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The seven requested parabolas are $y=c{x}^2$ for $c=-3,-2,-1,0,1,2,3$; the case $c=0$ is the horizontal axis. Positive $c$ opens upward, negative $c$ downward, and larger $|c|$ gives a narrower parabola. All pass through the origin with a horizontal tangent line.

To find orthogonal trajectories, use the continuous family $y=c{x}^2$ containing these seven representative members. Its slope is $y^{\prime }=2cx=2y/x$ away from $x=0$. Two finite slopes of perpendicular tangents multiply to $-1$, so the orthogonal slope isThis separable differential equation gives $2ydy=-xdx$, henceThese ellipses are the regular real trajectories; $C=0$ is a single point, not a regular curve, and $C<0$ has no real points. One example is $x^2+2y^2=2$, with semiaxes $\sqrt{2}$ horizontally and $1$ vertically.

The implicit tangent relation $2xdx+4ydy=0$ also describes points with vertical tangents, where the slope quotient is unavailable. At intersections with the member $y=0$, the ellipse has a vertical tangent line and is perpendicular to that horizontal axis. The common origin of the parabolas is not on any regular trajectory with $C>0$. The sketch shows the requested seven members and the example ellipse.

The region lies above the parabola $y=x^2$ and below the upper semicircle $y=\sqrt{1-x^2}$. Their intersections satisfy $x^4+x^2-1=0$, hence $x=\pm a$ with $a^2=(\sqrt{5}-1)/2$. In particular, $\sqrt{1-a^2}=a^2$. The region isIt is symmetric about the $y$ axis, with endpoints $(\pm a,a^2)$; at $x=0$ it extends from $y=0$ to $y=1$.

Put $\mathbf{v}={\mathbf{r}}_0-\mathbf{a}$. The squared distance to a point on the line is $|\mathbf{v}-\lambda \hat{\mathbf{t}}{|}^2$. Since $|\hat{\mathbf{t}}|=1$, completing the square in $\lambda$ gives its minimum at $\lambda =\hat{\mathbf{t}}\cdot \mathbf{v}$, with value $|\mathbf{v}{|}^2-(\hat{\mathbf{t}}\cdot \mathbf{v}{)}^2$. The cross product identity $|\hat{\mathbf{t}}\times \mathbf{v}{|}^2=|\mathbf{v}{|}^2-(\hat{\mathbf{t}}\cdot \mathbf{v}{)}^2$ therefore provesFor the given line, $\hat{\mathbf{t}}=(1,1,0)/\sqrt{2}$, so its squared distance to $(x,y,z)$ is $(x-y{)}^2/2+z^2$. Substituting the parabola givesThe global minimum occurs at $\mu =\pm 1$, henceThe corresponding points on the parabola are $\pm (1,1+2\sqrt{3},0)+(0,0,-3)$; their nearest points on the line are $\pm (1+\sqrt{3},1+\sqrt{3},0)$. The connecting vectors are orthogonal to the line, as the projection calculation requires.

In the $x,z$ plane, the embankment lies above the parabola $z=H{x}^2/L^2$ and below $z=H$. The bore has circular cross-sectionIt touches the valley floor at the origin. To check that this entire disk lies inside the projected embankment, its equation gives $x^2\le Dz-z^2\le Dz$. Since $HD/L^2<1$, it follows that $H{x}^2/L^2\le z$, with strict inequality away from the bottom contact. Also $z\le D<H$. Thus no circular segment needs to be subtracted from the tunnel section.

For each $(x,z)$ in the disk, the length of bore inside the material is the local width $2b(2-z/H)$. The required volume integral is thereforeThe disk has area $A=\pi D^2/4$ and centroid height $D/2$, so ${\iint }_{\mathcal{D}}zdxdz=AD/2$. Applying the volume of a solid with affine thickness formula givesThe width varies along the circular section; multiplying its area by the top width alone would miss this variation.

Write $d=b-a$ and $h=d/n$. The required Riemann sum isUsing $\sum k=n(n+1)/2$ and the given sum of squares,An equivalent expression separates its limit and finite-$n$ corrections:ConsequentlyThe function is continuous, so the Riemann sums converge to its definite integral, the desired area under the parabola. Both corrections are positive when $0<a<b$, agreeing with the geometric fact that these right-endpoint rectangles overestimate an increasing curve.

