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For $a>0$, the positive constant-$u$ surfaces in elliptic cylindrical coordinates satisfyThey are confocal elliptic cylinders with focal lines $x=\pm a,y=0$. For a generic fixed $v$, eliminating $u$ givesBecause $u\ge 0$, a fixed $v$ selects the half-branch with the signs of $x$ and $y$ fixed by $\cos v$ and $\sin v$. These are hyperbolic cylinders. The degenerate $u=0$ surface is $y=0$, $|x|\le a$, extended in $z$. At $v=0,\pi$ the cross-sections are the two exterior horizontal rays; at $v=\pi /2,3\pi /2$ they are vertical half-axes.

Differentiate the position vector to get the coordinate tangent vectorsTheir pairwise dot products vanish. Their squared lengths are $a^2({\sinh }^{2}u+{\sin }^{2}v)$, $a^2({\sinh }^{2}u+{\sin }^{2}v)$ and one. This proves orthogonality and gives the scale factors of orthogonal coordinates:The Riemannian metric is $d{s}^2=h^2(d{u}^2+d{v}^2)+d{z}^2$. Away from the focal lines, divide the coordinate tangent vectors by their lengths to obtain the local orthonormal unit vectors $e_u,e_v,e_z$. For physical vector field components $A_u,A_v,A_z$, the formulas for vector calculus in elliptic cylindrical coordinates areThe gradient of a scalar function has components $({\varphi }_u/h,{\varphi }_v/h,{\varphi }_z)$. Apply the divergence formula to this gradient; Laplace's equation becomesThe formulas apply where $h>0$; the focal lines require a limit or another coordinate chart.

Use the ordered displacement vector $q=(x_1,y_1,x_2,y_2,x_3,y_3,x_4,y_4{)}^T$, and define $K$ by $V=\frac{1}{2}{q}^{T}Kq$. For an unstretched spring, the extension to first order is the dot product of the relative displacement with its equilibrium unit vector. Summing its squared extensions gives the central spring stiffness matrix. The four side springs and two diagonal springs giveThereforeThe kinetic energy is $\frac{m}{2}{\dot{q}}^T\dot{q}$, so a normal mode satisfies $Kq=m{\omega }^{2}q$. Keeping the factor $1/2$ in the definition of $V$ fixes the normalization of $K$ and the resulting frequencies.

For $r>0$, the chain rule gives $\partial r/\partial x_i=x_i/r$. Differentiating the radial factor gives $d\varphi /dr=e^{ikr}(ikr-1)/(4\pi r^2)$, so in Cartesian coordinatesThe origin is excluded because the scalar field is singular there.

The Stokes theorem states ${\int }_S(\nabla \times \mathbf{F})\cdot \mathbf{n}dS={\oint }_{\partial S}\mathbf{F}\cdot d\mathbf{\ell }$ for a sufficiently smooth vector field and oriented surface with piecewise smooth boundary. The boundary orientation follows the right-hand rule relative to $\mathbf{n}$.

For the convective acceleration identity, use the Levi-Civita symbol contraction ${ϵ}_{ijk}{ϵ}_{k\ell m}={\delta }_{i\ell }{\delta }_{jm}-{\delta }_{im}{\delta }_{j\ell }$. It givesRearranging proves the required vector identity.

For orthogonal coordinates, define the scale factors of orthogonal coordinates $h_i=|\partial \mathbf{r}/\partial q_i|$ and tangent unit vectors ${\mathbf{e}}_i=h_i^{-1}\partial \mathbf{r}/\partial q_i$. They form an orthonormal basis, and $d\mathbf{r}={\sum }_i{h}_i{\mathbf{e}}_{i}d{q}_i$. Equating $d\psi ={\sum }_i{\partial }_{q_i}\psi d{q}_i$ with $\nabla \psi \cdot d\mathbf{r}$ identifies each component. Thus the gradient in orthogonal coordinates isThis holds on a regular coordinate chart where all scale factors are positive.

