A plane shock moving at speed $V_s$ through initially stationary gas has upstream speed $u_1=V_s$ and downstream speed $u_2=V_s/\chi$ in the shock frame, where $\chi$ is the shock compression ratio. Transforming back to the initial slab frame gives $v_2=V_s-u_2$. At the breakout time $L/V_s$, a uniform downstream slab has thickness $L-v_{2}L/V_s=L/\chi$. The strong monatomic limit has $\chi =4$, hence $L_2=L/4$ and $v_2=3V_s/4$. This uses a maintained planar post-shock state; boundary rarefactions and variations of shock speed can change the actual thickness of a freely evolving finite cloud.

Define $f(0,0)=0$ and, elsewhere,All first and second partial derivatives exist at the origin: $f_x=f_y=f_{xy}=f_{yx}=0$ and $f_{xx}=f_{yy}=2$. Thus its Hessian matrix there is the positive-definite matrix $2I$. Nevertheless $f(t,0)=t^2>0$ and $f(t,t)=-t^2<0$ for $t\ne 0$, so the origin is a saddle point of a scalar function. A valid second-derivative test needs a second-order Taylor expansion with a remainder $o(x^2+y^2)$; continuous second partial derivatives in a neighborhood are a sufficient hypothesis. Existence of those derivatives at the point alone does not supply this hypothesis.

Here is a short proof that uses only the row space. For a vector subspace $W\subseteq {\mathbb{F}}^n$, let $d_j$ be the dimension of its projection map onto the first $j$ coordinates, with $d_0=0$. In any reduced row echelon form with row space $W$, $d_j$ counts the pivot columns among those first $j$ columns: later-pivot rows project to zero, while earlier-pivot rows remain linearly independent. Thus the positions where $d_j-d_{j-1}=1$ determine all pivot positions from $W$ alone.

If those positions are $p_1,\dots ,p_r$, the map sending $w\in W$ to $(w_{p_1},\dots ,w_{p_r})$ is an isomorphism $W\to {\mathbb{F}}^r$: the reduced nonzero rows map to the standard basis. Its $i$th row is therefore the unique vector of $W$ with pivot-coordinate values equal to the $i$th standard basis vector. Any other reduced row echelon form with that row space has identical nonzero rows in identical order. At a fixed matrix size, the remaining rows must be zero. Together with existence by Gaussian elimination, this proves uniqueness.

For row reduction with a parameterized right-hand side at $\tau =5$, the nonzero reduced equations areColumn 3 has no pivot element, so $x_3$ is a free variable in a linear system. Taking $x_3=t$ gives every solution:There is no further restriction on $t$. Independent substitution into the original linear system givesConversely, the two reduced equations force exactly these coordinate formulas, so no solutions have been omitted. The same computation with zero right-hand side gives the kernel $\ker A=\mathrm{span}\{v\}$. Its dimension is one. The rank-nullity theorem appears here as $2+1=3$: two pivot columns and one freely chosen coordinate account for the three unknowns.^[ref]

In general a consistent linear system has solution set $x_0+\ker A$. Subtracting $A{x}_0=b$ from $Ax=b$ proves this description in both directions. The particular solution places the affine subspace, and the kernel supplies its directions.^[ref] Over the real numbers this example is an entire line; over the rational numbers the same formula uses rational $t$. A free variable in a linear system over a finite field instead has only finitely many choices.

The pivot columns identify a different object: the original column space. Here they selectwhich form its basis, while $a_3=-7a_1+3a_2$. The corresponding reduced columns are $(1,0,0,0{)}^T$ and $(0,1,0,0{)}^T$; their linear span is a different plane. For example, every original output satisfies $y_3=y_1+y_2$, but $(1,0,0,0{)}^T$ does not. Since $E$ is an invertible matrix, column linear relations survive $EA=R$ even though the output vectors move. This is why choosing original columns at the reduced pivot positions works.

The reciprocal lattice of a body-centered cubic lattice is face-centred cubic, and that of a face-centered cubic lattice is body-centred cubic. In both cases its conventional cubic side is $4\pi /a$ if the direct conventional side is $a$. Thus a direct BCC-to-FCC phase transition reverses the reciprocal lattice types. The reciprocal primitive cell volume is $(2\pi {)}^3/\Omega$, so thermal expansion contracts reciprocal spacings, while a discontinuous change in direct primitive unit cell volume gives an inverse change in reciprocal cell volume.

