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The printed point is not on the printed surface, so the requested tangent plane does not exist. At $(2,-2,9)$, the left side of the defining equation is $x^{2}y=-8$, whereas the right side is $4z{e}^{x+y}-11=25$. This is a direct counterexample to the premise that the stated point belongs to the surface.

To make the obstruction precise, set $G(x,y,z)=x^{2}y-4z{e}^{x+y}+11$. The specified surface is the level set $G=0$, but $G(2,-2,9)=-33$. Although $\nabla G(2,-2,9)=(-44,-32,-4)$, the resulting plane $11x+8y+z=15$ is tangent to the different level set $G=-33$ through that point. It is not an answer for the printed surface. A tangent plane requires the point to lie on the surface.

If the intended correction preserves $x=2,y=-2$, the surface equation instead forces $z=3/4$. At the corrected point $p_{\ast }=(2,-2,3/4)$,The corresponding conditional tangent plane isThis conditional calculation identifies a possible repair; it does not silently replace the source's coordinate $9$.

An integrating factor for a differential one-form is a nonzero function $\mu (x,y)$ for which $\mu (dy+fdx)$ is an exact differential, say $dF$. Then solutions lie on level sets $F=\text{constant}$. Locally, the compatibility condition is ${\partial }_x\mu ={\partial }_y(\mu f)$.

Here multiplication by $2y{e}^x$, which is nonzero for $y>0$, producesThe initial condition makes this first integral equal to $a^2$. Taking the positive branch yields the solutionon the connected interval containing zero where the radicand is positive. This domain restriction matters because the original differential equation is singular at $y=0$.

Completing the square gives $2x+x^2=(x+1{)}^2-1\ge -1$. As $y>0$, the differential equation consequently impliesAt $x=0$, $y^{\prime }=-a/2<0$. Thus, moving a little to the left from $y(0)=a\ge 1$, the curve rises above $a$. Throughout $y>1$ every segment of the direction field points downwards as $x$ increases, by the inequality above. Tracing backwards therefore makes $y$ rise further, so it cannot return to $y=1$. This graphical barrier argument proves $y>a\ge 1$ and $y^{\prime }<0$ for every $x<0$ on the solution. The explicit first integral also shows this branch extends throughout $x<0$: writing $t=-x>0$, the minimum of $e^t/t^2$ is $e^2/4>1$, so $a^2{e}^t-t^2>0$ for $a\ge 1$. Including the initial point, $y^{\prime }<0$ for all $x\le 0$.

For $a=1$ the curve is $y=\sqrt{e^{-x}-x^2}$ on $(-\infty ,b)$, where the strictly increasing function $x^2{e}^x$ has the unique positive root $x^2{e}^x=1$ at $b\simeq 0.703467$. The curve decreases through $(0,1)$ with slope $-1/2$ and reaches $(b,0)$ only as a limiting endpoint. Its derivative isAs $x\to -\infty$, $y\sim e^{-x/2}$ and $y^{\prime }\sim -\frac{1}{2}{e}^{-x/2}\to -\infty$. As $x\uparrow b$, the numerator tends to $-b^2-2b<0$ and the denominator to zero from above, so again $y^{\prime }\to -\infty$. More precisely, $y\sim \sqrt{b(b+2)(b-x)}$ at the endpoint. Both ends have the requested unbounded negative slope.

Introduce $H=x^2-y^2+y^4/2$. Along a solution,At $\alpha =0$, $H$ is a first integral. The identity $H+1/2=x^2+(y^2-1{)}^2/2$ shows its level sets are bounded: neither $x$ nor $y$ can escape to infinity on a fixed level. Consequently trajectories remain bounded as $t\to \infty$, including those starting far from the origin. This bound also prevents finite-time escape; no explicit solution is needed.

The equilibrium points are $(0,0)$ and $(0,\pm 1)$. The Jacobian matrix at $(0,y_{\ast })$ isAt the origin its eigenvalues are $(\alpha \pm \sqrt{{\alpha }^2+4})/2$, with opposite signs, so the origin is a saddle point in all three cases. An eigenvector for $\lambda$ lies on $y=-x/\lambda$; the stable direction has positive slope and the unstable direction negative slope. At $(0,\pm 1)$ the eigenvalues are $(\alpha \pm \sqrt{{\alpha }^2-8})/2$.

