Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 16

A directional derivative measures change per unit distance, so first normalize the direction: $\hat{v}=(4,3)/5$. The gradient isTaking the dot product with the unit vector,Using the unnormalized vector would instead give $-2$, the change per unit parameter along $(1,0)+tv$; it is not the rate per unit distance requested here.

The product rule gives the two partial derivativesThey both vanish at the origin. The exponential function is never zero, so any further stationary point must satisfy $x=0$ or $x^2-y^2=2$, and $y=0$ or $y^2-x^2=2$. Having both coordinates nonzero would require both opposite differences to be $2$, which is impossible. If $y=0$, the nonzero choices are $x=\pm \sqrt{2}$; if $x=0$, they are $y=\pm \sqrt{2}$. Hence the full list is

At the origin the Hessian matrix isThus the second-derivative test gives a saddle point. Its opposite signs can also be checked without a test formula: $g(x,0)=x^2{e}^{-x^2/2}>0$ for $x\ne 0$, whereas $g(0,y)=-y^2{e}^{-y^2/2}<0$ for $y\ne 0$. Both types of values occur arbitrarily close to the zero value at the origin, excluding either a local maximum or a local minimum.