Put $s=t^2-t^{-2}$. Part (b)(ii) gives $u=s+2i$, soThe parameter $s$ increases strictly from $-15/4$ to $15/4$. Therefore the imaginary part $Y=4s$ increases from $-15$ to $15$, while the real part is $X=s^2-4$. The endpoints areThe real axis crossing is at $s=0$, giving $\boxed{(-4,0)}$ at $t=1$. The two imaginary axis crossings satisfy $s^2=4$, givingIn order, the trajectory passes from its lower endpoint through $(0,-8)$, then the leftmost point $(-4,0)$, then $(0,8)$, to its upper endpoint. It is a right-opening parabola segment, traversed continuously upward as shown in the right panel.

The level set $f=c$ can be rearranged, without changing its domain, asThe requested special contours are thereforeThe first is a parabola, the second a single point, and the last two intersecting straight lines. Moreoverwith equality only at $(0,1)$. Hence $(0,1)$ is a strict global maximum. At $(0,-1)$ the crossing critical contour separates sectors with values above and below $-1$, so it is a saddle point of a scalar function.

As a Hessian matrix check, $H_f(0,1)=\mathrm{diag}(-3,-1)$ is negative definite, while $H_f(0,-1)=\mathrm{diag}(-1,1)$ is indefinite. There is no stationary minimum.

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.

At height $z$, the horizontal triangle has half-width $L\sqrt{1-z/(2H)}$ along $x$ and depth $D(1-z/H)$ along $y$. These shrink with $z$, so its orthogonal projection onto the $(x,y)$ plane is the base triangle with vertices $(-L,0)$, $(L,0)$ and $(0,D)$:The orthogonal projection onto the $(y,z)$ plane is the triangle with vertices $(0,0)$, $(D,0)$ and $(0,H)$:The orthogonal projection onto the $(x,z)$ plane is bounded by $z=0$, $z=H$ and the two curvesEquivalently, these curves lie on the parabola $z=2H(1-x^2/L^2)$, truncated at height $H$. In particular the top width is $\sqrt{2}L$, so this projection has a horizontal upper edge from $-L/\sqrt{2}$ to $L/\sqrt{2}$; it is not triangular. The top horizontal cross-section has zero depth, consistent with this edge.

From the factorisation in part (a), $x=0$ or $x=1$. Substitution into the parabola equation gives respectively $y=1$ or $y=0$. Thus the two intersections of the parabola and the unit circle are

The gradient is horizontal when $h_y=0$ and vertical when $h_x=0$. Factoring those partial derivatives gives the two loci:Thus each locus consists of an axis and two parabolas. Their common points are the five stationary points found below. At those points the gradient is zero and has no direction, so they are included in both algebraic zero-component conditions but excluded if “parallel” is required to mean a nonzero directed vector.

Differentiate the flux:On $-1<z_0<1$, $\Phi$ is strictly increasing. Its limiting endpoint values are $\Phi (-1)=0$ and $\Phi (1)=4\pi$, and $\Phi (0)=2\pi$. It has horizontal limiting tangents at the endpoints, curves upward for $z_0<0$, curves downward for $z_0>0$, and has an inflection point at zero.

The derivative graph is a downward-opening parabola, approaching zero at both endpoints and attaining its maximum $3\pi$ at $z_0=0$.

Coordinate surfaces and sketch. Adding and subtracting the defining relations gives $s=u^2+v^2$ and $x=u^2-v^2$. Since $y^2=s^2-x^2$, we have $y^2=4u^2{v}^2$. For $u^2=c>0$, eliminate $v^2$ to obtainThis is a parabolic cylinder parallel to the $z$ axis, opening towards negative $x$, with cross-sectional vertex $(c,0)$ and focus at the origin. Similarly, $v^2=d>0$ givesa parabolic cylinder opening towards positive $x$, with vertex $(-d,0)$ and the same focus. The degenerate $u^2=0$ surface is the half-plane $y=0,x\le 0$; $v^2=0$ gives $y=0,x\ge 0$. Constant $w$ gives the horizontal planes $z=w$.

Thus the requested cross-sections consist of $x=1-y^2/4$, $x=2-y^2/8$, $x=y^2/4-1$, $x=y^2/8-2$ and the two horizontal half-axes. The parabolas of the two families meet orthogonally.

Signed coordinates. Take $u\ge 0$ and choose the sign of $v$ to agree with $y$ away from $y=0$. Then the Cartesian coordinates areEvery coordinate triple gives exactly one point. For $u>0$, $v$ and $y$ have the same sign. A fixed signed $v\ne 0$ selects the upper or lower half of its parabolic cylinder; the entire cylinder corresponds to fixed $v^2$. On the cut $u=0$, the two values $v=\pm \sqrt{-x}$ represent the same point of the negative $x$ half-axis. This boundary identification is unavoidable: there $y=0$ even when $v\ne 0$, so the sign prescription is understood by approaching from either side of the cut. The interior $u>0$ gives a unique coordinate chart off that half-axis.