When $\alpha =0$, these latter eigenvalues are $\pm i\sqrt{2}$. The nonlinear classification follows from the strict minima of $H$ at the two points: nearby level sets are closed curves without other equilibria, and the nonvanishing vector field makes them periodic orbits. Thus the two points are center equilibria. For $-1/2<H<0$, there are two separate closed ovals, one around each center equilibrium. The level $H=0$ consists of the saddle point and two homoclinic orbits forming a figure eight, explicitlyFor $H>0$, a single outer closed orbit encloses all three equilibrium points. Motion is clockwise, as the vector points downwards wherever $x>0$. The upper homoclinic orbit leaves the origin into $x<0,y>0$ and returns through $x>0,y>0$; the lower one has the reflected orientation.

When $\alpha =0.1$, the two nonzero equilibrium points have eigenvalues $0.05\pm i\sqrt{7.99}/2$ and are unstable foci. Nearby paths spiral outward clockwise. The first integral becomes a strictly increasing energy along every nonconstant trajectory: although $\dot{H}$ can momentarily vanish when $x=0$, it cannot vanish on an interval unless the solution is an equilibrium. Thus there are no periodic orbits or homoclinic orbits. The two stable separatrices of the saddle point, traced backwards, spiral towards the corresponding unstable foci. Its unstable separatrices move onto positive energy and escape outwards. All other nonequilibrium trajectories outside the stable manifold are unbounded forwards: a bounded forward limit set would have to lie in $x=0$, whose only invariant points are the three equilibria; the two unstable foci cannot attract a nonconstant trajectory, and convergence to the saddle occurs only on its stable manifold. Escape is not finite-time blow-up, since $\dot{H}\le 2\alpha (H+1/2)$ bounds the energy on every finite interval.

When $\alpha =-0.1$, the eigenvalues at the nonzero points are $-0.05\pm i\sqrt{7.99}/2$, so they are stable foci. The energy decreases, keeping all forward trajectories in compact sublevel sets. A bounded limit set must lie in the invariant part of $x=0$, so trajectories converge to equilibria: generically to one of the two stable foci, exceptionally to the saddle along its stable manifold. The saddle's unstable separatrices spiral into the two stable foci; its stable manifold separates their attraction basins. Again, energy monotonicity excludes periodic orbits and homoclinic orbits. These three cases give the quartic double-well phase portrait with linear damping below.

Here the autonomous planar system is $\dot{x}=v$, $\dot{v}=-\cos x$. Its equilibrium points areThe Jacobian matrix isFor even $n$, the eigenvalues are $1,-1$, so these points are unstable saddle equilibria. Writing $\xi =x-x_{\ast }$, their local unstable and stable directions are $v=\xi$ and $v=-\xi$ respectively. For odd $n$, the eigenvalues are $i,-i$.

To classify the latter points nonlinearly, differentiate the first integralAt $x_{\ast }=3\pi /2+2m\pi$ this energy has a strict local minimum. Its nearby level sets are closed curves, proving that the odd-$n$ points are stable center equilibria, with no attraction to the center. The phase portrait consists of closed oscillatory orbits for $-1<H<1$, heteroclinic orbits at $H=1$ joining consecutive saddle equilibria, and open rotating orbits for $H>1$. The separating curves haveAbove the horizontal axis $\dot{x}=v>0$, so flow is rightward; below it flow is leftward. The closed orbits are clockwise, and these same signs orient every separating and rotating orbit.

The image is a ring torus: a circle of radius $b$ is rotated about the vertical axis at distance $a$ from that axis. Put $r=\sqrt{x^2+y^2}$. Since $a>b$, every image point has $r>0$, and the image is preciselyConversely, on this level set the pairs $((r-a)/b,z/b)$ and $(x/r,y/r)$ are unit vectors, so choosing their angles gives a preimage.

The specified angular rectangle removes two closed circles from the torus. More explicitly its image isHere $C_u$ is the outer equatorial circle and $C_v$ is a meridian. The given openness property makes $U$ open in $T$. On $U$, the angles of $(r-a)+iz$ and $x+iy$ are uniquely chosen in $(0,2\pi )$; their continuous argument branches give the inverse $(x,y,z)\mapsto (u,v)$. Thus the restriction is a homeomorphism onto $U$, and the inverse is locally smooth.