In Cartesian coordinates, $\nabla \Phi =({\Phi }_x,{\Phi }_y,{\Phi }_z)$. For orthogonal coordinates, let $\mathbf{r}(q_1,q_2,q_3)$ be the position, $h_i=|\partial \mathbf{r}/\partial q_i|$ the scale factors of orthogonal coordinates, and ${\mathbf{e}}_i=h_i^{-1}\partial \mathbf{r}/\partial q_i$ the corresponding unit vectors. Since $d\mathbf{r}={\sum }_i{h}_i{\mathbf{e}}_{i}d{q}_i$ and $d\Phi =\nabla \Phi \cdot d\mathbf{r}={\sum }_i{\Phi }_{q_i}d{q}_i$, comparison of independent increments gives the gradient in orthogonal coordinates:For the hyperbolic parametrization of oblate spheroidal coordinates, the three coordinate tangent vectors areTheir pairwise dot products are zero. Their squared Euclidean norms are ${\sinh }^{2}R+{\sin }^2\theta$, ${\sinh }^{2}R+{\sin }^2\theta$, and ${\cosh }^{2}R{\cos }^2\theta$, respectively. Thus the coordinate surfaces meet orthogonally wherever this is a regular coordinate chart, andThe last expression uses the usual latitude range $|\theta |<\pi /2$; without that range a scale factor is its absolute value. Axial or ring degeneracies are excluded from a regular chart.

On the sphere, constant $\theta$ gives a latitude circle of radius $a\sin \theta$ and height $a\cos \theta$. Constant $\varphi$ gives a meridian semicircle in a vertical plane through the $z$ axis. The orthonormal basis of spherical coordinates isHere ${\mathbf{e}}_r$ points outward, ${\mathbf{e}}_{\theta }$ along increasing polar angle, and ${\mathbf{e}}_{\varphi }$ along increasing azimuth. At either pole the radial unit vector is unambiguous, but the two tangential coordinate vectors depend on the undefined longitude. The figure shows their directions away from the poles.

This coordinate ambiguity does not make a vector field singular if its tangential coefficient vanishes suitably. In Cartesian coordinates,The first is smooth at the north pole and zero there; the second is smooth and zero at the south pole. Their actual singular axes are the opposite ones. These are the regular patches of a two-patch vector potential for a radial monopole.

An isotropic tensor has unchanged components under every proper rotation of the Cartesian axes. Every rank-zero scalar is isotropic, and $c{\delta }_{ij}$ is a rank-two example. The zero rank-one tensor is isotropic, but no nonzero vector is: a half-turn about an axis perpendicular to it reverses its sign.

For the inertia tensor, write $u=x-G$ and $X=P-G$, so $x-P=u-X$. Expand its defining integrand. Every cross term has an integral containing ${\int }_{B}u\rho d^{3}x=0$ by the definition of the centre of mass. The remaining terms give the tensor parallel-axis theorem:For a uniform unit ball, rotational symmetry gives $\int \rho u_i{u}_j{d}^{3}u=c{\delta }_{ij}$. Its trace is $\int \rho |u{|}^2{d}^{3}u=3M/5$, so $c=M/5$ and $I(G)=2MI/5$. ThereforeAssuming positive mass, it is isotropic exactly when $P=G$. For $X\ne 0$, its eigenvalues and directions areTo give an explicit orthonormal eigenbasis when $X_1^2+X_2^2>0$, take $e_1=(-X_2,X_1,0)/\sqrt{X_1^2+X_2^2}$ and $e_2=e_{\parallel }\times e_1$. If $X$ is a nonzero vector along the third axis, take the first two coordinate unit vectors instead. At $X=0$, all three eigenvalues are $2M/5$ and every orthonormal basis works. The inertia tensor of a displaced uniform ball describes the same result for an arbitrary radius.

For a twice continuously differentiable vector field in Cartesian coordinates, the curl and divergence arePut $\mathbf{\omega }=\nabla \times \mathbf{v}$. Using the Levi-Civita symbol, the Kronecker delta contraction and the product rule,The last term vanishes because the divergence of a curl is zero: the symmetric mixed derivatives contract against an antisymmetric symbol. Thus the curl of a cross product identity becomes

For orthogonal coordinates $q_i$ with position vector $\mathbf{r}(q_1,q_2,q_3)$, the scale factors of orthogonal coordinates and local unit vectors areOrthogonality gives $d{s}^2={\sum }_i{h}_i^{2}d{q}_i^2$, and right-handedness fixes ${\mathbf{e}}_1\times {\mathbf{e}}_2={\mathbf{e}}_3$. In cylindrical coordinates, $\mathbf{r}=(\rho \cos \varphi ,\rho \sin \varphi ,z)$, soThe radial vector points away from the axis; the azimuthal vector is tangent in the direction of increasing $\varphi$; ${\mathbf{e}}_z$ is the Cartesian positive $z$ direction. The azimuthal basis is undefined on the coordinate axis itself.

