Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

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: Compute dx/ds

We are given the parametric equation for x: \(x = 6s^2\). To find \(\frac{dx}{ds}\), differentiate x with respect to s:\[\frac{dx}{ds} = \frac{d}{ds}(6s^2) = 12s.\]

Step 2: Compute dy/ds

Similarly, for the parametric equation for y: \(y = -2s^3\), differentiate y with respect to s to find \(\frac{dy}{ds}\):\[\frac{dy}{ds} = \frac{d}{ds}(-2s^3) = -6s^2.\]

Step 3: Find dy/dx using dy/ds and dx/ds

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

Step 4: Calculate d²y/dx² using dy/dx

To find \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx} = -\frac{s}{2}\) with respect to s first, then divide by \(\frac{dx}{ds}\):\[\frac{d}{ds}\left(-\frac{s}{2}\right) = -\frac{1}{2},\]\[\frac{d^2y}{dx^2} = \frac{-\frac{1}{2}}{12s} = -\frac{1}{24s}.\]

Key Concepts

Parametric Equations

Parametric equations are a powerful tool in mathematics, especially in calculus and geometry. They involve expressing variables such as \( x \) and \( y \) as functions of another parameter, usually \( t \) or \( s \). In this exercise, the variables are represented as functions of \( s \): \( x = 6s^2 \) and \( y = -2s^3 \). This form allows us to describe curves that might be difficult to express explicitly as \( y = f(x) \).

So, what's the advantage of using parametric equations? They can model motion and curves that change direction or loop back on themselves, providing a simpler, more efficient way to track position over time or other parameters.

• Useful in physics to represent particle trajectories.
• Allows modeling of spirals or other complex curves easily.

Understanding how to differentiate parametric equations helps us analyze these paths and determine velocities and accelerations along them.

Chain Rule

The chain rule is a fundamental concept in calculus, used to differentiate compositions of functions. When dealing with parametric equations, it becomes a valuable tool when computing derivatives, such as \( \frac{dy}{dx} \).

Given that \( y \) and \( x \) are both functions of a third parameter \( s \), the derivative \( \frac{dy}{dx} \) can be found using the chain rule as the ratio of \( \frac{dy}{ds} \) to \( \frac{dx}{ds} \). This is because both \( dy \) and \( dx \) arise from their respective parametric forms:

• First, compute \( \frac{dy}{ds} \) by differentiating the \( y \)-function with respect to \( s \)
• Then, compute \( \frac{dx}{ds} \) by differentiating the \( x \)-function with respect to \( s \)
• The chain rule gives us \( \frac{dy}{dx} = \frac{dy/ds}{dx/ds} \)

This technique allows for a relatively straightforward approach to finding slopes and rates of change along curves described by parametric equations.

Second Derivative

The second derivative measures the curvature or concavity of a function's graph, providing insight into its acceleration or rate of change of slope. When working with parametric equations, computing the second derivative \( \frac{d^2y}{dx^2} \) requires careful application of implicit differentiation.

After finding the first derivative \( \frac{dy}{dx} \) in terms of \( s \), we differentiate it with respect to the parameter \( s \). Then, we divide this result by \( \frac{dx}{ds} \) again to find \( \frac{d^2y}{dx^2} \). This additional differentiation step helps analyze changes in the curve's slope, indicating points of inflection or extreme curvature. Here's the general process:

• Differentiation of \( \frac{dy}{ds} \) with respect to \( s \)
• Adjusting by \( \frac{dx/ds} \) results in final form \( \frac{d^2y}{dx^2} = \frac{d}{ds}(\frac{dy}{dx}) \div \frac{dx}{ds} \)

Understanding this helps in visualizing how a curve bends and twists as \( s \) varies.

Implicit Differentiation

Implicit differentiation is another calculus technique useful when dealing with equations that define \( y \) and \( x \) without explicit definitions. However, in parametric equations, functions are already in a differentiated form, so our focus is on utilizing these differentiated forms to find \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \).

Implicit differentiation helps us to:

• Differentiate functions that are difficult or impossible to solve explicitly
• Move through steps of differentiation by treating \( y \) as a function of \( x \), requiring derivatives of both sides of an equation
• Use relations when direct function forms aren't available

In parametric contexts, implicit differentiation simplifies tracking changes in curves and is inherently linked with the chain rule to transition from derivatives in parameter forms to regular Cartesian forms, like \( \frac{dy}{dx} \) and further derivatives.