For the immersion condition, the two derivatives have inner productsThey are linearly independent everywhere, since $a+b\cos u\ge a-b>0$. Hence this is a smooth embedded surface parametrization.

The angular cuts do not represent singularities of the embedded torus of revolution. To give a direct proof covering all points, define $F=(\sqrt{x^2+y^2}-a{)}^2+z^2-b^2$ on the open region $r>0$. On $T$,The implicit function theorem therefore describes $T$ near every point as a smooth graph of two coordinates. The whole torus is a smooth embedded surface.

A nonnegative sublinear operator satisfies $S(f+g)\le S(f)+S(g)$ and $S(cf)=|c|S(f)$ almost everywhere. It has weak type (1,1) if there is a finite constant $C$, independent of the function and threshold, such thatFor a signed or complex output use $|S(f)|$ in the level set. The bound controls the distribution of large values; it does not assert that $S(f)$ is integrable.

The principal real arctangent is defined here on $x\ne 0$, so work on either of the two open half-planes. Put $r=\sqrt{x^2+y^2}$. Direct differentiation givesHence ${\varphi }_{xx}+{\varphi }_{yy}=0$: $\varphi$ is harmonic on its domain. The Cauchy-Riemann equations for $f=\varphi +i\psi$ requireIntegrating these partial derivatives gives the harmonic conjugateThe constants may be chosen independently on the two components. The harmonic angle and logarithmic radius representation supplies the holomorphic functionwhere $\mathrm{Log}$ is the branch of the complex logarithm on the right half-plane. On the left half-plane it is the angle of $-z$, rather than the principal angle of $z$, that equals $\arctan (y/x)$. This distinguishes the literal arctangent from a globally defined argument; no single-valued complex logarithm exists on the whole punctured plane.

With $C_0=0$, the level sets $\varphi =C$, for $-\pi /2<C<\pi /2$, are the two opposite rays of $y=x\tan C$ with the origin removed. On each half-plane only one ray remains. The level sets $\psi =K$ are circles $r=e^{-K}$, with their two points on $x=0$ excluded from the literal domain. The rays and circles meet at right angles, as expected for harmonic conjugates.

The derivative matrixis symmetric in $j,k$. Contracting it with the antisymmetric Levi-Civita symbol gives the curlDirect differentiation of the scalar potential givesso $\mathbf{F}=\nabla \varphi$. To identify the level sets, use the cross product identityIf $\mathbf{\omega }\ne 0$, write $\mathbf{x}={\mathbf{x}}_{\parallel }+{\mathbf{x}}_{\perp }$ relative to its direction. Then $\varphi =-|\mathbf{\omega }{|}^2|{\mathbf{x}}_{\perp }{|}^2/2$. Thus each negative level $\varphi =c$ is a circular cylinder with axis $\mathbb{R}\mathbf{\omega }$ and radius $\sqrt{-2c}/|\mathbf{\omega }|$. The zero level is the axis itself; positive levels are empty. If $\mathbf{\omega }=0$, the scalar potential is identically zero, so its zero level is all of space and every other level is empty.

For $k=0$, the first integralhas $\dot{E}=0$. The equilibrium points are $(0,0)$ and $(1,0)$. The linearization has eigenvalues $\pm i$ at the origin and $\pm 1$ at $(1,0)$. The origin is a center equilibrium: the sufficiently small positive level sets of $E$ are closed curves, since its Hessian matrix there is positive-definite. The other equilibrium is a saddle equilibrium.

At the saddle, $E=1/6$, and the separatrix satisfiesIts two branches with $-1/2\le x\le 1$ form a homoclinic orbit through the saddle; its remaining branches extend to $x>1$. Inside the loop, the closed trajectories run clockwise: above the horizontal axis $\dot{x}=y>0$. The saddle directions have slopes $+1$ and $-1$. Outside the loop there are unbounded trajectories. The left panel shows the closed orbits and all saddle branches.

The gradient components areIf $x=0$ or $y=0$, their simultaneous zeros give $(0,0),(0,1),(1,0)$. If $xy\ne 0$, solving $2x+y=1$, $x+2y=1$ gives $(1/3,1/3)$. These are all the critical points. The Hessian matrix isAt each of $(0,0),(1,0),(0,1)$ its determinant is $-1$, so its two eigenvalues have opposite signs. All three are saddle points of a scalar function. At $(1/3,1/3)$,This point is a strict local maximum, of value $\boxed{1/27}$.

