Problem 41 - Calculus - AP Edition

Question

A particle is moving along the curve \(y=x /\left(x^{2}+1\right) .\) Find all values of \(x\) at which the rate of change of \(x\) with respect to time is three times that of \(y .\) [Assume that \(d x / d t\) is never zero.J

Step-by-Step Solution

Answer

The values of \( x \) are \( \pm \sqrt{\frac{-5 + \sqrt{33}}{2}} \)."

Step 1: Understand the problem

We are given a curve with the equation \( y = \frac{x}{x^2 + 1} \). We need to find values of \(x\) where the rate of change of \(x\) with respect to time \( \frac{dx}{dt} \) is three times the rate of change of \(y\) with respect to time \( \frac{dy}{dt} \).

Step 2: Differentiate y with respect to x

To find \( \frac{dy}{dt} \), we first need \( \frac{dy}{dx} \). So, differentiate \( y = \frac{x}{x^2 + 1} \) using the quotient rule:\[\frac{dy}{dx} = \frac{(x^2 + 1) \cdot (1) - x \cdot (2x)}{(x^2 + 1)^2} = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2}.\]

Step 3: Relate rates of change

Using the chain rule, \( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \). According to the problem, \( \frac{dx}{dt} = 3\frac{dy}{dt} \). Substitute \( \frac{dy}{dt} \) into this equation:\[\frac{dx}{dt} = 3 \left(\frac{1 - x^2}{(x^2 + 1)^2} \right) \frac{dx}{dt}.\]

Step 4: Solve for x

Since \( \frac{dx}{dt} eq 0 \), we can divide both sides of the equation by \( \frac{dx}{dt} \) and simplify:\[1 = 3 \left(\frac{1 - x^2}{(x^2 + 1)^2} \right).\]Multiply both sides by \( (x^2 + 1)^2 \) and solve for \(x\):\[(x^2 + 1)^2 = 3(1 - x^2).\]Expand and solve:\[ x^4 + 2x^2 + 1 = 3 - 3x^2 \] \[ x^4 + 5x^2 - 2 = 0. \]

Step 5: Solve the quadratic form

Now, let \( u = x^2 \), transforming the equation to \( u^2 + 5u - 2 = 0 \). Use the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):\[ u = \frac{-5 \pm \sqrt{25 + 8}}{2} = \frac{-5 \pm \sqrt{33}}{2}.\]Hence, \( u = \frac{-5 + \sqrt{33}}{2} \) and \( u = \frac{-5 - \sqrt{33}}{2} \). Only the first solution is valid as \( u = x^2 \geq 0 \), so \( x^2 = \frac{-5 + \sqrt{33}}{2} \).

Step 6: Find x

Solve for \( x \) by taking the square root of \( x^2 = \frac{-5 + \sqrt{33}}{2} \). Therefore, the values for \( x \) are \( \pm \sqrt{\frac{-5 + \sqrt{33}}{2}} \).

Key Concepts

Quotient Rule

The quotient rule is an essential tool in calculus, particularly when dealing with the differentiation of functions expressed as a ratio of two differentiable functions. If you have a function, say \( y = \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the derivative \( \frac{dy}{dx} \) is derived by applying the quotient rule. The formula for the quotient rule is:

• \( \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \)

In our exercise, \( y = \frac{x}{x^2 + 1} \). Here, \( u = x \) and \( v = x^2 + 1 \). When using the quotient rule, follow these steps:

• Differentiate the numerator \( u = x \) to get \( \frac{du}{dx} = 1 \).
• Differentiate the denominator \( v = x^2 + 1 \) to get \( \frac{dv}{dx} = 2x \).
• Substitute into the formula: \( \frac{dy}{dx} = \frac{(x^2 + 1) \cdot 1 - x \cdot 2x}{(x^2 + 1)^2} \).
• Simplify to get \( \frac{dy}{dx} = \frac{1 - x^2}{(x^2 + 1)^2} \).

Understanding this rule is key as it allows us to navigate through the differentiation of complex rational functions.

Chain Rule

The chain rule is another pivotal concept in calculus, required when differentiating composite functions. If a variable \( y \) is a function of \( u \), and \( u \) itself is a function of \( x \), the chain rule helps find the derivative of \( y \) with respect to \( x \). It's expressed as:

• \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)

Within the context of the given exercise, we use the chain rule to relate the rates of change of \( y \) and \( x \) with respect to time. That is, the rate \( \frac{dy}{dt} \) can be determined using:

• \( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \)

The exercise stipulates that the rate of \( x \) with respect to time, \( \frac{dx}{dt} \), is three times the rate of \( y \) with respect to time: \( \frac{dx}{dt} = 3 \cdot \frac{dy}{dt} \). This leads us to substitute and rearrange terms to solve for \( x \) accordingly. This application of the chain rule connects the understanding of derivatives with the concepts of rates of change across different variables.

Quadratic Formula

The quadratic formula is a fundamental algebraic solution technique for quadratic equations of the form \( ax^2 + bx + c = 0 \). The formula itself is given by:

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

In step 5 of our problem, we reformulate the expression \( x^4 + 5x^2 - 2 = 0 \) by setting \( u = x^2 \), thus reducing it to a simpler quadratic form \( u^2 + 5u - 2 = 0 \). Here, the quadratic formula is applied:

• Substitute \( a = 1 \), \( b = 5 \), and \( c = -2 \).
• Calculate: \( u = \frac{-5 \pm \sqrt{25 + 8}}{2} = \frac{-5 \pm \sqrt{33}}{2} \).

This resolves to potential values of \( u \), which can be further analyzed to find viable solutions for \( x \). Only the non-negative solution is considered due to the natural constraint \( x^2 \geq 0 \). This approach illustrates the quadratic formula's utility in transforming complex algebraic problems into solvable linear tasks.