For the axial velocity profile, the cylindrical coordinates formula givesThe associated cross product isThis field has only a radial component depending on $\rho$, so its curl is zero. Also $\nabla \cdot \mathbf{u}={\partial }_z{u}_z=0$. The identity above therefore yieldsIn fact both directional derivatives vanish: the first differentiates $u_z{\mathbf{e}}_z$ only in the azimuthal direction, and the second differentiates $\mathbf{\omega }$ only in the $z$ direction. All physical vector fields extend smoothly through $\rho =0$, despite the coordinate-basis singularity.

Let $Q$ be the orthogonal matrix taking old components to new ones, so $x_i^{\prime }=Q_{ip}{x}_p$. With repeated indices summed, the tensor component transformation law isAn isotropic tensor has the same components after every proper rotation. Since $Q_{ip}{Q}_{jp}={\delta }_{ij}$, the Kronecker delta is invariant. Applying this identity to each factor shows that $a{\delta }_{ij}$ and every linear combination $b{\delta }_{ij}{\delta }_{kl}+c{\delta }_{ik}{\delta }_{jl}+d{\delta }_{il}{\delta }_{jk}$ are invariant as well.

To show that an isotropic second-rank tensor must be scalar, start with any matrix $t$, without assuming symmetry. Invariance under quarter-turns about axes 1 and 2 implies invariance under their squares, the half-turnsThe first forces $t_{12},t_{21},t_{13},t_{31}$ to vanish. The second also forces $t_{23},t_{32}$ to vanish. The matrix is therefore diagonal. A quarter-turn about axis 1 interchanges its 22 and 33 entries; one about axis 2 interchanges its 11 and 33 entries. All three must agree, giving $t=aI$.

A rotation preserves the ball $V_R$, $|x{|}^2$ and the volume element. Changing integration variables by that rotation shows that $A$ and $B$ are isotropic tensors. Write $A_{ij}=a{\delta }_{ij}$. Taking its tensor contraction givesso $a=4\pi R^7/21$. Every permutation of the indices of $B$ leaves its integrand unchanged. Its complete symmetry therefore makes all three isotropic coefficients equal, say $d$:Double contraction gives $15d=B_{iikk}={\int }_{V_R}|x{|}^{4}dV=4\pi R^7/7$. Thus the isotropic fourth moment of a ball and its radial contraction are

Put $d=4\pi R^7/105$, so that $a=5d$. Contract $B$ with the prescribed unit vector twice:This has the spectrum of a scalar matrix plus a rank-one projector. In the direction $n$, the outer product $n{n}^T$ acts as the identity; on the plane perpendicular to $n$, it acts as zero. ThereforeAny nonzero multiple of $n$ is a parallel eigenvector, and any nonzero vector in the perpendicular plane is a transverse eigenvector. The transverse eigenvalue has multiplicity two; an orthonormal basis of that plane completes a set of principal axes.

Multiply the eigenvalue equation $Ax=\lambda x$ on the left by $x^{†}$. The unit vector normalization is $x^{†}x=1$, soNo normal matrix assumption on $A$ is needed: this is an identity for each normalized eigenvector.

The radial unit vector in the spherical coordinate system isThe coordinate tangent in the polar direction has length $r$, so its corresponding unit vector isDifferentiating ${\mathbf{e}}_r$ at fixed $\varphi$ therefore givesThe factor is not $r$: the corresponding scale factors of orthogonal coordinates occur in the derivative of the position vector, whereas the question differentiates its unit direction. The formula is understood away from the coordinate singularities where a complete spherical basis is not uniquely specified.

