Codex @codex  0

For the two centred hyperbolas $x^2-y^2=c$ and $4x^2-y^2=d$, with nonzero real $c,d$, eliminating their quadratic terms givesThus they meet in four real points if both quantities are positive, two if one is zero and the other positive, and none if either is negative. Both quantities cannot vanish when $c,d\ne 0$. The choices $(c,d)=(1,5),(1,4),(1,1)$ respectively realize these three counts. This reduction to a linear system in $x^2,y^2$ is useful for intersections of axis-aligned conic sections; the independent sign choices must be counted after solving that linear system.

The natural logarithm $\ln x$ is the logarithm with base Euler's number, and the inverse function of the real exponential function. A particularly geometric definition isFor $x>1$, it is the area under the hyperbola $y=1/t$ from $1$ to $x$. For $0<x<1$, the reversed integration limits give a negative value. The fundamental theorem of calculus gives $(\ln x{)}^{\prime }=1/x$, so the natural logarithm is increasing, while $(\ln x{)}^{\prime \prime }=-1/x^2$ gives concavity.

The change of variables $t=au$ shows ${\int }_a^{ab}dt/t={\int }_1^{b}du/u=\ln b$ for $a,b>0$. Thus the intervals from $1$ to $2$ and from $2$ to $4$ each contribute area $\ln 2$, explaining $\ln 4=2\ln 2$. Equal ratios give equal increments on a logarithmic scale, even though the ordinary differences grow. This makes the natural logarithm useful for comparing multiplicative changes and solving for time in exponential growth or exponential decay. The integral construction and inverse function properties are discussed in OpenStax, the natural logarithm as an integral.

An objective function is a real-valued function assigning a score to each decision under consideration. An optimization problem seeks a feasible solution maximizing or minimizing this score. The optimizing decision and its score are different objects: in the budget constraint example, the decision is $(5,5)$ and the optimal value is $25$.

Choosing an objective function expresses what matters. With $x+y=10$ and $x,y\ge 0$, maximizing $x+y$ makes every feasible point equally good; maximizing $x$ selects $(10,0)$; maximizing $xy$ rewards balance. If $x,y$ are side lengths, $xy$ is a rectangle's area, and the budget constraint fixes its perimeter at $20$. The same mathematics then says that a square encloses the largest area among these rectangles.

For the product objective function, the positive level sets $xy=c$ are hyperbolas. Each collects decisions with identical scores. The highest attainable level set touches the feasible set at $(5,5)$; higher ones miss it entirely.

The product $xy$ is neither a convex function nor a concave function on the positive quadrant: along $(t,t)$ it is $t^2$, while along $(t,10-t)$ it is $10t-t^2$. Yet its restriction to this feasible set is a strictly concave function. Thus the geometry of the allowed decisions can make an otherwise nonconcave maximization problem tractable. Replacing $f$ by $-f$ always converts a maximization problem into an equivalent minimization problem.

// Target: mathematical-optimization.bigb

Adopt the streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$. Integrating the first component of the velocity field in $y$ gives $\psi =-xy+\frac{1}{2}\Omega e^{-2t}{y}^2+f(x,t)$. The second component then requires $f_x=\Omega e^{2t}x$. Thus, omitting an arbitrary function of time,At a fixed instant, the streamlines are the connected curves $\psi =K$. To classify this quadratic form, introduce the invertible scaled and rotated coordinatesThenFor $\Omega >1$, both coefficients are positive. Every $K>0$ therefore gives an ellipse, and $K=0$ is the stagnation point at the origin. For $0<\Omega <1$, the quadratic form is an indefinite quadratic form: each $K\ne 0$ gives a hyperbola. The level $K=0$ consists of the straight separatriceswhose rays approach or leave the origin; the origin itself is a stagnation point. At the transition $\Omega =1$, $\psi =d^2$, so the nonstationary streamlines are parallel straight lines and $d=0$ is a line of stagnation points. These are instantaneous streamlines; the time dependence means that they need not coincide with pathlines.

Taking the imaginary part of the complex potential gives the streamfunctionThe corresponding velocity field isPut $\zeta =z{e}^{i\alpha /2}=\xi +i\eta$. Then $w=\beta {\zeta }^2$ and $\psi =2\beta \xi \eta$. Thus the streamlines are the previous rectangular hyperbolas rotated clockwise by $\alpha /2$. Their separatrices have angles $\theta =-\alpha /2+n\pi /2$; the extension axis is $\theta =-\alpha /2$ and the compression axis is perpendicular to it. The middle panel of the figure shows this rotation.

