Problem 18 - Advanced Engineering Mathematics

Question

Fill in the blank or answer true/false. Where appropriate, assume continuity of \(P, O\), and their first partial derivatives. If \(\int_{C} P d x+Q\) dy isindependent of the path, then \(F=P i+Q \mathbf{j}\) is the gradient of some function \(\phi\).___

Step-by-Step Solution

Answer

If \(P_dy + Q_dy\) is path-independent, \(F=P\mathbf{i}+Q\mathbf{j}\) is \(\nabla\phi\).

Step 1: Understand the Problem

We need to determine when a vector field \(F = P\mathbf{i} + Q\mathbf{j}\) is the gradient of some scalar function \(\phi\). This occurs when the line integral \(\int_{C} P\,dx + Q\,dy\) is independent of the path. We want to understand what conditions make the integral independent of the path.

Step 2: Recall the Necessary Condition

For a vector field \(F = P\mathbf{i} + Q\mathbf{j}\) to be a gradient field, the line integral \(\int_C P\,dx + Q\,dy\) is independent of the path if and only if \(\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}\). These are the conditions of equality of mixed partial derivatives that must be satisfied.

Step 3: Relate to Exact Differential

The differential form \(P\,dx + Q\,dy\) should be exact, implying that it is the differential of a scalar function \(\phi(x,y)\). This means \(F\) is the gradient field of \(\phi\) if the condition \(\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}\) holds throughout the region.

Key Concepts

Path Independence

In calculus, especially in vector calculus, path independence is a crucial concept when dealing with line integrals. When you hear the term "path independence," it generally means that the value of the line integral is the same, irrespective of the path taken between two points. This applies specifically to vector fields and their associated line integrals.
For a line integral \(\int_{C} P\,dx + Q\,dy\) to be path-independent, there are specific conditions that need to be met. Primarily, we are looking for the equality of mixed partial derivatives: \(\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}\).
This condition ensures that the vector field is conservative, meaning it can be expressed as the gradient of a scalar potential function. A conservative vector field guarantees path independence because the integral's value depends only on the endpoints of the path, not the actual path taken.

Exact Differential Form

An exact differential form is closely tied to the concept of path independence in vector calculus. In simple terms, a differential form \(P\,dx + Q\,dy\) is called exact if it is the differential of some scalar function, usually denoted \(\phi(x, y)\).
When a differential form is exact, it automatically implies that the line integral over any path is independent of the path taken. This follows from the fundamental theorem of line integrals, which tells us that if \(\phi(x, y)\) exists such that its gradient corresponds to the vector field \(F = P\mathbf{i} + Q\mathbf{j}\), then \(\int_{C} P\,dx + Q\,dy = \phi(B) - \phi(A)\), where \(A\) and \(B\) are the endpoints of the path \(C\).
The key condition for a form to be exact is that the mixed partial derivatives are equal, i.e., \(\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}\). If this holds throughout a region, the differential form is exact within that region.

Mixed Partial Derivatives

Mixed partial derivatives come into play when analyzing the conditions under which a vector field is conservative, and therefore, when the line integral is path independent. Mixed partial derivatives are the derivatives of functions that involve multiple variables, and they are taken with respect to different variables.
In practice, for a vector field \(F = P\mathbf{i} + Q\mathbf{j}\) to be the gradient of some function \(\phi(x, y)\), a crucial requirement is that the mixed partial derivatives must be equal: \(\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}\). This equality ensures the path independence of the line integral \(\int_{C} P\,dx + Q\,dy\).
Understanding mixed partial derivatives as being equal is not just an arbitrary condition but a reflection of the smoothness and symmetry of the function \(\phi\) from which our vector field originates. It is the mathematical assurance required to guarantee that the vector field can be derived from a single-valued scalar potential function and that the differential form is exact.