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For two observed layers $z_2>z_1$, define the positive logarithmic count slopeSince $k_B=R/N_A$, where $R$ is the molar gas constant, the sedimentation scale height givesSeveral heights check whether one slope fits the whole profile. A mixture of grain radii generally gives a sum of exponential functions, with the larger grains depleted faster aloft; one apparent radius need not describe every height.

An independent readout uses horizontal displacements at lag $\tau$, avoiding the vertical gravitational drift. Define the one-coordinate mean squared displacement $M_x(\tau )=⟨[x(t+\tau )-x(t){]}^2⟩$. In a stationary liquid without horizontal advection, translational diffusion predictsA camera sees two horizontal coordinates, so its squared radial displacement has twice the one-coordinate mean. Neither observable is path length or instantaneous velocity. The bulk Stokes–Einstein relation, with dynamic viscosity $\eta$, then gives

The 1926 lecture compares displacement-based and equilibrium determinations of the Avogadro constant.^[ref]

Rotational diffusion supplies a further observable when an identifiable grain orientation can be followed; centre positions alone cannot measure it.

For the chosen model above and $\eta =1.00\times 10^{-3}\mathrm{Pa}\mathrm{s}$, the bulk prediction is $D\simeq 0.8731\mu {\mathrm{m}}^2{\mathrm{s}}^{-1}$. At $\tau =10\mathrm{s}$, $M_x\simeq 17.46\mu {\mathrm{m}}^2$ and the planar value is $34.92\mu {\mathrm{m}}^2$. Reconstructing $N_A$ from these stipulated model outputs returns $6.02214076\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$ by construction; this arithmetic consistency check is not experimental confirmation. Historical force comes from comparing separately measured profiles and displacements using independently checked material parameters.

The diffusive flux and gravitational particle flux provide another useful connection. With constant drag coefficient $\zeta$, zero vertical net flux readswhere $v_s$ is the downward settling velocity magnitude. With $\zeta =6\pi \eta a$, the chosen model gives $v_s\simeq 0.02725\mu \mathrm{m}{\mathrm{s}}^{-1}$. The equilibrium profile forgets how rapidly it was established, while displacement measurements retain that rate information.

Choose an ideal absorbing chemical species with $\epsilon =8000\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}\mathrm{c}{\mathrm{m}}^{-1}$ at a fixed wavelength, and a cuvette with geometric length $l=1\mathrm{cm}$. These are stipulated model parameters. A micromolar molar concentration is $1\mu \mathrm{M}=10^{-6}\mathrm{mol}{\mathrm{L}}^{-1}$, so each additional $25\mu \mathrm{M}$ gives $\Delta A=8000\times 1\times 25\times 10^{-6}=0.2$. Five chosen standards therefore predictThe absorbance graph is a straight line of slope $0.008(\mu \mathrm{M}{)}^{-1}$, while the transmittance graph shows exponential decay. Equal increments of molar concentration multiply transmittance by $10^{-0.2}\approx 0.63096$; they do not subtract a fixed transmitted fraction.

Suppose a measurement returns the stipulated blank-normalized reading $T=0.40$ under those same conditions. The inferred absorbance and molar concentration areThis lies slightly below the $50\mu \mathrm{M}$ standard because $0.40$ is slightly above $10^{-0.4}\approx 0.39811$. A calibration determined experimentally by linear regression may instead have $A=mc+b$. Then use $c=(A-b)/m$, keeping the slope $m$ and intercept $b$ in their measured units. A nonzero intercept calls for checking the reference blank and sample preparation; it does not uniquely identify the cause.

The capacitance of a linear two-terminal capacitor is the constant relating its separated electric charge to its terminal electric potential difference:Here $Q$ is the electric charge on one electrical conductor, the other carries $-Q$, and $V$ is their electric potential difference with the same ordering. Reversing the terminal ordering reverses both $Q$ and $V$ and leaves capacitance unchanged. An ideal passive capacitor has $C\ge 0$; an ordinary pair of separated electrical conductors with positive permittivity has $C>0$. At $V=0$, use $Q=CV$ rather than the undefined ratio $0/0$.

For example, electric charge magnitude $Q=0.8854\mathrm{nC}$ at terminal electric potential difference $V=10\mathrm{V}$ gives capacitance $C=88.54\mathrm{pF}$. At $20\mathrm{V}$, the same linear capacitor has $Q=1.7708\mathrm{nC}$ and the same capacitance. The plots in capacitor energy show that capacitance is the slope of the electric charge versus terminal electric potential difference graph.

