Problem 228 - Advanced Calculus

Question

Let \(\mathrm{F}^{-7}\) be the following vector fields a) \(\mathrm{F}^{\rightarrow}{\underline{\phantom{xx}}}_{1}(\mathrm{x}, \mathrm{y}, \mathrm{z})=(1-\mathrm{yz}, 1-\mathrm{zx},-\mathrm{xy})\) b) \(\mathrm{F}^{\rightarrow}{\underline{\phantom{xx}}}_{2}(\mathrm{x}, \mathrm{y}, \mathrm{z})=(\mathrm{yz}, \mathrm{xz}, \mathrm{xy})\) c) \(\mathrm{F}^{\rightarrow}{\underline{\phantom{xx}}}_{3}(\mathrm{x}, \mathrm{y}, \mathrm{z})=(\log (\mathrm{xy}), \mathrm{x}, \mathrm{y})\) Determine, for each vector field, if \(\int \mathrm{F}^{\rightarrow} \cdot \mathrm{dc}^{\rightarrow}\) is independent of the path and if so integrate the vector field over the curve from the point \(\mathrm{P}=(1,6,5)\) to the point \(\mathrm{Q}=(4,3,2)\).

Step-by-Step Solution

Answer

The line integral for the vector field \(\mathrm{F}_{1}(x, y, z) = (1 - yz, 1 - zx, -xy)\) is path-independent, and its value is 24. The line integrals for the vector fields \(\mathrm{F}_{2}(x, y, z) = (yz, xz, xy)\) and \(\mathrm{F}_{3}(x, y, z) = (\log(xy), x, y)\) are path-dependent, so we cannot evaluate them using our method.

Step 1: Determine if the line integral is path-independent

For a vector field to have a path-independent line integral, it must be conservative. To check if the given vector field F1 is conservative, we need to verify if it satisfies the following conditions: 1. The partial derivative of its components match each other: \(\frac{\partial F_{1x}}{\partial y} = \frac{\partial F_{1y}}{\partial x}\), \(\frac{\partial F_{1x}}{\partial z} = \frac{\partial F_{1z}}{\partial x}\), and \(\frac{\partial F_{1y}}{\partial z} = \frac{\partial F_{1z}}{\partial y}\). 2. The vector field has continuous first-order partial derivatives. F1(x, y, z) = (1 - yz, 1 - zx, -xy) Calculating the partial derivatives: \(\frac{\partial F_{1x}}{\partial y} = -z\) \(\frac{\partial F_{1y}}{\partial x} = -z\) \(\frac{\partial F_{1x}}{\partial z} = -y\) \(\frac{\partial F_{1z}}{\partial x} = -y\) \(\frac{\partial F_{1y}}{\partial z} = -x\) \(\frac{\partial F_{1z}}{\partial y} = -x\) All partial derivatives exist and are continuous. We can see that the conditions above are satisfied. Therefore, F1 is conservative and its line integral is path-independent.

Step 2: Integrate the vector field over the curve

Since F1 is conservative, we can find the potential function G such that F1(x, y, z) = ∇G. Integrate each component of F1 to find G: G(x, y, z) = ∫(1 - yz) dx + ∫(1 - zx) dy + ∫(-xy) dz G(x, y, z) = x - xyz + f(y, z) + y - zxy + g(x, z) - xyz + h(x, y) Now we can simplify: G(x, y, z) = x + y + (-2xyz) + f(y, z) + g(x, z) + h(x, y) Since F1 is conservative and path-independent, we can find the line integral from P(1, 6, 5) to Q(4, 3, 2) by finding the difference in G: int(F1) = G(Q) - G(P) G(4, 3, 2) = 4 + 3 - 2(4)(3)(2) G(1, 6, 5) = 1 + 6 - 2(1)(6)(5) int(F1) = (4 + 3 - 48) - (1 + 6 - 60) int(F1) = -41 + 65 int(F1) = 24 So, for vector field F1, the line integral is path-independent and its value is 24. #b) Vector Field F2# We can follow the same steps as we did for F1. After checking the conditions for the vector field F2, it turns out that it is not conservative and the line integral is path-dependent. Thus, we cannot evaluate the line integral for this vector field. #c) Vector Field F3# Similarly, we follow the same steps for the vector field F3. After checking the conditions for vector field F3, it turns out that it is not conservative and the line integral is path-dependent. Thus, we cannot evaluate the line integral for this vector field. In conclusion, the line integral for the vector field F1 is path-independent and has a value of 24. The line integrals for the vector fields F2 and F3 are path-dependent, so we cannot evaluate them using our method.

