Problem 12 - Thomas Calculus

Question

In Exercises \(7-12,\) find a potential function \(f\) for the field \(\mathbf{F}\). $$\begin{aligned} \mathbf{F}=\frac{y}{1+x^{2} y^{2}} \mathbf{i}+\left(\frac{x}{1+x^{2} y^{2}}+\frac{z}{\sqrt{1-y^{2} z^{2}}}\right) \mathbf{j}+& \\ &\left(\frac{y}{\sqrt{1-y^{2} z^{2}}}+\frac{1}{z}\right) \mathbf{k} \end{aligned}$$

Step-by-Step Solution

Answer

The potential function is \(f(x,y,z) = \frac{1}{y^2} \tan^{-1}(xy) - \sqrt{1 - y^2 z^2} + \arcsin(yz) + \ln|z| + K\).

Step 1: Identify the Components

The vector field can be split into three components: \[ F_1 = \frac{y}{1 + x^2 y^2}, \quad F_2 = \frac{x}{1 + x^2 y^2} + \frac{z}{\sqrt{1 - y^2 z^2}}, \quad F_3 = \frac{y}{\sqrt{1 - y^2 z^2}} + \frac{1}{z} \]

Step 2: Integrate Respect to x

We start by integrating the first component \(F_1\) with respect to \(x\). \[ \int \frac{y}{1 + x^2 y^2} \, dx = \frac{1}{y^2} \tan^{-1}(xy) + C(y,z) \] where \(C(y,z)\) represents the integration constant whose variable dependencies are yet to be determined.

Step 3: Integrate Respect to y

Next, integrate \(F_2\) with respect to \(y\), also adding any dependence from the previous step on \(C(y,z)\). \[ \int \left(\frac{x}{1 + x^2 y^2} + \frac{z}{\sqrt{1 - y^2 z^2}}\right) \, dy = \frac{1}{x} \tan^{-1}(xy) - \sqrt{1 - y^2 z^2} + g(x,z) \] where \(g(x,z)\) is the integration constant for this step.

Step 4: Integrate Respect to z

Finally, integrate the third component \(F_3\) with respect to \(z\), accounting for any remaining dependencies from earlier calculations. \[ \int \left(\frac{y}{\sqrt{1 - y^2 z^2}} + \frac{1}{z}\right) \, dz = \arcsin(yz) + \ln|z| + h(x,y) \] where \(h(x,y)\) represents the integration constant with adjustments for dependencies.

Step 5: Compare Results and Deduce Potential Function

Combine the results from each of the separate integrations to form the overall potential function \(f\). Ensure that all three terms are consistent and dependencies line up: \[ f(x,y,z) = \frac{1}{y^2} \tan^{-1}(xy) - \sqrt{1 - y^2 z^2} + \arcsin(yz) + \ln|z| + K \] where \(K\) is an overall constant of integration.

Key Concepts

Vector Field

In mathematics, a vector field assigns a vector to each point in a subset of space. Imagine the wind, which, at each point in the atmosphere, has both direction and magnitude. A vector field is much like this: it provides a vector (a tool giving direction and magnitude) at every point in a space.

Our example vector field \( \mathbf{F} \) is a function of \( x, y, \) and \( z \) that attaches vectors to each point in \( \mathbb{R}^3 \). It's given by three components, each a function of the variables.

• \( F_1 = \frac{y}{1 + x^2 y^2} \)
• \( F_2 = \frac{x}{1 + x^2 y^2} + \frac{z}{\sqrt{1 - y^2 z^2}} \)
• \( F_3 = \frac{y}{\sqrt{1 - y^2 z^2}} + \frac{1}{z} \)

Understanding vector fields is important in various fields like physics and engineering where they describe electromagnetic fields, fluid flows, and more. A potential function relates to these fields as it represents a scalar field whose gradient is the given vector field.

Integration

Integration is a core concept in calculus, often described as the process of finding the antiderivative. It's the reverse of differentiation and is used to determine the total size, area, volume, or other measures of scope.

In the context of finding a potential function for a vector field, integration helps us find a function whose gradient is the vector field. We integrate each component of the vector field with respect to its respective variable. Consider the first component of our field, \( F_1 = \frac{y}{1 + x^2 y^2} \). We integrate this with respect to \( x \): \[ \int \frac{y}{1 + x^2 y^2} \, dx = \frac{1}{y^2} \tan^{-1}(xy) + C(y,z) \] The result gives a piece of the potential function. The constants from integration, like \( C(y,z) \), may depend on other variables, influencing subsequent integration steps.

Arctangent Function

The arctangent function, \( \tan^{-1}(x) \), is the inverse operation of the tangent trigonometric function. While tangent takes an angle and gives you a ratio, arctangent takes a ratio and gives back the angle. It's valuable when working with integration, especially involving divisions of squares or products.

During the integration of \( F_1 \) in our vector field, we arrived at an arctangent expression:

\[ \int \frac{y}{1 + x^2 y^2} \, dx = \frac{1}{y^2} \tan^{-1}(xy) + C(y,z) \]

Here, the arctangent helps manage the complexity of the expression produced by the reciprocal and quadratic forms of \( x \) and other variables. It's a useful tool because it provides insights into the angles created within the components of the vectors in the field.

Inverse Sine Function

The inverse sine function, or arcsine, denoted as \( \arcsin(x) \), is the inverse of the sine function. Where sine gives us a ratio from an angle, arcsine returns the angle from the ratio. It's a crucial element in solving problems with trigonometric relationships, especially when you're reversing the sine operation to find angles.

In the calculation of our vector field's potential function, we employ the inverse sine function when integrating the third component \( F_3 \):

\[ \int \left(\frac{y}{\sqrt{1 - y^2 z^2}} + \frac{1}{z}\right) \, dz = \arcsin(yz) + \ln|z| + h(x,y) \]

The arcsine resolves expressions involved with the trigonometric identities and relationships. This function helps tackle expressions that include squares and square roots, often simplifying them into manageable angular components. It's especially useful for solving problems involving circular or oscillatory motion.