Problem 18 - 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 \(Q .\) $$ \mathbf{F}(x, y)=e^{-y} \cos x \mathbf{i}-e^{-y} \sin x \mathbf{j} ; \quad P(\pi / 2,1), Q(-\pi / 2,0) $$

Step-by-Step Solution

Answer

The force field is conservative; work done from \(P\) to \(Q\) is \(-1 - \frac{1}{e}\).

Step 1: Check for a Conservative Field

A force field is conservative if it can be expressed as the gradient of a scalar potential function, meaning its curl is zero. For a 2D force field \( \mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \), we can check if it is conservative by seeing if \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \). Here, \( P(x, y) = e^{-y} \cos x \) and \( Q(x, y) = -e^{-y} \sin x \). Compute:\\[\frac{\partial P}{\partial y} = -e^{-y} \cos x \quad \text{and} \quad \frac{\partial Q}{\partial x} = -e^{-y} \cos x \]Since \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \), the field is conservative.

Step 2: Find the Scalar Potential Function

Since the field is conservative, there exists a scalar potential function \( \phi(x, y) \) such that \( abla \phi = \mathbf{F} \). Thus, \( \frac{\partial \phi}{\partial x} = P(x, y) = e^{-y} \cos x \) and \( \frac{\partial \phi}{\partial y} = Q(x, y) = -e^{-y} \sin x \). Integrate \( \frac{\partial \phi}{\partial x} \) with respect to \( x \):\[\phi(x, y) = \int e^{-y} \cos x \, dx = e^{-y} \sin x + g(y)\] Differentiate the result with respect to \( y \) to find \( g'(y) \):\[\frac{\partial \phi}{\partial y} = -e^{-y} \sin x + g'(y) = -e^{-y} \sin x\] Solve for \( g'(y) \):\[g'(y) = 0\], which implies \( g(y) = C \) where \( C \) is a constant.

Step 3: Determine the Potential Function

Combine the results to find the potential function:\[\phi(x, y) = e^{-y} \sin x + C\]Since the constant \( C \) doesn't affect the work calculation over a path, we can set it to zero for simplicity. So, the potential function is:\[\phi(x, y) = e^{-y} \sin x\]

Step 4: Calculate the Work Done by the Field

For a conservative force field, the work done moving from point \( P \) to \( Q \) is given by the difference in the potential function values at those points:\[ W = \phi(Q) - \phi(P) = \phi(-\pi/2, 0) - \phi(\pi/2, 1) \]Calculate \( \phi(-\pi/2, 0) \):\[\phi(-\pi/2, 0) = e^0 \sin(-\pi/2) = -1\]Calculate \( \phi(\pi/2, 1) \):\[\phi(\pi/2, 1) = e^{-1} \sin(\pi/2) = e^{-1}\]Thus, the work done is:\[ W = -1 - e^{-1} = -1 - \frac{1}{e}\]

Key Concepts

Potential Function

A potential function is a vital part of understanding conservative force fields. Think of it as a scalar field, representing a sort of 'energy landscape' for the force. For any location within the field, the value of the potential function indicates the potential energy at that point. In other words, it tells you how much work a force can do when moving an object to or from that point.
In this exercise, we sought to find a potential function \( \phi(x, y) \) associated with the given vector field \( \mathbf{F} \). This vector field is expressed in terms of components \( P(x, y) \) and \( Q(x, y) \). Because the field is conservative, we know it is the gradient of some scalar function. Hence, we integrated the given components to find \( \phi(x, y) = e^{-y} \sin x \). This function represents the potential energy landscape for the field.

Work Done

Work done by a force field on a particle as it moves from one point to another is equivalent to the change in potential energy between those points. In conservative fields, this makes calculations straightforward; you merely subtract the potential function values at the endpoints.
In our problem, we moved from \( P(\pi / 2, 1) \) to \( Q(-\pi / 2, 0) \). We measured the change in potential energy via the potential function \( \phi(x, y) \) previously determined. This gives us the work done as \( W = \phi(Q) - \phi(P) = -1 - \frac{1}{e} \). This negative value suggests the force did work on the particle as it moved 'downhill' in potential.

Gradient

The concept of a gradient is central to understanding how force fields and potential functions are related. Mathematically, the gradient of a scalar field indicates the direction and rate of the steepest ascent in that field. For vector fields with two components, it combines their partial derivatives.
When a force field \( \mathbf{F} \) is conservative, it can be expressed as the gradient of its potential function. This was checked by verifying the condition \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \). This equality showed that our vector field has no 'curl', i.e., it does not loop back on itself, reinforcing its conservativity. The scalar potential function we derived is indeed valid because its gradient matches the original force field.