Problem 16 - 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 \(Q .\) $$ \mathbf{F}(x, y)=2 x y^{3} \mathbf{i}+3 x^{2} y^{2} \mathbf{j} ; P(-3,0), Q(4,1) $$

Step-by-Step Solution

Answer

The work done by the force field is 16.

Step 1: Check if the Force Field is Conservative

A force field \( \mathbf{F}(x, y) = M(x,y) \mathbf{i} + N(x,y) \mathbf{j} \) is conservative if there exists a potential function \( f \) such that \( abla f = \mathbf{F} \). This is equivalent to checking if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) in some open and connected region. For the given force field, where \( M(x, y) = 2xy^3 \) and \( N(x, y) = 3x^2y^2 \), compute the partial derivatives: - \( \frac{\partial M}{\partial y} = 6xy^2 \) - \( \frac{\partial N}{\partial x} = 6xy^2 \) Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the field is conservative.

Step 2: Find the Potential Function

Since the force field is conservative, there exists a potential function \( f(x, y) \) with \( f_x = M(x, y) \) and \( f_y = N(x, y) \). Integrate \( f_x = 2xy^3 \) with respect to \( x \) to get \[ f(x, y) = x^2y^3 + g(y) \] where \( g(y) \) is a function of \( y \) only. Differentiate this with respect to \( y \) to get: \[ f_y = 3x^2y^2 + g'(y) \] Set equal to \( N(x, y) \): \[ 3x^2y^2 + g'(y) = 3x^2y^2 \] which implies \( g'(y) = 0 \). Thus, \( g(y) \) is a constant and the potential function is \( f(x, y) = x^2y^3 + C \).

Step 3: Calculate the Work Done

The work done by a conservative force field \( \mathbf{F} \) in moving a particle from point \( P \) to \( Q \) is given by the difference in the potential function at these points. Calculate \( f(Q) - f(P) \): - \( f(4, 1) = 4^2 \cdot 1^3 + C = 16 + C \) - \( f(-3, 0) = (-3)^2 \cdot 0^3 + C = 0 + C \) Thus, the work done is \( (16 + C) - (0 + C) = 16 \).

Key Concepts

Potential Function

In vector calculus, a potential function is a scalar field whose gradient is equal to a given vector field. This means that for the vector field \( \mathbf{F} \), there exists a function \( f(x, y) \) such that \( abla f = \mathbf{F} \). If you find such a function, it indicates that the vector field is conservative.

In our exercise, we began by verifying whether the force field \( \mathbf{F}(x, y) = 2xy^3 \mathbf{i} + 3x^2y^2 \mathbf{j} \) is conservative. After establishing this, we then derived the potential function. The key was integrating the components of \( \mathbf{F} \) as partial derivatives of the potential function. We integrated the term \( 2xy^3 \) with respect to \( x \), giving us \( f(x, y) = x^2y^3 + g(y) \), adding an unknown function \( g(y) \). This step is common when integrating with respect to just one variable.

Next, by differentiating this expression with respect to \( y \) and equating it to the term in \( \mathbf{F} \) involving \( j \)—namely \( 3x^2y^2 \)—we determined \( g'(y) = 0 \), leading to the conclusion \( g(y) \) must be a constant. This confirmed our potential function as \( f(x, y) = x^2y^3 + C \).

Work Done

The concept of work done by a conservative force field involves the change in the potential function between two points. For a conservative force field, the path taken between these two points doesn't matter—only the initial and final positions are important.

In our problem, we were asked to calculate the work done as a particle moves from point \( P(-3, 0) \) to \( Q(4, 1) \) within the field. Once we had the potential function \( f(x, y) = x^2y^3 + C \), calculating the work was straightforward.

• Compute \( f(Q) = f(4, 1) = 4^2 \cdot 1^3 + C = 16 + C \).
• Compute \( f(P) = f(-3, 0) = (-3)^2 \cdot 0^3 + C = 0 + C \).

The work done is the difference \( f(Q) - f(P) = 16 \). This calculation reinforces that work in a conservative field is path independent and only relies on the difference in potential energy between points \( P \) and \( Q \).

Partial Derivatives

Partial derivatives are used to find the rate at which a function changes with respect to one of its variables while keeping the others constant. In the context of vector fields and potential functions, they help in proving whether a force field is conservative.

For our exercise, the conditions for a force field to be conservative are centered around comparing the derivative of its components. Specifically, for a field \( \mathbf{F}(x, y) = M(x,y) \mathbf{i} + N(x,y) \mathbf{j} \), we need to confirm that the mixed partial derivatives from the potential function condition hold. This requires \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).

• Calculate \( \frac{\partial M}{\partial y} \) for \( M(x, y) = 2xy^3 \), resulting in \( 6xy^2 \).
• Calculate \( \frac{\partial N}{\partial x} \) for \( N(x, y) = 3x^2y^2 \), also resulting in \( 6xy^2 \).

Since these are equal, it confirms that the field is conservative. Understanding and using partial derivatives effectively is crucial in problems involving vector fields and ensures you can derive potential functions accurately.