Problem 34 - Thomas Calculus

Question

Gradient of a line integral Suppose that \(\mathbf{F}=\nabla f\) is a conservative vector field and $$g(x, y, z)=\int_{(0,0,0)}^{(x, y, z)} \mathbf{F} \cdot d \mathbf{r}$$ Show that \(\nabla g=\mathbf{F}.\)

Step-by-Step Solution

Answer

\(\nabla g = \mathbf{F}\) because \(g(x, y, z) = f(x, y, z) - f(0,0,0)\).

Step 1: Understanding the Problem

We start with the given vector field \(\mathbf{F} = abla f\), which is the gradient of some scalar field \(f(x, y, z)\). The line integral \(g(x, y, z) = \int_{(0,0,0)}^{(x,y,z)} \mathbf{F} \cdot d\mathbf{r}\) signifies the work done by \(\mathbf{F}\) in moving from the origin to the point \((x, y, z)\). We need to show that \(abla g = \mathbf{F}\).

Step 2: Apply the Fundamental Theorem for Line Integrals

The fundamental theorem for line integrals states that if \(\mathbf{F} = abla f\), then the line integral \(\int_C \mathbf{F} \cdot d\mathbf{r}\) over a curve \(C\) from a point \(A\) to \(B\) is simply \(f(B) - f(A)\). For our given function \(g(x, y, z)\), this becomes \(f(x, y, z) - f(0, 0, 0)\).

Step 3: Differentiate \(g\) to Find \(\nabla g\)

Now, we compute the gradient of \(g(x, y, z)\). Since \(g(x, y, z) = f(x, y, z) - f(0, 0, 0)\), the constant \(f(0, 0, 0)\) vanishes and the gradient is simply the gradient of \(f\). Thus, \(abla g = abla f\).

Step 4: Substitute the Given \(\mathbf{F}\)

We know that \(\mathbf{F} = abla f\). Therefore, substituting into the equation \(abla g = abla f\), we get \(abla g = \mathbf{F}\).

Key Concepts

Line Integrals

Line integrals are a way to integrate functions along a curve, accumulating values as you move through the field. In the context of vector fields, a line integral measures the total effect of a vector field along a path. Imagine an ant walking along a line drawn over a topographic map; the line integral provides the total "climb" made by the ant, integrating the field values along its path.

Line integrals are important when dealing with physical phenomena such as work done by a force field in moving along a path. For example, if \( \mathbf{F} \) is a force field, then the work done by the field when moving an object along a path \( C \) is captured by the line integral \( \int_C \mathbf{F} \cdot d\mathbf{r} \). This involves:

• Breaking the path into small segments
• Calculating the vector field's influence on each segment
• Summing up these influences

When the field \( \mathbf{F} \) is conservative, the work depends only on the starting and ending points, not on the path taken.

Conservative Vector Field

A conservative vector field is one where the line integral between two points is path-independent. This means the work done by the field in moving along any path between these two points remains the same, relying solely on the location of these points.

Such fields can be expressed as the gradient of a scalar function, noted as \( \mathbf{F} = abla f \). This implies several properties:

• The vector field \( \mathbf{F} \) is irrotational, meaning its curl is zero: \( abla \times \mathbf{F} = 0 \)
• Potential energy can be assigned at each point in the field, represented by the scalar function \( f \)

In practice, if you can find a scalar function \( f \) such that \( \mathbf{F} \) equals the gradient \( abla f \), and the field's curl is zero, you have a conservative vector field. These fields simplify many calculations as they depend only on "potentials" rather than complex path integrations.

Fundamental Theorem for Line Integrals

The fundamental theorem for line integrals connects line integrals to scalar functions. Particularly, it states if a vector field \( \mathbf{F} \) is conservative (i.e., \( \mathbf{F} = abla f \)), then the line integral of \( \mathbf{F} \) along a curve \( C \) from a point \( A \) to a point \( B \) equals the difference in the scalar function \( f \) at these points.

Mathematically, this is expressed as: \[ \int_C \mathbf{F} \cdot d\mathbf{r} = f(B) - f(A) \] This theorem highlights the remarkable simplification when dealing with conservative fields, transforming potentially complicated integrals into simple evaluations of a function at points:

• No need to calculate the integral over each path segment
• Focus solely on scalar function differences at endpoints

For solving problems, it reduces workload drastically by leveraging the "path independence" property of conservative fields. Understanding and applying this property can streamline solving physics and engineering problems involving complex force fields.