Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 12 / e

The double-angle formula gives $r^2\sin 2\varphi =2(r\cos \varphi )(r\sin \varphi )=2xy$. Therefore

In the positive quadrant, both components of the gradient are positive. Along $y=x$ the arrows point along $(1,1)$; near the horizontal axis they point mainly upwards, and near the vertical axis mainly rightwards. The level sets are the hyperbolas $xy=f/2$, including the coordinate axes for level zero. The gradient is orthogonal to each regular level set and points towards increasing values of $f$.

At $(1,2)$, the gradient is $(4,2)$. The requested direction must be normalized to the unit vector $(1,1)/\sqrt{2}$. The directional derivative is thereforeUsing $(1,1)$ without normalization would instead measure change per unit of a differently scaled path parameter.