dt} \).
In the given exercise, we are tasked with finding the values of \( x \) where the rate of change of \( x \) is three times that of \( y \). This equation is expressed as \( \frac{dx}{dt} = 3 \times \frac{dy}{dt} \). Understanding these relationships allows us to better analyze the dynamics of the particle's motion.

Derivatives are a vital tool in calculus, used to determine the rate of change of a function. By finding the derivative of a function, we can understand how one variable changes in relation to another. In the context of our problem, we have the function \( y = \frac{x}{x^2 + 1} \).
To find \( \frac{dy}{dx} \), which represents how \( y \) changes concerning \( x \), we use calculus differentiation rules. Specifically, the quotient rule is applied here due to the structure of the function as a ratio of two expressions.
Calculating the derivative \( \frac{dy}{dx} \) provides a way to connect \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \), as \( \frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} \). This connection is pivotal to solving for the required value of \( x \) in our problem.

The quotient rule is a method for differentiating functions that are the ratio of two differentiable functions. In this context, our function \( y = \frac{x}{x^2 + 1} \) can be represented as the quotient of \( u = x \) and \( v = x^2 + 1 \).
The quotient rule states:

• \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \)

By applying this rule, we found:

• \( \frac{du}{dx} = 1 \) since the derivative of \( u = x \) is simply 1
• \( \frac{dv}{dx} = 2x \) from differentiating \( v = x^2 + 1 \)

Substituting into the quotient rule, we get \( \frac{dy}{dx} = \frac{1 - x^2}{(x^2 + 1)^2} \). This expression is crucial for solving the remaining parts of the problem.

The quadratic formula is a reliable way to find the roots of quadratic equations. Quadratic equations have the form \( ax^2 + bx + c = 0 \). The standard formula for finding solutions is:

• \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In our exercise, after determining the necessary relationship between \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), we derive a quadratic equation in terms of \( z = x^2 \), leading to \( z^2 + 5z - 2 = 0 \).
Applying the quadratic formula, and considering positive values of \( x^2 \), enables us to find that \( x^2 = \frac{-5 + \sqrt{33}}{2} \). Subsequently, the possible values for \( x \) are \( x = \pm \sqrt{\frac{-5 + \sqrt{33}}{2}} \). This step is crucial because it allows us to extract the actual \( x \)-values where the conditions of the problem are satisfied.