Problem 24 - 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. $$ x=\sqrt{3} \theta^{2}, y=-\sqrt{3} \theta^{3} ; \theta \neq 0 $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = -\frac{3}{2}\theta \), \( \frac{d^2y}{dx^2} = -\frac{\sqrt{3}}{4\theta} \).

Step 1: Differentiate x and y with respect to θ

First, find the derivatives of the parametric equations with respect to the parameter \( \theta \). Obtain \( \frac{d x}{d \theta} \) from the equation \( x = \sqrt{3} \theta^2 \):\[ \frac{d x}{d \theta} = 2 \sqrt{3} \theta \]Now, differentiate \( y = -\sqrt{3} \theta^3 \):\[ \frac{d y}{d \theta} = -3 \sqrt{3} \theta^2 \]

Step 2: Calculate \( \frac{dy}{dx} \)

Use the chain rule for parametric equations: \( \frac{dy}{dx} = \frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} \). Substitute the derivatives:\[ \frac{dy}{dx} = \frac{-3 \sqrt{3} \theta^2}{2 \sqrt{3} \theta} \]Simplify the expression:\[ \frac{dy}{dx} = \frac{-3 \theta^2}{2 \theta} = -\frac{3}{2} \theta \]

Step 3: Differentiate \( \frac{dy}{dx} \) with respect to θ to find \( \frac{d^2y}{dx^2} \)

Differentiate the expression for \( \frac{dy}{dx} \) with respect to \( \theta \):\[ \frac{d}{d \theta} \left( -\frac{3}{2} \theta \right) = -\frac{3}{2} \]Now, use the chain rule again: \( \frac{d^2 y}{d x^2} = \frac{\frac{d}{d \theta} \left( \frac{d y}{d x} \right)}{\frac{d x}{d \theta}} \). Substitute the known derivatives:\[ \frac{d^2 y}{d x^2} = \frac{-\frac{3}{2}}{2 \sqrt{3} \theta} \]Simplify:\[ \frac{d^2 y}{d x^2} = \frac{-3}{4 \sqrt{3} \theta} = -\frac{\sqrt{3}}{4 \theta} \]

Key Concepts

Understanding Parametric Equations

When dealing with parametric equations, we have separate expressions for each coordinate in terms of a third variable, usually called a parameter. This means:

• For the x-coordinate: we have a function such as \( x = f(\theta) \).
• For the y-coordinate: we have another function like \( y = g(\theta) \).
• The parameter, in this case \( \theta \), allows us to explore the path or curve traced out as the parameter changes.

Parametric equations are incredibly useful for tracing complex curves that are cumbersome to handle with standard y in terms of x equations.
For instance, in this exercise, we dealt with:

• \( x = \sqrt{3} \theta^2 \)
• \( y = -\sqrt{3} \theta^3 \)

These equations tell us how x and y vary with \( \theta \), and we can navigate the paths they describe by changing \( \theta \). This makes parametric equations handy in fields like physics and engineering where paths, not just positions, are essential.

Applying the Chain Rule in Derivatives

The chain rule is a powerful tool in calculus that helps us differentiate composite functions by relating their derivatives. In the context of parametric equations, it allows the computation of the derivative \( \frac{dy}{dx} \) without explicitly solving for y as a function of x.
Here's how it works:

• First, find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \) by differentiating the parametric equations of x and y with respect to \( \theta \).
• Then, we use the formula: \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \).

In this exercise, differentiating gives us:

• \( \frac{dx}{d\theta} = \frac{d}{d\theta}(\sqrt{3} \theta^2) = 2\sqrt{3} \theta \)
• \( \frac{dy}{d\theta} = \frac{d}{d\theta}(-\sqrt{3} \theta^3) = -3\sqrt{3} \theta^2 \)

Substituting these results into the chain rule formula results in \( \frac{dy}{dx} = -\frac{3}{2} \theta \).
This showcases the beauty of the chain rule, making complex differentiation accessible without losing track of each variable's role.

Calculating the Second Derivative

To analyze the curvature or the concavity of the path in a parametric curve, finding the second derivative \( \frac{d^2y}{dx^2} \) is crucial. It measures how \( \frac{dy}{dx} \) changes with x, giving insight into how the slope itself is changing.
Here are the steps for calculating the second derivative:

• First, differentiate \( \frac{dy}{dx} \) with respect to \( \theta \) to find \( \frac{d}{d\theta}(\frac{dy}{dx}) \).
• Next, divide this derivative by \( \frac{dx}{d\theta} \) to apply the second part of the chain rule: \( \frac{d^2y}{dx^2} = \frac{\frac{d}{d\theta}(\frac{dy}{dx})}{\frac{dx}{d\theta}} \).

For this problem, differentiating \( -\frac{3}{2} \theta \) gives us \( -\frac{3}{2} \), and substituting into the formula, we have:

• \( \frac{d^2y}{dx^2} = \frac{-\frac{3}{2}}{2\sqrt{3}\theta} = -\frac{\sqrt{3}}{4\theta} \)

This result tells us more about the behavior of the curve. A negative second derivative indicates the curve is concave down in the regions we've calculated, offering more detailed insights into the nature of the path.