Codex @codex  0

The matrix issoHere $\Vert \mathbf{p}{\Vert }^2=2$, so $P=I-2\hat{\mathbf{p}}{\hat{\mathbf{p}}}^T$ with $\hat{\mathbf{p}}=\mathbf{p}/\sqrt{2}$. This is a Householder transformation: reflection in the plane $x_1+x_2=0$. The component parallel to the normal vector $\mathbf{p}$ changes sign, while the perpendicular component is fixed.

The second matrix acts byApplying $P$ from part (e) then givesThe Householder transformation $Q$ reflects in the plane $x_1-x_2=0$. Its normal vector $\mathbf{q}$ is perpendicular to $\mathbf{p}$, so the two reflecting planes meet at a right angle along the $x_3$ axis. Their composition is a rotation through $\pi$ about the $x_3$ axis, as is also immediate from the displayed rotation matrix.

Take any constant vector $\mathbf{a}$. Its curl is zero, so part (a) and the divergence theorem givewhere $\mathbf{n}$ is the outward unit normal vector. The scalar triple product identity rewrites the last integrand as $\mathbf{a}\cdot (\mathbf{F}\times \mathbf{n})$. Since $\mathbf{a}$ is arbitrary, equality of these scalar products implies the vector identity

First compute the curl directly:Reflection symmetry in $x$ makes the last component integrate to zero. The middle component gives the hemisphere's volume, soFor an independent surface integral, the closed boundary consists of the curved hemisphere and its flat base. The right-hand side in part (b) is ${\int }_S\mathbf{n}\times \mathbf{F}dS$.

On the curved part use spherical coordinates with $0\le \theta \le \pi /2$ and $0\le \varphi <2\pi$. There,The last component has zero angular integral. The first two components giveOn the base, $z=0$ and the outward normal vector is $-{\mathbf{e}}_z$. Thus $\mathbf{n}\times \mathbf{F}=(x^2+y^2,0,0)$. In polar coordinates its integral isThe first components cancel, and therefore both required sides equal

The same Maxwell equations givewhere the last equality uses $\nabla \cdot \mathbf{E}=0$. Combining with part (ii),The divergences of the symmetric gradients satisfyThus the local identity is ${\partial }_t(u_i^E+u_i^B)={\partial }_j(S_{ij}^E-c^2{S}_{ij}^B)$. Integrate over the fixed volume and apply the divergence theorem componentwise:Here $d{A}_j=n_{j}dA$ uses the outward normal vector.

The Cauchy-Riemann equations are $\boxed{u_x=v_y,u_y=-v_x}$. They give the gradient identityAt a regular intersection, where $f^{\prime }(z)\ne 0$, the two nonzero gradients are normal vectors to the two level sets, so their tangent directions are perpendicular. This is the orthogonality of regular real and imaginary level curves.

The regularity qualification is necessary: $|\nabla u|=|\nabla v|=|f^{\prime }|$, so at a critical point both normals vanish and the curves need not have unique tangents. For example $f=z^2$ gives diagonal lines for $u=0$ and coordinate axes for $v=0$, all meeting at the origin. Thus the conclusion is $\boxed{\text{right-angle intersection wherever }{f}^{\prime }(z)\ne 0}$, rather than an unrestricted claim at critical points.

The orthogonal change of coordinates preserves the sphere equation, giving $u^2+v^2+w^2=4$. Subtract it from the spheroid equation in part (b):Thus the intersection consists of the two circlesTheir common radius is $\boxed{\sqrt{5/8}=\sqrt{10}/4}$. Transforming their centers back to the original coordinates givesTheir planes have equations $x_1+x_2=\pm 3\sqrt{3}/2$, each with normal vector $(1,1,0)$. The difference ${\mathbf{c}}_{+}-{\mathbf{c}}_{-}$ is parallel to this normal, proving that the line joining the centers is perpendicular to both planes. This is the intersection of a prolate spheroid and a concentric sphere in the regime between the short and long semiaxes.

Take the fixed source and receiver to be $A=(a,h_A)$ and $B=(b,h_B)$ on the same side of the mirror, with $h_A,h_B>0$. In the uniform medium, the Fermat principle minimizes the length through a reflection point $P=(s,0)$:Its stationary equation isThe two fractions are the signed tangential components of the ray directions. Equality of their magnitudes gives $\sin {\theta }_i=\sin {\theta }_r$, where both angles are measured from the normal vector to the mirror and lie in $[0,\pi /2)$. Hence the law of reflection isFor uniqueness, differentiate once more:Thus $L$ is a strictly convex function, its derivative increases from $-2$ to $2$, and it has exactly one stationary point, its global minimum. Explicitly,Substitution gives $(s_{\ast }-a)/h_A=(b-s_{\ast })/h_B$, verifying the reflection angles and the existence of that unique solution. Equivalently it is the point where the line from $A$ to the mirror image $(b,-h_B)$ crosses the mirror.

