Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 14 / d

The gradient is orthogonal to each regular level set and points towards increasing $f$. Near either minimum, the linear approximation is $(8\delta x,\frac{7}{3}\delta y)$, so arrows point away from the point. Near the maximum at the origin it is $(-4\delta x,-\frac{5}{3}\delta y)$, so arrows point inwards. Near either saddle point it is $(-2\delta x,\frac{24}{7}\delta y)$: arrows point towards the vertical axis horizontally and away from the saddle vertically.

On the horizontal axis the gradient is $(4x(x^2-1),0)$. Its arrows point left on $x<-1$, right on $-1<x<0$, left on $0<x<1$, and right on $x>1$. On the vertical axis it isConsequently its arrows point down on $y<-1/\sqrt{2}$, up on $-1/\sqrt{2}<y<0$, down on $0<y<1/\sqrt{2}$, and up on $y>1/\sqrt{2}$. These directions, and representative arrows near each stationary point, appear in the right-hand panel. The gradient itself is zero at all five stationary points.