Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that $$ \nabla\left(f^{r}\right)=r f^{r-1} \nabla f $$

Step-by-Step Solution

Answer

The expression is verified: \( \nabla(f^r) = r f^{r-1} \nabla f \).

Step 1: Understand the Gradient Notation

The gradient operator, denoted \( abla \), when applied to a scalar function \( f \), produces a vector of partial derivatives. If \( f \) is a function of variables \( x, y, z \), then:\[abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)\]

Step 2: Apply the Power Rule for Derivatives

For a function of the form \( f^r \), where \( f \) is a function of coordinates, we can apply the power rule for derivatives. The power rule states that:\[\frac{d}{dx}(f^r) = r f^{r-1} \frac{df}{dx}\]This same rule applies when taking the gradient of \( f^r \).

Step 3: Calculate the Gradient of \( f^r \)

The gradient of \( f^r \) can be calculated as:\[abla(f^r) = \left( \frac{\partial}{\partial x}(f^r), \frac{\partial}{\partial y}(f^r), \frac{\partial}{\partial z}(f^r) \right)\]Using the power rule, each component becomes:\[\frac{\partial}{\partial x}(f^r) = r f^{r-1} \frac{\partial f}{\partial x}, \quad \text{and similarly for } \frac{\partial}{\partial y} \text{ and } \frac{\partial}{\partial z}\]

Step 4: Combine the Results

Substituting the results from the power rule, the gradient of \( f^r \) is:\[abla(f^r) = \left( r f^{r-1} \frac{\partial f}{\partial x}, r f^{r-1} \frac{\partial f}{\partial y}, r f^{r-1} \frac{\partial f}{\partial z} \right)\]This can be factored as:\[r f^{r-1} \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = r f^{r-1} abla f\]

Step 5: Verify the Final Expression Matches

We have shown that:\[abla(f^r) = r f^{r-1} abla f\]This confirms that the provided expression is correct and verified step-by-step.

Key Concepts

Vector Calculus

Vector Calculus is a branch of mathematics focusing on vector-valued functions. It plays a crucial role in understanding physical phenomena, such as electromagnetic fields and fluid flow. In vector calculus, we often deal with vector fields and perform operations like differentiation and integration with respect to these fields. The gradient, denoted by \(abla\), is a fundamental operator used to determine the rate and direction of change of scalar fields. When you take the gradient of a scalar function \(f\), it results in a vector field. This vector field points in the direction where the function increases most rapidly. The magnitude of this vector provides the maximum rate of change at a point.

• Gradient: \(abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)\)
• Vector fields can represent a range of physical quantities, such as velocity or force fields.

Power Rule for Derivatives

The power rule is a crucial principle in calculus which helps in finding derivatives of functions in the form \(f^r\). This rule is particularly useful in both single-variable and multivariable calculi.When applied to scalar functions, the power rule for derivatives states: if \(f(x)\) is a differentiable function of \(x\), then the derivative of \(f(x)\) raised to the power \(r\) is:\[ \frac{d}{dx} (f^r) = r f^{r-1} \frac{df}{dx} \]In vector calculus, this rule is expanded to include gradients. When taking the gradient of \(f^r\), each partial derivative is modified by the power rule:

• \( \frac{\partial}{\partial x} (f^r) = r f^{r-1} \frac{\partial f}{\partial x} \)
• Similarly applied for \(y\) and \(z\) components

This application simplifies computations and is instrumental in solving diverse mathematical problems, providing deeper insight into how functions behave with respect to each of their variables.

Partial Derivatives

Partial Derivatives are derivatives of multivariable functions taken with respect to one variable at a time, all other variables being considered constant. These derivatives provide essential information about the function's behavior in relation to each coordinate axis independently.For a function \(f(x, y, z)\), the partial derivative with respect to \(x\) is expressed as:\[ \frac{\partial f}{\partial x} \]Calculating partial derivatives is a fundamental step when working with the gradient of a function. Each component of the gradient vector is essentially a partial derivative of the function concerning its respective variable:

• The \(x\)-component: \(\frac{\partial f}{\partial x} \)
• The \(y\)-component: \(\frac{\partial f}{\partial y} \)
• The \(z\)-component: \(\frac{\partial f}{\partial z} \)

Understanding how to compute these derivatives is crucial for effectively applying the gradient and interpreting the results. They help determine how a small change in each variable affects the function's outcome.

Scalar Functions

Scalar Functions are mathematical maps that assign a single real number to every point in a space, often denoted as \(f(x, y, z)\). These functions are fundamental in various branches of mathematics and physics, modeling quantities like temperature, pressure, and potential energy over a region.Scalar functions are characterized by:

• Providing a single value (a scalar) for each coordinate point in the domain
• Being a key input for operators like the gradient in vector calculus

The gradient of a scalar function transforms the scalar-valued function into a vector field. This transformation helps reveal how the function changes over the region, specifically indicating the direction and magnitude of the steepest ascent.In the context of the exercise, \(f\) is a scalar function, and the use of the gradient operator \(abla\) on \(f\) assists in exploring how changes in \(f\) relate to the spatial variables \(x, y, z\). Understanding scalar functions is foundational for grasping complex concepts in multivariable calculus and their real-world applications.