Problem 11 - 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 \((2 x \cos y-1) \mathbf{i}-x^{2} \sin y \mathbf{j}\)

Step-by-Step Solution

Answer

The vector is a gradient and the function is \(f(x, y) = x^2 \cos y - x + C\).

Step 1 - Understand the Gradient

To determine if the given vector is a gradient, we must check if it is conservative. A vector field \(abla \textbf{F} = P(x, y) \textbf{i} + Q(x, y) \textbf{j}\) is conservative if there exists a scalar potential function \(f(x, y)\) such that \(abla f = \textbf{F}\).

Step 2 - Check Consistency Condition

For the vector to be a gradient, the mixed partial derivatives must be equal, \(P_y = Q_x\). Given \(P(x, y) = 2x \cos y - 1\) and \(Q(x, y) = -x^2 \sin y\), we compute \(P_y = \frac{\text{d}}{\text{d}y} (2x \cos y - 1) = -2x \sin y\) and \(Q_x = \frac{\text{d}}{\text{d}x} (-x^2 \sin y) = -2x \sin y\). Since \(P_y = Q_x\), the given vector field is a gradient.

Step 3 - Determine Potential Function

Find the potential function \(f(x, y)\) by integrating \(P(x, y)\) with respect to \(x\). This gives \(f(x, y) = \frac{\text{d}f}{\text{d}x} = 2x \cos y - 1 \rightarrow f(x, y) = x^2 \cos y - x + g(y)\).

Step 4 - Find the Correct Form of f(x, y)

Differentiate \(f(x, y)\) with respect to \(y\): \(abla_y f = \frac{\text{d}}{\text{d}y}(x^2 \cos y - x + g(y)) = -x^2 \sin y + g'(y)\). Since this must equal \(Q(x, y) = -x^2 \sin y\), we have \(-x^2 \sin y + g'(y) = -x^2 \sin y\). So, \(g'(y) = 0\), implying \(g(y)\) is a constant, \(C\).

Step 5 - Write the Final Function

Now we have the potential function: \(f(x, y) = x^2 \cos y - x + C\). Thus, the function having the given gradient is \(f(x, y) = x^2 \cos y - x + C\).

Key Concepts

gradient vector field

A gradient vector field is a vector field that can be expressed as the gradient of a scalar function. Essentially, this means every vector in the field points in the direction of the steepest increase of some function.
This scalar function is known as the potential function.
The notation for gradient is typically represented as \(abla f\), indicating the rate and direction of change in all directions.

• The gradient vector field for a function \(f(x, y)\) is \(abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\).
• Each component of the gradient is a partial derivative of \(f\).
• If we have a vector field \( \mathbf{F} = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \), we must check whether \( \mathbf{F} = abla f\) for some scalar function \( f \).

conservative vector field

A vector field is termed conservative if it is equal to the gradient of some potential function. Checking this involves determining if certain conditions are met.
Key conditions for a gradient vector field to be conservative include:

• The mixed partial derivatives must be identical: \( P_y = Q_x \).
• If \( P(x, y) \) and \( Q(x, y) \) meet this criterion, then there exists a scalar potential function \( f(x, y) \).

In the given exercise:

• We start with \( P(x, y) = 2x \cos y - 1 \) and \( Q(x, y) = -x^2 \sin y \).
• Compute the partial derivatives. Find \( P_y = \frac{\partial}{\partial y}(2x \cos y - 1) = -2x \sin y \) and \( Q_x = \frac{\partial}{\partial x}(-x^2 \sin y) = -2x \sin y \).
• Since \( P_y = Q_x \), the vector field is conservative.

potential function

The potential function is the scalar function whose gradient corresponds to a given vector field.
To find this function, we first integrate each component of the vector field:

• Consider the gradient: \(\mathbf{F} = (2x \cos y - 1) \mathbf{i} - x^2 \sin y \mathbf{j} \).
• We find the potential function by integrating \( P(x, y) \) with respect to \( x \), giving us \( f(x, y) = \int (2x \cos y - 1) \, dx = x^2 \cos y - x + g(y) \).
• To determine \( g(y) \), we differentiate \( f(x, y) \) with respect to \( y \) and set it equal to \( Q(x, y) \).

This process gives us:
\[ \frac{\partial}{\partial y}(x^2 \cos y - x + g(y)) = -x^2 \sin y + g'(y) \] Since this must equal \( -x^2 \sin y \), it follows that \( g'(y) = 0 \). Thus, \( g(y) \) is a constant, let’s call it \( C \).
In the end, we have:
\( f(x, y) = x^2 \cos y - x + C \). Hence, the potential function for the given gradient vector field is \( f(x, y) = x^2 \cos y - x + C \).