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!

We can also think about this problem analytically (as opposed to geometrically). Notice that if
\mathbf{F} = \langle F_1, F_2 \rangle = \left\langle \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} \right\rangle,
then by Clairaut’s theorem, we must have
\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.

Will Rosenbaum

Saarbrücken, Germany

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