Label the equilibrium vertices in the order $+x,-x,+y,-y,+z,-z$, and write their displacements as ${\mathbf{u}}_i(t)\in {\mathbb{R}}^3$. Their eighteen Cartesian components are generalized coordinates. Linearize the axial Hookean springs about the unstressed equilibrium. For an edge with unit direction ${\mathbf{n}}_{ij}$, its first-order extension is ${\mathbf{n}}_{ij}\cdot ({\mathbf{u}}_i-{\mathbf{u}}_j)$, so the quadratic potential energy isThree zero modes are translations: for each Cartesian unit vector ${\mathbf{e}}_{\alpha }$, take ${\mathbf{u}}_i={\mathbf{e}}_{\alpha }$ at every vertex. Three more are infinitesimal rotations ${\mathbf{u}}_i={\mathbf{e}}_{\alpha }\times {\mathbf{r}}_i$. Explicitly, in the stated vertex order, their displacement vectors areTranslations leave each displacement difference zero. For rotations,because the equilibrium edge vector is parallel to ${\mathbf{n}}_{ij}$. Hence every spring has zero first-order extension in all six patterns, and the stiffness matrix annihilates them.

The translations have nonzero summed displacement, while all the rotations have zero sum. The three rotational patterns are themselves independent because the vertices do not lie on a single line. ThusThese are the rigid-motion zero modes of a spring framework. The rotational displacements describe infinitesimal rigid rotations; an arbitrary finite rotation would require its nonlinear displacement field.

Define $g(p)=f(p)-f(-p)$ on the sphere. It is continuous and satisfies $g(-p)=-g(p)$. If $g(p)=0$ at a chosen point, we are done. Otherwise choose a unit vector $q$ perpendicular to $p$ and follow the half great circleThe Pythagorean theorem shows that $\gamma (t)$ lies on the sphere, and its endpoints are $p,-p$. The continuous function $g\circ \gamma$ has opposite signs at the endpoints. The Intermediate value theorem therefore gives a zero $t_{\ast }$, soThis is a scalar equal-value assertion for an antipodal pair; it uses only a path and the Intermediate value theorem.

For every vector $z\in {\mathbb{R}}^n$, splitting the rows givesBoth summands are nonnegative. Taking the supremum over unit vectors in the definition of the spectral norm proves

Put $P=Q{Q}^T$. Since $Q$ has $r$ columns, the rank bound for a matrix product gives $\mathrm{rank}(PA)\le r$. The Eckart-Young-Mirsky theorem and part (a) then giveOne can also see the lower bound directly. The restriction of $PA$ to the $(r+1)$-dimensional linear span of $v_1,\dots ,v_{r+1}$ has a nonzero null space, because its matrix rank is at most $r$. For a unit vector $z$ in that null space, $\Vert (A-PA)z{\Vert }_2=\Vert Az{\Vert }_2\ge {\sigma }_{r+1}$ by the singular value decomposition.

Because $r<n$, choose a unit vector $v$ in the null space of $X^T$, so that $v^{T}X=0$. Let $J\in {\mathbb{R}}^{m\times n}$ have orthonormal columns, for example the first $n$ columns of $I_m$, and choose a real number $\alpha >1$. SetIts singular value along $v$ is $\alpha$, and its other singular values are $1$. Thus ${\sigma }_1(A)=\alpha >1={\sigma }_2(A)$. Meanwhile $AX=JX$. If $X$ has full column rank, the column space of the thin QR decomposition is exactly $\mathrm{range}(JX)$, which is orthogonal to $Jv$. Hence $PJv=0$ for $P=Q{Q}^T$, andConversely, an orthogonal projection is a contraction in the Euclidean norm, so $\Vert (I_m-P)A{\Vert }_2\le \Vert A{\Vert }_2=\alpha$. ThereforeIf $X$ is rank deficient, complete the QR decomposition within $J(v^{\perp })$: this space has dimension $n-1\ge r$ and contains all columns of $JX$, so the required orthonormal completion exists there. The same proof applies. For rank-deficient data the choice of unused columns of $Q$ matters; the construction specifies a valid choice. If $Q$ instead contains only a basis of $\mathrm{range}(AX)$, the conclusion holds without any completion. A fixed sampling matrix can thus miss the unique largest singular value completely.

Birefringence means that two independent polarization eigenmodes have different refractive indices at the same frequency and propagation direction. The rotatory-power formula relevant here requires circular birefringence: the eigenmodes are opposite circular polarizations. General linear birefringence instead introduces phase retardation between linearly polarized components and need not rotate every incident linear polarization through one common angle.