A first-order structural transition need not continuously deform one perfect reciprocal array into another. Phase coexistence, new orientation variants, domains and loss of single-crystal alignment can produce two sets of diffraction peaks or many differently oriented sets. The two endpoint lattice types alone do not specify the transformation path or orientation relationship.

For a simple random walk on an infinite regular tree of degree $z\ge 2$, every vertex other than the chosen root has one neighbour closer to the root and $z-1$ farther away. The graph distance from the root is therefore a nearest-neighbour Markov chain on the nonnegative integers, with the displayed bulk transition probabilities and a certain outward step from zero. For shell probabilities $P_r(N)$,Shell probabilities include all vertices at the given distance. They differ from single-vertex probabilities by the shell size, $1$ at the root and $z(z-1{)}^{r-1}$ for $r\ge 1$.

Past-paper archive: www.sp.phy.cam.ac.uk/drp2/home. The archive lists papers for 2004–2018; worked solutions for these exams have not been verified. Lecture-problem solutions are not treated as past-exam solutions.

The binary-pulsar Einstein delay is the periodic clock-rate contribution to pulse arrival times from a pulsar moving in an eccentric binary. Near closest approach, the pulsar's orbital speed is larger and the companion's gravitational potential is deeper. Time dilation and gravitational redshift therefore make its clock run at a different rate relative to a distant time coordinate.

At leading order the variable timing term is ${\Delta }_E=\gamma \sin E$, with eccentric anomaly $E$ and time amplitude $\gamma$. Constant clock-rate offsets are absorbed into the fitted pulsar spin. This term is distinct from the geometric binary-pulsar Roemer delay and the propagation Shapiro time delay. Fitting it with other orbital observables constrains the component masses and tests general relativity.^[ref]

The difference between an observed arrival time and its model prediction. Positive pulsar timing residuals mean late arrivals in this convention. A periodic residual can expose an omitted orbit; a smooth long-term residual can also come from a wrong spin model or clock correction. Fitting and checking alternative causes are part of the inference, rather than interpreting every residual as gravitational radiation.

Pulsar dedispersion aligns radio wave arrival times in separate frequency channels before combining them. For an original illustrative model chooseBoth bracketed ratios are dimensionless, so $D$ supplies all the time units. The reference is $150\mathrm{MHz}$: delays are $0$ there, $0.13333\mathrm{s}$ at $100\mathrm{MHz}$, and $0.26833\mathrm{s}$ at $80\mathrm{MHz}$. These are chosen model values, not measurements of the first pulsar. Subtracting the delay aligns the peaks; summing channels first spreads a short pulse over a longer interval and reduces its peak. Finite channel width and incorrect delay parameters leave residual smearing.

The integrated electron number density along the line of sight, conventionally quoted in $\mathrm{pc}\mathrm{c}{\mathrm{m}}^{-3}$. A dispersion measure is a column, not a unique distance: converting it into distance requires a model of the interstellar medium. A clumpy nearby column and a longer dilute column can give the same pulsar dispersion delay.

For fixed strict preferences, the same output of the Gale-Shapley algorithm gives each receiver its worst partner obtainable in any stable matching, treating unmatched as last. "Worst" is restricted to stable possibilities: a receiver might still receive its favourite acceptable partner if every stable matching assigns it that partner. Reversing the proposing side gives receivers their best stable partners instead.

For a short proof, let $M$ be a proposer-optimal stable matching and $S$ any other stable matching. If a receiver $r$ preferred $M(r)=\ell$ to $S(r)$, then proposer optimality would make $\ell$ prefer $r=M(\ell )$ to $S(\ell )$, since the partners differ and preferences are strict. The pair $(\ell ,r)$ would block $S$. Therefore no receiver prefers its partner in $M$ to its partner in $S$. In the worked example, right-side ranks in $M_L$ are $2,2,3$ for $X,Y,Z$, whereas $M_R$ gives them $1,1,2$. Neither outcome is universally best for both sides.

For a random variable $X$ with probability density function $f_X$ on an interval, let $Y=h(X)$, where $h$ is a diffeomorphism onto another interval. The change of variables formula givesIndeed, integrating the right-hand side over an outcome interval gives the probability that $X$ lies in its inverse image. The absolute derivative preserves positive probability when $h$ reverses orientation. A gravity deflection of an atomic beam is a physical example: its speed-to-height transformation changes the probability density and its mode.

