Problem 18 - The Calculus with Analytic Geometry

Question

In Exercises 13 through 24 , find the indicated partial derivatives by holding all but one of the variables constant and applying theorems for ordinary differentiation. $$ r=e^{-\theta} \cos (\theta+\phi) ; \frac{\partial r}{\partial \theta} $$

Step-by-Step Solution

Answer

\( \frac{\partial r}{\partial \theta} = -e^{-\theta} (\sin(\theta + \phi) + \cos(\theta + \phi)) \)

Step 1: Identify the function and variables

Identify the function given: \( r = e^{-\theta} \cos(\theta + \phi) \). Recognize that our goal is to find the partial derivative of \( r \) with respect to \( \theta \).

Step 2: Apply the product rule

The function \( r \) is a product of \( e^{-\theta} \) and \( \cos(\theta + \phi) \). According to the product rule, \( \frac{d}{d\theta} [u \cdot v] = u'v + uv' \), where \( u = e^{-\theta} \) and \( v = \cos(\theta + \phi) \).

Step 3: Differentiate the first part, \( u = e^{-\theta} \)

Differentiate \( e^{-\theta} \) with respect to \( \theta \): \( \frac{d}{d\theta} (e^{-\theta}) = -e^{-\theta} \).

Step 4: Differentiate the second part, \( v = \cos(\theta + \phi) \)

Differentiate \( \cos(\theta + \phi) \) with respect to \( \theta \): \( \frac{d}{d\theta} (\cos(\theta + \phi)) = -\sin(\theta + \phi) \).

Step 5: Combine the results using the product rule

Combine the differentiated parts using the product rule: \[ \frac{\partial r}{\partial \theta} = (e^{-\theta})(-\sin(\theta + \phi)) + (-e^{-\theta})\cos(\theta + \phi) \].

Step 6: Simplify the expression

Simplify the expression: \[ \frac{\partial r}{\partial \theta} = -e^{-\theta} \sin(\theta + \phi) - e^{-\theta} \cos(\theta + \phi) \].

Step 7: Final simplified form

Combine like terms to get the final simplified form: \[ \frac{\partial r}{\partial \theta} = -e^{-\theta} (\sin(\theta + \phi) + \cos(\theta + \phi)) \].

Key Concepts

Product Rule

The product rule is essential when you're dealing with the differentiation of a product of two functions. In simple terms, if you have a function that's the product of two other functions, say \( u \) and \( v \), the product rule states that the derivative of this product with respect to a variable is: \( \frac{d}{d\theta} [u \times v] = u'v + uv' \). Here, you take the derivative of the first function and multiply it by the second function, then add the product of the first function and the derivative of the second function. This rule tries to manage the change in both components of the product independently and then combines them. This rule is extremely useful and frequently used in many calculus problems, especially when functions of multiple variables come into play.

Ordinary Differentiation

Ordinary differentiation involves finding the derivative of a function with respect to one variable. This is essential when you need to understand how the function changes as the variable changes. For example, when we differentiate \( e^{-\theta} \) with respect to \( \theta \), we get \( -e^{-\theta} \). Similarly, differentiating \( \cos(\theta + \phi) \) with respect to \( \theta \) yields \( -\sin(\theta + \phi) \). These individual derivatives are then combined using rules like the product rule to find the overall derivative in more complex scenarios.

Simplification of Expressions

Simplification is a critical step in differentiating complex expressions. Once we've applied the product rule and found the derivatives of the individual parts, we combine these results. However, the resulting expression is often messy and not in its simplest form. For instance, after applying the product rule to \( e^{-\theta} \cos(\theta + \phi) \), we initially obtain: \[ \frac{\partial r}{\partial \theta} = (e^{-\theta})(-\sin(\theta + \phi)) + (-e^{-\theta})\cos(\theta + \phi) \]. This can be messy to work with. Combining like terms and simplifying leads us to: \[ \frac{\partial r}{\partial \theta} = -e^{-\theta} (\sin(\theta + \phi) + \cos(\theta + \phi)) \]. This final form is much easier to interpret and use.

Functions of Multiple Variables

In calculus, functions often depend on more than one variable. These are called functions of multiple variables. For instance, the function \( r = e^{-\theta} \cos(\theta + \phi) \) depends on both \( \theta \) and \( \phi \). When we find the partial derivative of \( r \) with respect to \( \theta \), we treat \( \phi \) as a constant. This ability to focus on how the function changes with respect to one variable while holding the other variables constant is a crucial skill in multivariable calculus. Partial derivatives are the tools we use to achieve this, allowing us to understand the behavior of complex systems where multiple factors are at play.