Chapter 3: Problem 27
In Exercises \(27-32,\) find \(d p / d q\) $$ p=5+\frac{1}{\cot q} $$
Short Answer
Expert verified
\( \frac{d p}{d q} = \sec^2 q \)
Step by step solution
01
Understand the Given Function
We are given the function \( p = 5 + \frac{1}{\cot q} \). The expression \( \frac{1}{\cot q} \) can be rewritten as \( \tan q \), because \( \tan q = \frac{1}{\cot q} \). So, the function can be rewritten as \( p = 5 + \tan q \).
02
Calculate the Derivative of \(p\) with Respect to \(q\)
To find \( \frac{d p}{d q} \), we need to differentiate \( p = 5 + \tan q \) with respect to \( q \). The derivative of a constant, \( 5 \), is \( 0 \), and the derivative of \( \tan q \) is \( \sec^2 q \). Therefore, \( \frac{d p}{d q} = 0 + \sec^2 q = \sec^2 q \).
03
Interpret the Result
The derivative \( \frac{d p}{d q} = \sec^2 q \) tells us how \( p \) changes with respect to \( q \). The function \( p \) increases at a rate of \( \sec^2 q \) with any change in \( q \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometric Functions
Trigonometric functions are mathematical functions that relate the angles and sides of a triangle. In calculus, they play an important role in solving a variety of problems. At the heart of these functions are the six main trigonometric functions: sine (\( \sin \)), cosine (\( \cos \)), tangent (\( \tan \)), cotangent (\( \cot \)), secant (\( \sec \)), and cosecant (\( \csc \)). Each of these functions has a specific meaning and mathematical relationship with one another.
- For example, tangent is defined as the ratio of the opposite side to the adjacent side in a right triangle, and it can also be expressed as \( \tan q = \frac{\sin q}{\cos q} \).
- Similarly, cotangent is the reciprocal of tangent, written as \( \cot q = \frac{1}{\tan q} = \frac{\cos q}{\sin q} \).
Differentiation Rules
Differentiation is the process of finding the derivative of a function, which measures how the function value changes as its input changes. There are specific rules that make finding derivatives straightforward, especially when dealing with trigonometric functions.
- The Power Rule is one of the most fundamental differentiation rules, used for functions of the form \( x^n \). However, it doesn't directly apply here since we need trigonometric rules.
- For trigonometric functions, we use specific differentiation rules. For example, the derivative of \( \tan q \) is \( \sec^2 q \).
Rate of Change
The derivative of a function at a point essentially gives the rate of change of the function's value with respect to its input variable. In other words, it tells us how quickly or slowly a function's values are changing at a given point.
- For a linear function, the rate of change is constant. However, for non-linear functions, like \( p = 5 + \tan q \), the rate of change can vary greatly depending on the value of \( q \).
- In calculus, the derivative \( \frac{d p}{d q} = \sec^2 q \) informs us that as \( q \) changes, \( p \) changes at a rate of \( \sec^2 q \). This rate is dependent on the value of \( q \).