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

The gradient \(\nabla g\) equals the vector field \(\mathbf{F}\).

Step 1: Understand the Given Problem

We are given a vector field \(\mathbf{F} = abla f\), which means it is a conservative field. The task is to prove that the gradient of the line integral \(g(x, y, z)\) is the vector field \(\mathbf{F}\) itself.

Step 2: Represent the Line Integral

Write the given line integral as \(g(x, y, z) = \int_{(0,0,0)}^{(x, y, z)} \mathbf{F} \cdot d\mathbf{r}\). This represents the line integral of \(\mathbf{F}\) along a path from the origin to the point \((x, y, z)\).

Step 3: Recall the Fundamental Theorem of Line Integrals

The Fundamental Theorem for Line Integrals states that for a conservative vector field \(\mathbf{F} = abla f\), the line integral \(\int_{C} \mathbf{F} \cdot d\mathbf{r}\) from point \(A\) to \(B\) is \(f(B) - f(A)\). For our case, \(C\) is the path from \((0, 0, 0)\) to \((x, y, z)\), so \(g(x, y, z) = f(x, y, z) - f(0, 0, 0)\).

Step 4: Compute the Gradient \(\nabla g\)

Calculate the gradient of \(g\) with respect to \(x, y, z\). Since \(g(x, y, z) = f(x, y, z) - f(0, 0, 0)\), the gradient \(abla g = \left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z}\right)\) will be \(abla f = \mathbf{F}\).

Step 5: Derive the Gradient Components

Calculate each component of the gradient: \(\frac{\partial g}{\partial x} = \frac{\partial f}{\partial x}\), \(\frac{\partial g}{\partial y} = \frac{\partial f}{\partial y}\), and \(\frac{\partial g}{\partial z} = \frac{\partial f}{\partial z}\). Thus, \(abla g = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}) = \mathbf{F}\).

Step 6: Conclusion

By computing the gradient \(abla g\) and finding it to equal \(\mathbf{F}\), we have shown that \(abla g = \mathbf{F}\) as required by the problem.

Key Concepts

Conservative Vector Field

A vector field \( \mathbf{F} \) is called a **conservative vector field** if it can be expressed as the gradient of some scalar function \( f \), written as \( \mathbf{F} = abla f \). This property means that the vector field has no circulation around any closed path, and it is often related to the notion of potential energy in physics.
For any conservative field, the work done by the vector field along a path depends only on the starting and ending points, not on the specific path taken. This characteristic allows us to express the vector field in terms of a scalar potential function, and thus simplifies many problems involving energy or work calculations.

• Conservative fields are irrotational, meaning their curl is zero: \( abla \times \mathbf{F} = \mathbf{0} \).
• These fields are often associated with gravitational and electrostatic fields.
• Knowing a vector field is conservative allows us to use powerful theorems, like the Fundamental Theorem of Line Integrals, to evaluate path integrals easily.

Fundamental Theorem of Line Integrals

The **Fundamental Theorem of Line Integrals** is a crucial concept when dealing with conservative vector fields. It states that if \( \mathbf{F} = abla f \) is a conservative vector field, then the line integral of \( \mathbf{F} \) from point \( A \) to point \( B \) is simply: \[\int_{C} \mathbf{F} \cdot d\mathbf{r} = f(B) - f(A)\]Here, \( C \) represents the path from \( A \) to \( B \). This theorem simplifies the computation of line integrals considerably because it transforms a potentially challenging problem involving a path integral into a simple evaluation of a scalar function at two points.
For the exercise given, this means the line integral from the origin to a point \((x, y, z)\) is computed as:\[g(x, y, z) = f(x, y, z) - f(0, 0, 0)\]Some key takeaways include:

• The path taken does not impact the integral's value, reinforcing the path independence of conservative fields.
• This theorem is akin to the fundamental theorem of calculus but applies to multi-dimensional fields.
• It allows us to find potential functions which describe the vector field completely.

Gradient Computation

In the context of this exercise, **gradient computation** involves finding the gradient of the line integral function \( g(x, y, z) \). The function \( g(x, y, z) \) is derived from the path integral and is expressed as:\[g(x, y, z) = f(x, y, z) - f(0, 0, 0)\]As \( g(x, y, z) \) is rooted in the scalar function \( f \), calculating the gradient \( abla g \) means taking partial derivatives with respect to each variable:\[abla g = \left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z}\right)\]Given \( g(x, y, z) = f(x, y, z) - f(0, 0, 0) \), the derivatives are:

• \( \frac{\partial g}{\partial x} = \frac{\partial f}{\partial x} \)
• \( \frac{\partial g}{\partial y} = \frac{\partial f}{\partial y} \)
• \( \frac{\partial g}{\partial z} = \frac{\partial f}{\partial z} \)

Thus, we find \( abla g = \mathbf{F} \), showing that the gradient of the integral confirms our vector field \( \mathbf{F} \). This calculation verifies that the function \( g \) serves as a potential for the field, and leverages the properties of calculus to assert the nature of the vector field.