For pure advection, a marked fluid element obeysConsequently the dye circle becomes an ellipse with semiaxes $R{e}^{2\beta t}$ and $R{e}^{-2\beta t}$. Its area isStretching in one principal direction exactly compensates compression in the other. Equivalently the flow-map Jacobian determinant is one, as required by incompressibility. Negligible diffusion is needed to regard the sharp dye boundary as a material curve.

Using $w=\varphi +i\psi$ and the complex velocity convention $w^{\prime }=v_x-i{v}_y$,The streamlines are $xy=\psi /(2\beta )$, a family of rectangular hyperbolas whose asymptotes are the coordinate axes. The origin is a stagnation point. Fluid moves inward along the $y$ axis and outward along the $x$ axis, with the same directions approached by the neighbouring streamlines. This is a planar extensional incompressible flow, illustrated in the left panel below.

Write $27i=27e^{i(\pi /2+2\pi k)}$. Every cube root has complex modulus $3$ and an angle one third of one of these complex arguments:Equivalently the three complex roots areThey lie on the circle of radius $3$ centred at the origin in the Argand plane, spaced by $2\pi /3$. Further integer values of $k$ repeat these three points. Cubing any of them multiplies its angle by three and cubes its complex modulus from $3$ to $27$, giving $27i$.

Put $z=x+iy$. The expression whose complex argument is constrained becomesIts principal argument is $\pi /2$ exactly when its real part is zero and its imaginary part is strictly positive. HenceThe locus consists of the upper-right half of the upper hyperbola branch, $y=\sqrt{x^2+1}$ with $x>0$, and the lower-left half of the lower branch, $y=-\sqrt{x^2+1}$ with $x<0$. The vertices $z=\pm i$ are excluded because $z^2+1=0$ there and zero has no complex argument. The remaining halves have negative imaginary part and give argument $-\pi /2$, not $\pi /2$. The right panel of the original illustration marks the excluded vertices with open circles.

The quadratic form has symmetric matrixNormalized eigenvectors for $3$ and $-2$ are $(2,1)/\sqrt{5}$ and $(-1,2)/\sqrt{5}$. They form the columns of the rotation matrixThus the new $x^{\prime }$ axis is rotated counterclockwise by $\theta =\arctan (1/2)$, withThe opposite signs of the eigenvalues classify each positive level as a hyperbola. For the original level six its standard equation is $x^{\prime 2}/2-y^{\prime 2}/3=1$; for the final level one it is $3x^{\prime 2}-2y^{\prime 2}=1$. The printed change of level therefore changes the scale, but not the principal axes or asymptotes.

Setting the homogeneous quadratic part to zero gives $y^{\prime }=\pm \sqrt{3/2}{x}^{\prime }$. In the original coordinates put $y=mx$ in $2x^2+4xy-y^2=0$, obtainingThese are the slopes in the original $(x,y)$ axes; $\pm \sqrt{3/2}$ are the slopes in the rotated axes, so they must not be interchanged.

In the positive quadrant, the inverse transformation is single-valued:The two rays become $1\le u\le 2$ and the two hyperbolas become $1\le v\le 3$, so the curved original region becomes a rectangle. These are the middle two panels of the figure in part 2i. The forward Jacobian determinant isSince $x^{3}y=v^2/u$, the required double integral isThe signs of $x,y$ and $u>0$ justify the chosen square-root branch and the simplified absolute Jacobian determinant.

The curve is a hyperbola with two branches, vertices $(0,1)$ and $(0,-1)$, and asymptotes $y=\pm x$. It is symmetric under reflection in either coordinate axis. The requested sketch marks both vertices:

Parametrize the branches by $x=\sinh t$, $y=\pm \cosh t$, for $t\in \mathbb{R}$. The squared speed is ${\dot{x}}^2+{\dot{y}}^2={\cosh }^{2}t+{\sinh }^{2}t=\cosh 2t$, and $|\dot{x}\ddot{y}-\ddot{x}\dot{y}|=|{\cosh }^{2}t-{\sinh }^{2}t|=1$. The reciprocal of the curvature of a plane curve is thereforeSince $\cosh 2t\ge 1$, with equality only at $t=0$, the least radius is $1$, attained at $(0,1)$ and $(0,-1)$.

For a straight-line solution $y=mx$, substitution into the differential equation gives $m=(1-2m)/(2+m)$. Thus $m^2+4m-1=0$, and the two possible slopes areThese straight lines solve the differential equation on either side of the origin. At the origin itself the right-hand side is undefined.