Orthogonality and metric. Differentiate the position vector:The pairwise dot products vanish. Their lengths give the scale factors of orthogonal coordinates and hence the Riemannian metric:These are parabolic cylindrical coordinates. The differential calculation is regular for $u^2+v^2>0$; the $z$ axis is the coordinate singularity where the two planar tangent vectors vanish.

Laplacian. Put $h=2\sqrt{u^2+v^2}$. The physical components of the gradient of a $w$-independent scalar field are ${\varphi }_u/h,{\varphi }_v/h,0$. Applying the divergence formula cancels the two equal scale factors of orthogonal coordinates inside the derivatives:Therefore, away from the coordinate singularity,

For a repulsive inverse-square force, $V(r)=+\kappa /r$. The effective potential is positive and decreases from infinity to zero, so a trajectory reaching a finite radius has $E>0$. For nonzero angular momentum, it is the accessible branch of a hyperbola, with $e^2=1+2E{J}^2/(m{\kappa }^2)>1$. There are no bound ellipses or finite-distance zero-energy parabolas. At zero angular momentum, the motion is straight inward followed by straight outward after reflection at a radial turning point.

For an attractive inverse-square force, write $V(r)=-\kappa /r$ with $\kappa >0$, and take nonzero conserved angular momentum $J$. The conic section orbit hasThe physically allowed cases are ellipse for $-m{\kappa }^2/(2J^2)\le E<0$, parabola for $E=0$, and hyperbola for $E>0$. The minimum allowed energy gives the circular special case $e=0$. Here $E$ is measured relative to infinity. Zero angular momentum instead gives radial collision or escape trajectories, interpreted as degenerate limits rather than regular closed ellipses.

Use the Euler-Bernoulli beam equation for a uniform slender elastic beam, small deflection and small slope, neglecting axial tension, shear deformation, rotational inertia of a cross-section and damping. Let $E$ be Young's modulus, $I={\int }_A{z}^{2}dA$ the second moment of area about the neutral axis, and $\rho$ the mass per unit length. If density instead denotes mass per volume, replace this $\rho$ by ${\rho }_{\mathrm{bulk}}A$. The bending stiffness is $EI$; the small-slope curvature is $y_{xx}$. The bending and kinetic energies areTwice integrating the variation of $U$ by parts gives the restoring load $-EI{y}_{xxxx}$ per unit length. Newton's second law therefore givesAt a pivot, displacement is fixed but slope is free; zero end bending moment gives $EI{y}_{xx}=0$. This also follows from the natural boundary term $EI{y}_{xx}\delta y_x$ in the variation.

With separation of variables, $y=XT$ givesChoose the time normalization $T(0)=1$. Initial rest removes the sine time term, so $T(t)=\cos \omega t$. The spatial characteristic roots are $\pm ik,\pm k$, givingThe two conditions at $x=0$ give $B+D=0$ and $-B+D=0$, hence $B=D=0$. At $L$, addition and subtraction of $X(L)=X^{\prime \prime }(L)=0$ give $C\sinh kL=0$ and $A\sin kL=0$. Nontrivial positive-frequency modes therefore have $C=0$ and $k_n=n\pi /L$. Thus the simply supported beam normal modes areThe zero-frequency cubic solution also vanishes under these four endpoint conditions, so $n=0$ is not an additional mode.

For the specified release, small slope requires $|a|L\ll 1$. Expand the initial deflection in the Fourier sine series. Its coefficients areWriting ${\omega }_1={\pi }^2\sqrt{EI/\rho }/L^2$, the resulting motion isOnly odd normal modes occur because the initial profile is symmetric about the midpoint, and their amplitudes decrease as $n^{-3}$. The frequencies grow as $n^2$, so the motion is dispersive and is not a single shape multiplied by one cosine. Nevertheless the ideal undamped solution is exactly periodic with period $2\pi /{\omega }_1$. At $t=\pi /(2{\omega }_1)$ all odd-mode cosines vanish, so the bar is instantaneously straight; at $t=\pi /{\omega }_1$ it has the opposite initial deflection. The first mode dominates displacement, with third-mode amplitude only $1/27$ of the first.