The zero level set is exactlyThese three lines form a triangle containing the maximum. Inside it $f>0$; near the maximum positive level sets are nested closed curves, with their leading shapes elliptical because the Hessian matrix is negative definite. At each vertex the zero lines cross, separating alternating signs, and nearby nonzero level sets have hyperbolic branches. For example, the leading quadratic terms are $xy$ at $(0,0)$, $-\eta (\xi +\eta )$ for $\xi =x-1,\eta =y$ at $(1,0)$, and $-\xi (\xi +\eta )$ for $\xi =x,\eta =y-1$ at $(0,1)$.

Work locally where the needed derivatives of the level set can be represented by the implicit function theorem. Differentiating while holding the indicated coordinate fixed givesMultiplying the three implicit differentiation formulas provesThis cyclic partial derivative identity is asserted where all three finite coordinate derivatives exist, in particular when $f_x{f}_y{f}_z\ne 0$. It is not a globally defined product at points where one of the necessary coordinate parametrizations is singular.

In the $(x,y)$-plane take $z=0$; holding another fixed $z$ merely shifts all contour labels by $z^2$. The level sets satisfyFor $a=1$, the origin is a local minimum and $(0,\pm 1)$ are saddle points of a scalar function at contour value $1$. For $0<c<1$, a closed contour around the origin is accompanied by unbounded outer components. At $c=1$ the contour satisfies $x=\pm (y^2-1)$ and crosses at the two saddle points of a scalar function. For $c>1$, two unbounded components remain; negative contours have only outer components.

For $a=-1$, the origin is a saddle point of a scalar function, and the zero contour is $x=\pm y\sqrt{y^2+2}$, with tangent lines $x=\pm \sqrt{2}y$. Positive contours form two components extending on opposite sides of the $y$-axis; negative contours form upper and lower components. These level sets are illustrated below.

Here both the divergence and vorticity vanish:The flow is incompressible and irrotational. With $\mathbf{u}=\nabla \varphi$ and $(u_x,u_y)=({\psi }_y,-{\psi }_x)$,The streamlines are the level sets $x^2-y^2=\text{constant}$. To determine the arrows, put $r=x+y$ and $s=x-y$. A particle satisfies $\dot{r}=-2r$, $\dot{s}=2s$. Thus the separatrix $y=x$ approaches the stagnation point at the origin, whereas $y=-x$ leaves it; the other streamlines are hyperbolas following those asymptotes. The origin itself is a stationary particle.

Write $\mathbf{g}=-g{\mathbf{e}}_z$. The vertical component of the balance gives hydrostatic pressure,Its horizontal gradient is ${\nabla }_{h}p=\rho g{\nabla }_{h}h$. The horizontal balance is consequentlyThus $\mathbf{u}\cdot {\nabla }_{h}h=0$: the velocity is tangent to every level set of $h$, so those contours are streamlines of the geostrophic flow.

Across a vertical section joining the contours $h_0$ and $h_0+\Delta h$, an element of horizontal distance normal to the contours is $dn=dh/|{\nabla }_{h}h|$, while $|\mathbf{u}|=(g/f)|{\nabla }_{h}h|$. The volumetric flow rate is thereforeFor dimensional analysis, $[g]=L{T}^{-2}$, $[f]=T^{-1}$ and $[h_0]=[\Delta h]=L$, so the result has dimension $L^3{T}^{-1}$, exactly that of a volume flux.

Along a trajectory, the chain rule givesThus $\boxed{H=x^2+y^2-y^4}$ is a first integral, and every trajectory lies on one of its level sets.

For $a\in U$, the function $f:U\to {\mathbb{R}}^n$ is differentiable at $a$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe linear map is unique and is the Fréchet derivative $D{f}_a=A$. The function is continuously differentiable at $a$ if it is differentiable on a neighbourhood of $a$ and the map $x\mapsto D{f}_x$, with values in the space of linear maps equipped with the operator norm, is continuous at $a$.