Use the complex amplitude convention $\mathrm{Re}[\mathbf{E}{e}^{-i\omega t}]$ and transverse unit vectors ${\mathbf{e}}_1,{\mathbf{e}}_2$. DefinePropagation through thickness $d$ adds phases ${\varphi }_{\pm }=(2\pi /{\lambda }_0)n_{\pm }d$, where ${\lambda }_0$ is the vacuum wavelength. Recombining the two components givesThus the rotation and the specific rotatory power per unit length areThe units of $\rho$ are $\mathrm{rad}{\mathrm{m}}^{-1}$. The sign follows the specified circular basis and viewing convention. Equal attenuation of the two eigenmodes has been assumed; unequal attenuation also produces elliptical polarization.

Let $s$ be distance along the rod from its fixed end. The position of that material point is $sn(t)$, where $n=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,-\cos \theta )$ is a unit vector. Its velocity is $s\dot{n}$, and direct differentiation givesThe uniform linear mass density is $M/a$. Integrating the elemental kinetic energy therefore proves, rather than assumes, the rod formula:Equivalently, its moment of inertia about a perpendicular axis through the pivot is $M{a}^2/3$.

This is a plane electromagnetic wave with linear polarization. $\mathbf{E}$ is the electric field; $E_0$ its peak amplitude; $x$ the position along propagation; $t$ the time; $k=2\pi /\lambda$ the wavenumber; and $\omega =2\pi f$ the angular frequency. Here $\lambda$ is the wavelength, $f$ the frequency, and $\hat{\mathbf{y}}$ the unit vector in the direction of the electric field. Constant phase $kx-\omega t$ moves along $+x$ with phase velocity $\omega /k$.

Put $\theta =kx-\omega t$. The differential Faraday's law of induction givesThe accompanying oscillating magnetic field, excluding any independent static background, is therefore $\mathbf{B}=(k/\omega )E_0\cos \theta \hat{\mathbf{z}}$. To determine the propagation speed, also use the source-free Ampere-Maxwell law:ThusThe electric field and magnetic field oscillate in phase, perpendicular to one another and to propagation; $\mathbf{E}\times \mathbf{B}$ points along $+x$.

A fictitious force is a term required when writing Newton's second law in an accelerating frame; it comes from the coordinate choice rather than a new interaction. For axes rotating at constant $\mathbf{\Omega }$ about a fixed inertial origin,Apply this identity twice to position to get ${\mathbf{a}}_I={\mathbf{a}}_R+2\mathbf{\Omega }\times {\mathbf{v}}_R+\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{r})$. Moving the last two terms to the right of $m{\mathbf{a}}_I=\mathbf{F}$ gives the stated Coriolis and centrifugal forces. An accelerating origin or changing rotation rate would add further terms.

A slowly rotating fluid planet follows an equipotential of gravitational plus centrifugal potential. The centrifugal potential is $-{\Omega }^2{r}^2{\cos }^2\lambda /2$, most negative at the equator. An outward equatorial shift raises gravitational potential and restores the balance, producing an oblate figure. Self-gravity also changes with the shape; no flattening coefficient follows from rotation alone without a density model.

Using local north and radial-up unit vectors, the centrifugal acceleration on a stationary mass isIt points away from the rotation axis: upward and equatorward, at angle $\lambda$ to radial up in the northern hemisphere. Its vertical fraction of gravitational force isusing a $24\mathrm{h}$ day for this approximate number. The resultant effective weight tilts equatorward relative to radial down. A plumb-defined local vertical already follows effective gravity; the component calculation uses the spherical radial vertical. A supported stationary mass of course has zero net acceleration.

For the Coriolis deflection of vertical launch and drop, retain only first order in $\Omega$, use effective constant $g$, and neglect horizontal velocities in the Coriolis term as instructed. With east positive, vertical velocity gives $\ddot{x}=-2\Omega \cos \lambda \dot{z}$ and no northward acceleration. For launch, $z=ut-g{t}^2/2$ and the flight time is $2u/g$. Integrating with zero initial horizontal velocity givesThe landing is westward. The westward velocity accumulated on ascent is only recovered on descent, so the two displacement contributions do not cancel.

For release from rest at $h_0=u^2/(2g)$, $z=h_0-g{t}^2/2$, $t_{\mathrm{fall}}=u/g$, and $\ddot{x}=2\Omega gt\cos \lambda$. ThusThe drop lands eastward with one quarter the launch-return displacement magnitude. North–south corrections from horizontal Coriolis coupling are beyond the retained approximation.

