I have to show that a line integral is path independent between two points, and while I know how to check if one is, I have no idea where to begin proving that one is. The equation looks very simple to integrate, but I don't even know where to start. The line integral is
(3x-5y)dx+(7y-5x)dy, and it is over the line segment from (1,3) to (5,2). The only thing I have figured out is that the path is (-t,4t) with t ranging from 0 to one, but after that I'm lost.
[Math] Show that a line integral is path independent
line-integrals
Best Answer
Hints:
To show path independence, suppose you have a closed curve $C$ through any two points. Then, applying Green's theorem, we have:
$$\int_{C} Ldx + Mdy = \iint_D \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right)dxdy$$
Where $D$ is the region enclosed by $C$ and $L = 3x - 5y$ and $M = 7y-5x$. So what happens to the integrand on the RHS above, and what can we conclude?
Based on this work, what is a sufficient condition for showing that a general vector field is path-independent?
To compute the specific path integral for the line segment connecting $(1, 3)$ and $(5, 2)$, we first need to get a correct parametrization. Pedro is correct in the comments. You've gone astray since at $t = 0$, your parametrization is putting you at the origin.
All lines can be parametrized in the form $\alpha(t) = \mathbf{p_0} + \mathbf{c}t$. Given that we want to be at $(1, 3)$ at time $t = 0$, then what is a good choice for $\mathbf{p_0}$? What, then, must $\mathbf{c}$ be if we wish to be at $(5, 2)$ when $t = 1$?
Of course, that is the most difficult part. Once you have a good parametrization, the rest is just plug-and-chug.