There is initial curvature incompatibility at a moment-free beam end: the prescribed parabola has $y_{xx}(x,0)=2a$, whereas a moment-free pivot requires zero curvature at an endpoint. Therefore the data do not define a classical solution with curvature continuous through the two end-time corners. They do define a finite-energy weak solution after release of the forces or end moments that prepared the initial shape. The sine expansion represents that solution: its displacement reproduces the initial parabola, each mode satisfies the released endpoint conditions, and ${\sum }_n{n}^4|A_n{|}^2<\infty$ gives finite bending energy. One must not differentiate the initial series twice at its endpoints and assign the interior parabola's curvature there. This qualification leaves the modal displacement prediction intact.

For two isolated point masses, introduce total mass $M_T=M+m$, reduced mass $\mu =Mm/M_T$ and relative separation $r$. The relative acceleration is $\ddot{\mathbf{r}}=-G{M}_T\mathbf{r}/r^3$. Conserved specific angular momentum $h=|\mathbf{r}\times \dot{\mathbf{r}}|$ fixes a plane and a constant areal velocity $h/2$. The specific orbital energy isFor $h\ne 0$, the trajectory is a conic:Negative energy gives an ellipse, including the circular limit $e=0$; zero energy gives a parabola; positive energy gives a hyperbola. Bound ellipses obey $\epsilon =-G{M}_T/(2a_o)$ andThus Kepler's laws of planetary motion follow from angular momentum conservation and the inverse-square force. For a circular orbit of radius $r$, $v_c=\sqrt{G{M}_T/r}$, while escape from that radius requires $v_{\mathrm{esc}}=\sqrt{2G{M}_T/r}$. Energy determines whether the orbit is bound; angular momentum also determines its shape. These ideal results neglect extended-body perturbations, drag and relativistic corrections.

The Lorentz force determines the trajectory:Thus a magnetic field changes momentum direction but does no work; an electric field can change the particle's kinetic energy. In a uniform electric field with negligible magnetic field and nonrelativistic speed, $\mathbf{a}=q\mathbf{E}/m$ is constant. A transverse initial velocity then gives a parabola.

In a uniform magnetic field, the parallel velocity is unchanged and the perpendicular motion is circular. Nonrelativistically,the combined trajectory is a helix, with circulation reversed when the charge sign reverses. Relativistically $r_L=p_{\perp }/(|q|B)$ and the gyrofrequency is $|q|B/(\gamma m)$.

For perpendicular uniform fields in the magnetic-dominated case, the drift velocity is ${\mathbf{v}}_d=\mathbf{E}\times \mathbf{B}/B^2$, which obeys $\mathbf{E}+{\mathbf{v}}_d\times \mathbf{B}=0$. A particle at this velocity is undeflected: the crossed-field velocity selector selects speed $E/B$ in the appropriate direction, independent of mass and charge. Radiation reaction and collisions are neglected in these ideal trajectories.

For a test mass moving around a fixed central mass $M$, gravity is central and conservative. Specific angular momentum $h=|\mathbf{r}\times \mathbf{v}|$ and specific orbital energy $ϵ=v^2/2-GM/r$ are constant. The motion is planar and sweeps equal areas in equal times, $dA/dt=h/2$.

For a nonradial orbit, the inverse-square force gives a conic section,Negative energy gives an ellipse, zero energy a parabola and positive energy a hyperbola. For an ellipse of semimajor axis $a$,A circular orbit is the $e=0$ case, with $v=\sqrt{GM/r}$. Escape from radius $r$ requires $v\ge \sqrt{2GM/r}$ in the absence of further forces. For two comparable masses, apply the relative two-body equation with $GM$ replaced by $G(M+m)$, rather than treating one body as fixed.

In a fixed point-mass Newtonian gravitational field, a test mass $m$ experiencesBecause this is a central force, its torque about the attracting centre vanishes and angular momentum is conserved. The orbit lies in a fixed plane and sweeps out equal areas in equal times. Since the force is also conservative, the specific orbital energy is constant:The last two terms form an effective potential. They show how the angular-momentum barrier competes with attraction and determines the turning radii. For nonradial motion the orbit is a conic section:Negative energy gives an ellipse, with $e<1$; zero energy gives a parabola; positive energy gives a hyperbola. The circular limit has $e=0$, $v_c=\sqrt{GM/r}$ and $\epsilon =-GM/(2r)$. For a bound ellipse of semimajor axis $a$, $\epsilon =-GM/(2a)$ and the orbital period is $T=2\pi \sqrt{a^3/(GM)}$. The local escape speed is $\sqrt{2GM/r}$. These are the point-mass, test-particle limits; extended mass distributions or noncentral perturbations change the orbit laws.