Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

find \(d y / d x\) and \(d^{2} y / d x^{2}\) without eliminating the parameter. $$ x=2 \theta^{2}, y=\sqrt{5} \theta^{3} ; \theta \neq 0 $$

Step-by-Step Solution

Answer

\(dy/dx = \frac{3\sqrt{5}}{4}\theta\), \(d^2y/dx^2 = \frac{3\sqrt{5}}{16\theta}\).

Step 1: Differentiate x with respect to θ

To find \(dx/d\theta\), differentiate the given parametric equation \(x = 2\theta^2\) with respect to \(\theta\). This yields \(dx/d\theta = 4\theta\).

Step 2: Differentiate y with respect to θ

Differentiate the given parametric equation \(y = \sqrt{5}\theta^3\) with respect to \(\theta\) to find \(dy/d\theta\). This yields \(dy/d\theta = 3\sqrt{5}\theta^2\).

Step 3: Find dy/dx using parametric derivatives

Using the chain rule for parametric equations, \(dy/dx = (dy/d\theta) / (dx/d\theta)\). Substitute \(dy/d\theta = 3\sqrt{5}\theta^2\) and \(dx/d\theta = 4\theta\) to get \(dy/dx = (3\sqrt{5}\theta^2) / (4\theta) = (3\sqrt{5}/4)\theta\).

Step 4: Differentiate dy/dx with respect to θ

To find \(d^2y/dx^2\), first differentiate \(dy/dx = (3\sqrt{5}/4)\theta\) with respect to \(\theta\) to get the second derivative in terms of \(\theta\): \(d(dy/dx)/d\theta = 3\sqrt{5}/4\).

Step 5: Find d²y/dx² using the chain rule

To express \(d^2y/dx^2\) in terms of \(\theta\), use \(d^2y/dx^2 = (d(dy/dx)/d\theta) / (dx/d\theta)\). Substitute \(d(dy/dx)/d\theta = 3\sqrt{5}/4\) and \(dx/d\theta = 4\theta\) to get \(d^2y/dx^2 = (3\sqrt{5}/4) / (4\theta) = 3\sqrt{5}/(16\theta)\).

Key Concepts

Chain Rule

The Chain Rule is a fundamental tool in calculus that allows us to take derivatives of composite functions. Essentially, it's used when you have a function inside another function, so you need to "chain" the derivatives together.

In the context of parametric differentiation, as seen in our exercise, we're dealing with functions that are expressed in terms of a parameter like \( \theta \) instead of directly in terms of \( x \). To find \( dy/dx \), which is the derivative of \( y \) with respect to \( x \), we need to use the Chain Rule.

This is because \( dy/dx \) is found by dividing \( dy/d\theta \) by \( dx/d\theta \).

• First, find the derivative of \( y \) with respect to the parameter \( \theta \).
• Next, find the derivative of \( x \) with respect to the parameter \( \theta \).
• Finally, use the Chain Rule to divide these two derivatives, \( dy/dx = (dy/d\theta) / (dx/d\theta) \).

This method is extremely useful when working with curves defined by parametric equations since it simplifies the differentiation process without the need to directly eliminate the parameter.

Second Derivative

The second derivative, denoted as \( d^2y/dx^2 \), describes the curvature or the concavity of a graph. It tells us how the rate of change of the slope is itself changing.

In our exercise, after finding \( dy/dx \) using the Chain Rule, we move forward to find the second derivative. This involves differentiating \( dy/dx \) with respect to \( \theta \).

• Begin by taking the derivative of \( dy/dx \) with respect to \( \theta \). In this case, it's \( 3\sqrt{5}/4 \).
• Use the Chain Rule again to then divide this result by \( dx/d\theta \).
• The final expression for \( d^2y/dx^2 \) is \( 3\sqrt{5}/(16\theta) \).

This tells you that the curvature of the graph changes with respect to \( \theta \), and you can analyze the nature of these changes to understand the behavior of the function. This step is crucial for assessing the graph's inflection points or analyzing stability in dynamics.

Parametric Equations

Parametric equations represent a different way to express curves and functions, using a third variable called a parameter instead of directly relating \( x \) and \( y \). This method is particularly useful for describing more complex curves beyond simple Cartesian equations.

In the given exercise, the parameter \( \theta \) is used to express both \( x \) and \( y \).

• \( x = 2\theta^2 \), which expresses our horizontal component.
• \( y = \sqrt{5}\theta^3 \), which represents our vertical component.

By letting \( \theta \) vary, you can trace out the entire curve defined by these relations.

Parametric equations can be particularly handy for modeling real-world phenomena where one can track the position of an object over time or where curves intersect unusually through motion constraints. The key advantage here is the separation of \( x \) and \( y \) into simpler parts, simplifying differentiation tasks like finding \( dy/dx \) using the Chain Rule.

Calculus Problem Solving

In calculus, problem-solving requires understanding fundamental concepts and applying them appropriately. Our exercise demonstrates essential steps in working with parametric equations and differentiation.

This exercise involves finding both the first and the second derivatives without eliminating the parameter \( \theta \). By breaking the process into clear steps using parametric equations:

• Start by differentiating \( x \) and \( y \) separately with respect to \( \theta \).
• Apply the Chain Rule to find \( dy/dx \).
• Continue by differentiating \( dy/dx \) with respect to \( \theta \) to get the second derivative \( d^2y/dx^2 \).

Breaking down complex problems into simpler, more manageable parts can significantly ease the solving process. Thus, a structured, step-by-step approach is key to tackling calculus problems effectively, enabling a clear understanding of underlying mathematical concepts.