Question

If \(w=(r \cos \theta)^{r \sin \theta},\) find \(\partial w / \partial \theta.\)

Short answer

\( \frac{\partial w}{\partial \theta} = (r \, \text{cos} \, \theta)^{r \, \text{sin} \, \theta} [ r \, \text{cos} \, \theta \, \text{ln} (r \, \text{cos} \, \theta) - r \, \frac{\text{sin}^2 \, \theta}{\text{cos} \, \theta} ] \)

Step-by-step solution

- Express the problem in logarithmic form

Given \(w = (r \, \text{cos} \, \theta)^{r \, \text{sin} \, \theta}\), take the natural logarithm of both sides to simplify the differentiation process. So, \( \text{ln} \, w = \text{ln} \, [(r \, \text{cos} \, \theta)^{r \, \text{sin} \, \theta}] \). Using the logarithm power rule, rewrite it as: \( \text{ln} \, w = r \, \text{sin} \, \theta \, \text{ln} \, (r \, \text{cos} \, \theta) \).

- Differentiate using the chain rule

Differentiate both sides with respect to \( \theta \). On the left side, use the chain rule: \( \frac{d}{d \theta} \text{ln} \, w = \frac{1}{w} \cdot \frac{\partial w}{\partial \theta} \). On the right side, use the product rule:\( \frac{d}{d \theta} [r \, \text{sin} \, \theta \, \text{ln}(r \, \text{cos} \, \theta)] \).

- Apply the product rule

Apply the product rule to \( r \, \text{sin} \, \theta \, \text{ln}(r \, \text{cos} \, \theta) \): \( \frac{d}{d \theta}[r \, \text{sin} \, \theta] \, \text{ln}(r \, \text{cos} \, \theta) + r \, \text{sin} \, \theta \frac{d}{d \theta} [\text{ln} (r \, \text{cos} \, \theta)] \).

- Differentiate each component

Differentiate individually: \( \frac{d}{d \theta} [ r \, \text{sin} \, \theta] = r \, \text{cos} \, \theta \) and \( \frac{d}{d \theta} [\text{ln} (r \, \text{cos} \, \theta)] = - \text{tan} \, \theta \). Substituting these back, get: \( r \, \text{cos} \, \theta \, \text{ln} (r \, \text{cos} \, \theta) + r \, \text{sin} \, \theta \cdot [- \text{tan} \, \theta] \).

- Simplify the expression

Combine the terms to simplify: \( r \, \text{cos} \, \theta \, \text{ln} (r \, \text{cos} \, \theta) - r \, \text{sin} \, \theta \, \text{tan} \, \theta \). Since \( \text{tan} \, \theta = \frac{\text{sin} \, \theta}{\text{cos} \, \theta} \), we have: \( r \, \text{cos} \, \theta \, \text{ln} (r \, \text{cos} \, \theta) - r \, \frac{\text{sin}^2 \, \theta}{\text{cos} \, \theta} \).

- Multiply through by \( w \)

Recall \( \frac{1}{w} \cdot \frac{\partial w}{\partial \theta} = r \, \text{cos} \, \theta \, \text{ln} (r \, \text{cos} \, \theta) - r \, \frac{\text{sin}^2 \, \theta}{\text{cos} \, \theta} \). Multiply through by \( w \) (i.e., \( (r \, \text{cos} \, \theta)^{r \, \text{sin} \, \theta} \)): \( \frac{\partial w}{\partial \theta} = (r \, \text{cos} \, \theta)^{r \, \text{sin} \, \theta} [ r \, \text{cos} \, \theta \, \text{ln} (r \, \text{cos} \, \theta) - r \, \frac{\text{sin}^2 \, \theta}{\text{cos} \, \theta} ] \).

Key concepts

Chain Rule

When it comes to calculus, the chain rule is a powerful tool. It helps us differentiate composite functions. For instance, if we have a function of another function, like with the form \[ h(x) = f(g(x)) \]we use the chain rule to differentiate. The chain rule states that: \[ \frac{d}{dx} h(x) = f'(g(x)) \times g'(x) \]This means we first differentiate the outer function and then multiply it by the derivative of the inner function. An application of this can be seen in the differentiation of \[ \text{ln} \text{w} \]in our exercise, where we encountered \[ \frac{d}{d \theta} \text{ln} \text{w} = \frac{1}{w} \frac{\text{∂} w}{\text{∂} \theta} \] Understanding the chain rule is critical as it simplifies differentiating more complex expressions!

Logarithmic Differentiation

Logarithmic differentiation is incredibly useful in handling functions that involve products, quotients, or powers. It particularly shines when dealing with complex expressions like \[ w = (r \text{cos} \theta)^{r \text{sin} \theta} \] As taking the natural logarithm of both sides leverages the properties of logarithms to simplify differentiation. The properties that come in handy are:

• \[ \text{ln}(a^b) = b \text{ln}(a) \]
• \[ \text{ln}(ab) = \text{ln}(a) + \text{ln}(b) \]

These allow us to rewrite our equation, making it easier to apply differentiation techniques. For example, transforming \[ w = (r \text{cos} \theta)^{r \text{sin} \theta} \] into \[ \text{ln} \text{w} = r \text{sin} \theta \text{ln}(r \text{cos} \theta) \] This approach is vital for breaking down otherwise unwieldy expressions.

Product Rule

The product rule is another key concept in calculus that's essential for differentiating products of two functions. When given a product of two differentiable functions, for example, \[ h(x) = f(x) \times g(x) \] we use the product rule, which states: \[ h'(x) = f'(x) \times g(x) + f(x) \times g'(x) \] In our exercise, this rule was applied to \[ r \text{sin} \theta \text{ln} (r \text{cos} \theta) \] Differentiating this, we split it into parts, differentiating each component individually before combining them:

• Differentiate \[ r \text{sin} \theta \] which is \[ r \text{cos} \theta \]
• Differentiate \[ \text{ln} (r \text{cos} \theta) \] resulting in \[ -\text{tan} \theta \]

Combining both, we ensure we accurately capture the effect of both parts of the product.

Trigonometric Functions

Trigonometric functions like \[ \text{sin}(\theta) \] and \[ \text{cos}(\theta) \] are fundamental in calculus. Both have specific differentiation rules:

• \[ \frac{d}{d \theta} \text{sin} (\theta) = \text{cos} (\theta) \]
• \[ \frac{d}{d \theta} \text{cos} (\theta) = -\text{sin} (\theta) \]

These functions often show up in more complex expressions, and understanding their properties is vital. For example, we used: \[ \text{tan} (\theta) = \frac{\text{sin} (\theta)}{\text{cos} (\theta)} \] This relationship helped simplify our expression towards the end of the exercise. By substituting and rearranging terms, we made the differentiation process more manageable. Recognizing and applying these properties and relationships will strengthen your calculus skills.