Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 2 / 6A / b

The partial derivatives of $f$ are $f_x=2x$, $f_y=4y(a-y^2)$ and $f_z=2z$. Therefore the three implicit differentiation formulas areTheir product is $-1$ wherever $xyz(y^2-a)\ne 0$. The excluded factors identify exactly where these quotient formulas cannot all represent finite partial derivatives.

At $(x,0,z)$ the gradient is $(2x,0,2z)$. The outward radial unit vector is $(x,0,z)/\sqrt{x^2+z^2}$ when $(x,z)\ne (0,0)$. Taking its scalar product with the gradient gives the radial directional derivativeEquivalently, on this plane $f=r^2$. At the origin the radial unit vector is undefined, but every unit directional derivative is zero. Differentiating along the unnormalized scaling path $(tx,0,tz)$ instead gives $2(x^2+z^2)$ at $t=1$; that is a rate per scaling parameter, not per unit radial distance.

A stationary point requires $x=z=0$ and $y(a-y^2)=0$. The Hessian matrix is diagonal:For $a>0$, the origin has positive definite Hessian matrix, hence is a strict local minimum, with value zero. The two additional stationary points $(0,\pm \sqrt{a},0)$ have value $a^2$ and Hessian matrix $\mathrm{diag}(2,-8a,2)$; both are saddle points.

For $a<0$, the origin is the only stationary point and has an indefinite Hessian matrix, so it is a saddle point. For $a=0$, the Hessian matrix test is inconclusive, but $f(x,y,z)=x^2+z^2-y^4$ takes positive values along the $x$-axis and negative values along the $y$-axis arbitrarily near zero. Thus the origin is a degenerate saddle point. There are no local maxima in any case: varying $x$ or $z$ away from a stationary point increases $f$.

In the $(x,y)$-plane take $z=0$; holding another fixed $z$ merely shifts all contour labels by $z^2$. The level sets satisfyFor $a=1$, the origin is a local minimum and $(0,\pm 1)$ are saddle points of a scalar function at contour value $1$. For $0<c<1$, a closed contour around the origin is accompanied by unbounded outer components. At $c=1$ the contour satisfies $x=\pm (y^2-1)$ and crosses at the two saddle points of a scalar function. For $c>1$, two unbounded components remain; negative contours have only outer components.

For $a=-1$, the origin is a saddle point of a scalar function, and the zero contour is $x=\pm y\sqrt{y^2+2}$, with tangent lines $x=\pm \sqrt{2}y$. Positive contours form two components extending on opposite sides of the $y$-axis; negative contours form upper and lower components. These level sets are illustrated below.