Past exam of the mathematics course of the University of Oxford / Undated / Prelims / Mathematics III resit / 3

The middle component suggests a scalar potential with $-{\varphi }_y=2y{\cosh }^2(xz)$, so $\varphi =-y^2{\cosh }^2(xz)+h(x,z)$. The chain rule and $2\sinh t\cosh t=\sinh (2t)$ giveMatching the other two components requires $h_x=h_z=0$. ThereforeIts negative gradient is the given vector field in all three components; the additive constant has no effect on that gradient.

A parametrization of the straight line segment is $\mathbf{r}(t)=(t,t,t)$, $0\le t\le 1$. Then $\mathbf{F}(\mathbf{r}(t))=(t,t^2,t)$ and ${\mathbf{r}}^{\prime }(t)=(1,1,1)$. By the line integral formula,

Evaluate the line integral separately on the three oriented line segments. Their parametrizations and contributions areThusThe different value in i demonstrates path dependence. The vector field in this part is different from the conservative vector field in part (a).

Write $A=a^2>0$ and $E=e^{-(x^2+y^2)/A}>0$. The partial derivatives factor asA critical point has both partial derivatives zero. If $x,y\ne 0$, this would require $x^2-2y^2=A$ and $x^2-2y^2=-2A$, which is impossible. On the axes the remaining conditions giveAt either critical point with $x\ne 0$, the Hessian matrix isIt is negative definite, so the second-derivative test shows that both $(\pm |a|,0)$ are strict local maxima, of value $a^2/e$. They also attain the global maximum: $f\le x^2{e}^{-x^2/A}\le A/e$. For completeness, the Hessian matrix at $(0,\pm |a|)$ is $\mathrm{diag}(6/e,8/e)$, giving strict local minima, while at the origin it is $\mathrm{diag}(2,-4)$, giving a saddle point.