Problem 15 - The Calculus with Analytic Geometry

Question

In the following exercises, determine if the vector is a gradient. If it is, find a function having the given gradient \((2 y-5 z) \mathbf{i}+(2 x+8 z) \mathbf{j}-(5 x-8 y) \mathbf{k}\)

Step-by-Step Solution

Answer

The vector is a gradient. The scalar potential function is \(f(x, y, z) = (2 y - 5 z)x + 8 z y + C\).

Step 1: Understand the Problem

The task is to determine if the given vector \(2 y-5 z\mathbf{i}+(2 x+8 z)\mathbf{j}-(5 x-8 y)\mathbf{k}\) is a gradient and, if it is, find the corresponding scalar potential function.

Step 2: Recall the Definition of a Gradient

A vector field \mathbf{F} is a gradient vector field if there exists a scalar function \(f\) such that \(\mathbf{F} = abla f\). This means that the vector components must be partial derivatives of a single scalar function.

Step 3: Assign the Vector Components to Partial Derivatives

Identify the vector components that correspond to the partial derivatives: \(\frac{\partial f}{\partial x} = 2 y - 5 z, \ \frac{\partial f}{\partial y} = 2 x + 8 z, \ \frac{\partial f}{\partial z} = -5 x - 8 y\).

Step 4: Integrate Partial Derivative with Respect to x

Integrate \(\frac{\partial f}{\partial x} = 2 y - 5 z\) with respect to \(x\): \[ f(x, y, z) = (2 y - 5 z)x + g(y, z) \] where \(g(y, z)\) is an arbitrary function.

Step 5: Partial Derivation with Respect to y

Take the partial derivative of the result from Step 4 with respect to \(y\): \[ \frac{\partial f}{\partial y} = 2 x + \frac{\partial g}{\partial y} \] and equate it to the given \(\frac{\partial f}{\partial y} = 2 x + 8 z\). This implies: \[ \frac{\partial g}{\partial y} = 8 z \]

Step 6: Solve for g(y, z)

Integrate \(\frac{\partial g}{\partial y} = 8 z\) with respect to \(y\): \[ g(y, z) = 8 z y + h(z) \] where \(h(z)\) is an arbitrary function.

Step 7: Partial Derivation with Respect to z

Take the partial derivative of the complete expression \( f(x, y, z) = (2 y - 5 z)x + 8 z y + h(z) \) with respect to \(z\): \[ \frac{\partial f}{\partial z} = -5 x + 8 y + h'(z) \] and equate it to the given \(\frac{\partial f}{\partial z} = -5 x - 8 y\). This implies: \[ h'(z) = 0 \]

Step 8: Determine h(z)

Since \(h'(z) = 0\), \(h(z)\) must be a constant function. Let \(h(z) = C\) where \(C\) is a constant.

Step 9: State the Scalar Potential Function

Combining all the parts, the scalar potential function \(f(x, y, z)\) is: \[ f(x, y, z) = (2 y - 5 z)x + 8 z y + C \]

Key Concepts

scalar potential function

In mathematics and physics, the concept of a scalar potential function is integral when dealing with vector fields. A scalar potential function, often denoted as \(f(x, y, z)\), is a function whose gradient (denoted as \(abla f\)) gives us the vector field we are interested in. The gradient of \(f\) is a vector field composed of the partial derivatives of \(f\) with respect to its variables (x, y, z). For example, if we have a gradient vector field given by \(\mathbf{F} = 2 y - 5 z\mathbf{i} + 2 x + 8 z\mathbf{j} - 5 x - 8 y\mathbf{k}\), we aim to find a function \(f(x, y, z)\) such that \(\mathbf{F} = abla f\). This means that each component of the vector field should correspond to a partial derivative of the scalar potential function. Understanding the relationship between the vector field and the scalar potential function helps us solve problems related to conservative fields, which are fields where the line integral is path-independent.

partial derivatives

Partial derivatives are the building blocks to understanding how a function changes as its variables change. When working with scalar potential functions and vector fields, partial derivatives allow us to dissect each component of a function's rate of change independently with respect to its input variables. Consider a scalar function \(f(x, y, z)\). Its partial derivative with respect to \(x\), represented as \(\frac{\partial f}{\partial x}\), tells us how the function changes as only \(x\) changes, keeping \(y\) and \(z\) constant. Given our vector field \(ablaf\), we find the partial derivatives:

• \(\frac{\partial f}{\partial x} = 2 y - 5 z\)
• \(\frac{\partial f}{\partial y} = 2 x + 8 z\)
• \(\frac{\partial f}{\partial z} = -5 x - 8 y\)

These partial derivatives are individually matched to components of our given vector field. The concept of partial derivatives simplifies the multi-variable functions into manageable pieces that reveal crucial information about the overall function’s nature.

integration of partial derivatives

Integration of partial derivatives is the reverse process of differentiation. In this context, we aim to reconstruct the original scalar potential function \(f(x, y, z)\) from its partial derivatives. Here’s how we do this step by step:1. **Integrate with respect to \(x\)**: Starting with \(\frac{\partial f}{\partial x} = 2 y - 5 z\), we find the integral with respect to \(x\): \[f(x, y, z) = (2 y - 5 z)x + g(y, z)\]. Here, \(g(y, z)\) is an arbitrary function of \(y\) and \(z\).2. **Differentiate with respect to \(y\)**: Next, we take the partial derivative of this function with respect to \(y\): \[\frac{\partial f}{\partial y} = 2 x + \frac{\partial g}{\partial y}\] and set it equal to the given \(\frac{\partial f}{\partial y} = 2 x + 8 z\), meaning \(\frac{\partial g}{\partial y} = 8 z\). Upon integrating with respect to \(y\), we get \(g(y, z) = 8 z y + h(z)\), where \(h(z)\) is another arbitrary function.3. **Differentiate with respect to \(z\)**: Finally, we differentiate with respect to \(z\) using \(f(x, y, z)\) and equate it to the partial derivative given \(\frac{\partial f}{\partial z} = -5 x - 8 y\). This integration will usually reveal any remaining constants/functions, finalizing our scalar potential: \[f(x, y, z) = (2 y - 5 z)x + 8 z y + C\], where C is a constant. Understanding this process helps us derive the scalar potential functions from a vector field, offering deeper insights into its properties.