Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the gradient \(\nabla f\). $$ f(x, y, z)=x^{2} y+y^{2} z+z^{2} x $$

Step-by-Step Solution

Answer

\( \nabla f = (2xy + z^2, x^2 + 2yz, y^2 + 2zx) \)

Step 1: Identify the partial derivatives

To find the gradient \( abla f \), we must determine the partial derivatives of \( f(x, y, z) = x^2y + y^2z + z^2x \) with respect to each variable, \( x \), \( y \), and \( z \).

Step 2: Compute the partial derivative with respect to \( x \)

Differentiate \( f(x, y, z) \) with respect to \( x \). The term \( x^2y \) differentiates to \( 2xy \), \( y^2z \) is zero since it does not contain \( x \), and \( z^2x \) differentiates to \( z^2 \). So, \( \frac{\partial f}{\partial x} = 2xy + z^2 \).

Step 3: Compute the partial derivative with respect to \( y \)

Differentiate \( f(x, y, z) \) with respect to \( y \). The term \( x^2y \) becomes \( x^2 \), the term \( y^2z \) differentiates to \( 2yz \), and \( z^2x \) is zero since it does not contain \( y \). Therefore, \( \frac{\partial f}{\partial y} = x^2 + 2yz \).

Step 4: Compute the partial derivative with respect to \( z \)

Differentiate \( f(x, y, z) \) with respect to \( z \). The term \( x^2y \) is zero since it does not contain \( z \), the term \( y^2z \) becomes \( y^2 \), and \( z^2x \) differentiates to \( 2zx \). Therefore, \( \frac{\partial f}{\partial z} = y^2 + 2zx \).

Step 5: Combine the partial derivatives to form the gradient

The gradient \( abla f \) is a vector consisting of all the partial derivatives: \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (2xy + z^2, x^2 + 2yz, y^2 + 2zx) \).

Key Concepts

Understanding Partial Derivatives

Partial derivatives are a fundamental tool in calculus, especially when dealing with functions of multiple variables. It involves taking the derivative of a function with respect to one variable while keeping the others constant. For example:

• When finding the partial derivative of a function with respect to the variable \( x \), treat \( y \) and \( z \) as constants.
• This means that you will differentiate each term involving \( x \), ignoring any parts of the function that do not contain \( x \).

This is a key step in multivariable calculus when investigating how a variable uniquely affects a function's behavior. In our exercise, the function \( f(x, y, z) = x^2y + y^2z + z^2x \) was differentiated to find its partials. Partial derivatives help in constructing the gradient of a function, guiding us on how the function changes in different directions in its input space.

Exploring Multivariable Calculus

Multivariable calculus expands calculus to functions with more than one variable. This branch is critical in various fields, including physics and engineering. The primary idea is to deal with functions like \( f(x, y, z) \) that depend on multiple inputs and consider the rate of change with respect to each input.
In our exercise, we encountered a multivariable function: \( f(x, y, z) = x^2y + y^2z + z^2x \). This function exists in a three-dimensional space, which means it is influenced by three variables. By applying partial derivatives, we analyze how changes in one variable influence the full function.

• Each partial derivative represents a tiny change in the output related to a tiny change in one input, keeping other inputs constant.
• This leads to the gradient, a vector that indicates the direction and rate of steepest ascent of the function.

Multivariable calculus allows us to tackle complex problems that single-variable calculus cannot solve, giving us comprehensive insights into functions' behavior in multi-dimensional spaces.

Grasping Vector Calculus and the Gradient

Vector calculus combines elements of calculus and linear algebra to understand vector fields and functions. One important concept is the gradient. The gradient is a vector that points in the direction of the greatest rate of increase of a scalar field.

• The magnitude of the gradient indicates how steep the incline of the function is along that direction.
• For instance, in our exercise, the gradient is \( abla f = (2xy + z^2, x^2 + 2yz, y^2 + 2zx) \).

This vector helps in multiple applications, such as optimizing functions in machine learning and understanding electrostatic fields in physics.
When computing a gradient, each component corresponds to the partial derivative of the function with respect to a given variable. Therefore, understanding and calculating partial derivatives is crucial, as they directly contribute to the formation of the gradient in vector calculus contexts.