Problem 26 - University Calculus: Early Transcendentals

Question

Find \(d r / d \theta\). $$r=(1+\sec \theta) \sin \theta$$

Step-by-Step Solution

Answer

\( \frac{dr}{d\theta} = 1 + \sec \theta \)

Step 1: Identify the Differentiation Technique

To find \( \frac{dr}{d\theta} \), we need to differentiate the given expression \( r = (1 + \sec\theta) \sin\theta \) with respect to \( \theta \). Since this expression is a product of two functions, we will use the product rule for differentiation.

Step 2: Apply the Product Rule

The product rule states that if \( u = f(x)g(x) \), then \( \frac{du}{dx} = f'(x)g(x) + f(x)g'(x) \). Here, let \( u = (1+ \sec \theta) \) and \( v = \sin \theta \). So, \( \frac{du}{d\theta} = \sec \theta \tan \theta \) and \( \frac{dv}{d\theta} = \cos \theta \).

Step 3: Differentiate Using Product Rule

Using the product rule, we have: \[ \frac{dr}{d\theta} = \frac{d}{d\theta}[(1 + \sec\theta) \sin\theta] = (1+\sec\theta) \frac{d}{d\theta}(\sin\theta) + \sin\theta \frac{d}{d\theta}(1+\sec\theta) \]Substitute the derivatives we found earlier:\[ = (1+\sec\theta)(\cos\theta) + \sin\theta(\sec\theta \tan\theta) \]

Step 4: Simplify the Expression

Combine the terms:\[ \frac{dr}{d\theta} = \cos \theta + \sec \theta \cos \theta + \sin \theta \sec \theta \tan \theta \]Since \( \sec \theta = \frac{1}{\cos \theta} \), the term \( \sec \theta \cos \theta = 1 \). Also, the term \( \sec \theta \tan \theta = \frac{\sin \theta}{\cos^2 \theta} \), so: \[ \frac{dr}{d\theta} = \cos \theta + 1 + \frac{\sin^2 \theta}{\cos \theta} \]

Step 5: Final Simplification

To simplify further, note that \( \sin^2 \theta = 1 - \cos^2 \theta \). Using this identity:\[ \frac{dr}{d\theta} = \cos \theta + 1 + \frac{1 - \cos^2 \theta}{\cos \theta} \] Combine the terms in the fraction:\[ = \cos \theta + 1 + \frac{1}{\cos \theta} - \cos \theta \]\[ = 1 + \frac{1}{\cos \theta} \]Thus, \[ \frac{dr}{d\theta} = 1 + \sec \theta \].

Key Concepts

Product Rule

The product rule is an essential differentiation technique when dealing with products of two functions. In mathematics, when you need to find the derivative of a function that is the product of two other functions, you'll apply the product rule. This rule helps in neatly breaking down complex problems into manageable pieces.
For a function described as the product of two functions, say \( f(x) = u(x) \cdot v(x) \), the product rule states that the derivative \( \frac{d}{dx}[u(x) \cdot v(x)] \) is calculated as:

• \( u'(x) \cdot v(x) \)
• plus \( u(x) \cdot v'(x) \)

This simplifies to \( u'(x) v(x) + u(x) v'(x) \). Grasping the product rule is crucial for success in calculus, and it provides a reliable method when dealing with many real-world applications. Understanding the product rule through practical exercises helps build a strong foundation for tackling more complex problems in calculus.

Trigonometric Functions

Trigonometric functions play a significant role in various mathematical computations, especially when calculating the derivatives of functions involving angles. They are made up of the functions sine, cosine, tangent, and their respective reciprocals (cosecant, secant, and cotangent). These functions are essential in calculus for dealing with periodic functions and angular measurements.
In this exercise, trigonometric functions such as \( \sin \theta \), \( \cos \theta \), and \( \sec \theta \) are featured as components of the expression that requires differentiation. Knowing how to differentiate these trigonometric functions is crucial:

• The derivative of \( \sin \theta \) is \( \cos \theta \).
• The derivative of \( \cos \theta \) is \( -\sin \theta \).
• The derivative of \( \sec \theta \) involves both \( \sec \theta \) and \( \tan \theta \), given by \( \sec \theta \tan \theta \).

It's important to be familiar with these derivatives to effectively apply them in differentiation problems involving trigonometric functions.

Derivative Calculation

Calculating derivatives is a fundamental process in calculus used to determine the rate of change of variables. It can be particularly challenging when involving complex expressions, such as products of trigonometric functions. In this exercise, we are required to calculate the derivative of a product compounded by trigonometric expressions: \( r = (1 + \sec \theta) \sin \theta \).
To calculate \( \frac{dr}{d\theta} \), one must appropriately apply the product rule in conjunction with trigonometric derivatives. We begin by identifying each part of the product:

• First function \( u = (1 + \sec \theta) \)
• Second function \( v = \sin \theta \)

After obtaining the derivatives of these parts, \( \sec \theta \tan \theta \) for \( u \) and \( \cos \theta \) for \( v \), use these to apply the product rule:

• \( (1+\sec\theta) \cdot \cos \theta \)
• plus \( \sin\theta \cdot \sec \theta \tan \theta \)

Simplifying this results in the simplified derivative, \( \frac{dr}{d\theta} = 1 + \sec \theta \). Mastery of such calculations forms the stepping stone to understanding and handling more advanced calculus challenges.