Problem 17 - 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 \(Q .\) $$ \mathbf{F}(x, y)=y e^{x y} \mathbf{i}+x e^{x y} \mathbf{j} ; P(-1,1), Q(2,0) $$

Step-by-Step Solution

Answer

The work done by the force field is \(1 - e^{-1}\).

Step 1: Verify the Force Field is Conservative

A force field is conservative if the curl of the force field is zero. The force field given is \( \mathbf{F}(x, y) = y e^{xy} \mathbf{i} + x e^{xy} \mathbf{j} \). To calculate the curl, \( abla \times \mathbf{F} \), we set up the components as follows: \( \mathbf{F}_1 = y e^{xy} \) and \( \mathbf{F}_2 = x e^{xy} \). The partial derivative of \( \mathbf{F}_1 \) with respect to \( y \) is:\( \frac{\partial}{\partial y} (y e^{xy}) = e^{xy} + x y e^{xy} \).The partial derivative of \( \mathbf{F}_2 \) with respect to \( x \) is: \( \frac{\partial}{\partial x} (x e^{xy}) = e^{xy} + x y e^{xy} \).Since these are equal, \( abla \times \mathbf{F} = 0 \), so \( \mathbf{F} \) is conservative.

Step 2: Use Potential Function Method for Work

Since the force field is conservative, there exists a potential function \( \phi \) such that \( \mathbf{F} = abla \phi \). We find \( \phi \) by integrating \( \mathbf{F}_1 \) with respect to \( x \):\[ \phi(x, y) = \int y e^{xy} \, dx = e^{xy} + g(y), \]where \( g(y) \) is an arbitrary function of \( y \). Next, differentiate \( \phi \) with respect to \( y \) and set it equal to \( \mathbf{F}_2 \):\[ \frac{\partial \phi}{\partial y} = x e^{xy} + g'(y) = x e^{xy}. \]This implies \( g'(y) = 0 \), so \( g(y) = C \), a constant. Thus, \( \phi(x, y) = e^{xy} + C \).

Step 3: Compute the Work Done

The work done by a conservative force field from a point \( P \) to a point \( Q \) can be determined by the difference in the potential function at these points:\[ W = \phi(Q) - \phi(P). \]Substitute point \( Q \) as \( (2, 0) \) and point \( P \) as \( (-1, 1) \):\[ \phi(2, 0) = e^{2 \cdot 0} + C = e^0 + C = 1 + C, \]\[ \phi(-1, 1) = e^{-1 \cdot 1} + C = e^{-1} + C. \]Thus, the work done is:\[ W = (1+C) - (e^{-1} + C) = 1 - e^{-1}. \]

Key Concepts

Potential Function

In a conservative force field, such as the one defined by \( \mathbf{F}(x, y) = y e^{xy} \mathbf{i} + x e^{xy} \mathbf{j} \), we can find a potential function \( \phi \). This function plays a crucial role since it allows us to determine the work done by the field simply by evaluating the potential at two points. The beauty of a potential function is that it is related directly to the force field by the gradient operation. Specifically, for a conservative force field, the force \( \mathbf{F} \) is equal to the gradient of the potential function, expressed as \( \mathbf{F} = abla \phi \).

To find \( \phi \), we begin by integrating the force field's components. For example, integrating \( y e^{xy} \) with respect to \( x \) gives \( e^{xy} + g(y) \). Here, \( g(y) \) is a function solely of \( y \), accounting for any missing terms from the integration with respect to \( x \).

The next step involves ensuring this potential matches both components of \( \mathbf{F} \). By differentiating \( \phi \) with respect to \( y \) and matching it to the \( y \)-component of the vector field, we can determine \( g'(y) \). This ensures that the potential function correctly represents the force field. In this case, since \( g'(y) = 0 \), our potential function simplifies to \( \phi(x, y) = e^{xy} + C \), providing a clear pathway to calculate work done.

Work Done

The concept of work done by a force field on a particle moving along a path is greatly simplified for conservative force fields. With a potential function \( \phi \) established, the work done depends only on the values of this potential at the starting and ending points instead of the specific path taken.

In our exercise, the work done as the particle moves from point \( P(-1, 1) \) to \( Q(2, 0) \) is given by the expression \( W = \phi(Q) - \phi(P) \). Here, \( \phi(Q) \) and \( \phi(P) \) are found by plugging the coordinates of \( Q \) and \( P \) into the potential function \( \phi(x, y) = e^{xy} + C \).

At point \( Q(2, 0) \), the potential function evaluates to \( \phi(2, 0) = 1 + C \) because \( e^{2 \cdot 0} = e^0 = 1 \). Meanwhile, at point \( P(-1, 1) \), it evaluates to \( \phi(-1, 1) = e^{-1} + C \), yielding the work as \( W = 1 - e^{-1} \).

Notice that the work calculated is independent of path and requires no direct integration along the path, underscoring the efficiency of using a potential function in conservative force fields.

Curl of a Vector Field

To determine if a force field is conservative, we look at the curl of the vector field. A force field is considered conservative if the curl is zero everywhere in the region considered. The curl operation in two dimensions is used to test the absence of rotational forces.

For our force field \( \mathbf{F}(x, y) \), the curl \( abla \times \mathbf{F} \) involves checking the partial derivatives of the vector field's components. Specifically, we calculate \( \frac{\partial \mathbf{F}_1}{\partial y} \) and \( \frac{\partial \mathbf{F}_2}{\partial x} \) separately and compare them.

In this case, both derivatives equal \( e^{xy} + x y e^{xy} \). Since these are equal, the curl effectively equals zero, confirming that \( \mathbf{F} \) is conservative. This result allows us to conclude that there exists a potential function \( \phi \) and that work done can be calculated directly from this potential function.

The significance of the curl being zero is that it implies a path-independent property of work done by the force field, a hallmark of conservative forces in physics.