Problem 16 - Calculus Early Transcendentals

Question

Confirm that \(\phi\) is a potential function for \(\mathbf{F}(\mathbf{r})\) on some region, and state the region. $$ \begin{array}{l}{\text { (a) } \phi(x, y)=2 y^{2}+3 x^{2} y-x y^{3}} \\\ {\quad \mathbf{F}(x, y)=\left(6 x y-y^{3}\right) \mathbf{i}+\left(4 y+3 x^{2}-3 x y^{2}\right) \mathbf{j}}\end{array} $$ $$ \begin{aligned} \text { (b) } \phi(x, y, z)=x \sin z+y \sin x+z \sin y & \\\ \mathbf{F}(x, y, z)=(\sin z+y \cos x) \mathbf{i}+(\sin x+z \cos y) \mathbf{j} \\\\+(\sin y+x \cos z) \mathbf{k} \end{aligned} $$

Step-by-Step Solution

Answer

Both \( \phi(x, y) \) and \( \phi(x, y, z) \) are potential functions for their respective \( \mathbf{F} \) on \( \mathbb{R}^2 \) and \( \mathbb{R}^3 \).

Step 1: Compute Gradient of \( \phi(x, y) \)

Calculate the partial derivatives of \( \phi \) with respect to \( x \) and \( y \). \( \phi(x, y) = 2y^2 + 3x^2y - xy^3 \). The partial derivative with respect to \( x \) is \( \frac{\partial \phi}{\partial x} = 6xy - y^3 \), and with respect to \( y \) is \( \frac{\partial \phi}{\partial y} = 4y + 3x^2 - 3xy^2 \).

Step 2: Compare Gradient with \( \mathbf{F}(x,y) \)

The vector field \( \mathbf{F}(x,y) = (6xy - y^3) \mathbf{i} + (4y + 3x^2 - 3xy^2) \mathbf{j} \) matches the computed gradient of \( \phi(x, y) \). This confirms that \( \phi \) is a potential function for \( \mathbf{F}(x, y) \) on the whole \( \mathbb{R}^2 \).

Step 3: Compute Gradient of \( \phi(x, y, z) \)

Calculate the partial derivatives of \( \phi \) with respect to \( x \), \( y \), and \( z \). \( \phi(x, y, z) = x \sin z + y \sin x + z \sin y \). The partial derivatives are \( \frac{\partial \phi}{\partial x} = \sin z + y \cos x \), \( \frac{\partial \phi}{\partial y} = \sin x + z \cos y \), and \( \frac{\partial \phi}{\partial z} = \sin y + x \cos z \).

Step 4: Compare Gradient with \( \mathbf{F}(x,y,z) \)

The vector field \( \mathbf{F}(x,y,z) = (\sin z + y \cos x) \mathbf{i} + (\sin x + z \cos y) \mathbf{j} + (\sin y + x \cos z) \mathbf{k} \) matches the computed gradient of \( \phi(x, y, z) \). This confirms that \( \phi \) is a potential function for \( \mathbf{F}(x, y, z) \) on the whole \( \mathbb{R}^3 \).

Key Concepts

Potential Function

A potential function, typically denoted as \( \phi \), is a scalar function that helps us define a vector field. Think of it as a tool to describe the direction and strength of a field using a simpler form. To identify \( \phi \) as a potential function, the gradient of \( \phi \) must match the given vector field. This means that applying partial derivatives to \( \phi \) results in vectors that are exactly the same as the ones defined by the vector field. For instance, in the exercise, we confirmed \( \phi \) as a potential function for \( \mathbf{F} \) by checking their gradients matched. This match implies that the vector field is derived from the potential function across the regions specified, which are \( \mathbb{R}^2 \) for two dimensions and \( \mathbb{R}^3 \) for three dimensions.

Gradient

The gradient of a scalar function \( \phi \), denoted as \( abla \phi \), is a vector field that points in the direction of the maximum rate of increase of the function. The components of this gradient are the partial derivatives of \( \phi \) with respect to each variable.

• For a function \( \phi(x, y) \), the gradient is \( \left( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y} \right) \).
• For a function \( \phi(x, y, z) \), it's \( \left( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z} \right) \).

In our exercise, we calculated these gradients and confirmed that they matched the components of the vector field \( \mathbf{F} \), proving that \( \phi \) could act as its potential function. The gradient not only helps in understanding the change rate but is crucial in fields like physics, where it translates to fields like electric and gravitational.

Vector Field

A vector field assigns a vector to every point in a space, often used to represent things like magnetic fields, fluid flow, or force fields. Each vector in a vector field has both magnitude and direction, offering a powerful way to visualize complex patterns and phenomena.

• In the exercise, we examined vector fields defined as \( \mathbf{F}(x, y) = (6xy - y^3) \mathbf{i} + (4y + 3x^2 - 3xy^2) \mathbf{j} \).
• And \( \mathbf{F}(x, y, z) = (\sin z + y \cos x) \mathbf{i} + (\sin x + z \cos y) \mathbf{j} + (\sin y + x \cos z) \mathbf{k} \).

These functions assign a distinct vector to every point based on the input coordinates \( x \), \( y \), and \( z \). By confirming the alignment of vector fields with the gradient of \( \phi \), we demonstrated that these fields could indeed be expressed or derived from potential functions.

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They represent the rate at which a function changes as all other variables are held constant, effectively showcasing how much a function is varying in one specific direction.

• The notation \( \frac{\partial \phi}{\partial x} \) implies the partial derivative of \( \phi \) with respect to \( x \), focusing on changes along the \( x \)-axis.
• Similarly, \( \frac{\partial \phi}{\partial y} \) and \( \frac{\partial \phi}{\partial z} \) denote changes along the \( y \)-axis and \( z \)-axis, respectively.

In our exercise, by calculating these partial derivatives, we obtained components of the gradient. Checking these against the vector field revealed that the original field could indeed be described as the gradient of a potential function. Partial derivatives help us elongate complex surfaces into understandable directions, making them crucial in multivariable and vector calculus.