Problem 24 - Advanced Engineering Mathematics

Question

Find the first partial derivatives of the given function. $$ f(\theta, \phi)=\phi^{2} \sin \frac{\theta}{\phi} $$

Step-by-Step Solution

Answer

\( \frac{\partial f}{\partial \theta} = \phi \cos \left( \frac{\theta}{\phi} \right) \); \( \frac{\partial f}{\partial \phi} = 2\phi \sin \left( \frac{\theta}{\phi} \right) - \theta \cos \left( \frac{\theta}{\phi} \right) \).

Step 1: Determine the function to differentiate

We need to find the first partial derivatives of the function \( f(\theta, \phi) = \phi^2 \sin \frac{\theta}{\phi} \). This means we'll find \( \frac{\partial f}{\partial \theta} \) and \( \frac{\partial f}{\partial \phi} \).

Step 2: Find the partial derivative with respect to \(\theta\)

To find \( \frac{\partial f}{\partial \theta} \), treat \( \phi \) as a constant. The derivative of \( \sin \left( \frac{\theta}{\phi} \right) \) with respect to \( \theta \) is \( \cos \left( \frac{\theta}{\phi} \right) \times \frac{1}{\phi} \) using the chain rule. So, \[ \frac{\partial f}{\partial \theta} = \phi^2 \times \cos \left( \frac{\theta}{\phi} \right) \times \frac{1}{\phi} = \phi \cos \left( \frac{\theta}{\phi} \right). \]

Step 3: Find the partial derivative with respect to \(\phi\)

To find \( \frac{\partial f}{\partial \phi} \), use the product rule since \( f(\theta, \phi) = \phi^2 \cdot \sin \left( \frac{\theta}{\phi} \right) \). Let \( u = \phi^2 \) and \( v = \sin \left( \frac{\theta}{\phi} \right) \). Then, \( \frac{du}{d\phi} = 2\phi \) and \( \frac{dv}{d\phi} = \cos \left( \frac{\theta}{\phi} \right) \times -\frac{\theta}{\phi^2} \), again using the chain rule. Applying the product rule \( \frac{d(uv)}{d\phi} = u'v + uv' \), we get: \[ \frac{\partial f}{\partial \phi} = 2\phi \cdot \sin \left( \frac{\theta}{\phi} \right) + \phi^2 \cdot \left(-\frac{\theta}{\phi^2}\cos \left( \frac{\theta}{\phi} \right) \right) \] Thus, \[ \frac{\partial f}{\partial \phi} = 2\phi \sin \left( \frac{\theta}{\phi} \right) - \theta \cos \left( \frac{\theta}{\phi} \right). \]

Step 4: Verify and interpret results

The first partial derivatives of \( f \) with respect to \( \theta \) and \( \phi \) are: - \( \frac{\partial f}{\partial \theta} = \phi \cos \left( \frac{\theta}{\phi} \right) \) which describes how \( f \) changes with \( \theta \) holding \( \phi \) constant.- \( \frac{\partial f}{\partial \phi} = 2\phi \sin \left( \frac{\theta}{\phi} \right) - \theta \cos \left( \frac{\theta}{\phi} \right) \) which describes how \( f \) changes with \( \phi \) holding \( \theta \) constant.

Key Concepts

First Partial Derivative

Partial derivatives are a foundational concept in the study of multivariable calculus. Essentially, they help us understand how a function changes with respect to one variable while keeping others constant. Here, in the case of the function \( f(\theta, \phi) = \phi^2 \sin \frac{\theta}{\phi} \), the task is to find its first partial derivatives.

• For \( \frac{\partial f}{\partial \theta} \), treat \( \phi \) as a constant. This means we observe how the function responds if only \( \theta \) is varied.
• For \( \frac{\partial f}{\partial \phi} \), treat \( \theta \) as a constant, observing the function's response to changes in \( \phi \) alone.

Calculating these derivatives provides valuable insights into the behavior of functions across surfaces or spaces, which is essential in optimizing and modeling real-world scenarios.

Chain Rule

The chain rule is a critical tool when dealing with derivatives, especially when functions are composed of other functions. It allows us to compute the derivative of such composite functions efficiently. For instance, given a function like \( \sin \left( \frac{\theta}{\phi} \right) \), both \( \theta \) and \( \phi \) are involved in a nested manner.In practice, the chain rule tells us that the derivative of \( \sin \left( \frac{\theta}{\phi} \right) \) with respect to \( \theta \) is \( \cos \left( \frac{\theta}{\phi} \right) \times \frac{1}{\phi} \). The inner derivative, \( \frac{\theta}{\phi} \), multiplies the derivative of the outer function, \( \sin \), which is \( \cos \). This rule simplifies the process of finding derivatives involving intricate layers of variables, which often appear in real-world modeling.

Product Rule

The product rule is indispensable when differentiating expressions that are products of two or more functions. In our example, the function \( f(\theta, \phi) = \phi^2 \cdot \sin \left( \frac{\theta}{\phi} \right) \) is a perfect candidate for the product rule because it's the product of \( \phi^2 \) and \( \sin \left( \frac{\theta}{\phi} \right) \).To apply:

• Identify the two functions: \( u = \phi^2 \) and \( v = \sin \left( \frac{\theta}{\phi} \right) \).
• Differentiate each function separately: \( \frac{du}{d\phi} = 2\phi \) and \( \frac{dv}{d\phi} \) using the chain rule.
• Apply the product rule: \( \frac{d(uv)}{d\phi} = u'v + uv' \).

This gives us insight into how the interplay between multiplying functions affects the rate of change and is frequently used in physics and engineering problems.

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in mathematics, especially when dealing with periodic phenomena. They have unique properties that make them crucial in calculus, and understanding their derivatives is key.For example, the function \( f(\theta, \phi) = \phi^2 \sin \left( \frac{\theta}{\phi} \right) \) involves the sine function. When differentiating the sine function with respect to any variable, the result is a cosine function, multiplies by the derivative of the inside function if it involves multiple variables.Key points about trigonometric derivatives:

• \( \frac{d}{dx} \sin(x) = \cos(x) \).
• Trigonometric identities can simplify complex derivatives.

Grasping these concepts aids in solving a wide range of problems, from wave behavior in physics to oscillations in engineering.