Problem 28 - Thomas Calculus

Question

Work Find the work done by the gradient of \(f(x, y)=(x+y)^{2}\) counterclockwise around the circle \(x^{2}+y^{2}=4\) from \((2,0)\) to itself.

Step-by-Step Solution

Answer

The work done by the gradient field is zero.

Step 1: Understanding the Problem

We need to find the work done by the vector field which is the gradient of the function \(f(x, y) = (x + y)^2\) around the circle given by \(x^2 + y^2 = 4\), starting and ending at the point \((2,0)\). Since the path is a closed curve, the fundamental theorem of line integrals for conservative fields may apply.

Step 2: Find the Gradient

The gradient of \(f(x, y) = (x+y)^2\) is the vector field \( abla f = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \right)(x+y)^2 = (2(x+y), 2(x+y))\). So, \( abla f = (2(x+y), 2(x+y)).\)

Step 3: Use Properties of Conservative Fields

For conservative vector fields, and when the path is closed, the work done (line integral of the vector field along the closed path) is zero. Since we are moving counterclockwise around a circle, which is a closed path, and the vector field is the gradient of a scalar function, it is conservative.

Step 4: Calculate the Line Integral

Even without directly calculating the line integral, we apply the property that the line integral of a gradient over a closed path is zero since \(abla f\) is a conservative vector field. Thus, the work done is zero.

Key Concepts

Gradient Vector Field

A gradient vector field is essentially a representation of how a scalar function changes at any given point in space. To find the gradient of a function, we take the derivative, or rate of change, with respect to each variable in the function. Here, the function is given as \( f(x, y) = (x + y)^2 \). To find its gradient, we denote it as \( abla f \) which is composed of partial derivatives with respect to \( x \) and \( y \).

Let's calculate these partial derivatives:

• With respect to \( x \), the derivative is \( 2(x + y) \).
• With respect to \( y \), the derivative is also \( 2(x + y) \).

Therefore, the gradient vector field \( abla f \) is \( (2(x + y), 2(x + y)) \). This vector field tells us how the value of \( f \) changes as we move in the plane. It always points in the direction of greatest increase of the function, and its magnitude reflects how steep that increase is.

Conservative Vector Fields

A vector field is termed conservative if it is the gradient of some scalar function. This implies certain properties that are useful, especially in calculating line integrals.

• The most notable property is that the work done by a conservative vector field over a closed path is zero.
• This occurs because, with conservative fields, the line integral is path-independent and solely dependent on the start and end points.

In the context of our example, since the vector field \( abla f = (2(x+y), 2(x+y)) \) arises from the gradient of the function \( f(x, y) \), it's conservative. So, regardless of the path taken, as long as it forms a closed loop, the work done by this vector field is zero.

Fundamental Theorem of Line Integrals

The Fundamental Theorem of Line Integrals bridges the relationship between the integration of a conservative vector field over a curve and the evaluation of its scalar potential function. This theorem particularly simplifies calculations by stating that the line integral of a vector field that is the gradient of a function depends only on the endpoints of the path and not on the specific path taken—this is known as path independence.

Mathematically, if \( \mathbf{F} = abla f \), and there’s a curve \( C \) with endpoints \( A \) and \( B \), the line integral of \( \mathbf{F} \) along \( C \) is given by:\[ \int_C \mathbf{F} \cdot d\mathbf{r} = f(B) - f(A) \]In our case, since the path is a circle returning to its start point, \( A = B \). Thus, the line integral or the work done around a closed path is \( f(B) - f(A) = 0 \). This straightforward property leads directly to the conclusion that the work done by a conservative field over a closed path is zero, perfectly applying to our gradient vector field example.