Problem 20 - 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 \(e^{z} \cos x \mathbf{i}+z \sin y \mathbf{j}+\left(e^{z} \sin x-\cos y\right) \mathbf{k}\)

Step-by-Step Solution

Answer

The vector is a gradient. The potential function is \( f(x, y, z) = e^{z} \, \text{sin} \, x - \, \text{cos} \, y + C \).

Step 1: Identify components of the vector

The given vector \(e^{z} \, \text{cos} \, x \mathbf{i}+z \, \text{sin} \, y \mathbf{j}+(e^{z} \, \text{sin} \, x- \, \text{cos} \, y) \mathbf{k}\) can be written as its component functions: \(F_1 = e^{z} \, \text{cos} \, x\), \(F_2 = z \, \text{sin} \, y\), \(F_3 = e^{z} \, \text{sin} \, x - \, \text{cos} \, y\).

Step 2: Check if the vector is a gradient

A vector \(\mathbf{F} = F_1\mathbf{i} + F_2 \mathbf{j} + F_3 \mathbf{k}\) is a gradient vector if all mixed partial derivatives are equal, i.e., \(\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}\), \(\frac{\partial F_1}{\partial z} = \frac{\partial F_3}{\partial x}\), and \(\frac{\partial F_2}{\partial z} = \frac{\partial F_3}{\partial y}\).

Step 3: Compute the partial derivatives

\(\frac{\partial F_1}{\partial y} = 0 \) and \(\frac{\partial F_2}{\partial x} = 0\)Therefore, the first condition: \(\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}\) holds.Now check the second condition:\(\frac{\partial F_1}{\partial z} = e^{z} \, \text{cos} \, x \) and \(\frac{\partial F_3}{\partial x} = e^{z} \, \text{cos} \, x\)So, the second condition also holds.

Step 4: Verify the last condition for mixed partial derivatives

Check the third condition: \(\frac{\partial F_2}{\partial z} = \, \text{sin} \, y \) and \(\frac{\partial F_3}{\partial y} = \, \text{sin} \, y\)Thus, the third condition is also satisfied.

Step 5: Conclude the vector is a gradient

Since all mixed partial derivatives are equal, the vector \( \mathbf{F} \) is indeed a gradient vector.

Step 6: Find a potential function

To find the function \( f \) such that \( abla f = \mathbf{F} \), integrate each component:From \(F_1 = e^{z} \, \text{cos} \, x\), integrate with respect to \( x \):\( f = e^{z} \, \text{sin} \, x + g(y, z)\).From \(F_2 = z \, \text{sin} \, y\), integrate with respect to \( y \):\( f = - z \, \text{cos} \, y + h(x, z)\).Compare: \( f = e^{z} \, \text{sin} \, x - \, \text{cos} \, y + C\).

Key Concepts

Gradient

The gradient of a function represents the direction of the steepest ascent. If you think of a function as representing a hill, the gradient tells you which way to go to start climbing up the fastest. Mathematically, the gradient of a function f(x, y, z) is a vector: \(abla f = \frac{\text{d}f}{\text{d}x}\textbf{i} + \frac{\text{d}f}{\text{d}y}\textbf{j} + \frac{\text{d}f}{\text{d}z}\textbf{k}\). So, in our given problem, determining whether the vector given is a gradient involves verifying if it can be expressed as \(abla f\) for some function f. Checking the mixed partial derivatives ensures that the vector can indeed be a gradient.

Partial Derivatives

Partial derivatives measure how a function changes as one of its input variables changes while the other variables are held constant. For example, the partial derivative of a function f with respect to x, denoted as \(\frac{\text{d}f}{\text{d}x}\), shows how f changes when only x changes. In the exercise, we needed to calculate partial derivatives like \(\frac{\text{d}F_1}{\text{d}z}\) and \(\frac{\text{d}F_2}{\text{d}x}\) to verify conditions for the vector to be a gradient. Here are the partial derivatives computed:

Potential Function

A potential function f for a vector field \(\textbf{F}\) is a scalar function whose gradient is \(\textbf{F}\). This means \(abla f = \textbf{F}\). Determining if a vector field is a gradient involves checking if there exists such an f. In the given exercise, after verifying the vector is a gradient, we integrated each component function to find the potential function. This is how we got \( f = e^z \text{sin} x - \text{cos} y + C \), indicating that \(abla f\) indeed matches the given vector.

Vector Calculus

Vector calculus extends calculus to vector fields. It includes operations like grad (gradient), div (divergence), and curl. In this problem, primarily, the gradient operation was used. Vector fields have broad applications in physics and engineering, describing various forces and phenomena. The gradient specifically is widely used in fields like electromagnetism and fluid dynamics. This exercise reinforced understanding the gradient and finding potential functions, which are foundational concepts in vector calculus.