Problem 19 - Thomas Calculus

Question

In Exercises \(17-20,\) find \(d r / d \theta\) $$ r=\sec \theta \csc \theta $$

Step-by-Step Solution

Answer

\( \frac{dr}{d\theta} = \frac{\sin^2 \theta - \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} \).

Step 1: Differentiate with Respect to θ

Start by finding the derivative of the given expression with respect to \( \theta \). The function is \( r = \sec \theta \csc \theta \). Use the product rule for differentiation, which is if you have two functions \( u \) and \( v \), the derivative of their product \( uv \) is \( u'v + uv' \). In this case, \( u = \sec \theta \) and \( v = \csc \theta \).

Step 2: Differentiate \( u = \sec \theta \)

The derivative of \( \sec \theta \) with respect to \( \theta \) is \( \sec \theta \tan \theta \). So, \( u' = \sec \theta \tan \theta \).

Step 3: Differentiate \( v = \csc \theta \)

The derivative of \( \csc \theta \) with respect to \( \theta \) is \(-\csc \theta \cot \theta \). So, \( v' = -\csc \theta \cot \theta \).

Step 4: Apply Product Rule

Now apply the product rule: \( \frac{dr}{d\theta} = u'v + uv' \).Substitute \( u' = \sec \theta \tan \theta \), \( v = \csc \theta \), \( u = \sec \theta \), and \( v' = -\csc \theta \cot \theta \).

Step 5: Simplify the Expression

Compute the expression: \[ \frac{dr}{d\theta} = (\sec \theta \tan \theta)(\csc \theta) + (\sec \theta)(-\csc \theta \cot \theta) \]This simplifies to:\[ \frac{dr}{d\theta} = \sec \theta \csc \theta \tan \theta - \sec \theta \csc \theta \cot \theta \]Factor out \( \sec \theta \csc \theta \):\[ \frac{dr}{d\theta} = \sec \theta \csc \theta (\tan \theta - \cot \theta) \].

Step 6: Simplify Trigonometric Terms

Since \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), the expression for \( \tan \theta - \cot \theta \) becomes:\[ \tan \theta - \cot \theta = \frac{\sin \theta}{\cos \theta} - \frac{\cos \theta}{\sin \theta} \]The expression then further simplifies to:\[ \frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta \cos \theta} \]Thus,\[ \frac{dr}{d\theta} = \sec \theta \csc \theta \frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta \cos \theta} \].

Step 7: Finalize the Result

Now, substitute back the definition of \( \sec \theta \) and \( \csc \theta \), \( \sec \theta = \frac{1}{\cos \theta} \) and \( \csc \theta = \frac{1}{\sin \theta} \), into the expression:\[ \frac{dr}{d\theta} = \frac{1}{\cos \theta \sin \theta} \cdot \frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta \cos \theta} \]This simplifies to:\[ \frac{dr}{d\theta} = \frac{\sin^2 \theta - \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} \]

Key Concepts

Product Rule in Differentiation

In calculus, the product rule is essential when dealing with derivatives of multiplied functions. Imagine you have two functions, \( u(x) \) and \( v(x) \). When you want to find the derivative of their product \( u(x)v(x) \), you can't differentiate each term separately like in addition or subtraction. This is where the product rule comes into play. The formal statement of the product rule is: \[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \]. This means you take the derivative of the first function \( u \) while leaving the second function \( v \) alone, then add it to the original \( u \) and the derivative of \( v \).
This rule helps in breaking down the differentiation into simpler parts. When addressing more complicated expressions, especially when each component function is complex or involves another rule like chain rule, it becomes beneficial. It's a building block for using calculus effectively.
In our original exercise, the product rule is applied to differentiate \( \sec \theta \csc \theta \). With \( u = \sec \theta \) and \( v = \csc \theta \), applying the product rule helps in finding the derivative efficiently, giving us an organized way to separate each function's contribution to the derivative.

Understanding Trigonometric Functions

Trigonometric functions are fundamental in calculus and are often seen in differentiation problems. The main functions include sine \( \sin \), cosine \( \cos \), tangent \( \tan \), cosecant \( \csc \), secant \( \sec \), and cotangent \( \cot \). Each has distinct properties and plays specific roles in trigonometry and calculus.

• \( \sin \theta \) and \( \cos \theta \) are the basic functions, defined as the opposite and adjacent over the hypotenuse in a right triangle.
• \( \sec \theta = \frac{1}{\cos \theta} \) and \( \csc \theta = \frac{1}{\sin \theta} \) provide reciprocal relationships, essential for more complex trigonometric integration and differentiation.

Trigonometric identities are powerful tools. For example, \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \cot \theta = \frac{\cos \theta}{\sin \theta} \) are useful in simplifying expressions. In the given exercise, these functions were used to simplify and complete the differentiation.
The derivatives of these functions are also fundamental:

• \( \frac{d}{d\theta} \sec \theta = \sec \theta \tan \theta \)
• \( \frac{d}{d\theta} \csc \theta = -\csc \theta \cot \theta \)

Understanding how these derivatives behave is crucial in manipulating expressions and simplifying complex calculations.

Detailed Differentiation Steps

Differentiation involves finding how a function changes as its input changes. In this exercise, we practiced differentiating \( r = \sec \theta \csc \theta \). This involves several structured steps, utilizing known mathematical rules to reach an answer.**Step 1: Identifying Functions**Recognize the functions involved, setting \( u = \sec \theta \) and \( v = \csc \theta \). This helps apply differentiation rules effectively.**Step 2: Applying the Product Rule**Using the product rule, \( \frac{dr}{d\theta} = u'v + uv' \), we individually differentiate \( u \) and \( v \), then combine them accordingly.

• For \( u = \sec \theta \), \( u' = \sec \theta \tan \theta \)
• For \( v = \csc \theta \), \( v' = -\csc \theta \cot \theta \)

Substitute these back into the product rule equation.**Step 3: Simplifying the Expression**After substituting the derivatives, simplify by factoring common terms. Here, factor out \( \sec \theta \csc \theta \) to condense and clarify the result.**Step 4: Dealing with Trigonometric Simplification**Simplify trigonometric terms like \( \tan \theta - \cot \theta \). Convert into common terms and simplify, i.e., \( \frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta \cos \theta} \).These steps showcase a systematic approach to differentiation, built on understanding trigonometric derivatives and algebra simplification. Such skills are essential for solving more complex calculus problems.