Codex @codex  0

The gradient vanishes whenIf $x=0$, the second equation gives $y=0$ or $y=1$. If $y=-1/2$, it gives $x^2=3/4$. These exhaust the possibilities. The stationary points and their values are

At each regular point, the gradientis normal to the level set and points towards increasing $f$. The arrows in the contour plot above show this direction; their lengths have been normalized to keep the outer arrows legible. Near the origin they point outwards, as expected for a local minimum. Near $(0,1)$, write $\eta =y-1$; the linear approximation is $\nabla f\simeq (3x,-\eta )$. Thus the arrows point away along the horizontal direction and towards the point along the vertical direction. The other two saddle points have the same behaviour after rotations through $2\pi /3$. All four stationary points have zero gradient, so no direction arrow is assigned at the points themselves.

It is helpful first to separate an exact gradient:The fundamental theorem for line integrals makes the gradient contribution $2\alpha +{\alpha }^2$ for either path. On the curved path, $dy=2tdt$, $dz=3t^{2}dt$, $y=t^2$ and $z=t^3$, so the remaining contribution isThus

A stationary point has zero gradient. The two partial derivatives giveSolving this system of linear equations gives

Put $s=x^2+y^2$. The gradient components areA stationary point requires both components to vanish. If $x\ne 0$, then $s=1$; the second component subsequently forces $y=0$, giving $(\pm 1,0)$. If $x=0$, either $y=0$ or, writing $u=y^2$,Only $u=1/2$ is nonnegative. Thus there are five stationary points, with valuesThe cases $x=0$ and $s=1$ exhaust the first partial derivative equation, so no other stationary points have been lost by division.

The contour plot is symmetric under either $x\mapsto -x$ or $y\mapsto -y$, because $f$ depends only on their squares. Denote the three critical heights by $m=\frac{7}{2}\ln 3$, $h=\frac{1}{4}+\frac{7}{2}\ln (7/2)$ and $M=1+\frac{7}{2}\ln 3$, where $m<h<M$.

For $m<c<h$, the level set $f=c$ consists of separate closed curves around the two minima. At $c=h$ the branches meet at the two saddle points. For $h<c<M$, there is an inner closed curve surrounding the local maximum and an outer closed curve surrounding all five stationary points. At $c=M$ the inner curve shrinks to the origin; for $c>M$ only the outer closed curve remains. At each saddle point the critical level set crosses in the two directions determined byThe figure plots representative level sets, including the critical saddle height; its second panel adds the gradient directions requested below.

The gradient is orthogonal to each regular level set and points towards increasing $f$. Near either minimum, the linear approximation is $(8\delta x,\frac{7}{3}\delta y)$, so arrows point away from the point. Near the maximum at the origin it is $(-4\delta x,-\frac{5}{3}\delta y)$, so arrows point inwards. Near either saddle point it is $(-2\delta x,\frac{24}{7}\delta y)$: arrows point towards the vertical axis horizontally and away from the saddle vertically.

On the horizontal axis the gradient is $(4x(x^2-1),0)$. Its arrows point left on $x<-1$, right on $-1<x<0$, left on $0<x<1$, and right on $x>1$. On the vertical axis it isConsequently its arrows point down on $y<-1/\sqrt{2}$, up on $-1/\sqrt{2}<y<0$, down on $0<y<1/\sqrt{2}$, and up on $y>1/\sqrt{2}$. These directions, and representative arrows near each stationary point, appear in the right-hand panel. The gradient itself is zero at all five stationary points.

For the second vector field, the curl isIt is nonzero, for example $(0,-2,2)$ at $(1,0,0)$. A differentiable scalar potential would have zero curl of its gradient, so $\mathbf{G}$ is not conservative. The fact that its curl vanishes on the line $x=y=z$ does not make it vanish throughout the domain.

By implicit differentiation of the ellipse equation,The tangent line through $(1,1)$ is thereforeIts normal vector is $(9,16)$, parallel to the gradient $(18,32)$ of $9x^2+16y^2$ at that point.

The original PDF specifies evaluation at $(1,1,1)$; the local TeX incorrectly gives $(1,1,0)$. Write the scalar field explicitly as $f=c/(x^2+y^2+z^2)$. Its gradient isA directional derivative measures rate per unit distance, so normalize the direction to the unit vector $\mathbf{n}=(1,1,0)/\sqrt{2}$. ThenUsing the unnormalized direction vector would instead differentiate with respect to a path parameter, not distance, and would give a different scale.

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.

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.

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

For $a=0$, differentiate the proposed scalar potential directly:ThusFor $a\ne 0$, the curl is the nonzero vector $(0,0,2a)$. A gradient of a twice differentiable scalar field has zero curl, so no such scalar field exists. Equivalently the closed line integral in part (d) is nonzero, contradicting the path independence of a conservative vector field. The field is smooth on all of ${\mathbb{R}}^3$, so no excluded singular set complicates this conclusion.

The gradient is

At $A=(1/2,1/4)$, the gradient is $(-3/64,-9/32{)}^T$. The position vector has length $\sqrt{5}/4$, so its normalized direction is $(2,1{)}^T/\sqrt{5}$. Hence the directional derivative is

The zero level set is especially simple:The two axes and the unit circle divide the plane into regions where the signs alternate. Inside the circle, $f$ has the opposite sign to $xy$; outside it has the same sign. Closed level curves surround the two interior local minima and two interior local maxima; the zero curves cross at the five saddle points of a scalar function. Farther out, the nonzero level curves are open within the displayed square.

Near each local minimum, the gradient points outward; near each local maximum, it points inward. The five remaining points have saddle directions, determined by their indefinite Hessian matrices.

On the $x$-axis the exact gradient is $(0,x(x^2-1){)}^T$, and on the $y$-axis it is $(y(y^2-1),0{)}^T$. Thus arrows are perpendicular to the axes and reverse direction at the origin and the unit-axis stationary points.

On the unit circle, substitute $x^2+y^2=1$ into the gradient to obtainIt is radial outward in quadrants I and III, radial inward in quadrants II and IV, and zero at the four axis intersections. These arrows, together with the nearby saddle and extremum patterns, are shown in the right panel of the contour figure. The gradient is normal to every regular level curve and points toward increasing values of $f$.

Use $g=x^2+y^2+z^2$. The Lagrange multiplier equations areTaking the dot product with $(x,y,z{)}^T$ gives $2T=2\lambda$, so $\lambda =T$.

If $\lambda \ne 0$, the first and third equations give $x=z$. The middle equation then gives $x=\lambda y$, so $2{\lambda }^2=1$. The norm constraint gives $x^2=z^2=1/4$ and $y^2=1/2$. Thus the hottest points and their temperature areThe coldest points areFor $\lambda =0$, the equations give $y=0$, $z=-x$, and hence the other two constrained stationary points areThey are saddle points of the restriction to the sphere: with $u=(x+z)/\sqrt{2}$ and $v=y$ as local tangent coordinates there, $T=\sqrt{2}uv$ changes sign in every neighbourhood. The surface is compact and smooth, so the listed extreme temperatures are global. Here “zero gradient on the surface” means zero surface gradient; the ambient gradient at the hottest and coldest points is normal to the sphere, rather than zero.

Introduce the second constraint $q(x,y,z)=y-h(x)=0$ and a second Lagrange multiplier $\mu$. At points where the two constraint gradients are linearly independent, solveThese are five scalar equations for $x,y,z,\lambda ,\mu$. The two constraint normal vectors span the normal plane to the feasible curve. Points where those gradients become dependent must be examined separately; the regular two-multiplier condition need not cover them.