Columbia University's institutional laboratory manual, Experiment 1, Glassware Calibration, Primary and Secondary Standards, and Manual Titrations, Part 3, specifies two weighed $0.2$–$0.3\mathrm{g}$ portions of potassium hydrogen phthalate, transferred completely into separate $250\mathrm{mL}$ Erlenmeyer flasks. Dissolve each in approximately $100\mathrm{mL}$ distilled water; any warmed sample cools to room temperature before titration. Add two or three drops of phenolphthalein. Rinse and condition a clamped burette with the diluted sodium hydroxide, then titrate while swirling to faint pink lasting about $20$ seconds. Calculate concentration from each portion's own mass and delivered titre; repeat if results scatter. Part 1 identifies a Mettler balance, and Part 2 includes paired weighings of exposed sodium hydroxide and potassium hydrogen phthalate. These are prescribed procedures, not reported measurements.^[ref]

The same manual directs keeping the diluted sodium hydroxide covered and limiting air exposure after titrant standardization.^[ref] The MIT handout explains the associated carbon dioxide problem: carbonate contamination of sodium hydroxide is controlled during preparation, bottles are promptly capped, and atmospheric carbon dioxide makes the pink phenolphthalein endpoint fade.^[ref] Consequently, a standardized concentration applies to the stored chemical solution under its handling conditions; continued uptake of carbon dioxide can change the reacting inventory per delivered volume.

The color is the readout, and stoichiometry supplies the inference. Phenolphthalein indicates a pH range, so its first reproducible pale pink is an operational titration endpoint, rather than a direct observation of equal reacting inventories. A potentiometric titration can locate an endpoint from a pH versus volume curve, but glass electrode calibration alone still does not determine sodium hydroxide concentration. Repeat portions test scatter; incomplete dissolution, an unmatched titration blank, or a shared endpoint error can bias all repeats together.

Let $m$ be the transferred sample mass, $p$ the known mass fraction of the intended chemical compound, and $M$ its molar mass. In a model with complete dissolution and chemically inactive impurities,The purity fraction is dimensionless: $99.95\%$ would mean $p=0.9995$, not $99.95$. These equations assume that the impurity contributes no titrant demand; an impurity that also reacts needs its own inventory equation. The value of $p$ must refer to the same preparation state as $m$, for example the specified dry material. A mass fraction assigned before unaccounted uptake of water cannot simply be used after it.

For an original worked example, stipulate pure potassium hydrogen phthalate, abbreviated KHP, so $p=1$, a selected mass $m=0.51055\mathrm{g}$, and rounded molar mass $M=204.22\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. The NIST Chemistry WebBook lists formula ${\mathrm{C}}_8{\mathrm{H}}_5\mathrm{K}{\mathrm{O}}_4$ and molecular weight $204.2212$, consistent with the rounded molar mass used here.^[ref] The selected mass, purity assumption and volumes in this example are chosen inputs, not measured results or a claim about a real batch's purity.

The acid-bearing anion of potassium hydrogen phthalate loses one proton to hydroxide. The net chemical equation isThus sodium hydroxide has one-to-one stoichiometry with potassium hydrogen phthalate in this neutralization, givingChoose a delivered endpoint titre of $24.85\mathrm{mL}$ and matched titration blank of $0.05\mathrm{mL}$. ThenThe extra displayed digits show the arithmetic of the stipulated model; they do not establish experimental accuracy. A compact rounded result is $0.1008\mathrm{mol}{\mathrm{L}}^{-1}$.

When the entire weighed portion is titrated, added water changes concentration and the pH path but preserves the initial acid inventory. The dissolution volume therefore does not appear in $n_s=mp/M$. If instead a volumetric flask prepares a stock and a volumetric pipette selects an aliquot, the selected inventory is $n_s{V}_a/V_f$ for homogeneous stock, with delivered aliquot volume $V_a$ and final stock volume $V_f$. The same bookkeeping explains why complete transfer matters even when the amount of dissolution water is approximate.

A resistance pressure gauge is a pressure sensor whose electrical resistance is calibrated against pressure. Even a simple local response needs both pressure and temperature. For an original illustrative model definewhere $R_0$ is the reference electrical resistance at $p=0$, $T=T_0$, and zero history offset. The derivatives $\alpha$ and $\beta$ are local sensitivities with units $\mathrm{P}{\mathrm{a}}^{-1}$ and ${\mathrm{K}}^{-1}$, respectively; $h$ is a dimensionless history offset. Constant sensitivities are an approximation over a chosen range, not material constants valid at every pressure.