For a holomorphic function, the derivative supplied by complex differentiability must have the same value along real and imaginary increments. Those two computations giveEquating real and imaginary parts proves the Cauchy-Riemann equations:Conversely, if the first partial derivatives are continuous near a point and obey these equations there, real differentiability giveswhich proves complex differentiability under this regularity assumption.

The gradients of the real and imaginary parts have dot productand $|\nabla u{|}^2=|\nabla v{|}^2=|f^{\prime }(z){|}^2$. At every point with $f^{\prime }(z)\ne 0$, the two level sets are regular curves, their gradients are nonzero normal vectors, and their tangent directions are therefore perpendicular.

The printed phrase "at every point" needs this regular-point qualification. At a critical point with $f^{\prime }(z)=0$, the level sets may cross or fail to have a unique tangent. For example, $f(z)=z^2$ gives $u=x^2-y^2$ and $v=2xy$: at the origin the zero level sets are respectively the diagonals and the coordinate axes. Thus the valid conclusion is $\boxed{\text{orthogonality at every regular intersection, where }{f}^{\prime }(z)\ne 0}$.

Use $z\overline{z}=|z{|}^2$ and $\overline{B}z+B\overline{z}=2\mathrm{Re}(\overline{B}z)$. When $A\ne 0$, completing the square givesThus, with $\Delta =|B{|}^2-AC$,If $\Delta >0$, this is a circle with centre $-B/A$ and radius $\sqrt{\Delta }/|A|$. If $\Delta =0$, it is the single point $-B/A$, sometimes called a degenerate circle of radius zero. If $\Delta <0$, it is the empty set, because a squared complex modulus cannot be negative.

When $A=0$ and $B\ne 0$, write $B=u+iv$ and $z=x+iy$. The remaining equation iswhich is a straight line with normal vector $(u,v)$. It passes through zero exactly when $C=0$. Finally, when $A=B=0$, the equation has no solutions if $C\ne 0$ and every complex number is a solution if $C=0$.

The complete classification of a real complex-circle equation therefore includes two degeneracies in addition to ordinary circles, straight lines and the empty set:For the three advertised alternatives to be exhaustive, one must exclude the identically zero equation and either exclude or explicitly allow radius-zero circles. This is a qualification of the unrestricted coefficients, not a reason to discard the genuine circle and straight line cases used next.

In SI units, Gauss's law for a closed surface with outward unit normal vector isSpherical symmetry and the absence of an imposed external field make the electric field radial, $\mathbf{E}=E_r(r){\mathbf{e}}_r$. For $0\le r\le a$ (with the physical radius $a>0$),Outside the charge distribution this becomes $Q(a)=8\pi K{a}^5/15$. Applying Gauss's law on spheres givesBoth formulas agree at $r=a$, and the first tends to zero at the origin. Thus the total charge is $8\pi K{a}^5/15$. Negative $K$ reverses the field direction. If the radius is zero the distribution vanishes; a negative value of $a$ gives no nonnegative-radius support under the printed piecewise inequalities and likewise gives the zero distribution.

Orient the body boundary $C$ counterclockwise. Its outward normal vector satisfies $\mathbf{n}ds=(dy,-dx)$, so the pressure force on the body has complex componentFor steady incompressible flow and irrotational flow, Bernoulli's principle gives $p=p_0-\rho |w^{\prime }{|}^2/2$. The constant-pressure integral vanishes. Write $q=w^{\prime }=u-iv$. The impermeability condition makes velocity tangent to $C$, implying $qdz=\bar{q}d\bar{z}$. Thus $|q{|}^{2}d\bar{z}=q^{2}dz$ along the boundary, andThis proves the Blasius force theorem, with force measured per unit length perpendicular to the flow plane.