For fixed geometry and linear material response, capacitance depends on dimensions and permittivity, rather than the source's terminal electric potential difference. This is why capacitance is a useful property of a component: once it is known, $Q=CV$ predicts the separated electric charge for each operating terminal electric potential difference. The SI unit is the farad, $1\mathrm{F}=1C/V$. ^[ref].

For combinations of capacitors, the same terminal definition gives an equivalent capacitance. Two identical capacitors of capacitance $C$ connected in parallel share the same terminal electric potential difference, and their electric charges add, giving $C_{\mathrm{eq}}=2C$. Connected in series with an initially neutral isolated junction, they carry equal plate electric charge magnitudes and their terminal electric potential differences add, giving $C_{\mathrm{eq}}=C/2$. ^[ref].

A force gauge reads the pulling force. For horizontal straight-line sliding at constant velocity, its reading equals the opposing kinetic friction magnitude, provided these are the only horizontal forces. For total supported mass $m$, including added loads, $N=mg$, so a mean sliding reading $\overline{F}$ gives ${\mu }_k=\overline{F}/(mg)$. The important readout is the sustained sliding force after the starting peak; the peak belongs to the static friction threshold.

The University of Mississippi's Experiment 6 uses a wooden block on a wooden plane, a triple-beam balance, added masses, and a dual-range 50 N force probe connected through LabPro to Logger Pro. Its procedure calibrates the probe with a $0.5\mathrm{kg}$ load on a hanger, entering the corresponding force in newtons, and measures the block's mass separately. With the plane horizontal, it starts with $0.5\mathrm{kg}$ on the block and pulls horizontally at constant speed. The recorded force trace has a starting peak; the procedure selects the subsequent constant-force interval and takes its mean. It repeats with $0.5\mathrm{kg}$ increments up to $3\mathrm{kg}$ added, then plots sliding force against normal force to estimate the coefficient of kinetic friction from the slope.^[ref]

For a replication, check the gauge's zero with the string slack and its response to known gravitational weights; zeroing alone does not check its scale. Keep the plane level, the pull parallel to it, and the block face and track region consistent. Specify the contacting materials and whether the surfaces are dry, cleaned or lubricated; repeat runs to see whether debris or wear changes the sliding force. Verify constant velocity independently from position versus time, rather than assuming that a flat force trace proves zero acceleration. These are proposed measurement controls, not additional observations from the laboratory writeup. Record the selected interval, variation between runs and gauge resolution as sources of measurement error.

The derivative measures the instantaneous rate at which a function changes. For a real function $f$ defined near $a$, it exists when the difference quotient has a finite two-sided limit:The quotient is the slope of the straight line joining two nearby points of the graph. Its limit is the slope of the tangent line. Thus a derivative describes a local rate without choosing a particular finite measurement interval.

For $f(x)=x^2$, expanding the numerator givesAt $a=1$, the increments $h=1$, $1/2$ and $1/10$ give secant line slopes $3$, $2.5$ and $2.1$. The limit is $2$, and the tangent line is $y=1+2(x-1)$. The picture makes the limiting process visible.

The same calculation shows why the derivative is useful for approximation:This is a local linear approximation. Near $a=1$, replacing $(1+h{)}^2$ by $1+2h$ makes the exact error $h^2$, smaller than the leading change when $h$ is small. A curved graph can therefore be studied locally using a straight line.

If the function records position against time, its derivative is velocity. If it records an objective function, its derivative tests local changes that matter in optimization; in several variables, the gradient collects these directional rates and guides gradient descent. The interpretation comes from the function and its units, rather than from a new rule for each application.

Existence matters. For the absolute value function $f(x)=|x|$, the one-sided derivatives at zero are $-1$ and $1$, so there is no single tangent line described by a finite derivative there. Nor does a zero derivative alone identify a maximum or minimum: $f(x)=x^3$ has $f^{\prime }(0)=0$ but keeps increasing through zero.

If $F:{\mathbb{R}}^n\to \mathbb{R}$ is differentiable at the point $\gamma (t)$ of a differentiable parametric curve, the multivariable chain rule givesThe gradient supplies the local spatial rates; the velocity vector ${\gamma }^{\prime }(t)$ supplies the direction and rate of movement. Their dot product is the change in the sampled scalar field per unit parameter. If the parameter is time, this is a rate per unit time; a directional derivative per unit distance instead uses a unit vector.

