Question

Find all first partial derivatives of each function. \(f(r, \theta)=3 r^{3} \cos 2 \theta\)

Short answer

The partial derivatives are \( \frac{\partial f}{\partial r} = 9r^2 \cos 2\theta \) and \( \frac{\partial f}{\partial \theta} = -6r^3 \sin 2\theta \).

Step-by-step solution

Understanding Partial Derivatives

To find the first partial derivatives of a function with multiple variables, differentiate the function with respect to each variable while keeping the other variables constant. Here we have the function \( f(r, \theta) = 3r^3 \cos 2\theta \), and we need to find \( \frac{\partial f}{\partial r} \) and \( \frac{\partial f}{\partial \theta} \).

Finding Partial Derivative with respect to r

To find \( \frac{\partial f}{\partial r} \), differentiate \( f(r, \theta) = 3r^3 \cos 2\theta \) with respect to \( r \). Treat \( \cos 2\theta \) as a constant.\[\frac{\partial f}{\partial r} = \frac{\partial}{\partial r}(3r^3 \cos 2\theta) = 3 \cos 2\theta \cdot \frac{\partial}{\partial r}(r^3)\]Apply the power rule:\[\frac{\partial}{\partial r}(r^3) = 3r^2\]So,\[\frac{\partial f}{\partial r} = 3 \cos 2\theta \cdot 3r^2 = 9r^2 \cos 2\theta\]

Finding Partial Derivative with respect to θ

To find \( \frac{\partial f}{\partial \theta} \), differentiate \( f(r, \theta) = 3r^3 \cos 2\theta \) with respect to \( \theta \). Treat \( 3r^3 \) as a constant.\[\frac{\partial f}{\partial \theta} = \frac{\partial}{\partial \theta}(3r^3 \cos 2\theta) = 3r^3 \cdot \frac{\partial}{\partial \theta}(\cos 2\theta)\]Apply the chain rule:\[\frac{\partial}{\partial \theta}(\cos 2\theta) = -\sin 2\theta \cdot (2)\]So,\[\frac{\partial f}{\partial \theta} = 3r^3 \cdot (-2\sin 2\theta) = -6r^3 \sin 2\theta\]

Summarizing the Results

The first partial derivative with respect to \( r \) is \( 9r^2 \cos 2\theta \) and the first partial derivative with respect to \( \theta \) is \( -6r^3 \sin 2\theta \).

Key concepts

Calculus

Calculus is a branch of mathematics that studies how things change. It provides a framework for describing the way quantities vary over time or space. At its core, calculus deals with two main concepts: differentiation and integration.

• Differentiation: The process of finding the rate at which something changes. This includes finding slopes of curves, which are often represented as derivatives.
• Integration: The reverse process of differentiation. It involves finding the total accumulation of a quantity, like the area under a curve.

In this exercise, we particularly focus on differentiation, though in the broader context of multivariable calculus.

Multivariable Functions

Multivariable functions are functions that have more than one input variable. For example, in our function, we have two variables: \( r \) and \( \theta \). Since real-world scenarios often involve multiple changing factors, studying multivariable functions is crucial.

• Such functions are often used in physics, economics, and engineering to model complex systems.
• The relationship between the variables can be examined to understand how a change in one might influence the others.

Understanding these functions allows us to tackle challenges involving multiple variables with ease, just like the one seen in finding the partial derivatives in this exercise.

Differentiation

Differentiation is the mathematical process of finding the derivative of a function. The derivative gives you the rate at which a function is changing at any given point.

When dealing with multivariable functions, we compute partial derivatives. Each partial derivative represents how the function changes as one specific variable changes, while others remain constant.

To differentiate \( f(r, \theta) = 3r^3 \cos 2\theta \):

• Partial Derivative with respect to \( r \): Here, treat \( \theta \) (and thus \( \cos 2\theta \)) as a constant. Use the power rule to differentiate \( r^3 \).
• Partial Derivative with respect to \( \theta \): Here, treat \( r \) as a constant. Use the chain rule to differentiate \( \cos 2\theta \).

These procedures correctly identify how the function \( f \) changes in relation to each individual variable.

Chain Rule

The chain rule is a fundamental technique in calculus used for differentiating compositions of functions. When a function is composed of two or more functions, the chain rule comes into play.

In our exercise, we use the chain rule to find the partial derivative of \( f(r, \theta) \) with respect to \( \theta \):

• The function inside the cosine function is \( 2\theta \). When differentiating \( \cos 2\theta \), the chain rule allows us to deal with this inner function.
• We take the derivative of \( \cos \) (which is \(-\sin\)) and then multiply it by the derivative of \( 2\theta \), which is \( 2 \).

By applying the chain rule, we effectively measure how deeply nested functions within our larger function contribute to its overall rate of change.