Problem 64 - Thomas Calculus in SI Units

Question

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=\frac{\theta \sin \theta}{\sqrt{\sec \theta}}$$

Step-by-Step Solution

Answer

The derivative of \(y\) with respect to \(\theta\) is \(\frac{dy}{d\theta} = y \left( \frac{1}{\theta} + \cot \theta - \frac{1}{2} \sec \theta \tan \theta \right)\).

Step 1: Take the Natural Logarithm

Start by taking the natural logarithm of both sides of the equation to make differentiation easier. We have:\[\ln y = \ln \left( \frac{\theta \sin \theta}{\sqrt{\sec \theta}} \right)\] This can be simplified using logarithmic properties to:\[\ln y = \ln \theta + \ln \sin \theta - \frac{1}{2} \ln \sec \theta\]

Step 2: Differentiate Implicitly

Differentiate both sides with respect to \( \theta \). For the left side, use the chain rule:\[\frac{d}{d\theta}(\ln y) = \frac{1}{y} \cdot \frac{dy}{d\theta}\]For the right side differentiate each term:\[\frac{d}{d\theta}(\ln \theta) = \frac{1}{\theta}, \quad \frac{d}{d\theta}(\ln \sin \theta) = \cot \theta, \quad \frac{d}{d\theta} \left( \frac{1}{2} \ln \sec \theta \right) = \frac{1}{2} \cdot \sec \theta \tan \theta \cdot \sec \theta = \frac{1}{2} \sec^2 \theta \tan \theta \]Therefore:\[\frac{1}{y} \cdot \frac{dy}{d\theta} = \frac{1}{\theta} + \cot \theta - \frac{1}{2} \sec \theta \tan \theta\]

Step 3: Solve for \( \frac{dy}{d\theta} \)

Multiply through by \( y \) to solve for \( \frac{dy}{d\theta} \):\[\frac{dy}{d\theta} = y \left( \frac{1}{\theta} + \cot \theta - \frac{1}{2} \sec \theta \tan \theta \right)\]Substitute back the expression for \( y \):\[\frac{dy}{d\theta} = \frac{\theta \sin \theta}{\sqrt{\sec \theta}} \left( \frac{1}{\theta} + \cot \theta - \frac{1}{2} \sec \theta \tan \theta \right)\]

Step 4: Final Simplification

Simplify the expression, if necessary. However, due to the nature of trigonometric simplifications, the form may not simplify significantly beyond factorization. The derivative is:\[\frac{dy}{d\theta} = \sin \theta \sqrt{\cos \theta} + \theta \cos \theta \sqrt{\cos \theta} - \frac{ \theta \sin \theta \tan \theta \cos \theta}{2} \]

Key Concepts

Derivative of Trigonometric Functions

Trigonometric functions like sine, cosine, and secant are fundamental in calculus. Knowing their derivatives is essential for solving calculus problems effectively.
When dealing with trigonometric functions, each function has a specific derivative:

• The derivative of \(\sin(\theta)\) is \(\cos(\theta)\).
• The derivative of \(\cos(\theta)\) is \(-\sin(\theta)\).
• For \(\sec(\theta)\), the derivative is \(\sec(\theta)\tan(\theta)\).

These derivatives are essential in the problem-solving process, as they provide the foundation for finding the rate of change of trigonometric expressions.
In our exercise, we differentiate \(\sin(\theta)\) and \(\sec(\theta)\) as part of finding the derivative of the given function. Understanding these basic derivatives simplifies the entire differentiation process.

Implicit Differentiation

Implicit differentiation is particularly useful when dealing with complex relationships that aren't solely dependent on one variable.
This technique involves differentiating both sides of an equation with respect to one variable, often introducing other variables in the process.
In this exercise, logarithmic differentiation simplifies handling the problem's complexity. We start with the natural logarithm to allow for easier manipulation of the function's various elements.
With implicit differentiation, we applied the chain rule to find \( \frac{d}{d\theta}(\ln y) \). This led to differentiating the combination of functions inside the logarithm:

• \(\ln(\theta)\) differentiates to \(\frac{1}{\theta}\),
• \(\ln(\sin(\theta))\) becomes \(\cot(\theta)\),
• and \(\frac{1}{2} \ln(\sec(\theta))\) becomes \(\frac{1}{2} \sec^2(\theta) \tan(\theta)\).

This detailed breakdown helps manage functions that involve multiple trigonometric components, ultimately leading to the derivative.

Calculus Problem-Solving

In calculus, problem-solving often involves several techniques combined to handle complex functions and expressions.
This exercise illustrates how using logarithmic differentiation can unravel seemingly complicated relationships within a function. Here’s a breakdown of the strategic problem-solving steps we took:

• Logarithmic Transformation: Applying the natural logarithm simplified the original expression, thus making it easier to differentiate.
• Utilizing Implicit Differentiation: This allowed us to differentiate without having to simplify the function entirely beforehand.
• Combination of Trigonometric Derivatives: Handling multiple derivatives of trigonometric functions required careful application of their known rules.
• Re-substitution: After differentiating, substituting back the original expression for \(y\) was key to simplifying the derivative in terms of \(\theta\).

Problem-solving in calculus is often about recognizing which tools and techniques will unlock the solution most efficiently.
This example highlights the integration of knowledge about derivatives, logarithms, and algebraic manipulation.