Take $F(x,y)=x^2+2y^2$ and the unit circle path $\gamma (t)=(\cos t,\sin t)$. Along the path,At $t=\pi /4$, the gradient is $(\sqrt{2},2\sqrt{2})$, the velocity vector is $(-1/\sqrt{2},1/\sqrt{2})$, and their dot product is $-1+2=1$. The sampled value is $3/2$ and increases at rate one, even though one coordinate is decreasing. This is why a moving observation must combine every coordinate's contribution.

For a level curve, $F(\gamma (t))$ is constant, so the derivative along a curve is zero and its velocity vector is orthogonal to the gradient. Along other paths, the same formula combines a field's geometry with the observer's motion. ^[ref].

A motivating laboratory realization is The Hall Effect and Conductivity in Germanium by Rex Tayloe. The handout specifies a germanium sample holder with temperature and magnetic field control, a low-voltage supply, digital meters, and a current-reversing switch. Measure the sample dimensions, map the contacts, and establish the directions of electric current, magnetic field, and transverse voltage. At fixed temperature and zero magnetic field, sweep positive and negative electric currents and record longitudinal electric potential differences. A straight voltage-versus-electric current relation tests Ohm's law; sample geometry converts its slope to electrical resistivity. For the Hall effect, minimize the zero-field transverse offset and measure transverse voltage at all four combinations of $\pm I$ and $\pm B$, repeating for several electric currents. The readouts test longitudinal proportionality, transverse polarity, and transverse proportionality to electric current. These are protocol instructions, not reported measurement outcomes.^[ref]

For interpreting such measurements, an ideal uniform bar of wire cross-sectional area $A$ with longitudinal sensing contacts separated by $L$ gives $\sigma =LI/(A{V}_L)$. Four-terminal sensing suppresses voltage drops in current contacts when the voltage meter draws negligible electric current; it does not fix uncertain dimensions or a nonuniform current density. Reverse electric current and use $[V_L(+I)-V_L(-I)]/2$ to reject a constant voltage offset. Keep temperature stable, allow the current density to settle, and reduce electric current to test for Joule heating. Those are controls for interpreting the material coefficient, rather than reasons to assume every measured curve is linear.

For the transverse measurement define $\mathbf{J}=J_x\hat{\mathbf{x}}$, $\mathbf{B}=B_z\hat{\mathbf{z}}$, width $w$ along $y$, and thickness $d$ along $z$. Define $V_H=V(y=-w/2)-V(y=+w/2)=E_{y}w$, so $J_x=I/(wd)$ givesExtract the part odd in both electric current and magnetic field usingThis combination rejects a constant offset and longitudinal pickup even in magnetic field. It cannot reject another signal with the same two reversal symmetries. Check field polarity and magnitude independently; a sign error in the contact map changes the inferred Hall coefficient.

For one Drude model species in transverse open circuit, the steady Lorentz force balance is $0=q(\mathbf{E}+{\mathbf{v}}_d\times \mathbf{B})-m{\mathbf{v}}_d/\tau$ with $v_{d,y}=0$. Its $y$ component gives $E_y=v_{d,x}{B}_z$. ConsequentlyLongitudinal electrical conductivity and the Hall coefficient therefore separate number density from charge carrier mobility in this single-species model. Inferring a momentum relaxation time additionally requires an appropriate mass. A constant-field longitudinal measurement alone cannot supply all three.

A Hall coefficient is not universally the reciprocal of signed number density. For example, adding independent electron and electron hole contributions with densities $n,p$ and positive charge carrier mobilities ${\mu }_e,{\mu }_h$ in the weak-magnetic field limit givesThe Hall coefficient can vanish even while electrical conductivity is positive, and its sign weights charge carrier mobility as well as number density. The germanium handout explicitly asks students to track possible transverse sign changes with temperature; it does not establish a universal single-carrier interpretation. Real electronic band structure and unequal scattering also limit that interpretation.^[ref]

Use the Malthusian delay differential equation with time measured in units of its constant delay:Here $a$ is a real constant. On the first interval $y(t-1)=1$, hence $y(t)=1+at$. On the next interval the delayed input is $1+a(t-1)$, already known, so the method of steps for a delay differential equation givesEach polynomial integrates the preceding interval's delayed polynomial, and its integration constant ensures continuity. At $a=1$, the step endpoints are $y(1)=2$, $y(2)=7/2$, and $y(3)=37/6$. At $t=3/2$, the value is $21/8$, but its derivative is the earlier value $y(1/2)=3/2$. This is the geometric distinction from the ordinary differential equation $y^{\prime }=y$, whose derivative would equal its current value.

