Problem 27 - University Calculus: Early Transcendentals

Question

Find \(d p / d q\). $$p=5+\frac{1}{\cot q}$$

Step-by-Step Solution

Answer

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

Step 1: Rewrite the Given Function

The given function is \(p = 5 + \frac{1}{\cot q}\). We know that \(\cot q = \frac{\cos q}{\sin q}\), so we can rewrite the function as \(p = 5 + \frac{\sin q}{\cos q}\) which simplifies to \(p = 5 + \tan q\).

Step 2: Differentiate Using the Sum Rule

To find \(\frac{dp}{dq}\), differentiate each term of the function separately: \(\frac{d}{dq}(5)\) and \(\frac{d}{dq}(\tan q)\).

Step 3: Differentiate Constant Term

The derivative of a constant is 0. Therefore, \(\frac{d}{dq}(5) = 0\).

Step 4: Differentiate Tangent Function

The derivative of \(\tan q\) is \(\sec^2 q\). Therefore, \(\frac{d}{dq}(\tan q) = \sec^2 q\).

Step 5: Combine the Derivatives

Add the derivatives from Steps 3 and 4 to get \(\frac{dp}{dq} = 0 + \sec^2 q = \sec^2 q\).

Key Concepts

Trigonometric Functions

Trigonometric functions are fundamental in understanding the relationships between the angles and sides of triangles. In this context, they help in expressing various real-world phenomena like waves and oscillations. The key trigonometric functions are sine, cosine, and tangent, each derived from right-angled triangles. When given an angle \( q \):

• \( \sin q \) gives the ratio of the opposite side to the hypotenuse.
• \( \cos q \) represents the adjacent side over the hypotenuse.
• \( \tan q \) signifies the opposite side to the adjacent side.

Additionally, functions like the cotangent (cot), which is the reciprocal of the tangent, are often involved. In this particular exercise, cotangent was converted to tangent using the identity \( \cot q = \frac{1}{\tan q} \), showing how these functions relate and can be rewritten for ease of differentiation.

Derivative Rules

Understanding how derivatives work is crucial in calculus as they help determine rates of change. The derivative of a function gives us information about the slope of the function at any given point. Basic derivative rules simplify finding these derivatives, making calculations more intuitive. Common rules include:

• **Power Rule:** For any function \( x^n \), its derivative is \( nx^{n-1} \).
• **Constant Rule:** The derivative of a constant is always zero.
• **Sum Rule:** The derivative of a sum is the sum of the derivatives.
• **Trig Functions:** \( \frac{d}{dq}(\tan q) = \sec^2 q \), as used in this exercise.

Employing these rules together allows one to differentiate complex expressions systematically. Here, the sum rule was particularly helpful in separating the constant term from the trigonometric term.

Calculus Problem Solving

Solving calculus problems often involves breaking down the problem into smaller, manageable steps. This is key to understanding not just the "how" but also the "why" behind each calculation. For instance, in the given problem, stepping through each part of the differentiation process was crucial. Let's break down the approach:

• **Rewriting:** Before differentiating, rewrite expressions in easily manageable forms, such as converting \( \cot q \) to \( \tan q \).
• **Applying Derivative Rules:** Use derivative rules like the sum and constant rules to handle different components separately.
• **Combining Results:** Combine the results to find the final derivative, as done with powering through \( dp \over dq \) to yield \( \sec^2 q \).

By following the structured approach, each component becomes clear, ensuring that solving calculus problems becomes a logical and methodical exercise, rather than a daunting challenge.