Problem 21 - Thomas Calculus

Question

In Exercises \(21-24,\) find \(d p / d q\) $$ p=5+\frac{1}{\cot q} $$

Step-by-Step Solution

Answer

\( \frac{dp}{dq} = \sec^2 q \)

Step 1: Understand the Function

The function given is \( p = 5 + \frac{1}{\cot q} \). For simplicity, recognize \( \frac{1}{\cot q} \) as \( \tan q \), since \( \tan q = \frac{1}{\cot q} \). Thus, the function becomes \( p = 5 + \tan q \).

Step 2: Apply Derivative Rules

To find \( \frac{dp}{dq} \), we need to differentiate \( p = 5 + \tan q \) with respect to \( q \). The derivative of a constant is 0, while the derivative of \( \tan q \) is \( \sec^2 q \).

Step 3: Calculating the Derivative

Apply the derivatives from Step 2: \( \frac{d}{dq}(5) = 0 \) and \( \frac{d}{dq}(\tan q) = \sec^2 q \). Therefore, \( \frac{dp}{dq} = 0 + \sec^2 q \).

Step 4: Final Expression

We conclude that the derivative of \( p \) with respect to \( q \) is \( \frac{dp}{dq} = \sec^2 q \).

Key Concepts

Trigonometric Functions

Trigonometric functions are essential mathematical tools used to relate angles to side lengths in right-angled triangles. They include basic functions such as sine (\( \sin \theta \)), cosine (\( \cos \theta \)), and tangent (\( \tan \theta \)). Each function has its unique properties and relationships. For example, the tangent of an angle is the ratio of the sine to the cosine:

\[\tan \theta = \frac{\sin \theta}{\cos \theta}\]These functions are periodic and have specific derivatives associated with their changes in angles. Understanding these derivatives is crucial for different kinds of problems in calculus and beyond.

A few derivatives related to trigonometric functions are:

• The derivative of \( \sin q \) is \( \cos q \).
• The derivative of \( \cos q \) is \(-\sin q \).
• The derivative of \( \tan q \) is \( \sec^2 q \).

When working with these functions, it's essential to understand not only their values but also their rates of change.

Chain Rule

The chain rule is a fundamental technique in calculus for differentiating composite functions. A composite function is one that is formed by one function inside another, such as \( f(g(x)) \). The chain rule states that to find the derivative of a composite function, you multiply the derivative of the outer function by the derivative of the inner function:

\[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\]This rule is incredibly useful when dealing with complex expressions where one function operates on another. In practice, it helps simplify the differentiation process for functions that aren't easily solvable using basic rules.

Let’s consider the trigonometric function scenario where you might need to apply the chain rule. Suppose you have \( h(x) = \tan(3x) \). Using the chain rule:

• The derivative of \( \tan u \) where \( u=3x \) is \( \sec^2(3x) \cdot \frac{d}{dx}(3x) \).
• This becomes \( 3 \sec^2(3x) \), highlighting how the chain rule manages the differentiation of nested functions.

Learning to use the chain rule effectively can make working with integrations and derivatives much more manageable.

Differentiation Techniques

Differentiation techniques are crucial to solving calculus problems efficiently. Simple functions are differentiated using the standard rules, but more complex functions require advanced techniques like the chain rule, product rule, or quotient rule. In the context of our initial exercise, we saw a straightforward instance of differentiation.

With the function \( p = 5 + \tan q \), we leveraged the trigonometric derivative directly:

• The derivative of the constant \( 5 \) is \( 0 \), simplifying the process as constants vanish under differentiation.
• The derivative of \( \tan q \) is \( \sec^2 q \), which we used to arrive at the final answer of \( \frac{dp}{dq} = \sec^2 q \).

Other differentiation techniques might involve breaking down complex expressions into simpler parts, switching functions to their more basic forms, or employing strategies to simplify the steps leading to the derivative.

Mastering multiple techniques allows flexibility and ensures you're well-equipped to tackle any calculus problem you encounter.