More generally, for any integer $n\ge 0$,At an integer join the newly introduced term vanishes. Inside a step, differentiation lowers each power by one and gives exactly $a$ times the preceding step evaluated at $t-1$. Together with $y(0)=1$, this verifies the formula by mathematical induction; it is an exact construction, not a truncated power series.

The history and the new equation need not prescribe matching derivatives at the starting time; their mismatch can reappear later in higher derivatives.^[ref] In this example, $y^{\prime }(0^{-})=0$ and $y^{\prime }(0^{+})=a$. For $a\ne 0$, $y^{\prime }$ is continuous at $t=1$ but $y^{\prime \prime }$ jumps there by $a^2$; at $t=2$, the first jump is in $y^{\prime \prime \prime }$, of size $a^3$. The solution of a differential equation is therefore continuously differentiable for positive time, although it need not have continuous derivatives of every order at step boundaries.

In a gas syringe, the graduated barrel reading $V_{\mathrm{read}}$ can omit a fixed connecting or sensor volume $V_d$ occupied by the same sealed gas. For a common settled temperature, Boyle's law then requiresThis second relation makes a graph of $1/p$ against $V_{\mathrm{read}}$ a straight line; its intercept divided by its slope estimates $V_d$ under these assumptions. Unequal temperatures between the barrel and the connecting space require a different correction, analogous to the gas thermometer dead-space correction.

For an original hypothetical example, let $V_d=1\mathrm{c}{\mathrm{m}}^3$, $V_{\mathrm{read},1}=20\mathrm{c}{\mathrm{m}}^3$, $p_1=100\mathrm{kPa}$, and $V_{\mathrm{read},2}=10\mathrm{c}{\mathrm{m}}^3$. The total volume falls from $21$ to $11\mathrm{c}{\mathrm{m}}^3$, so $p_2=100(21/11)\mathrm{kPa}\approx 190.91\mathrm{kPa}$, rather than $200\mathrm{kPa}$. The omitted volume makes the total compression ratio smaller than the barrel-reading ratio. These stipulated values are independent of the manufacturer's apparatus correction and are not measurements.

An implicit solution describes a solution of a differential equation through a relation $F(x,y)=C$, instead of isolating $y$ in an explicit formula. The relation describes a level set; an actual scalar solution of a differential equation must be a differentiable function $y(x)$ on an interval whose function graph lies in the equation's domain. A whole level set need not be one such function graph.

For example, considerImplicit differentiation of $x^2+y^2=C$ gives $2x+2y{y}^{\prime }=0$. For $C>0$, the circle contains two solutions of a differential equation,Both satisfy the equation on the stated open interval, and their function graphs stay in $D$. An initial condition $(x_0,y_0)\in D$ fixes $C=x_0^2+y_0^2$ and selects the sign of $y_0$. In particular, $y(0)=2$ gives $C=4$ and the upper branch on $(-2,2)$. The constant alone does not select the branch.^[ref]

The precise local criterion uses the implicit function theorem. Let $F$ be continuously differentiable on an open subset of the plane, with $F(x_0,y_0)=C$. If its partial derivative $F_y(x_0,y_0)\ne 0$, the nearby level set is a unique continuously differentiable function graph $y=g(x)$ through that point. The chain rule givesThis proves that the local function graph solves $y^{\prime }=-F_x/F_y$ there. For a different prescribed equation $y^{\prime }=f(x,y)$, one must also check $F_x+F_{y}f=0$ along that branch and that it stays in the domain of $f$. A level set by itself does not verify an arbitrary ordinary differential equation.^[ref]

For $F=x^2+y^2$, the upper branch through $(1,\sqrt{3})$ has $F_y=2\sqrt{3}$ and tangent line slope $-1/\sqrt{3}$. At $(2,0)$, however, $F_y=0$ and $F_x=4$. The circle remains a smooth regular level set with a vertical tangent line: it can be written locally as $x=\sqrt{4-y^2}$, but not as a single function graph $y=g(x)$ over an open interval containing $2$. Moreover, $-x/y$ is undefined there, so the scalar solution of a differential equation cannot include this point in its domain. Changing the parameter draws the curve through the point but does not extend that scalar initial value problem.

