Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(21-30\), find \(d y / d x\) and \(d^{2} y / d x^{2}\) without eliminating the parameter. $$ x=6 s^{2}, y=-2 s^{3} ; s \neq 0 $$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = -\frac{s}{2}\), \(\frac{d^2y}{dx^2} = -\frac{1}{24s}\).

Step 1: Find dx/ds

Given the parametric equations: - \(x = 6s^2\)Differentiate \(x\) with respect to \(s\):\[ \frac{dx}{ds} = \frac{d}{ds}(6s^2) = 12s \]

Step 2: Find dy/ds

Given the parametric equations: - \(y = -2s^3\)Differentiate \(y\) with respect to \(s\):\[ \frac{dy}{ds} = \frac{d}{ds}(-2s^3) = -6s^2 \]

Step 3: Find dy/dx

We use the chain rule to find \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{dy}{ds} \cdot \frac{ds}{dx} = \frac{-6s^2}{12s} = -\frac{s}{2} \]

Step 4: Differentiate dy/dx with respect to s

Differentiate \( -\frac{s}{2} \) with respect to \(s\):\[ \frac{d}{ds} \left( -\frac{s}{2} \right) = -\frac{1}{2} \]

Step 5: Find d²y/dx²

To find \( \frac{d^2y}{dx^2} \), use \[ \frac{d^2y}{dx^2} = \frac{d}{ds}\left(\frac{dy}{dx}\right) \cdot \frac{ds}{dx} \]From Step 4, \( \frac{d}{ds}\left(\frac{dy}{dx}\right) = -\frac{1}{2} \) and from Step 1, \( \frac{ds}{dx} = \frac{1}{12s}\):\[ \frac{d^2y}{dx^2} = -\frac{1}{2} \cdot \frac{1}{12s} = -\frac{1}{24s} \]

Key Concepts

Chain Rule

The chain rule is an essential part of calculus, particularly when working with parametric equations. It helps us find the derivative of one variable with respect to another by relating their derivatives through a parameter. In simpler terms, when you have two functions dependent on a third variable, the chain rule connects their rates of change.

In the original problem, we're given parametric equations:

• \( x = 6s^2 \)
• \( y = -2s^3 \)

To find \( \frac{dy}{dx} \), we cannot directly differentiate \( y \) with respect to \( x \) because both \( x \) and \( y \) are defined using a third variable \( s \). Instead, we differentiate each with respect to \( s \) and then apply the chain rule. Specifically:

• First, find \( \frac{dy}{ds} \) and \( \frac{dx}{ds} \)
• Then, use the chain rule formula: \( \frac{dy}{dx} = \frac{dy}{ds} \cdot \frac{ds}{dx} \)

This method allows us to calculate \( \frac{dy}{dx} \) without eliminating the parameter \( s \), maintaining the relationship between \( x \) and \( y \) through their dependence on \( s \). Thus, the chain rule provides a critical link between the parametric derivations.

Second Derivative

After finding the first derivative, \( \frac{dy}{dx} \), the next step is often finding the second derivative, \( \frac{d^2y}{dx^2} \), which offers insights into the concavity and curvature of the curve described by the parametric equations.

The second derivative measures how the rate of change, represented by the first derivative, itself changes. To find \( \frac{d^2y}{dx^2} \), you need to differentiate \( \frac{dy}{dx} \) with respect to \( s \), and then adjust by multiplying it by \( \frac{ds}{dx} \), as shown in the formula:

• \( \frac{d}{ds}\left(\frac{dy}{dx}\right) \cdot \frac{ds}{dx} \)

In our example, we first calculated \( \frac{dy}{dx} = -\frac{s}{2} \). Next, we differentiated \( \frac{dy}{dx} \) with respect to \( s \) to get \( -\frac{1}{2} \). We then used \( \frac{ds}{dx} = \frac{1}{12s} \), derived from the reciprocal of \( \frac{dx}{ds} \). Multiplying these together gives us the second derivative:

• \( \frac{d^2y}{dx^2} = -\frac{1}{2} \cdot \frac{1}{12s} = -\frac{1}{24s} \)

This expression tells us how the rate of change in \( y \) with respect to \( x \) is itself changing with respect to \( x \). It is crucial for determining the shape of the graph.

Parametric Equations

Parametric equations are a way to express mathematical curves and paths through parameters, which hide the direct dependency between the variables. This approach can simplify the representation and differentiation of more intricate equations.

In our specific case, the curve is described by the parametric equations:

• \( x = 6s^2 \)
• \( y = -2s^3 \)

Here, \( s \) is the parameter that independently controls both \( x \) and \( y \), indirectly establishing a relationship between them.

One of the benefits of using parametric equations is they can describe complex curves that are difficult to express with a single equation. For example:

• They allow for easier computation of derivatives using the chain rule.
• They provide a more flexible framework to represent complex geometric shapes or motions.

When you're asked not to eliminate the parameter, as in this exercise, it means retaining the parametric form. The challenge is to calculate derivatives while keeping the parameter \( s \) in play, resulting in potentially more insightful perspectives on the geometry of the curve. Parametric equations hence offer significant power in both calculus and geometry.