Problem 15 - Calculus Early Transcendentals

Question

Confirm that the force field \(\mathbf{F}\) is conservative in some open connected region containing the points \(P\) and \(Q,\) and then find the work done by the force field on a particle moving along an arbitrary smooth curve in the region from \(P\) to \(Q .\) $$ \mathbf{F}(x, y)=x y^{2} \mathbf{i}+x^{2} y \mathbf{j} ; P(1,1), Q(0,0) $$

Step-by-Step Solution

Answer

The work done by the force field from P to Q is \(-\frac{1}{2}\).

Step 1: Verify the Curl of the Force Field

To determine if the force field \( \mathbf{F}(x, y) = xy^2 \mathbf{i} + x^2y \mathbf{j} \) is conservative, we need to check if its curl is zero. The curl of a two-dimensional vector field \( \mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \) is given by \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \).For this vector field:- \( P(x,y) = xy^2 \), so \( \frac{\partial P}{\partial y} = 2xy \).- \( Q(x,y) = x^2y \), so \( \frac{\partial Q}{\partial x} = 2xy \).Since \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2xy - 2xy = 0 \), the curl of \( \mathbf{F} \) is zero, making the field conservative.

Step 2: Find the Potential Function

To find a potential function \( \phi(x, y) \) for \( \mathbf{F} \), solve \( abla \phi = \mathbf{F} \). This means \( \frac{\partial \phi}{\partial x} = xy^2 \) and \( \frac{\partial \phi}{\partial y} = x^2y \).1. Integrate \( \frac{\partial \phi}{\partial x} = xy^2 \) with respect to \( x \): \[ \phi(x, y) = \int xy^2 \, dx = \frac{x^2y^2}{2} + g(y) \] where \( g(y) \) is an arbitrary function of \( y \).2. Differentiate the result with respect to \( y \): \[ \frac{\partial \phi}{\partial y} = x^2y + g'(y) \]3. Set this equal to \( x^2y \): \[ x^2y + g'(y) = x^2y \] Thus, \( g'(y) = 0 \), which implies \( g(y) \) is a constant. The potential function is \( \phi(x, y) = \frac{x^2y^2}{2} + C \), where \( C \) is a constant.

Step 3: Calculate the Work Done on a Path from P to Q

For a conservative force field, the work done along a path from point \( P \) to point \( Q \) is the difference in the potential function \( \phi \) evaluated at these points.Calculate \( \phi \) at points \( P(1,1) \) and \( Q(0,0) \):- \( \phi(1, 1) = \frac{1^2 \times 1^2}{2} = \frac{1}{2} \)- \( \phi(0, 0) = \frac{0^2 \times 0^2}{2} = 0 \)The work done, \( W \), is:\[ W = \phi(Q) - \phi(P) = 0 - \frac{1}{2} = -\frac{1}{2} \]

Key Concepts

Potential Function

A potential function, often denoted as \( \phi(x, y) \), is a scalar function whose gradient is equal to the given vector field. In the context of conservative force fields, such a function helps us understand the energy stored in a system. If a force field is conservative, it means that it must have a potential function associated with it. To find a potential function, we need to ensure that the gradient of \( \phi \) is equal to the vector field \( \mathbf{F}(x, y) \). For the field \( \mathbf{F}(x, y) = xy^{2} \mathbf{i} + x^{2}y \mathbf{j} \), the equations are:

• \( \frac{\partial \phi}{\partial x} = xy^2 \)
• \( \frac{\partial \phi}{\partial y} = x^2y \)

We integrate these partial derivatives to find \( \phi(x, y) \). Since such fields are path-independent, their work only depends on the endpoints of the path. This makes potential functions crucial for calculating the work done by conservative forces.

Curl of a Vector Field

The curl of a vector field is a key concept in determining whether a force field is conservative. For a two-dimensional vector field \( \mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \), the curl can be calculated as: \[ \text{Curl} = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \]If the curl is zero, the vector field is conservative, implying that a potential function exists. In our problem, the field \( \mathbf{F}(x, y) = xy^{2} \mathbf{i} + x^{2}y \mathbf{j} \) yields:

• For \( P(x,y) = xy^{2} \), \( \frac{\partial P}{\partial y} = 2xy \)
• For \( Q(x,y) = x^{2}y \), \( \frac{\partial Q}{\partial x} = 2xy \)

Calculating these, we have \( 2xy - 2xy = 0 \), confirming that the field is conservative as its curl is zero. This tool of checking the curl is a powerful way to spot conservative fields without explicitly finding a potential function.

Work Done in Physics

In physics, the concept of 'work done' by a force acting on an object along a path is crucial. For conservative forces, this work can be found using the potential function. The work done by a conservative force field when moving a particle from point \( P \) to point \( Q \) is the difference in the potential function at these points:\[ W = \phi(Q) - \phi(P) \]This difference represents the energy transferred when moving between two points through a conservative field. For the given problem, the force field is found to be conservative, so:

• Calculate the potential function at point \( P(1,1) \), yielding \( \phi(1,1) = \frac{1}{2} \).
• Calculate at \( Q(0,0) \), yielding \( \phi(0,0) = 0 \).

Then the work done is \( W = 0 - \frac{1}{2} = -\frac{1}{2} \). This useful calculation not only shows the work done in terms of actual movement along a certain path, but also verifies the conservative nature by being able to perform this algebraic computation.