Key Concepts

Line Integrals

The concept of line integrals is essential in vector calculus, especially when dealing with vector fields. A line integral essentially measures the sum of a field's effects along a certain path. Imagine dragging a particle through a magnetic or gravitational field; the line integral accounts for the total influence the field exerts on the particle along its journey.

The general form of a line integral of a vector field \(\mathbf{F}\) along the piecewise smooth curve \(C\) is given by:
\[ \int_C \mathbf{F} \cdot d\mathbf{r} \]
Where \(\mathbf{r}\) represents the position vector of points along the curve, and \(d\mathbf{r}\) is the differential element along the path. The dot product \(\mathbf{F} \cdot d\mathbf{r}\) signifies that we are interested in the component of \(\mathbf{F}\) that is tangent to the curve.

• This integral accounts for how the vector field interacts with the path of integration.
• It might be dependent on the path taken or independent, depending on the nature of the vector field.

This leads us to explore the concept of path independence, which is intimately connected to whether a vector field is conservative.

Path Independence

Path independence is a fascinating property of certain line integrals. For a function to have a path-independent line integral, the simplification to a single evaluation of a difference of a potential function is possible. In simpler words, it doesn't matter which path you take between two points; the result is the same.

This property is a hallmark of a conservative vector field. A vector field is conservative if there exists a scalar potential function \(G(x, y, z)\) such that the field is the gradient of this potential: \(\mathbf{F} = abla G\).

• When a vector field is conservative, the line integral only depends on the endpoints of the curve.
• If you can find a potential function, the vector field is indeed conservative.

When solving exercises involving path independence, it means you can compute the line integral by finding this potential function and evaluating it at the endpoints. For example, in our exercise, vector field \(\mathbf{F}_1\) was conservative, making its line integral path-independent. Thus, it was evaluated using the potential function, \(G(x, y, z)\).

Vector Calculus

Vector calculus is an advanced field of mathematics that deals with vector fields and their derivatives and integrals. It forms the cornerstone for understanding physical phenomena through mathematics, ranging from electromagnetism to fluid flow.

Key operations in vector calculus include the gradient, divergence, and curl.

• The gradient \(abla G\) transforms a scalar field into a vector field.
• The divergence \(abla \cdot \mathbf{F}\) tells us how much "stuff" is emanating from a point in a vector field.
• The curl \(abla \times \mathbf{F}\) detects rotational aspects of a vector field.

These operations are vital in understanding whether a vector field can be conservative, which tremendously simplifies line integral evaluations. In our exercise, identifying the vector field as conservative stemmed from these calculations. Further exploration of these operations deepens understanding of many phenomena in physics and engineering.

Partial Derivatives

Partial derivatives are integral to understanding how vector fields work and are a key concept in vector calculus. These derivatives measure the rate at which a function changes as one of its variables varies, with all other variables held constant.

In three-dimensional space, for functions \(f(x, y, z)\), we have the partial derivatives, \(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\) and \(\frac{\partial f}{\partial z}\). Partial derivatives play a central role in fulfilling the conditions required for a vector field to be conservative:

• Each component of the vector field must have continuous partial derivatives.
• Certain partial derivatives must equal each other, a condition which confirms path independence in the vector field.

In the exercise given, calculating the partial derivatives of vector fields like \(\mathbf{F}_1(x, y, z)\) confirmed that the components of the vector field satisfied the symmetry condition necessary for it to be conservative. Understanding these calculations confidently takes one closer to mastering vector calculus.