Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

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

Step 1: Differentiate x with respect to theta

Given the function \(x = \sqrt{3} \theta^2\), differentiate \(x\) with respect to \(\theta\). Use the power rule \( \frac{d}{d\theta}(\theta^n) = n\theta^{n-1} \). \[ \frac{dx}{d\theta} = \sqrt{3} \cdot 2\theta = 2\sqrt{3}\theta \]

Step 2: Differentiate y with respect to theta

Given the function \(y = -\sqrt{3} \theta^3\), differentiate \(y\) with respect to \(\theta\). Use the power rule \( \frac{d}{d\theta}(\theta^n) = n\theta^{n-1} \). \[ \frac{dy}{d\theta} = -\sqrt{3} \cdot 3\theta^2 = -3\sqrt{3}\theta^2 \]

Step 3: Find dy/dx using chain rule

To find \(\frac{dy}{dx}\), use the chain rule: \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\). Substitute the expressions from Steps 1 and 2. \[ \frac{dy}{dx} = \frac{-3\sqrt{3}\theta^2}{2\sqrt{3}\theta} = \frac{-3\theta}{2} \]

Step 4: Differentiate dy/dx with respect to theta to find d^2y/dx^2

Start by differentiating \(\frac{dy}{dx} = \frac{-3\theta}{2}\) with respect to \(\theta\). \[ \frac{d}{d\theta}\left(\frac{-3\theta}{2}\right) = -\frac{3}{2} \]

Step 5: Differentiate dx/dtheta to use in second derivative

Recall \(\frac{dx}{d\theta} = 2\sqrt{3}\theta\). Differentiate with respect to \(\theta\). \[ \frac{d}{d\theta}(2\sqrt{3}\theta) = 2\sqrt{3} \]

Step 6: Find the second derivative using chain rule

To find \(\frac{d^2y}{dx^2}\), use the formula: \(\frac{d}{d\theta}\left(\frac{dy}{dx}\right) \bigg/ \frac{dx}{d\theta}\). Substitute the expressions from Steps 4 and 5. \[ \frac{d^2y}{dx^2} = \frac{-\frac{3}{2}}{2\sqrt{3}} = -\frac{3}{4\sqrt{3}} \]

Step 7: Simplify the second derivative

To simplify \(-\frac{3}{4\sqrt{3}}\), multiply the numerator and the denominator by \(\sqrt{3}\) to rationalize the denominator. \[ \frac{d^2y}{dx^2} = -\frac{3\sqrt{3}}{12} = -\frac{\sqrt{3}}{4} \]

Key Concepts

Understanding the Chain Rule in Parametric Differentiation

The chain rule is a fundamental concept in calculus used for finding the derivative of composite functions. In the context of parametric differentiation, it helps us compute derivatives when the equations are given in terms of a third variable, usually denoted as \( \theta \). This can be very useful particularly when \(x\) and \(y\) are expressed as functions of \(\theta\) instead of directly in terms of each other.

• To apply the chain rule, consider that we have two derivatives to manage: \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \).
• The derivative \( \frac{dy}{dx} \) can be found using the formula: \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \).

By utilizing the chain rule, we can navigate the interdependent changes in \(x\) and \(y\) as \(\theta\) varies, giving us powerful insights into the behavior of parametric curves. In the original solution, the chain rule is applied by leveraging the derivatives of both \(x\) and \(y\) with respect to \(\theta\), which allows us to calculate \( \frac{dy}{dx} \) efficiently.

Basics of Calculus in Differentiation

Calculus is the mathematical study of continuous change and is divided mainly into differential calculus and integral calculus. In the context of this exercise, differential calculus is our focus since it deals with the concept of the derivative, which represents the rate of change.

• In differential calculus, the derivative of a function gives us the slope of the function at any given point, effectively describing how the function is changing at that point.
• The power rule, which appears in this problem, states that if \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).

Differentiating parametric equations involves treating the parameter \(\theta\) as the independent variable and applying rules like the power rule to find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \). These derivatives are used in the chain rule to find rates of change like \( \frac{dy}{dx} \), which directly relate to calculus's central goal of understanding how functions change.

Finding and Understanding the Second Derivative

The second derivative is a critical concept in calculus that provides information on the 'concavity' of a function, or how the function's slope is changing. Finding \( \frac{d^2y}{dx^2} \) in parametric equations involves differentiating \( \frac{dy}{dx} \) with respect to \(\theta\) again, then using the chain rule once more.

• The derivative \( \frac{d}{d\theta} \left( \frac{dy}{dx} \right) \) is first calculated, indicating how \( \frac{dy}{dx} \) (the slope itself) changes as \( \theta \) changes.
• Then, we divide this new derivative by \( \frac{dx}{d\theta} \) to adjust for the change in \(x\), arriving at \( \frac{d^2y}{dx^2} \).

The significance of \( \frac{d^2y}{dx^2} \) is that it tells us how the curve 'bends' or 'curves', influencing the graphical behavior. For instance, if \( \frac{d^2y}{dx^2} \) is positive, the graph is concave up, while a negative value indicates concave down. Thus, by carefully calculating the second derivative, we gain deeper insights into the geometry of the parametric equation's graph.