Problem 23 - 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. $$ x=2 \theta^{2}, y=\sqrt{5} \theta^{3} ; \theta \neq 0 $$

Step-by-Step Solution

Answer

\( \frac{d y}{d x} = \frac{3\sqrt{5}}{4}\theta \); \( \frac{d^2 y}{d x^2} = \frac{3\sqrt{5}}{16\theta} \).

Step 1: Differentiate x with respect to θ

The given parameterization is \( x = 2 \theta^2 \). Differentiate \( x \) with respect to \( \theta \) using the power rule. The derivative \( \frac{d x}{d \theta} \) is \( 4\theta \).

Step 2: Differentiate y with respect to θ

The given parameterization is \( y = \sqrt{5} \theta^3 \). Differentiate \( y \) with respect to \( \theta \) using the power rule. The derivative \( \frac{d y}{d \theta} \) is \( 3\sqrt{5} \theta^2 \).

Step 3: Find \( \frac{d y}{d x} \)

Use the formula \( \frac{d y}{d x} = \frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} \). Substitute in the derivatives from Steps 1 and 2: \( \frac{d y}{d x} = \frac{3\sqrt{5} \theta^2}{4\theta} = \frac{3\sqrt{5}}{4}\theta \).

Step 4: Differentiate \( \frac{d y}{d x} \) with respect to θ

Differentiate \( \frac{d y}{d x} = \frac{3\sqrt{5}}{4} \theta \) with respect to \( \theta \). The derivative is \( \frac{3\sqrt{5}}{4} \).

Step 5: Find \( \frac{d^2 y}{d x^2} \)

Find \( \frac{d^2 y}{d x^2} \) by using the chain rule: \[ \frac{d^2 y}{d x^2} = \frac{d}{d\theta}\left(\frac{d y}{d x}\right) \times \frac{1}{\frac{d x}{d \theta}} \]. Substitute the derivatives: \( \frac{d^2 y}{d x^2} = \frac{3\sqrt{5}}{4} \times \frac{1}{4\theta} = \frac{3\sqrt{5}}{16\theta} \).

Key Concepts

Understanding Calculus

Calculus is a branch of mathematics that helps us understand changes between quantities. It consists of two main concepts: differentiation and integration. Here, we are focused on differentiation, which deals with how a function changes as its input changes.
One common application is finding rates of change — for instance, how fast a car is accelerating. It can also be used to find slopes of curves or predict rates of growth.
In this exercise, we deal with parametric equations, where we have expressions for both x and y in terms of a third variable, \(\theta\). Differentiating these equations sheds light on how changes in the parameter affect our coordinates. This is essential in physics and engineering, where many systems naturally involve parameters rather than single-variable functions.

Applying the Chain Rule

The chain rule is an essential tool in calculus used when differentiating a function composed of other functions. It allows us to find the derivative of these composite functions easily. The chain rule is expressed as:
\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\]
In our context, we needed the chain rule for finding the second derivative \( \frac{d^2y}{dx^2} \). After finding \( \frac{dy}{dx} \) using the parametric derivatives, the chain rule helps us connect back to \(\theta\) without directly eliminating the parameter.
This step ensures we accurately account for how changes in \(\theta\) influence both x and y, crucial for finding slopes and concavity in parametric curves.

Exploring the Second Derivative

The second derivative represents the rate at which the first derivative changes. It provides insights into the curvature or concavity of a function. If the first derivative tells you the slope of a tangent line to a curve, the second derivative describes whether the curve is "bending" upwards or downwards at a particular point.
In our problem, after finding \(\frac{dy}{dx}\), we used the chain rule to calculate the second derivative \(\frac{d^2y}{dx^2}\). The formula used was:
\[ \frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(\frac{dy}{dx}\right) \cdot \frac{1}{\frac{dx}{d\theta}}\]
This result helps us understand the behavior of our parametric curve more deeply and can indicate points of inflection where the curvature changes direction. Such knowledge is powerful in modeling dynamic systems like projectile motion or complex mechanical designs.