For speed $v=k{r}^2$ in the punctured plane, the travel-time element is $ds/(k{r}^2)$. Under the inversion in a circle with radial coordinate $u=1/r$, this becomes $k^{-1}\sqrt{d{u}^2+u^{2}d{\theta }^2}$, the arc length in Euclidean space divided by $k$. Thus a shortest travel-time route corresponds to a straight line segment in the inverted plane, whenever that segment avoids the origin and lies in the allowed region. Between $(a,0)$ and $(0,a)$ the inverted segment has equation $u(\cos \theta +\sin \theta )=1/a$. The route is $r=a(\cos \theta +\sin \theta )$, and its minimum travel time is $\sqrt{2}/(ka)$. The triangle inequality for arc length proves global minimality.

The brachistochrone problem asks for the curve along which a particle released from rest reaches a specified target fastest under a uniform gravitational field, without friction. With downward depth $u$, conservation of energy gives speed $\sqrt{2gu}$ and travel time $\int \sqrt{(1+u^{\prime 2})/(2gu)}dx$. The Beltrami identity makes $u(1+u^{\prime 2})$ constant, leading to a cycloid. If only the horizontal endpoint coordinate $x_0>0$ is fixed, the natural boundary conditions for a free endpoint select the half arch ending at its deepest point, with depth $2x_0/\pi$ and minimum time $\sqrt{\pi x_0/g}$.

For constant $K$, this quadratic combination of two-by-two gradient minors is a divergence. Indeed,because the term containing $u_{j,\beta \alpha }$ vanishes by antisymmetry of the Levi-Civita symbol. Its volume integral vanishes for homogeneous displacement boundary data. It also vanishes pointwise on rank-one matrices, since ${\nu }_{\alpha }{\nu }_{\beta }$ is symmetric. Consequently one can strengthen an elastic energy density by such a term without changing fixed-boundary incremental energies or acoustic tensors.

A cache miss on the first access to a cache block in an initially empty memory cache with no fetching ahead of requests. Larger capacity cannot eliminate these first fetches; subsequent cache misses can have other causes.

The additional memory access latency charged for a cache miss beyond a common lookup time in the average memory access time model. It includes obtaining the requested cache block and making the read complete; this model excludes extra costs from writes and overlapping requests.

A repeated cache block access that misses in a restricted design but hits in the comparison fully associative cache of the same capacity and cache line size using LRU replacement. Read 4 of the direct-mapped cache comparison under cache miss isolates this effect while another cache slot is unused.

A repeated cache block access that misses even in the comparison fully associative cache of the same capacity and cache line size using LRU replacement. In the memory cache example these are reads 6, 7 and 11. The replacement assumption matters to this classification.

The fraction of $R>0$ memory references yielding cache misses, with $M$ their count: $m=M/R$. When each reference is classified as exactly one cache hit or cache miss, $h+m=1$; the two-slot fully associative cache example has $m=7/12$.

A resident storage entry that can hold one cache block and metadata identifying its contents. Cache line identity belongs to backing computer memory; cache slot identity belongs to the consulted memory cache.

The fraction of $R>0$ memory references yielding cache hits, with $H$ their count: $h=H/R$. The two-slot fully associative cache example has $h=5/12$. Count the same reference unit and consulted level for both cache hit rate and cache miss rate.

An axisymmetric Gaussian Burgers vortex with constant circulation in axial strain $\alpha$ has core area parameter $s={\delta }^2$ satisfying $\dot{s}+\alpha s=4\nu$. Thus $s=4\nu /\alpha +(s_0-4\nu /\alpha )e^{-\alpha t}$, an attracting balance of transverse compression and viscous diffusion for every positive finite initial radius.

The swirl viscous dissipation of a steady Burgers vortex, integrated per unit length and per unit density, is $\nu \int {\omega }_z^{2}dA=\alpha {\Gamma }^2/(8\pi )$. It remains finite as kinematic viscosity tends to zero at fixed strain and circulation. This is excess over the background strain: the uniform strain itself has infinite total viscous dissipation over an infinite cross-section.

A planar strained shear layer is a Burgers vortex sheet. For strain $u_x=-\alpha x$, $u_z=\alpha z$ and transverse velocity $u_y=U\mathrm{erf}(x/{\delta }_s)$, viscous balance gives ${\delta }_s^2=2\nu /\alpha$. At fixed velocity jump its excess viscous dissipation per sheet area scales as $U^2\sqrt{\nu \alpha }$, which vanishes as kinematic viscosity tends to zero.

Neutral buoyancy means a fully immersed body's upward buoyancy equals its downward force from gravity. For a body of fixed volume in a constant-mass density fluid under a uniform gravitational field, this occurs when its mean mass density equals the surrounding mass density. A $0.10\mathrm{m}$ cube of mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$ in the stipulated water has mass $1.000\mathrm{kg}$, so both forces are $9.81\mathrm{N}$.