If $T:{\mathbb{R}}^m\to {\mathbb{R}}^n$ is linear, thenConsequently $D{T}_a=T$ at every $a$. The derivative is a constant function of $a$, hence is continuous, so every linear map is continuously differentiable everywhere.

The mean value inequality says that if the line segment $[x,y]$ lies in $U$ andthenTo prove it, put $v=f(y)-f(x)$. The claim is immediate if $v=0$. Otherwise set $e=v/\Vert v\Vert$ and apply the one-dimensional mean value theorem to the real-valued functionFor some $c\in (0,1)$, the chain rule givesas required.

Now suppose that $U$ is open and connected and that $D{f}_a=0$ for every $a\in U$. Every point has an open ball contained in $U$. The mean value inequality with $M=0$ shows that $f$ is constant on each such ball, so $f$ is locally constant. Fix $a_0\in U$. The level setis nonempty and open in $U$; its complement is also open because $f$ is locally constant. Since $U$ is connected, $E=U$. This proves the zero derivative on a connected open set result: $f$ is constant.

The inverse function theorem states that if $f:U\to {\mathbb{R}}^m$ is continuously differentiable, $a\in U$, and $D{f}_a$ is an invertible linear map, then there are open neighbourhoods $V$ of $a$ and $W$ of $f(a)$ such that $f{|}_V:V\to W$ is a bijection whose inverse is continuously differentiable.

For the curve in the question, define the continuously differentiable functionThenThe implicit function theorem, which follows from the inverse function theorem applied to $(x,y)\mapsto (x,F(x,y))$, therefore gives an open interval $I$ containing $0$, an open neighbourhood $U_0$ of $(0,0)$, and a continuously differentiable, hence continuous, function $g:I\to \mathbb{R}$ such that

At a regular point of the optimization constraint, meaning $\nabla g\ne 0$, the necessary Lagrange multiplier condition isIndeed, every allowed tangent vector $t$ to the level set $g=0$ satisfies $\nabla g\cdot t=0$. At a constrained extremum, the directional derivative $\nabla f\cdot t$ must vanish too. Thus both gradients are normal to the same tangent line and are parallel. This includes $\nabla f=0$. At a singular point with $\nabla g=0$, the displayed condition need not be necessary; those points require a separate examination.

The ellipsoid is the level set $F(x,y,z)=2x^2+y^2+z^2=1$. Its gradient is $(4x,2y,2z)$, which never vanishes on the ellipsoid. Half this gradient gives the outward normal vectorThis normal vector remains valid at $\theta =0,\pi$, where the given parametrization is singular.

The linearization at rest at $r=0$ isFor ${\omega }^2<g/a$, small disturbances oscillate. To verify nonlinear stability, multiply the full equation of motion by $\dot{r}/a^2$ and integrate to obtain the first integralFor ${\omega }^2<g/a$, $J$ is positive away from the equilibrium point in a sufficiently small neighborhood of $(r,\dot{r})=(0,0)$; its small level sets confine the motion. ThusAt equality, every resting location on the wire is an equilibrium point, so a displacement at zero relative velocity simply remains there: the equilibrium is neutral with respect to resting displacements. It is unstable with respect to general perturbations in phase space, however. An arbitrarily small outward initial velocity $v_0>0$ gives $\dot{r}=v_0/\sqrt{1+r^2/a^2}>0$ from the first integral, eventually taking the bead outside any prescribed small neighborhood. Thus strict phase space stability holds only for ${\omega }^2<g/a$.

For this homogeneous first-order differential equation, set $v=y/x$, with $x\ne 0$. The product rule gives $y^{\prime }=v+x{v}^{\prime }$, and substitution simplifies the equation toThe useful further substitution is $w=e^{-v}-v^2$. Its chain rule derivative satisfiesThus $(x^{2}w{)}^{\prime }=0$. Replacing $v$ and $w$ gives the implicit solutionNo division by $w$ was needed, so the level $C=0$ is included. A direct verification also identifies the integrating factor for a differential one-form:Setting this exact differential to zero recovers precisely the original equation after multiplication by the nonzero factor $e^{y/x}$. The coefficient $1+2v{e}^v$ never vanishes for real $v$, since $v{e}^v\ge -1/e$ implies $1+2v{e}^v\ge 1-2/e>0$. Hence these level sets give local function graphs $y(x)$ everywhere with $x\ne 0$.