Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 16

Translate the coordinates by $u=x-1$ and $v=y+1$, and write $E=e^{-(u^2+v^2)/2}>0$. The partial derivatives areIf $u=0$ or $v=0$, these equations together force $u=v=0$. Otherwise they require $u^2=v^2=1$. Thus the complete set of stationary points is

At the central stationary point, $u=v=0$, so $f=0$. At each of the other stationary points, $u^2+v^2=2$ and hence $f=uv/e$. Therefore

In the translated coordinates $u=x-1$, $v=y+1$, the zero level set is the pair of lines $u=0$ and $v=0$. The function is positive in the quadrants $uv>0$ and negative in those with $uv<0$, and tends to zero as $u^2+v^2\to \infty$. Thus its nonzero contours form closed curves around the four off-centre stationary points.

For an explicit description of the contour plot, use polar coordinates $u=r\cos \theta$, $v=r\sin \theta$:At fixed direction within a quadrant, the magnitude increases until $r=\sqrt{2}$ and then decreases towards zero. Also $|u|e^{-u^2/2}\le e^{-1/2}$, with equality exactly when $|u|=1$, and similarly for $v$. Consequently $|f|\le e^{-1}$; each positive level below $e^{-1}$ encloses a positive peak in each positive quadrant, with the analogous negative contours in the negative quadrants.

To classify the stationary points, the Hessian matrix isAt $u=v=0$ it is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, with eigenvalues $1$ and $-1$, so $(1,-1)$ is a saddle point. At $u,v\in \{-1,1\}$ it is $-2fI$. Therefore $(2,0)$ and $(0,-2)$ are strict local maxima, and $(2,-2)$ and $(0,0)$ are strict local minima. These are also global maxima and minima by the bound just proved.

The gradient isIt is perpendicular to each nonzero level set and points towards increasing $f$. Near the two positive local maxima, the Hessian matrix is $-2e^{-1}I$, so the gradient arrows point inward. Near the two negative local minima, the Hessian matrix is $2e^{-1}I$, so they point outward. These arrows are shown on the figure in part (c).

At the saddle point, $\nabla f\approx (v,u{)}^T$. Along $v=u$, the arrows point away from the centre on both sides; along $v=-u$, they point towards the centre on both sides. On the positive $u$ axis they point up, on the negative $u$ axis down, on the positive $v$ axis right, and on the negative $v$ axis left. This specifies the four intervening sectors of the gradient pattern.