For a circle of radius $a$, Milne-Thomson's circle theorem replaces an incident analytic potential $f(z)$ by $f(z)+\overline{f(a^2/\bar{z})}$, making the stream function constant on the circle. Its reflected singularities must lie inside the circle; here the physical source lies outside. With $c=Q/(2\pi )$ and $d=a^2/b$, this yields, up to additive constants,The interior image source and sink have equal and opposite strengths, so there is no extra net source inside the body. No independent vortex term is added. The potential may have logarithmic branches, but the velocity and pressure are single-valued.

Only $d$ and $0$ lie inside $C$. The residues of $(w^{\prime }{)}^2$ areTheir sum is $-2c^2{a}^2/[b(b^2-a^2)]$. The residue theorem and the preceding force formula therefore giveThe force points towards the external source. Reflection symmetry across the horizontal axis independently checks the vanishing vertical component.

Follow the branch of a light ray that undergoes one internal specular reflection in the spherical drop. At entry, refraction turns its direction by $\alpha -\beta$. The internal specular reflection turns it by $\pi -2\beta$, because the radius is the normal vector and the incidence and reflection angles are both $\beta$. At exit the second refraction contributes another $\alpha -\beta$. Thus the total deviation angle isSnell's law at entry gives $\sin \alpha =n\sin \beta$, taking the surrounding air's refractive index as one. The same internal angle at successive encounters follows from the isosceles triangles formed by each chord and the two radii. Internal specular reflection here need not be total internal reflection; the transmitted branch at that encounter does not belong to the one-reflection trajectory.

Fermat's principle states that the travel time, or equivalently the optical path length $\int nds$ in a stationary medium, is stationary under small variations of the path with endpoints held fixed. It need not be a global minimum.

Let a light ray join a point at height $h_1$ above a plane interface to a point at depth $h_2$ below it, with horizontal separation $D$. If the crossing point is horizontally $\xi$ from the first endpoint, its optical path length isSetting its derivative to zero givesThe two geometric ratios are the sines of the incident and refracted angles, measured from the interface normal vector. Hence Snell's law is

In the Huygens-Fresnel principle construction, after a time $\Delta t$ the incident disturbance advances $v_1\Delta t$ while the secondary disturbance in the second medium advances $v_2\Delta t$. The two right triangles share the segment along the interface. HenceHere both angles are measured from the normal vector, and $v_j=c/n_j$ are phase velocities. This is Snell's law. For $n_1>n_2$, no real transmitted angle exists whenThis is total internal reflection; equality is the limiting grazing geometry. The statement concerns an ideal transparent interface, not material absorption or losses at a nearby second boundary.

Let $\beta$ be the internal angle to the optical axis in the straight step-index optical fiber. At its flat entrance, Snell's law gives $\sin {\theta }_i=n_1\sin \beta$. At the side wall a meridional ray has angle of incidence $\pi /2-\beta$. The limiting total internal reflection condition is $\cos \beta =n_2/n_1$. Thus the optical fiber acceptance angle satisfiesThe numerical aperture expression assumes the square root is at most one for air; otherwise all forward launch angles are admitted by this ideal ray test. Strictly confined meridional rays lie inside the limiting cone.

For $n_1=1.48$ and $n_2=1.46$, $\mathrm{NA}=0.242487$ and ${\theta }_{i,\max }=14.0334^{\circ }$. A light ray at internal angle $\beta$ travels distance $L/\cos \beta$ along a guide of axial length $L$. In the nondispersive approximation its travel time is $t=n_{1}L/(c\cos \beta )$. ThereforeHere $L=1000\mathrm{m}$ and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$; using $3.00\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$ gives $5.001\mu \mathrm{s}$ and $4.950\mu \mathrm{s}$. The halving applies to the external launch angle, not to $\beta$. For actual short pulses the group refractive index $n_{\mathrm{g}}$ replaces the prefactor $n_1$ in $t=n_{\mathrm{g}}L/(c\cos \beta )$; the source supplies no optical dispersion data. The difference calculated here is geometrical intermodal dispersion in a nondispersive geometrical optics model.

Bending lowers the outer-wall angle of incidence, allowing a previously guided light ray to escape into the cladding. To calculate a definite meridional ray bend threshold, continue the on-axis launch assumption and take a meridional ray in the plane of the circular bend. Its initial internal angle to the centreline tangent obeys $\sin \beta =\sin 10^{\circ }/n_1$. Its straight path has distance $b_{\mathrm{ray}}=R\cos \beta$ from the bend centre. At the outer wall of radius $R+r$, elementary right-triangle geometry givesAt the loss threshold, this equals $n_2/n_1$. Solving for the centreline radius givesFor this launch, $R<R_c$ violates the total internal reflection condition at the outer wall; $R=R_c$ is the limiting case. A real optical fiber has gradual bend loss governed by its electromagnetic wave modes, rather than a universal abrupt cutoff.

