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Magazine Chapter 3: Problem 28
Find \(d p / d q\). $$p=(1+\csc q) \cos q$$
Short Answer
Expert verified
The derivative \( \frac{dp}{dq} = -\frac{\cos^2 q}{\sin^2 q} - \sin q - 1 \).
Step by step solution
01
Understand the Problem
You need to differentiate the function \(p = (1 + \csc q) \cos q\) with respect to \(q\). This means you will find \( \frac{dp}{dq} \).
02
Use the Product Rule
The function \(p = (1 + \csc q) \cos q\) is a product of two functions: \((1 + \csc q)\) and \(\cos q\). The product rule for differentiation states that \((uv)' = u'v + uv'\). Identify \(u = (1 + \csc q)\) and \(v = \cos q\).
03
Differentiate \(u = 1 + \csc q\)
Differentiate \(u = 1 + \csc q\). The derivative of a constant \(1\) is 0, and the derivative of \(\csc q\) is \(-\csc q \cot q\). Thus, \(u' = 0 - \csc q \cot q = -\csc q \cot q\).
04
Differentiate \(v = \cos q\)
Differentiate \(v = \cos q\). The derivative of \(\cos q\) is \(-\sin q\). Thus, \(v' = -\sin q\).
05
Apply the Product Rule
Using the product rule, \( \frac{dp}{dq} = u'v + uv' = (-\csc q \cot q)(\cos q) + (1 + \csc q)(-\sin q)\).
06
Simplify the Expression
Simplify the expression from Step 5:\[ (-\csc q \cot q)(\cos q) + (1 + \csc q)(-\sin q) \]Distribute each term:\[ -\csc q \cos q \cot q - \sin q - \csc q \sin q \]Use the identity \(\csc q = \frac{1}{\sin q}\) and \(\cos q \cot q = \frac{\cos^2 q}{\sin q}\) to further simplify:\[ -\frac{\cos^2 q}{\sin^2 q} - \sin q - 1 \]Then, combine like terms if possible.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Product Rule
The product rule is a fundamental concept in calculus that is essential when differentiating products of two functions. It helps you find the derivative of a function that can be expressed as the product of two or more other functions. To apply the product rule, remember the formula:
This clear separation allows us to individually differentiate \( u \) and \( v \), and then apply the formula.
This method makes handling complex expressions much more manageable.
- If you have two functions, say \( u(q) \) and \( v(q) \), then the derivative of their product \( (uv)' \) is given by: \( u'v + uv' \).
This clear separation allows us to individually differentiate \( u \) and \( v \), and then apply the formula.
This method makes handling complex expressions much more manageable.
Trigonometric Identities
Trigonometric identities are equations that are true for all angles and are especially invaluable when simplifying expressions in calculus.
In this particular problem, identities are useful for simplifying terms during the differentiation process.
Some common trigonometric identities that are frequently used include:
In this particular problem, identities are useful for simplifying terms during the differentiation process.
Some common trigonometric identities that are frequently used include:
- The reciprocal identity: \( \csc q = \frac{1}{\sin q} \).
- The quotient identity: \( \cot q = \frac{\cos q}{\sin q} \).
- \( \cos q \cot q \) becomes \( \frac{\cos^2 q}{\sin q} \).
- Now, \( \csc q \cos q \cot q \) simplifies to \( \frac{\cos^2 q}{\sin^2 q} \) when combined.
Derivatives
Derivatives are central to understanding rates of change within calculus. They measure how a function changes as its input changes.
A common basic derivative that is vital in trigonometric calculus involves functions like sine, cosine, or tangent.
For instance:
A common basic derivative that is vital in trigonometric calculus involves functions like sine, cosine, or tangent.
For instance:
- The derivative of \( \cos q \) is \(-\sin q \), and is a standard result.
- The derivative of \( \csc q \) is \(-\csc q \cot q \), which might be less familiar but equally important.
- For \( u' \), differentiating \( 1 + \csc q \) yields \( -\csc q \cot q \).
- For \( v' \), differentiating \( \cos q \) yields \(-\sin q \).
