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Here the stationary point is the origin, of value two. The Taylor expansion isThus nearby level curves are approximately concentric circles. Globally the level curves above two are closed around the unique minimum, symmetric under interchanging $x$ and $y$, but not under changing either coordinate's sign. The faster exponential growth for positive coordinates compresses the curves on their positive sides.

For case (i),Each component has the sign of its coordinate, so the gradient points outward from the minimum. Near the origin it is $(x,y{)}^T+O(x^2+y^2)$.

For case (ii),Its $x$ component has the sign of $x$, while its $y$ component has the opposite sign to $y$. Thus arrows point outward on the $x$-axis and inward on the $y$-axis, with local form $(x,-y{)}^T+O(x^2+y^2)$. These directions are included in the contour figure; in both cases the gradient is perpendicular to the level curves and points toward increasing $h$.

Along a level curve, $V_{x}dx+V_{y}dy=0$, with $V_x=V_0\theta /\pi$ and $V_y=V_0(1+\log r)/\pi$. ThereforeIn the intended upper half-plane $0<\theta <\pi$, the slope is positive for $0<r<e^{-1}$, vertical at $r=e^{-1}$, and negative for $r>e^{-1}$. At the positive real boundary $\theta =0$ the slope is zero except at the stationary point $(e^{-1},0)$, where both derivatives vanish and the quotient is indeterminate. More generally, outside the chosen half-plane its sign is the sign of $-\theta (1+\log r)$, with the argument branch held fixed. A convenient parametric description of the curves to sketch is $V\pi /V_0=r(\sin \theta \log r+\theta \cos \theta )$.