Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-16, find all first partial derivatives of each function. \(f(r, \theta)=3 r^{3} \cos 2 \theta\)

Step-by-Step Solution

Answer

\( \frac{\partial f}{\partial r} = 9r^2 \cos 2\theta, \quad \frac{\partial f}{\partial \theta} = -6r^3 \sin 2\theta \).

Step 1: Identify Variables and Function Components

The given function is \( f(r, \theta) = 3r^3 \cos 2\theta \). The variables involved are \( r \) and \( \theta \). We need to find the partial derivatives with respect to each of these variables.

Step 2: Differentiate with Respect to \( r \)

To find \( \frac{\partial f}{\partial r} \), treat \( \theta \) as a constant. Differentiate \( f(r, \theta) = 3r^3 \cos 2\theta \) with respect to \( r \). The derivative of \( r^3 \) is \( 3r^2 \), so: \[\frac{\partial f}{\partial r} = 3 \cdot 3r^2 \cos 2\theta = 9r^2 \cos 2\theta.\]

Step 3: Differentiate with Respect to \( \theta \)

To find \( \frac{\partial f}{\partial \theta} \), treat \( r \) as a constant. The derivative of \( \cos 2\theta \) is \( -2\sin 2\theta \) (using chain rule, since we need to multiply by the derivative of \( 2\theta \), which is 2):\[\frac{\partial f}{\partial \theta} = 3r^3 \cdot (-2 \sin 2\theta) = -6r^3 \sin 2\theta.\]

Step 4: Summarize Partial Derivatives

The partial derivative with respect to \( r \) is \( 9r^2 \cos 2\theta \), and the partial derivative with respect to \( \theta \) is \( -6r^3 \sin 2\theta \). These represent the rates of change of the function in each respective direction.

Key Concepts

Calculus

Calculus is a branch of mathematics focused on change through differentiation and accumulation through integration. One primary aim is to determine how functions change as their inputs change. In single-variable calculus, we mostly look at continuous functions of one variable. But calculus extends into multivariable functions, where we consider functions with more than one variable. Differentiation, in this context, forms the foundation for calculating derivatives.Derivatives:

• The derivative of a function represents its rate of change. For example, in the function's graph, the derivative is the slope of the tangent line at any point.
• In mathematical notation, if our variable is \( x \), the derivative of a function \( f(x) \) is denoted as \( f'(x) \) or \( \frac{df}{dx} \).

For multivariable functions, calculus involves partial derivatives, which examine how a function changes as only one of its several variables changes, keeping all other variables constant.

Multivariable Functions

In calculus, multivariable functions are functions with more than one input variable. In our example, the function is \( f(r, \theta) = 3r^3 \cos 2\theta \), and the input variables are \( r \) and \( \theta \).Understanding Multivariable Functions:

• The domain of multivariable functions involves all possible pairs of inputs that provide a meaningful output.
• The output values form a surface in a three-dimensional space when plotted graphically. For example, if \( f(x, y) \) is a common function, forming three dimensions with \(x\), \(y\), and \(f(x, y)\).
• One key concept is how each variable affects the overall function. This is where partial derivatives become essential.

Partial Derivatives: These derivatives examine the rate of change of the function with respect to one variable at a time by keeping the other variables constant. Calculating them helps us analyze the influence of each specific variable.

Chain Rule

The chain rule is a fundamental derivative rule in calculus used to differentiate compositions of functions. When dealing with multivariable functions, it becomes vital when variables themselves are functions of other variables.Consider two functions given by:

• An outer function \( f(u) \)
• An inner function \( u(g(x)) \)

The derivative of the composition \( f(g(x)) \) is obtained as:\[\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} \]Using the Chain Rule:

• In our example, to find \( \frac{\partial f}{\partial \theta} \), we work with \( \cos 2\theta \). The chain rule helps calculate the derivative of \( \cos \) while accounting for the inner function \( 2\theta \).
• Compute the derivative of the inner function, which is \( 2 \), and multiply it by the derivative of the outer function. Thus, getting \( -2\sin 2\theta \).

This detailed approach ensures accurate differentiation of functions within complex variable frameworks.