Ideally it can remain at rest at any depth for which it stays fully immersed and clear of boundaries. Raising or lowering it changes the absolute pressure, but neither its displaced volume nor the difference between the vertical pressure forces. There is therefore no vertical restoring force toward a preferred depth in this model. Neutral buoyancy is a vertical force balance, not a claim about resistance to tilting or about a compressible body whose volume changes with pressure.

Archimedes' principle states that buoyancy equals the downward gravitational weight, or weight, of the displaced fluid. In a resting fluid of constant mass density ${\rho }_f$ under uniform gravitational acceleration $g$, its magnitude isHere $V_{\mathrm{displaced}}$ is the volume from which the body excludes the fluid: the whole external volume for a fully immersed solid body, and just the immersed volume for a partly immersed one. The body's own mass does not enter this buoyancy formula. ^[ref]

A short proof outline for a fully immersed body is to imagine replacing it by exactly the fluid it displaces. The surrounding hydrostatic pressure is unchanged. That replacement fluid is at rest, so its upward pressure force must balance its downward force from gravity. The original body receives the same inward pressure forces on the same closed surface, hence the same upward buoyancy.

Equivalently, let $\Omega$ be the displaced region, $\partial \Omega$ its closed surface and ${\mathbf{e}}_z$ the upward unit direction. The divergence theorem, applied component by component to the surface integral, and hydrostatic pressure giveThis also explains why horizontal forces cancel even for an irregular shape. A spatially constant surface pressure contributes zero to the closed-surface integral. If the mass density varies with depth, replace ${\rho }_f{V}_{\mathrm{displaced}}$ by the mass ${\int }_{\Omega }{\rho }_{f}dV$ of the displaced fluid.

The closed-surface argument assumes that the fluid applies pressure over the entire boundary of a fully immersed body. A body sealing against a tank bottom may have no fluid beneath it; then the lower-face pressure force used in the proof is absent, and a supporting contact force must be treated separately. For partial immersion, the usual buoyancy formula assumes the exposed surface receives the common atmospheric pressure, neglects the much smaller buoyancy from the air, and neglects surface tension.

For the $0.10\mathrm{m}$ cube in the stipulated water, full immersion displaces $0.001{\mathrm{m}}^3$, so $F_b=1000\times 9.81\times 0.001=9.81\mathrm{N}$. If its uniform mass density is $600\mathrm{kg}{\mathrm{m}}^{-3}$, its mass is $m=600\times 0.001=0.600\mathrm{kg}$ and its downward force from gravity is $mg=5.886\mathrm{N}$. To hold it fully immersed at rest requires an additional downward forceOn release from rest it initially moves upward; its eventual floating equilibrium has a smaller displaced volume. A same-size cube of uniform mass density $1200\mathrm{kg}{\mathrm{m}}^{-3}$ has mass $1.200\mathrm{kg}$ and downward force from gravity $11.772\mathrm{N}$. Full immersion still gives $9.81\mathrm{N}$ of buoyancy, leaving an initial net downward force of $11.772-9.81=1.962\mathrm{N}$. Subsequent motion also involves drag.

Magnetic buoyancy is the tendency of magnetized fluid to rise when magnetic pressure support reduces its density relative to its surroundings. For a thin flux concentration in lateral balance, $p_{\mathrm{in}}+B^2/(2{\mu }_0)=p_{\mathrm{out}}$. At equal temperature and composition an ideal-gas equation of state then gives ${\rho }_{\mathrm{in}}<{\rho }_{\mathrm{out}}$, producing an upward buoyant force. Thermal exchange, magnetic tension and stable stratification modify this simple argument. In a solar dynamo, buoyant toroidal flux can emerge as tilted bipolar active regions.

The spatial gradient of buoyancy determines the buoyancy anomaly produced by a small parcel displacement: $b^{\prime }=-\mathbf{\xi }\cdot \nabla \bar{b}$. Its vertical component is the squared buoyancy frequency in a stably stratified Boussinesq approximation. A horizontal component enters thermal wind and the propagation of a Boundary Rossby wave.

A proposed modern measurement extends the weak acid dissociation measurement by estimating the local slope of an acid-base titration. Transfer a measured sample with a pipette into a stirred vessel; clamp a combination glass electrode/reference electrode clear of the stirring magnet and place a temperature probe nearby. A calibrated burette delivers a chemical base of independently standardized concentration $c_T$, such as sodium hydroxide. Record the initial sample volume $V_0$, cumulative delivered volume $v$, stable pH, and temperature after each small addition. Use a separate fresh portion for an acidward series. These are proposed measurement conditions, not reported experimental results.