When both partial derivatives vanish, even the geometry needs separate inspection. The level $x^2+y^2=0$ is an isolated point, not a function graph on any nonempty open interval. The zero level set of $G(x,y)=y^2-x^2$ instead consists of the two lines $y=x$ and $y=-x$ crossing at the origin, where $\nabla G=0$. Each line satisfies the implicit ordinary differential equation $y{y}^{\prime }=x$, including at the origin; the explicit form $y^{\prime }=x/y$ excludes the origin. Thus algebraically dividing by a vanishing coefficient changes the domain and can change which solutions of a differential equation are admissible. Vanishing $F_y$ alone is only failure of a sufficient test: $F(x,y)=y^3$ has the smooth zero function graph $y=0$ despite $F_y=0$ everywhere on it.

Implicit solutions are especially useful for exact first-order ordinary differential equations and for finding a first integral of an ordinary differential equation: a conserved relation can reveal allowed states without an explicit formula. For the normalized harmonic oscillator written as $\dot{x}=y$, $\dot{y}=-x$, the energy $H=(x^2+y^2)/2$ has $\dot{H}=x\dot{x}+y\dot{y}=0$. Its positive level sets are the same circles. Where $y\ne 0$, eliminating time gives $dy/dx=-x/y$ and the local implicit solution above. The harmonic oscillator's parametric curve continues through $y=0$ because its vector field governing evolution in time is defined there; treating $x$ as the independent variable fails at that turn. The origin is an equilibrium point of the harmonic oscillator, although it is excluded from the scalar equation $y^{\prime }=-x/y$.

For the first-order ordinary differential equation $y^{\prime }=f(x,y)$ defined on an open subset $D$ of the plane, an isocline of slope $k$ is the level setHere $k$ is a constant. At every point of this isocline, the direction field prescribes a tangent vector proportional to $(1,k)$ for a solution of a differential equation. It does not generally prescribe the tangent line of the isocline itself. The level set can have several components or fail to be a smooth curve, and points outside $D$ are excluded.

For example, take $y^{\prime }=x-y$ on the whole plane. Its isoclines are the parallel straight linesEach line has slope $1$, while the direction field along it has slope $k$. Thus the isocline $y=x$ carries horizontal segments even though the line itself rises. A function graph of a solution of a differential equation follows the segments, rather than following this line. Only the isocline with $k=1$, namely $y=x-1$, is also a solution of a differential equation.^[ref]

The first-order linear differential equation has the exact familyThe initial condition $y(x_0)=y_0$ selects $A=(y_0-x_0+1)e^{x_0}$ and hence a unique solution of a differential equation for every real $x$. The picture shows $y(0)=-1,0,1,2$. For $A=0$ the solution of a differential equation stays on the $k=1$ isocline; for other $A$ it moves through different isoclines. These formulas check the pictured trajectories by direct differentiation, independently of how the direction field was sampled.

The zero-slope isocline $f(x,y)=0$ is called a nullcline. Below the nullcline $y=x$ in this example, $y^{\prime }>0$; above it, $y^{\prime }<0$. For $A>0$, the solution of a differential equation meets it at $x=\log A$, where $y^{\prime }=0$ and $y^{\prime \prime }=1$, a strict local minimum. The nullcline is therefore a useful guide to turning points, although it is not itself a solution of a differential equation. In the equivalent planar autonomous differential equation $\dot{x}=1$, $\dot{y}=x-y$, it is the nullcline for the second component; there is no equilibrium point, because the first component never vanishes.

This makes isoclines useful even when no explicit solution of a differential equation is available: organize a direction field by common slopes, read where function graphs rise or fall, and identify possible turning points on nullclines.

A linearized plot of Boyle's law uses $x=1/V$ as its horizontal coordinate. Then absolute pressure satisfies $p=Cx$: the slope is $C=nRT$, and the ideal intercept is zero. In the figure, $(V,p)=(100,100)$ and $(50,200)$ in $\mathrm{c}{\mathrm{m}}^3$ and $\mathrm{kPa}$ become $(x,p)=(0.010,100)$ and $(0.020,200)$ in $\mathrm{c}{\mathrm{m}}^{-3}$ and $\mathrm{kPa}$. Both give slope $10000\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3$. The Vernier experiment includes this reciprocal-volume graph as its extension.^[ref]

At constant atmospheric pressure, gauge pressure instead obeys $p_g=Cx-p_{\mathrm{atm}}$, with the same slope and a negative intercept. A nonzero fitted intercept in an intended absolute pressure plot can indicate sensor bias or an incorrect volume correction, but is not by itself a unique diagnosis. The origin is a mathematical extrapolation to infinite volume and zero absolute pressure, rather than an additional measured state.