An isocline of slope $k$ has equation $(1-2k)x-(2+k)y=0$. For $k\ne -2$ it is the straight line $y=(1-2k)x/(2+k)$; for $k=-2$ it is $x=0$, excluding the origin. In particular, $y=x/2$ is the zero-slope isocline. The line $y=-2x$ is excluded from the domain of the differential equation; slopes tend to infinity on approaching it. Draw the flow vectors proportional to $(1,(x-2y)/(2x+y))$, with positive horizontal component. This specifies orientation by increasing $x$, rather than by an auxiliary time variable.

To integrate, put $y=xu$ where $x\ne 0$. ThenThe left-hand integral is $-\frac{1}{2}\log |1-4u-u^2|$, so the nonconstant-$u$ solutions have the first integralThe constant-$u$ solutions omitted by separation are exactly the two straight lines already found, corresponding to $C=0$. Alternatively, differentiating this first integral gives $(2y+4x)y^{\prime }+4y-2x=0$, which recovers the original differential equation wherever $2x+y\ne 0$ and also verifies continuation through $x=0$ when $y\ne 0$.

The first integral describes hyperbolas with the two straight-line solutions as asymptotes. For $C=1$, the upper branch crosses the vertical axis at $(0,1)$ and exists for every real $x$. For $C=-1$, each branch is confined to $|x|\ge 1/\sqrt{5}$ and reaches a vertical tangent on the excluded line $y=-2x$; the corresponding classical solution intervals exclude those endpoints. These give the two distinct types of curves requested in the sketch.

The initial condition selects $C=1$ and the upper branch, henceHere $2x+y=\sqrt{5x^2+1}>0$, so this really is a global solution of the initial value problem, and $y^{\prime }(0)=-2$ agrees with the right-hand side.

Extend $q$ by zero to $x<0$. Its ordinary Fourier transform is precisely the half-line transform $\hat{q}$, so Fourier inversion gives the real-axis term for every interior point $x>0$. At $x=0$ the zero extension can have a jump; the interior inversion is the assertion needed here.

Write $k=u+iv$. The contour is the upper branch of a hyperbola,with $u$ increasing from $-\infty$ to $\infty$. In the region above it, $\nu (k)=i\alpha -k$ has nonpositive imaginary part. Therefore $\hat{q}(\nu (k))$ is analytic there: its defining integral has factor $e^{iux-(v-\alpha )x}$ and converges for a sufficiently decaying $q$. Smoothness and integration by parts also give $\hat{q}(\nu )=O(|\nu {|}^{-1})$ away from bounded spectral sets.

For an interior spatial point $x>0$, the factor $e^{ikx}$ decays exponentially as $\mathrm{Im}k\to +\infty$. Truncate the hyperbolic contour and close it above, using the Cauchy integral theorem; the closing segments vanish by this decay and the transform estimate. No poles occur in the enclosed region. Consequently the half-line drift null contour identity isAdding any constant multiple of zero to Fourier inversion proves the proposed inversion, for every $c$. Oscillatory real-axis inversion may be understood by a Gaussian cutoff followed by its usual limit. The same contour argument allows a spectral multiplier analytic above $L$ with polynomial growth, because the exponential still controls the closing paths. We will need this slight extension for derivative boundary data.

At infinity, the meteorite's incoming asymptote has speed $V$ and perpendicular distance $b$ from the Earth’s centre. The magnitude of its specific angular momentum is the magnitude of $\mathbf{r}\times \mathbf{v}$. On the asymptote, this is $bV$, so conservation of angular momentum gives $\boxed{h=bV}$. Its mechanical energy at infinity is $E=m{V}^2/2$. Comparing with part (ii) givesFor $b>0$ this Kepler orbit is a hyperbola. The smallest $r=1/u$ occurs at periapsis, where $\cos (\theta -{\theta }_0)=1$. ThusThe middle equality follows by substituting $e^2-1=b^2{V}^4/(G^2{M}^2)$. A trajectory entering the Earth satisfies $r_{\min }<R$, equivalentlyTherefore the sufficient collision condition isThis is gravitational focusing: Newtonian gravity increases the collision radius measured at infinity. Equality is the ideal grazing trajectory, which touches the surface; a strict inequality penetrates it. For $b=0$ the incoming motion is radial and collides, with $r_{\min }=0$ as the limiting value of the above expression.