For the instrument calibration, use fresh standard buffer solutions bracketing the measured pH range, with values appropriate to the actual temperature. The Hanna manual specifies separate rinse and calibration portions, gentle stirring, and approximately $4\mathrm{cm}$ immersion of the probe tip. It calls for stable readings; the sensing bulb and reference electrode junction must remain immersed. Rinse between standards and samples and avoid carrying rinse water into the measured portion.^[ref]

The readout pairs each added amount of substance with the measured pH. With negligible added volume, neighboring readings give the finite difference estimateassigned to the interval's midpoint. High buffer capacity appears as a small change of pH for a measured reagent dose. Choose increments small enough to resolve changing buffer capacity, but large enough that the change of pH exceeds measurement noise. Repeat with a fresh portion and different increment sizes; remeasure a standard afterward to check drift. A water blank can reveal contamination, but its pH must not be subtracted from the sample pH. These controls are proposed tests of the measurement and model assumptions.

If additions appreciably increase volume, even the analytical concentration changes: $C(v)=C_0{V}_0/(V_0+v)$. Merely dividing the added moles by the current volume does not restore the constant-volume derivative. One alternative is explicitly to report a changing-volume convention. For $c(v)=c_{T}v/(V_0+v)$, defineThis definition includes the reagent's changing concentration and differs from the constant-volume buffer capacity defined above.^[ref]

To compare with the constant-volume theory, fit the changing-volume chemical equilibrium model instead of treating the observed slope as $\beta$. In the same ideal monoprotic acid system, define $S_0=[{\mathrm{A}}^{-}{]}_0+[{\mathrm{OH}}^{-}{]}_0-[{\mathrm{H}}^{+}{]}_0$. For baseward additions, the electric charge balance at each reading isUse a minus sign before $c_{T}v$ for the corresponding acidward series. From the model parameters, evaluate the constant-volume buffer capacity at the chosen analytical concentration and pH. The fit also tests whether one acid dissociation constant describes the observations.

Limitations include incomplete mixing, slow glass electrode response, uncertain reagent concentration, and changes in temperature or ionic strength. Dissolved atmospheric carbon dioxide introduces an additional chemical acid system; comparable exposure and freshly prepared samples help distinguish it from the intended monoprotic acid response. Additional reacting chemical species and nonideal activity coefficients invalidate the simple formula. Especially when pH increments are tiny, measurement noise can dominate the inferred derivative; reporting the dose, pH interval, volume convention, and repeatability is part of reporting a meaningful buffer capacity.

Prepare the same ideal monoprotic acid model at initial pH $5$, with appropriate accompanying ions to balance electric charge. To predict the new pH after a finite signed addition $\Delta b$, solve $Q({\mathrm{pH}}_1)-Q(5)=\Delta b$ using the chemical equilibrium expression above. For $\Delta b=0.002\mathrm{mol}{\mathrm{L}}^{-1}$ of added chemical base, the calculated final pH is $5.366798$ at $C=0.010\mathrm{mol}{\mathrm{L}}^{-1}$, but $5.034749$ at $C=0.100\mathrm{mol}{\mathrm{L}}^{-1}$. The initial local approximation gives changes of about $0.34605$ and $0.034730$, respectively; its accuracy decreases as the partner balance changes appreciably.

For the dilute example, additions beyond approximately $0.005\mathrm{mol}{\mathrm{L}}^{-1}$ in either direction exceed the initial inventory of one partner. The full chemical equilibrium calculation still produces a finite pH; an inventory-only Henderson-Hasselbalch equation calculation would attempt to use a negative partner quantity. At $+0.006\mathrm{mol}{\mathrm{L}}^{-1}$ the dilute example reaches pH $10.99564$, whereas the concentrated example reaches only $5.104698$. This demonstrates a finite operating range, not an abrupt cessation of all buffer capacity.

Stipulate pKa $5$ and $K_w=10^{-14}$; these are model inputs, not measurements of an identified substance. At pH $5$ the pair contribution is $0.575646\dots C$. For analytical concentrations $C=0.010$ and $0.100\mathrm{mol}{\mathrm{L}}^{-1}$, respectively, it isThe free hydrogen ion and hydroxide contribution is the same at this pH, ${\beta }_{\mathrm{water}}=(\ln 10)(10^{-5}+10^{-9})\mathrm{mol}{\mathrm{L}}^{-1}=0.0000230282\dots \mathrm{mol}{\mathrm{L}}^{-1}$ per pH. The full buffer capacities are therefore approximately $0.00577949$ and $0.0575877\mathrm{mol}{\mathrm{L}}^{-1}$ per pH. Identical pH fixes the free hydrogen ion thermodynamic activity; it does not fix the inventory of weak acid and conjugate base available for neutralization.

At one unit of pH on either side of pKa, $r=10$ or $0.1$, giving ${\beta }_{\mathrm{pair}}/{\beta }_{\mathrm{pair},\max }=40/121\approx 0.331$. The partners still form a buffer solution, but its local resistance is only about one third of the maximum for that analytical concentration. At fixed pH, the pair contribution scales exactly with $C$; the full buffer capacity contains the additional $C$-independent water contribution. Consequently, extreme dilution does not justify dropping self-ionization of water.