Problem 2 - 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 \(y^{2} \mathbf{i}+3 x^{2} \mathbf{j}\)

Step-by-Step Solution

Answer

The vector \( y^2 \mathbf{i} + 3x^2 \mathbf{j} \) is not a gradient.

Step 1 - Identify Components

Identify the components of the vector. The given vector is \( y^2 \mathbf{i} + 3x^2 \mathbf{j} \). Here, the components are \( F_x = y^2 \) and \( F_y = 3x^2 \).

Step 2 - Check for Gradient Condition

A vector field \( F = P \mathbf{i} + Q \mathbf{j} \) is a gradient if it is conservative. This implies that the partial derivatives of its components are equal, i.e., \( \frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y} \). Compute the partial derivatives: \( \frac{\partial Q}{\partial x} = \frac{\partial (3x^2)}{\partial x} = 6x \) and \( \frac{\partial P}{\partial y} = \frac{\partial (y^2)}{\partial y} = 2y \).

Step 3 - Verify Condition

Compare the computed partial derivatives: \( 6x \) and \( 2y \). Since these are not equal, \( \frac{\partial (3x^2)}{\partial x} eq \frac{\partial (y^2)}{\partial y} \), the given vector is not a gradient.

Key Concepts

Partial Derivatives

Partial derivatives help us understand how a function changes as we alter just one of its variables while keeping others constant. For instance, if we have a function like \( f(x, y) \), the partial derivative with respect to \( x \), written as \( \frac{\partial f}{\partial x} \), measures the rate of change of \( f \) as \( x \) changes, with \( y \) fixed. Similarly, \( \frac{\partial f}{\partial y} \) measures the rate of change with respect to \( y \). In this exercise, the vector field is given by \( F = y^2 \mathbf{i} + 3x^2 \mathbf{j} \), where we identified the components as \( F_x = y^2 \) and \( F_y = 3x^2 \). By taking partial derivatives of these components, we get \( \frac{\partial (3x^2)}{\partial x} = 6x \) and \( \frac{\partial (y^2)}{\partial y} = 2y \). These derivatives help us check whether the vector field satisfies certain conditions that would make it a gradient field.

Conservative Vector Fields

A vector field is called conservative if it can be represented as the gradient of some scalar function. In other words, for a vector field to be conservative, there must exist a function \( f \) such that its gradient \( abla f \) equals the vector field. For a 2-dimensional vector field \( F = P \mathbf{i} + Q \mathbf{j} \), it is conservative if and only if the mixed partial derivatives of \( P \) and \( Q \) are equal, i.e., \( \frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y} \). This condition helps ensure that the vector field has a potential function \( f \), making it conservative. In our example, since \( \frac{\partial (3x^2)}{\partial x} = 6x \) is not equal to \( \frac{\partial (y^2)}{\partial y} = 2y \), the vector field \( y^2 \mathbf{i} + 3x^2 \mathbf{j} \) is not conservative.

Gradient Conditions

To determine whether a vector field is a gradient (or conservative), it must satisfy specific gradient conditions. A key condition is that the partial derivatives of the vector components as mentioned above must be equal. For example, for \( F = y^2 \mathbf{i} + 3x^2 \mathbf{j} \) to be a gradient field, the condition \( \frac{\partial (3x^2)}{\partial x} = \frac{\partial (y^2)}{\partial y} \) must hold true at all points. Another important aspect of gradient fields is that they have path-independent line integrals. This means that the work done moving along a path in such a field depends only on the starting and ending points, not the path taken. By verifying these conditions, we can identify whether the given vector field is indeed a gradient field or not. In our specific problem, we verified the gradient condition, and it failed, confirming that the vector \( y^2 \mathbf{i} + 3x^2 \mathbf{j} \) is not a gradient.