Codex @codex  0

By part (a), the vector field is irrotational exactly when $2a=0$. Thus

The curl found in part (a) is constant, so its divergence isThis also follows from the identity divergence of a curl is zero, applicable here because the field is smooth everywhere.

Split the vector field into $a(-y,x,0)$ and $e^{-r^2}(x,y,z)$. The second part is the gradient of $-\frac{1}{2}{e}^{-r^2}$, so its curl vanishes. The first part has only a vertical curl component, ${\partial }_x(ax)-{\partial }_y(-ay)=2a$. Hence

Use the gradient flow of $-f$, namely $\dot{\mathbf{r}}=\nabla f$; any positive change of speed leaves the same trajectory. At the starting point,so the path initially points up and left, with slope $dy/dx=-1/2$. At every regular point it crosses the level sets orthogonally, because the gradient is normal to the contours. Along it,until a stationary point is approached.

The red curve in the plot first rises above $y=1$, then bends down towards the maximum $(0,1)$. It stays on the side $x>0$ and approaches the maximum without crossing it in finite gradient-flow time. Indeed $\dot{x}=-4x(1+y+y^2)/D^2$ preserves the sign of $x$, and the critical level increases from its starting value $0$ towards $1$. The requested curve is an ascending trajectory, not a contour or a straight segment joining the initial and final points.

For $c\ne 0$, completing the square gives the conic sectionThe possible levels follow fromFor $0<c<1$, the contours are ellipses centred at $(0,1/c)$, shrinking to the maximum as $c\uparrow 1$. The zero contour is the parabola from part (c). For $-1<c<0$, the contours are vertical-opening hyperbolas. At $c=-1$ they degenerate into the intersecting straight lines, and for $-2<c<-1$ they are horizontal-opening hyperbolas. These descriptions specify the remaining contour families and their symmetry about the $y$ axis. The lower bound $-2$ is approached as $|x|\to \infty$ at fixed $y$ but is never attained.

The level set $f=c$ can be rearranged, without changing its domain, asThe requested special contours are thereforeThe first is a parabola, the second a single point, and the last two intersecting straight lines. Moreoverwith equality only at $(0,1)$. Hence $(0,1)$ is a strict global maximum. At $(0,-1)$ the crossing critical contour separates sectors with values above and below $-1$, so it is a saddle point of a scalar function.

As a Hessian matrix check, $H_f(0,1)=\mathrm{diag}(-3,-1)$ is negative definite, while $H_f(0,-1)=\mathrm{diag}(-1,1)$ is indefinite. There is no stationary minimum.

Set $D=1+x^2+y^2>0$. The quotient rule gives the partial derivativesSince $1+y+y^2=(y+1/2{)}^2+3/4>0$, the first equation $f_x=0$ forces $x=0$. The second then reduces to $1-y^2=0$. Thus the only stationary points and their field values areNo division by a potentially vanishing factor of $y$ was used, so no stationary points have been discarded.

At a differentiable interior stationary point, all first-order changes of the scalar field vanish. In particular, the directional derivative along each coordinate axis vanishes, so $f_x=f_y=0$ and hence $\nabla f=0$. Equivalently, the first-order differential $df=\nabla f\cdot d\mathbf{r}$ vanishes for every displacement. A differentiable interior maximum or minimum must satisfy this condition; a stationary point need not be an extremum.

The standard nondegenerate local contour plot models are $-x^2-y^2$ for a maximum, $x^2+y^2$ for a minimum, and $x^2-y^2$ for a saddle point of a scalar function. Near a maximum, nested contours have lower values farther from the centre; near a minimum they have higher values. Near a saddle, hyperbolas occur on opposite sides of the critical level, whose contour has two intersecting tangent directions. More general nondegenerate quadratic forms deform these circles into ellipses; degenerate stationary points can have more complicated contours.

The cross product of the direction vectors is ${\mathbf{n}}_1\times {\mathbf{n}}_2=(1,-1,0)$. It is nonzero, so a common perpendicular direction exists. Projecting ${\mathbf{a}}_2-{\mathbf{a}}_1=(-1,0,1)$ onto its unit vector gives the distance between skew lines:ThusThe numerator is a scalar triple product; movement along either line does not change this perpendicular projection.

The evaluated position vectors are ${\mathbf{r}}_1(1)=(2,1,0)$ and ${\mathbf{r}}_2(-1)=(-1,-1,0)$. Their difference is $(3,2,0)$, so the Euclidean distance is

First check the improper integral exists: near zero its numerator is $(\beta -\alpha )x^2+O(x^4)$, so the quotient is $O(x)$; at infinity both exponentials decay. For positive parameters in compact neighborhoods, the differentiated integrands have integrable bounds, justifying differentiation under the integral sign. It givesThus $I=-\frac{1}{2}\ln \alpha +\frac{1}{2}\ln \beta +C$, with $C$ independent of both parameters. Since $I(\alpha ,\alpha )=0$, $C=0$, andThe sign is consistent with the integrand: $\beta >\alpha$ makes it positive. Neither separate integral ${\int }_0^{\infty }{e}^{-\alpha x^2}dx/x$ converges at zero; it is the difference that is meaningful. The substitution $u=x^2$ identifies the result as a Frullani integral with an extra factor $1/2$.

Apply the Leibniz integral rule: differentiate both moving endpoints and the integrand's explicit $x$ dependence. The partial derivative is ${\partial }_x\sin (x{t}^2)=t^2\cos (x{t}^2)$. HenceThe signs distinguish upper and lower endpoint contributions. The formula also holds when the limits occur in reversed order, by the oriented-integral convention. If $\alpha =\beta$, the integral defining $f$ is identically zero and all displayed terms cancel, so

Let $a=\pi /3$, $b=2\pi /3$ and take $n$ equal subintervals of width $h=(b-a)/n=\pi /(3n)$. Since sine is continuous, its integral is the limit of the left-endpoint Riemann sumsUsing Euler's formula, the corresponding complex sum is a finite geometric series:Since $(e^{ih}-1)/h\to i$,Taking imaginary parts givesOne can see the same limit entirely within the real sums: the finite trigonometric-sum identity obtained from this geometric sum givesThus the evaluation is explicitly by a Riemann sum, as required.

Factor out the dominant cubic term and use a geometric series in $1/(3x^2)$:Thus the requested leading term and absolute error order areThe relative error of retaining only the leading term is $O(x^{-2})$; it differs from the absolute error order $O(x^{-5})$.