Problem 12 - 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 \((\sin 2 x-\tan y) \mathbf{i}-x \sec ^{2} y \mathbf{j}\)

Step-by-Step Solution

Answer

The vector is a gradient. The function is \(f(x,y) = -\frac{1}{2}\cos 2x - x \tan y + C\).

Step 1 - Check if the vector is a gradient

To determine if the given vector \(\vec{F} = (\sin 2x - \tan y) \mathbf{i} - x \sec^2 y \mathbf{j}\) is a gradient of some scalar function \(f(x, y)\), we must check if the mixed partial derivatives are equal, i.e., \(\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}\). Compute both partial derivatives and compare.

Step 2 - Compute \(\frac{\partial F_1}{\partial y}\)

Calculate the partial derivative of \(F_1 = \sin 2x - \tan y\) with respect to \(y\): \(\frac{\partial}{\partial y}(\sin 2x - \tan y) = -\sec^2 y\).

Step 3 - Compute \(\frac{\partial F_2}{\partial x}\)

Calculate the partial derivative of \(F_2 = -x \sec^2 y\) with respect to \(x\): \(\frac{\partial}{\partial x}(-x \sec^2 y) = -\sec^2 y\).

Step 4 - Compare the partial derivatives

Both partial derivatives are \( -\sec^2 y\). Since \(\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}\), the vector is indeed a gradient of some function \(f(x, y)\).

Step 5 - Find the potential function \(f(x, y)\)

To find the function \(f(x, y)\) whose gradient gives the vector field, integrate \(F_1 = \sin 2x - \tan y\) with respect to \(x\): \(\int (\sin 2x - \tan y) \, dx = -\frac{1}{2} \cos 2x - x \tan y + g(y)\), where \(g(y)\) is a function of \(y\). Next, integrate \(F_2 = -x \sec^2 y\) with respect to \(y\): \(\int -x \sec^2 y \, dy = -x \tan y + h(x)\), where \(h(x)\) is a constant.

Step 6 - Combine the results

By comparing the integrations, we find the potential function \(f(x, y)\): \(f(x, y) = -\frac{1}{2} \cos 2x - x \tan y + C\), where \(C\) is a constant.

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and scalar fields. It is an essential tool in many areas of science and engineering. A vector field assigns a vector to each point in space. When we talk about gradient vector fields, we're looking at the steepest ascent directions of a scalar function.

In the context of our exercise, we want to determine if a given vector field is a gradient vector field. This means checking if the vector field is derived from some potential function. If it is, the vector field is conservative, and we can find a scalar function whose gradient matches it.

Partial Derivatives

Partial derivatives measure how a function changes as its input variables change, one at a time. For a function of two variables, say, \(f(x, y)\), the partial derivative with respect to \(x\) is written as \(\frac{\partial f}{\partial x}\) and with respect to \(y\) as \(\frac{\partial f}{\partial y}\).

In our exercise, we calculated partial derivatives to verify if the mixed derivatives are equal. For the vector \(\vec{F} = (\sin 2x - \tan y) \mathbf{i} - x \sec^2 y \mathbf{j}\), we computed:

• \(\frac{\partial F_1}{\partial y} = -\sec^2 y\)
• \(\frac{\partial F_2}{\partial x} = -\sec^2 y\)

Since these partial derivatives are equal, it confirms that our vector field has continuous mixed derivatives and thus is a gradient vector field.

Potential Function

A potential function, or scalar potential, is a scalar field whose gradient gives us the original vector field. Finding a potential function helps in simplifying problems in physics and engineering, like calculating work done by a force field.

In our exercise, we found the potential function by integrating the components of the given vector field. By integrating \(F_1 = \sin 2x - \tan y\) with respect to \(x\) and \(F_2 = -x \sec^2 y\) with respect to \(y\), we combined the results:

• \(f(x, y) = -\frac{1}{2} \cos 2x - x \tan y + C\)

This function gives us the scalar field that matches the gradient of our original vector field.

Integration

Integration in vector calculus involves finding the scalar function whose derivative (gradient) is the given vector field. It helps in identifying potential functions and solving problems like the given exercise.

To integrate \(F_1\) and \(F_2\), we integrated each component separately. For example, integrating \(\sin 2x - \tan y\) with respect to \(x\), we get:

• \(\int (\sin 2x - \tan y) \, dx = -\frac{1}{2} \cos 2x - x \tan y + g(y)\)

Similarly, integrating \(-x \sec^2 y\) with respect to \(y\) gives:

• \(\int -x \sec^2 y \, dy = -x \tan y + h(x)\)

By combining results from these integrations and consolidating terms, we obtained the potential function for the vector field. This integration method is crucial in solving many vector calculus problems.