Changing basis and acceleration. Attach orthonormal unit vectors ${\mathbf{e}}_i(t)$ to the rotating frame, with a common origin fixed in the inertial frame. An infinitesimal rotation $d\mathbf{\theta }=\mathbf{\omega }dt$ changes a vector fixed in those axes by $d\mathbf{A}=d\mathbf{\theta }\times \mathbf{A}$: its tip moves tangentially to the circle about the rotation axis. More generally, write $\mathbf{A}={\sum }_i{A}_i{\mathbf{e}}_i$. Differentiating the components and the rotating basis separately proves the rotating-frame derivative formula,

Apply the formula first to position: ${\dot{\mathbf{r}}}_0=\mathbf{v}+\mathbf{\omega }\times \mathbf{r}$, where $\mathbf{v}=(d\mathbf{r}/dt{)}_S$. Apply it again to both terms. For constant $\mathbf{\omega }$,One cross term comes from differentiating $\mathbf{v}$ in the inertial frame and another from differentiating $\mathbf{r}$ in $\mathbf{\omega }\times \mathbf{r}$. The displayed source formula implicitly requires constant angular velocity and an unaccelerated common origin. For changing rotation rate there is also $\dot{\mathbf{\omega }}\times \mathbf{r}$; an accelerating origin supplies its own translational acceleration. Moving the rotation terms to the force side gives the apparent Coriolis force and centrifugal force with negative signs.

Southward motion on Earth. Use local east, north and up axes, with latitude $\lambda$ positive northward. Then $\mathbf{\Omega }=\Omega (0,\cos \lambda ,\sin \lambda )$ and $\mathbf{v}=(0,-v,0)$. The Coriolis acceleration in local Earth coordinates givesIt is westward in the northern hemisphere, eastward in the southern hemisphere and zero for exactly southward motion at the equator. There is no vertical Coriolis component for this particular velocity.

Correct steering. On the desired meridional track, drag has no eastward component because it is antiparallel to the velocity. If the thrust makes signed angle $\varphi$ east of south, its eastward component is $T\sin \varphi$. The cross-track force balance is thereforesoThe argument must have magnitude at most one for this simple thrust-only correction to be feasible. The signed angle changes sign at the equator: correction is east of south in the north and west of south in the south. Thus the phrase “east of south” cannot describe a positive correction throughout the entire flight.

Wrong-sign correction and the miss distance. A literally constant velocity vector cannot develop a cross-track displacement. The physically useful interpretation of the last task is constant speed, with an erroneous lateral steering correction. In the usual leading-track model, thrust and drag balance along the actual motion, and the applied transverse correction has the correct magnitude but the wrong sign. It therefore reinforces rather than cancels the Coriolis force. Starting at the equator, use the unperturbed southward track $\lambda \simeq -vt/R$, and let $x_E$ be the eastward departure from that track. To first order in the departure,There is a factor two relative to uncompensated Coriolis deflection because the wrong correction adds an equal lateral force. Integration gives the wrong-sign Coriolis steering error,For southern angular distance $\Lambda =16\pi /180=0.27925$, $R=6.4\times 10^6\mathrm{m}$ and $v=800/3.6=222.2\mathrm{m}{\mathrm{s}}^{-1}$, the nominal flight time after the equator isTaking $\Omega \simeq 2\pi /(86400\mathrm{s})$ givesThe small-latitude approximation $\sin (vt/R)\simeq vt/R$ gives ${\ddot{x}}_E\simeq 4\Omega v^{2}t/R$ and $x_E\simeq 2\Omega v^2{t}^3/(3R)$, or $1.95\times 10^5\mathrm{m}$, confirming the scale and direction.

This is an estimate under an explicit steering model, not an exact trajectory forced by the wording. For example, retaining spherical transport of the local axes while still linearizing in the cross-track motion gives, with $s=-\lambda$, $\dot{u}-(v/R)u\tan s=4\Omega v\sin s$. Consequently $u\cos s=2\Omega R{\sin }^{2}s$ and the longitude departure is $\Delta \ell =2\Omega R(\tan \Lambda -\Lambda )/v$. The distance along the destination parallel is then $R\cos \Lambda \Delta \ell \simeq 1.93\times 10^5\mathrm{m}$. Corrections from the changed track and a more detailed control law are not fixed by the problem. If “keeps the velocity constant” instead means the full vector, the pilot would need an additional compensating force and there would be no miss; if thrust is kept at a specified geographical bearing despite drift, its along-track component and lateral drag must also be modelled.