Choose synthetic coefficients $\alpha =0.020\mathrm{GP}{\mathrm{a}}^{-1}$ and $\beta =2.0\times 10^{-4}{\mathrm{K}}^{-1}$, with $T-T_0=10\mathrm{K}$ and $h=0$. A measured ratio $R/R_0=1.022$ then gives $p=(0.022-0.002)/0.020=1.00\mathrm{GPa}$. Ignoring the temperature change gives $1.10\mathrm{GPa}$, a $0.10\mathrm{GPa}$ error. This model is deliberately stipulated; these numbers are not measured manganin coefficients or Bridgman's data.

If two histories differ by $\Delta h=0.0004$, their outputs at the same pressure differ by $0.0004R_0$; interpreting both with a single history-free curve gives an apparent pressure separation $\Delta h/\alpha =0.020\mathrm{GPa}$. The half separation is $0.010\mathrm{GPa}$. Reversing the drive can additionally cause mechanical friction hysteresis, independently of this sensor model. Record both the electrical calibration history and the loading direction.

For independent small standard measurement uncertainties $u_y,u_T,u_h$ and exact stipulated $\alpha ,\beta$, first-order propagation givesFor $u_y=10^{-4}$, $u_T=0.5\mathrm{K}$, and $u_h=10^{-4}$, this is $u_p=0.00866\mathrm{GPa}$. Real calibration also includes coefficient measurement uncertainties, correlations, reference-point measurement uncertainty, and any unresolved nonhydrostatic loading. A sensor calibrates its local environment; it cannot by itself certify the absence of deviatoric stress throughout the sample.

For polar coordinates $x=r\cos \theta$, $y=r\sin \theta$ with $r\ge 0$, the Jacobian determinant is $r$. The change of variables formula therefore gives the area element $dA=rdrd\theta$. Integrating over the radial interval $r_1(\theta )\le r\le r_2(\theta )$ gives $[r_2(\theta {)}^2-r_1(\theta {)}^2]/2$ per unit angle. Bounds must describe the intended region without counting it more than once.

For boundary value $c$ on $(-a,a)$ and zero outside, integrating the Poisson kernel for the upper half-plane gives the displayed bounded harmonic function for $y>0$. It approaches the assigned values at boundary continuity points and the midpoint value $c/2$ at either jump. Changing the value at an isolated endpoint does not alter the integral. This is the unique bounded harmonic extension; pointwise decay in the vertical direction alone, without a spatial growth restriction, does not give uniqueness.

Consider an ideal pressure-driven closure plug with a large chamber-facing projected area $A_p$. Its opposite side seats against a smaller annular packing surface area $A_{\mathrm{seal}}$. The remaining projected part $A_u=A_p-A_{\mathrm{seal}}>0$ is the unsupported area: it has no opposing packing contact in that plane. Here $p$ is the chamber pressure relative to the exterior. Let $q$ denote the mean axial compressive contact stress on the packing, not the chamber's hydrostatic stress.

Neglect friction, the plug's weight, and other axial reactions. The plug's Newton's second law in static balance isThe load is the same on both sides; its smaller supporting surface area raises the packing's mean contact stress. For a stipulated ratio $A_p/A_{\mathrm{seal}}=1.25$ and $p=1\mathrm{GPa}$, the ideal seating stress is $q=1.25\mathrm{GPa}$. Its excess over chamber pressure is $0.25\mathrm{GPa}$. Reducing $A_u$ to zero removes this amplification. This is the elementary force balance behind self-energizing seating, consistent with the lecture's Figure 1 caption.^[ref]

The relation is a model, not a universal sealing law. The mean axial contact stress need not equal the local contact stress along every possible leakage path. Confinement and plasticity in continuum mechanics determine whether the packing closes a gap or instead extrudes into it. Friction diverts load into other surfaces; material plasticity in continuum mechanics changes contact geometry; the vessel must carry its own stress. A finite initial seating load may also be needed before chamber pressure can develop. An arbitrarily small $A_{\mathrm{seal}}$ does not yield arbitrarily useful sealing: the assumptions fail when local deformation, extrusion or structural failure takes over.