Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 12

Assume $F$ is continuously differentiable near the point of interest on $F=0$, and that the relevant denominators below are nonzero. The total differential on the surface isHolding $z$ fixed gives $F_{x}dx+F_{y}dy=0$, so the implicit function theorem givesWhen $F_x{F}_y\ne 0$, their product proves the reciprocity rule for constrained partial derivatives:Similarly, holding $y$ or $x$ fixed gives $(\partial x/\partial z{)}_y=-F_z/F_x$ and $(\partial z/\partial y{)}_x=-F_y/F_z$. With $F_x{F}_y{F}_z\ne 0$, multiply the three ratios:This is the cyclic partial derivative identity. Nonzero denominators and local differentiable coordinate representations are substantive assumptions: at a singular coordinate representation, a formal cancellation of zero or infinite factors is not a valid proof.

Differentiate the defining function with the other coordinates independent. Put $C=\cosh (x+z)$ for brevity. ThenThe implicit differentiation ratios therefore giveEach expression applies on the implicit surface wherever its own denominator is nonzero. The hyperbolic cosine appears because the derivative of $\sinh (x+z)$ with respect to either $x$ or $z$ is $\cosh (x+z)$.

There is a global-domain issue in the printed question. Write $D=a^2{x}^2+b^2{y}^2>0$ away from the origin. Direct partial differentiation givesThus the differential one-form is a closed differential form for every real $a,b\ne 0$. The Poincaré lemma makes it locally an exact differential form on small neighborhoods avoiding the origin. There is no additional condition such as $a^2=b^2$.

However, the stated domain is the whole punctured plane, which is not simply connected. To test global exactness, use the closed ellipse $x=\cos t/a$, $y=\sin t/b$, $0\le t\le 2\pi$. Here $D=1$ andThe line integral of a globally exact differential form around any closed curve would be zero. ConsequentlyThe distinction is precisely that of the elliptically scaled angular one-form. The remaining parts can be answered consistently using a local angle branch, as follows.

On a chart with $x\ne 0$, setThe chain rule verifies both components:A coordinate-independent description of the local potential is a continuous branch of $\arg (ax+iby)$, the complex argument of the scaled coordinates. Such branches can also cover points where $x=0$, using a different angular chart. Continuing once around the missing origin changes that angle by a nonzero multiple of $2\pi$, so there is no global single-valued $f$ on the full stated domain.

The nonzero factor $ab/D$ can be cancelled in the local differential equation, leaving $xdy-ydx=0$. On an interval with $x\ne 0$,Hence the explicit general solution ison any interval that avoids the excluded origin. It also follows from fixing the local angle potential, with slope $C=(a/b)\tan (f-C_0)$. As parameterized curves there are also vertical rays $x=0$, which cannot be represented as single-valued functions $y(x)$. Geometrically the connected integral curves are rays, not lines continued through the undefined origin.