Changing the added mass while retaining the same contact pair and sliding speed tests whether a constant coefficient of kinetic friction is a useful model. Compute $N=(m_{\mathrm{block}}+m_{\mathrm{added}})g$ for each level horizontal run, and plot the mean sliding force vertically against normal force horizontally. The constant-coefficient prediction is a straight line through the origin; its slope is ${\mu }_k$. This extracts a ratio from several loads rather than relying on one reading.

The apparatus sketch identifies the contact pair and horizontal pull used by the cited laboratory procedure. The plotted line and points are the original stipulated ${\mu }_k=0.30$ model from coefficient of kinetic friction, not laboratory data. The two model points giveIn actual measurements, a linear regression $\overline{F}=b+{\mu }_{k}N$ can expose an unexplained intercept $b$. A nonzero intercept calls for checking the gauge zero, other resisting forces and the contact model; forcing the line through zero can hide those problems. Conversely, constant ratios within measurement error support the approximation only over the loads, speeds and surface conditions tested. A load sweep alone cannot establish independence from contact area or sliding speed; those require separate controlled variations.

A log-log plot uses logarithmic scales on both axes. For a positive power law $y=C{x}^p$ with $x,C>0$, positive references $x_0,y_0$ giveIts transformed slope is the exponent $p$ when both axes use the same logarithm base and the slope is measured in logarithmic coordinate units. For example, $y=2x^2$ takes the values $2,8,32$ at $x=1,2,4$. Every doubling of $x$ quadruples $y$, giving slope two on a log-log plot.

This differs from a semi-log plot: a power law has constant proportional response to a proportional change of input, whereas exponential growth has constant proportional response to an additive change of input. Both can appear straight with appropriate axes, so one must read the scale labels before interpreting a slope.

The mean value theorem says that an average rate of change is attained as an instantaneous rate somewhere between the endpoints. Let $a<b$ and let the real-valued function $f:[a,b]\to \mathbb{R}$ be continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. Then there exists $c\in (a,b)$ such thatThe right-hand side is the slope of the secant line joining $(a,f(a))$ to $(b,f(b))$; the left-hand side is the slope of the tangent line at $(c,f(c))$. The derivative need not be continuous, and the function need not be differentiable at the endpoints. ^[ref]

For the polynomial $f(x)=x^2$ on $[1,3]$, the secant line has slope $(9-1)/(3-1)=4$. Since the derivative is $f^{\prime }(x)=2x$, the required interior point is $c=2$. The tangent line there is $y=4+4(x-2)=4x-4$, whereas the secant line is $y=1+4(x-1)=4x-3$. Their equal slopes make them parallel, but they are distinct straight lines: the theorem equates rates, not heights.

The practical strength is that the unknown point $c$ can disappear from an estimate. Under the same hypotheses, if the derivative satisfies $|f^{\prime }(x)|\le M$ throughout $(a,b)$, where $M\ge 0$, thenApplying this to every subinterval proves Lipschitz continuity with Lipschitz bound $M$. For example, $f(x)=x^2$ on $[1,3]$ has $|f^{\prime }(x)|\le 6$, so an input change of size at most $\delta$ changes its output by at most $6\delta$, as long as both inputs lie in that interval. A local derivative bound therefore controls finite changes without computing an integral. The sign of the derivative similarly gives the derivative criterion for monotonicity.

Taking $M=0$ shows that a function with zero derivative throughout a real interval is a constant function, provided it is continuous at any included endpoints. The interval condition is essential: on the domain of a function $(-2,-1)\cup (1,2)$, the function equal to zero on the left interval and one on the right interval has zero derivative everywhere. It is a locally constant function, yet not a constant function on the whole domain of a function. The mean value theorem cannot cross the gap.

A concise proof subtracts the secant line: set $m=(f(b)-f(a))/(b-a)$ and $h(x)=f(x)-f(a)-m(x-a)$. Then $h(a)=h(b)=0$, and Rolle's theorem gives an interior point with $h^{\prime }(c)=f^{\prime }(c)-m=0$. The geometric idea is to remove the average trend and find a horizontal tangent line in what remains.