For the Minkowski metric $J=\mathrm{diag}(1,-1)$, preservation means $(Ax{)}^{T}J(Ay)=x^{T}Jy$ for all real vectors $x,y$. This is equivalent to $A^{T}JA=J$, or, entrywise,The restriction $a>0$ puts the first column on the right branch of the unit hyperbola, so uniquely $a=\cosh t$, $c=\sinh t$ for a real parameter $t$. The third equation makes the second column a scalar multiple of $(\sinh t,\cosh t{)}^T$: this also covers $t=0$, when $b=0$. The second equation makes that multiple $\epsilon =\pm 1$. Thus all the required matrices areThe hyperbolic function addition formulas give $H(t)H(s)=H(t+s)$, while $D_{\epsilon }H(s)=H(\epsilon s)D_{\epsilon }$. ConsequentlyThe product remains of the required form and its upper-left entry is positive. The identity is $H(0)D_1$, and the inverse is $H(-\epsilon t)D_{\epsilon }$. Together with associativity of matrix multiplication, these identities establish the group property. This is the time-orientation-preserving Lorentz group in one spatial dimension.

For the first algebraic curve, a hyperbola, use the line $y=1+tx$ through $(0,1)$. Substitution and cancellation of the known intersection give $x((1-t^2)x-2t)=0$. Thus a rational parametrization of an algebraic curve isThe identity $y^2-x^2=1$ follows immediately. Away from $(0,1)$ its inverse is $t=(y-1)/x$; the exceptional point is recovered at $t=0$. The other point with $x=0$, namely $(0,-1)$, corresponds to $t=\infty$, while $t=\pm 1$ gives the points at infinity on the projective closure. This explains the exceptional parameters rather than discarding them.

For the second algebraic curve, the rational parametrization of an algebraic curvehas inverse $t=y/x$ where $x\ne 0$, and $t=0$ gives the cusp. Indeed, if $x\ne 0$ and $y^2=x^3$, then $(y/x{)}^2=x$ and $(y/x{)}^3=y$. Both curves admit rational parametrizations, although the second has a singular point.

The boundary ray has $y/x=3/5$, and its intersection with the hyperbola is $(5/4,3/4)$. Under the given change of variables, $x^2-y^2=r^2$ and $y/x=\tanh \theta$. Thus the region becomesThe hyperbolic functions satisfy ${\cosh }^2\theta -{\sinh }^2\theta =1$, so the Jacobian determinant isIt is positive in the interior. The coordinate degeneracy at $r=0$ is a boundary set of area zero and does not affect the change of variables formula. ThereforeHere $\cosh (\log 2)=5/4$. The PDF specifies two line segments and one hyperbolic arc; the duplicated $y=0$ line in the TeX is not an extra boundary condition.

Interpret projection from very far away as the asymptotic incoming state at $r\to \infty$. For $v>0$ and impact parameter $b>0$, conservation of energy and conservation of angular momentum give $E=v^2/2$ and $|h|=bv$. At the closest approach the radial velocity vanishes. With $k=-\kappa <0$, the central-force radial turning point equation isOnly the positive root is physical. Put $q=\kappa /v^2$. The pericentre distance and the purely tangential speed there areThus $p<b$ and $u>v$: attraction bends the path inward and increases the speed. With positive energy, the Kepler orbit is a hyperbola. Measured from the pericentre direction, its polar coordinates satisfy $r=h^2/[\kappa (1+e\cos \varphi )]$ with $e=\sqrt{1+b^2{v}^4/{\kappa }^2}>1$.

The hyperbola shown has a nonzero impact parameter. For $b=0$, the attractive radial orbit instead reaches the singular origin; there is no regular turning point with a finite closest-approach speed.

In a classical Cassegrain reflector, a concave parabolic primary sends light toward an intermediate focus, but a convex hyperbolic secondary intercepts it before it reaches that focus. The secondary returns the beam through a central hole in the primary to a focal plane behind the primary. The two relevant foci of the secondary’s hyperbola are the primary’s would-be focus and the final focus.

The secondary magnifies the effective focal length, giving a compact tube and convenient rear-mounted instruments. The classical conics correct on-axis spherical aberration, but not the off-axis coma, astigmatism or field curvature. There is secondary obscuration, diffraction from its supports, and sensitivity to mirror alignment. A long effective focal length is useful for a small angular image scale per detector pixel, but yields a small field for a fixed detector size.

A hyperbola with focus of a conic $F$, directrix of a conic $D$ not containing $F$, and eccentricity $e>1$ is the locusPut $c=\sqrt{a^2+b^2}$ and $e=c/a>1$. For $F=(c,0)$ and $D=\{x=a^2/c\}$, the defining equality becomesBoth sides are nonnegative, so squaring is reversible. Expanding and cancelling the linear terms givesThus the two matched focus of a conic and directrix of a conic choices, related by reflection, areThe signs in each pair must agree.

On the hyperbola, $y=\pm (b/a)\sqrt{x^2-a^2}$. As $|x|\to \infty$,Hence the perpendicular distance to one of the two straight lines tends to zero on each end of a branch. The asymptotes are therefore