Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 7-12, find \(\nabla f\). $$ f(x, y, z)=y^{2} e^{-2 z} $$

Step-by-Step Solution

Answer

\( \nabla f = \left( 0, 2y e^{-2z}, -2y^2 e^{-2z} \right) \).

Step 1: Understand the Problem

The exercise requires us to find the gradient of the function \( f(x, y, z) = y^2 e^{-2z} \). The gradient \( abla f \) is a vector of partial derivatives of \( f \) with respect to its variables \( x \), \( y \), and \( z \). These partial derivatives are \( \frac{\partial f}{\partial x} \), \( \frac{\partial f}{\partial y} \), and \( \frac{\partial f}{\partial z} \).

Step 2: Partial Derivative with respect to x

Take the partial derivative of \( f(x, y, z) = y^2 e^{-2z} \) with respect to \( x \). Notice that \( f \) does not contain \( x \), so the derivative is 0:\[ \frac{\partial f}{\partial x} = 0 \].

Step 3: Partial Derivative with respect to y

Take the partial derivative of \( f(x, y, z) = y^2 e^{-2z} \) with respect to \( y \). Treat \( x \) and \( z \) as constants:\[ \frac{\partial f}{\partial y} = 2y e^{-2z} \].

Step 4: Partial Derivative with respect to z

Take the partial derivative of \( f(x, y, z) = y^2 e^{-2z} \) with respect to \( z \). Treat \( x \) and \( y \) as constants:\[ \frac{\partial f}{\partial z} = -2y^2 e^{-2z} \].

Step 5: Combine Partial Derivatives into Gradient

The gradient \( abla f \) is the vector of partial derivatives:\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = \left( 0, 2y e^{-2z}, -2y^2 e^{-2z} \right) \].

Key Concepts

Partial Derivatives

When working with functions of several variables, understanding partial derivatives is essential. A partial derivative measures how a function changes as one of its variables is varied while keeping the other variables constant. In our original exercise, we took the function \( f(x, y, z) = y^2 e^{-2z} \). We computed partial derivatives for each variable: \( x \), \( y \), and \( z \). To find the partial derivative with respect to \( x \), notice that \( f \) does not involve \( x \), so the derivative is zero: \( \frac{\partial f}{\partial x} = 0 \). For \( y \), treat \( z \) as a constant and derive accordingly: \( \frac{\partial f}{\partial y} = 2y e^{-2z} \). Similarly, for \( z \), \( y^{2} \) can be seen as a constant, leading to: \( \frac{\partial f}{\partial z} = -2y^2 e^{-2z} \). These partial derivatives form the building blocks of the gradient vector.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions of more than one variable. In this context, it becomes crucial to understand how changing more than one variable affects the function. Functions such as \( f(x, y, z) = y^2 e^{-2z} \) in our exercise, rely on inputs from multiple dimensions, which makes understanding their change rate tricky without the proper tools.

• In single-variable calculus, derivatives inform us how a function changes along the x-axis. In multivariable calculus, partial derivatives tell us how the function changes along each dimension individually.
• The gradient vector, \( abla f \), consolidates all these changes, showing the direction of the steepest ascent in the field of scalar or multivariable functions.

Multivariable calculus allows us to navigate complex systems by providing the tools to measure and predict changes across multiple variables.

Vector Calculus

Vector calculus is a branch of mathematics focused on vector fields and operations. In particular, it deals with vectors in multiple dimensions and how they interact, overlap, and map out spaces. When we talk about the gradient \( abla f \), we're using a concept from vector calculus that represents a vector field of all possible directional derivatives.

• Each component of the gradient vector is a partial derivative that indicates how \( f(x, y, z) \) changes across a particular direction.
• The resulting vector field provides a visual sense of the function's behavior throughout space, highlighting regions of steep or gentle inclines.
• The gradient is particularly useful in optimizations, helping to find local maxima or minima by offering insight into where the function increases the most.

By employing vector calculus, we gain a comprehensive method to analyze and visualize multivariable functions in action, making it an indispensable part of advanced mathematics and physics.