Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the gradient \(\nabla f\). $$ f(x, y, z)=x z \ln (x+y+z) $$

Step-by-Step Solution

Answer

The gradient is \( \nabla f = \left( z \ln(x+y+z) + \frac{xz}{x+y+z}, \frac{xz}{x+y+z}, x \ln(x+y+z) + \frac{xz}{x+y+z} \right) \).

Step 1: Understand the Gradient

The gradient of a function is a vector of its partial derivatives. If we have a function of three variables, as in this exercise, the gradient is given by: \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \). Our goal is to find each of these partial derivatives for the function \( f(x, y, z) = x z \ln(x+y+z) \).

Step 2: Compute \( \frac{\partial f}{\partial x} \)

For the partial derivative with respect to \( x \), we apply the product rule because \( f(x, y, z) = x z \ln(x+y+z) \) is a product of \( x z \) and \( \ln(x+y+z) \). This gives \( \frac{d}{dx} [x z] \cdot \ln(x+y+z) + x z \cdot \frac{d}{dx} [\ln(x+y+z)] \). Calculating, we have \( z \ln(x+y+z) + \frac{xz}{x+y+z} \).

Step 3: Compute \( \frac{\partial f}{\partial y} \)

For the partial derivative with respect to \( y \), \( x z \) is constant, and we only differentiate \( \ln(x+y+z) \). This derivative is \( \frac{d}{dy} [\ln(x+y+z)] = \frac{1}{x+y+z} \). Therefore, \( \frac{\partial f}{\partial y} = x z \cdot \frac{1}{x+y+z} = \frac{xz}{x+y+z} \).

Step 4: Compute \( \frac{\partial f}{\partial z} \)

For the partial derivative with respect to \( z \), we again apply the product rule: \( \frac{d}{dz} [x] \cdot z \ln(x+y+z) + x \cdot \ln(x+y+z) + x \cdot z \frac{d}{dz} [\ln(x+y+z)] \). Simplifying gives \( x \ln(x+y+z) + \frac{xz}{x+y+z} \).

Step 5: Write the Gradient Vector

Combine all the partial derivatives into the gradient vector: \[ abla f = \left( z \ln(x+y+z) + \frac{xz}{x+y+z}, \frac{xz}{x+y+z}, x \ln(x+y+z) + \frac{xz}{x+y+z} \right). \] This is the gradient of the function \( f(x, y, z) = x z \ln(x+y+z) \).

Key Concepts

Partial Derivatives

To find the rate of change of a function with multiple variables, we use partial derivatives. These derivatives only consider the variation of the function with respect to one variable, keeping the other variables constant. In our example function: - The partial derivative with respect to \( x \) is found by varying \( x \) while treating \( y \) and \( z \) as constants. - The partial derivative with respect to \( y \) considers changes in \( y \), keeping \( x \) and \( z \) constant. - Similarly, for \( z \), we vary \( z \) and treat \( x \) and \( y \) as constants.
This focus on one variable at a time makes it simpler to analyze complex functions involving several independent variables.

Product Rule

When computing derivatives, especially in vector calculus, the product rule is crucial. It helps us differentiate a product of two or more functions.
The rule states that if you have two functions, \( u \) and \( v \), their derivative is given by:\[ \frac{d}{dx}(uv) = u'v + uv' \] In our case, \( f(x, y, z) = xz \ln(x+y+z) \) means applying the product rule to a product \( u = xz \) and \( v = \ln(x+y+z) \).
This simplifies calculating partial derivatives of each component, structured and accurate, and manages complexity efficiently when functions are expressed as products.

Vector Calculus

Vector calculus is all about operations on vector fields, which are functions of several variables. In our exercise, the gradient is such an operation. A gradient of a scalar function is a vector field.
The gradient symbol, \( abla \), is used to capture the partial derivatives of each variable. For a three-variable function, the gradient is expressed as:

• The rate of change along the x-axis: \( \frac{\partial f}{\partial x} \)
• The rate of change along the y-axis: \( \frac{\partial f}{\partial y} \)
• The rate of change along the z-axis: \( \frac{\partial f}{\partial z} \)

Combining these partial derivatives gives us a vector that maps out the direction and speed of the greatest increase of the function, making vector calculus a powerful tool for analyzing multi-dimensional functions.

Mathematical Functions

Mathematical functions such as \( f(x, y, z) = xz \ln(x+y+z) \) allow us to model relationships between variables. Each part of this function – \( xz \) and \( \ln(x+y+z) \) – represents a component of a real-world relationship.
Functions can be:

• Simple linear or polynomial expressions.
• Complex combinations involving logarithmic or exponential terms.

They help us understand how changes in one or more variables influence outcomes. This is key in many scientific and engineering problems where variables interact in non-linear ways. Recognizing how these functions behave is vital for calculating changes and predicting outcomes through derivatives.