# Explanation of 17.1 #34

This problem asks you to argue that the depicted vector field is not conservative. Recall that a vector field $$\mathbf{F}(x, y)$$ is conservative if we can write
$\mathbf{F}(x, y) = \nabla V(x, y)$
where $$V(x, y)$$ is a scalar function. That is,
$\mathbf{F} = \left\langle \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} \right\rangle.$
The geometric interpretation of the gradient $$\nabla V$$ is that

1. $$\nabla V$$ is perpendicular to the level curves given by $$V(x, y) = c$$
2. $$\nabla V$$ points in the direction of greatest increase of $$V$$
3. $$|\nabla V|$$ is the slope of $$V$$ in the direction of greatest increase.

From the figure, the vector field $$\mathbf{F}$$ always points horizontally to the right. Therefore, by property 1 of the gradient, the level curves of $$V$$ must be vertical lines.

The figure also depicts the lengths of the vectors $$\mathbf{F}(x, y)$$ as decreasing as $$y$$ increases. By properties 2 and 3, this tells us that the supposed potential function $$V$$ is increasing as $$x$$ increases, but that this increase is slower for larger values $$y$$. However, this implies that the level curves are further apart for larger values $$y$$. This contradicts the previous observation that the level curves of $$V$$ must be vertical lines as vertical lines are a constant $$x$$–width apart!

$\mathbf{F} = \langle F_1, F_2 \rangle = \left\langle \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} \right\rangle,$
$\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}.$
In the depicted vector field, $$F_1$$ decreases as $$y$$ increases, but $$F_2$$ is constant (in fact, $$0$$). Therefore, $$\frac{\partial F_1}{\partial y}\neq 0$$, but $$\frac{\partial F_2}{\partial x} = 0$$. Thus the vector field cannot possibly be conservative, as these two partial derivatives would be equal for a conservative vector field.