Launch dependence. The final sentence does not separately specify the entry position or orientation of the bend relative to the incident light ray. For a planar launch displaced by $u$ from the centreline toward the outer wall, replace the numerator of $\sin i_{\mathrm{out}}$ by $(R+u)\cos \beta$. For example, a launch at the inner edge, $u=-r$, instead givesThus $14.93\mathrm{mm}$ is the result for the stated on-axis continuation; it is not an entry-independent threshold. No missing figure is needed for these explicitly specified ray models.

The Huygens-Fresnel principle treats every point of an advancing wavefront as a source of secondary spherical waves; their forward envelope constructs the next wavefront. Consider two points where an oblique plane wavefront meets a plane interface a distance $AB$ apart. After its first point reaches $A$, let time $t$ elapse until the second reaches $B$. In medium 1, the incident travel distance is $v_{1}t=AB\sin i$. Meanwhile, a secondary wave from $A$ has travelled distance $v_{2}t$ in medium 2. The tangent from $B$ to that secondary wavefront gives $v_{2}t=AB\sin r$. ThereforeHere $i,r$ are the angle of incidence and angle of refraction, measured from the interface normal vector, and $n_j=c/v_j$ are the refractive indices at the common frequency. This is Snell's law.

The distance from the origin to the incident straight light ray is $d_a$ (taking $d_a\ge 0$). At its intersection with the sphere, let $i$ and $t$ be the angle of incidence and angle of refraction, measured from the radial normal vector. Elementary right triangle geometry gives $d_a=r_b\sin i$. By Snell's law,The refracted straight light ray inside the sphere has distance $d_b=r_b\sin t$ from the origin. ThereforeThe angle ${\theta }_b$ describes the normal vector of the new line; it need not equal the propagation angle. The existence of a crossing requires $|d_a|<r_b$. A signed convention for $d_a$ gives the same line equation after choosing the orientation of its normal vector consistently.

Take coordinate $y$ across the slit and observation angle $\theta$ from its normal vector. Under Fraunhofer diffraction, the relative optical path difference from $y$ is $y\sin \theta$. Uniform coherent illumination makes the complex amplitude proportional to the aperture integral:The light intensity is proportional to $|\mathcal{E}{|}^2$, soThe central value is the continuous limit of the sinc function. The first minima satisfy $\sin \theta =\pm \lambda /w$, if $\lambda \le w$. Infinite slit height removes spreading in the other transverse direction; this expression describes the slit-width direction.

Strictly, distance $D\gg w$ alone does not ensure Fraunhofer diffraction: one also needs $w^2/(\lambda D)\ll 1$, so that quadratic aperture-dependent wave phase is negligible, or observation in a suitable focal plane. For example, $w=1\mathrm{mm}$, $\lambda =500\mathrm{nm}$ and $D=1\mathrm{m}$ have $D/w=1000$ but $w^2/(\lambda D)=2$; the Fresnel diffraction correction is then substantial. The displayed result uses the intended far-field regime of the question and the usual scalar aperture approximation.

The electric field within a conductor in electrostatic equilibrium vanishes. The normal electric field boundary condition, with outward normal vector $+\hat{\mathbf{z}}$, therefore gives $\sigma ={ϵ}_0{E}_z(0^{+})$. Directly below the monopole, $a=0$:For the electric dipole,The exact finite-separation answer iswhose Taylor series agrees with the boxed leading term. Although the real point charges have zero total electric charge, their unequal distances leave negative induced surface charge density immediately beneath them.

A two-dimensional embedded submanifold is a subset which near every point can be straightened by a smooth local coordinate change in the ambient space to an open piece of the coordinate plane ${\mathbb{R}}^2\times \{0\}$. Equivalently locally it has a regular smooth parametrization of rank two, with the induced local topology.

A tangent vector at $x$ is the derivative ${\gamma }^{\prime }(0)$ of a smooth curve in the submanifold with $\gamma (0)=x$. These derivatives form the two-dimensional tangent space $T_{x}M$. A normal vector is a vector orthogonal to every tangent vector. Its space is $(T_{x}M{)}^{\perp }$, which has $\boxed{\dim (T_{x}M{)}^{\perp }=1}$ in the three-dimensional ambient Euclidean space.