Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 12

The two coefficients are the corresponding partial derivatives of $e^{x+y}$. Therefore the differential form is exact, withThe arbitrary constant does not change the total differential.

For a smooth exact differential form $Mdx+Ndy$, its coefficients must satisfy $M_y=N_x$ by Clairaut's theorem. HereThese are not equal as functions on the plane, so the form is not exact. Equality at isolated points or along some lines is insufficient: an exactness test must hold throughout an open domain.

The coefficients follow by differentiating the same monomial:Thus this exact differential form hasExhibiting a global scalar potential proves exactness directly, without needing an additional domain argument.

The partial derivatives are $g_x=\sin x$ and $g_y=y$, so all stationary points are $(n\pi ,0)$ for integers $n$. Their Hessian matrix is $\mathrm{diag}(\cos x,1)$. At even multiples of $\pi$ it is a positive-definite matrix, whereas at odd multiples it has one positive and one negative eigenvalue. ConsequentlyThere are no local maxima. Since both $1-\cos x$ and $y^2/2$ are nonnegative, these are also global minima. In the requested open plotting window, the stationary points are $(0,0)$ and $(\pm \pi ,0)$; the local minima at $(\pm 2\pi ,0)$ lie on excluded edges.

For a level set $g=c$, solving for the vertical coordinate givesFor $0<c<2$ these level sets are closed ovals around each local minimum. At $c=0$ the level sets shrink to the local minima. The level set through the saddle points iswhere the two signed branches together describe the same set as $y=\pm 2|\cos (x/2)|$. They cross at $(\pm \pi ,0)$ and reach $(0,\pm 2)$. The crossing slopes at a saddle point are $\pm 1$, as also follows from the local quadratic form $(-\delta x^2+\delta y^2)/2$. For $c>2$, the upper and lower level sets remain separate and extend across the plotting window without meeting the horizontal axis.