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Independence of Path

Conservative Vector Fields

MathJax TeX Test Page Let's say you have a vector field $F(x,y) = < M(x,y), N(x,y)>$. That is conservative iff there exists a function f(x,y) such that $\nabla{}f = F(x,y)$.

Determining conservativeness  

MathJax TeX Test Page Let's say you have a vector field $$< y^3 \cos(x), 3y^2 \sin(x)>$$ Remember that the x term in that field is $f_x$, while they y term is $f_y$. If both parts are continuous, then by Clairaut's Theorem, $f_{xy} = f_{yx}$. So, if F = < M(x,y), N(x,y)>, $\boxed{M_y = N_x}$. $$\dfrac{\partial}{\partial y} y^3 \cos(x) = 3y^2 \cos(x) $$ $$\dfrac{\partial}{\partial x} 3y^2 \sin(x) = 3y^2 \cos(x)$$ They're the same! By Clauraut's Theorem, there exists a function f. Now, let's find that function! However, I want to demonstrate something, so I will slightly change the field to < y^3 \cos(x) + 1, 3y^2 \sin(x)> $$f_x = y^3 \cos(x) + 1 \rightarrow f = y^3 \sin(x) + x + g(y)$$ $$f_y = 3y^2 \sin(x) \rightarrow f = y^3 \sin(x) + h(x)$$ Notice how the functions are not the same, but there is a constant. When there are extra terms, that means they only have an impact in either x or y, so you must combine the two expressions to create one functions with all of the terms. So, here $f(x,y) = y^3 \sin(x) + x$.

TFAE

TFAE means "The Following Are Equivalent". So, the following things are either all true or all false.

MathJax TeX Test Page First, we have to assume that F is continuous on open, connected region R. $$1. \text{ F is conservative}$$ $$2. \exists f: \mathbb{R}^n \rightarrow \mathbb{R} \text{ such that } \nabla f = F$$ $$3. \text{ For all C contained in region R, }\int_C F \cdot dr \text{ is independent of path}$$ $$4. \text{ For all closed curves in region R, } \oint_C F \cdot dr = 0$$ $$5. \text{If M and N have cont. first partials and region R is simply connected, } \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$ $$6. \text{curl }F = 0$$

Independence of Path

If the entire phrase of part five is true, including continuous first partials and simply connected, then the integral is independent of path. This means that if you want to take the path integral from A to B, you can go along a straight line, a curve, split it up into two straight lines, it doesn't matter. Any possible path from two points has the same path integral. This is only true when it is conservative. That is crucial to understand. This is something specific to conservative vector fields.

Fundamental Theorem of Line Integrals

MathJax TeX Test Page The Fundamental Theorem of Line Integrals states that if $f \in \Omega^0(U)$ and C is a smooth curve in U from p to q, then $$\int_C df = f(q) - f(p)$$ Another way of writing it is if F = $\nabla f$, then $$\int_C F = f(q) - f(p)$$ This theorem greatly simplifies computation. Let's say you have the path following $g = \begin{bmatrix}\cos(t)\\ \sin(t)\end{bmatrix}$ from $0 \leq t \leq \pi$.

It turns out $\int_{g(t)} ydx + xdy$ is actually very easy to solve. The equation above means that the form above is exact. $(\dfrac{\partial}{\partial y} y = \dfrac{\partial}{\partial x} x = 1)$. In this equation, $f = xy$, $\nabla f = f_x dx + f_ydy = ydx + xdy$ So, we plug in f(t) at the end points. $$f(\pi) = 0 * 1 = 0. f(0) = 0 * 1 = 0. f(\pi) - f(0) = \boxed{0}$$
David Witten

Change of Variables

Line Integrals

Line Integrals