Answer
(a) Using Figure 17, we conclude that the level curves of $f$ must be vertical lines and ${\bf{F}}$ is a constant vector field.
(b) Using Figure 17, we conclude that the level curves of $f$ must grow farther apart as $y$ increases.
(c) The results in part (a) and part (b) are incompatible. Hence, $f$ cannot exist.
Work Step by Step
Let $f$ be the potential function for ${\bf{F}}$. Then $\nabla f = {\bf{F}}$.
(a) From Figure 17, we see that the vector fields ${\bf{F}}$ is pointing horizontally and perpendicular to the level curves. Therefore, if a potential function $f$ for ${\bf{F}}$ exists, then the level curves of $f$ must be vertical lines. Since the spacings between level curves are constant, we conclude that $\nabla f = {\bf{F}}$ is a constant vector field.
(b) We observe from Figure 17, the vector field ${\bf{F}}$ is decreasing as $y$ increases. This implies that $f$ is also decreasing. So, the level curves of $f$ must grow farther apart (it is getting less steeper from one level to the next one). Therefore, the level curves of $f$ must grow farther apart as $y$ increases.
(c) From part (a) we obtain the result that $\nabla f$ is a constant vector field. However, in part (b) we require that the level curves of $f$ must grow farther apart as $y$ increases, which means that $\nabla f$ is decreasing and $\nabla f$ is not constant. So, there is a contradiction. Therefore, part (a) and (b) are incompatible, and hence $f$ cannot exist.