In the branching model, mixed inhibition retains inhibitor binding to both free enzyme and the enzyme-substrate complex. Unequal $K_i$ and $K_i^{\prime }$ shift both apparent parameters; their equality gives pure noncompetitive inhibition. For the chosen unequal case, $I=30\mu \mathrm{M}$ gives $\alpha =4$ and ${\alpha }^{\prime }=2$, hence $V_{\max ,\mathrm{app}}=50\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$ and $K_{M,\mathrm{app}}=40\mu \mathrm{M}$. Because $K_i<K_i^{\prime }$, binding favours free enzyme, and the apparent Michaelis constant increases. Reversing that inequality decreases it.

The limiting behaviour provides a useful diagnostic within this model:Low substrate concentration isolates $\alpha$ through the initial slope; high substrate concentration isolates ${\alpha }^{\prime }$ through the plateau. At $I=30\mu \mathrm{M}$, the chosen competitive inhibition and unequal mixed inhibition examples have the same low-$s$ slope, $1.25\mathrm{mi}{\mathrm{n}}^{-1}$, but different plateaus, $100$ and $50\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$. Measuring only low substrate concentrations would miss that distinction.

The Pareto distribution describes a positive random variable whose tail probability falls as a power law. In the shape–minimum convention used here, $X\sim \mathrm{Pareto}(\alpha ,\beta )$ has shape parameter $\alpha >0$ and minimum $\beta >0$, with probability support $[\beta ,\infty )$. Its probability density function isThe minimum $\beta$ has the same units as $X$; the shape parameter $\alpha$ is dimensionless. At fixed $\alpha$, changing $\beta$ gives a scale family, since $X/\beta \sim \mathrm{Pareto}(\alpha ,1)$. This convention is also used in uniform-Pareto conjugacy.^[ref]

The survival function and cumulative distribution function areFor $0<p<1$, solving $F(x)=p$ gives the quantile functionIts right-hand limit at $p=0$ is $\beta$; it tends to infinity as $p\uparrow 1$.^[ref] There is no probability mass at $\beta$: changing a probability density function's value at that single point changes no probabilities.

Choose $\alpha =2$ and $\beta =10$ as an original theoretical example. Then $\mathbb{P}(X>20)=1/4$, $\mathbb{P}(X>100)=1/100$, the median is $10\sqrt{2}\simeq 14.142$, and the 0.99 quantile is $Q(0.99)=100$. These chosen values and all figures below describe mathematical probability distributions, not actual wealth observations.

On this log-log plot, write $r=x/\beta$ to obtain $\log \overline{F}(\beta r)=-\alpha \log r$. The slope is $-\alpha$, whereas the probability density function has power exponent $\alpha +1$. Multiplying a threshold by ten divides its tail probability by $10^{\alpha }$. A larger $\alpha$ therefore concentrates the Pareto distribution nearer $\beta$; a smaller $\alpha$ makes extreme values more plausible. This exact scaling is useful when modelling large positive losses or sizes over several orders of magnitude.

The Pareto distribution is a heavy-tailed distribution: for every $t>0$, the exponential moment $\mathbb{E}[e^{tX}]$ diverges because $e^{tx}{x}^{-\alpha -1}$ eventually grows without bound. In contrast, the Laplace distribution has exponentially decreasing tails and some finite positive exponential moments. The Student's t-distribution also has power tails, but on both sides of the real line, with a smooth central region rather than a positive minimum.

Solving the implicit equation for $y$ when $\Delta x\ne 0$ givesHere “gradient” means the slope $dy/dx$, rather than the gradient $(a,b)$ of an implicit defining function. If $\Delta x=0$ and $A\ne B$, the straight line is vertical and has no finite slope. If $\Delta y=0$ and $\Delta x\ne 0$, its slope is zero.

Take $z$ positive upwards, so the source's sign convention makes $p$ a downward applied load. The small-slope tension contribution on a short interval is $T{w}^{\prime \prime }$. Let $r\ge 0$ be the upward contact reaction. Force balance gives $T{w}^{\prime \prime }-p+r=0$. Nonpenetration is $w\ge f$, and reaction is absent where $w>f$. ThereforeThe contact boundary must have continuous $w$ to avoid a broken string, and continuous slope in this smooth-contact model with no concentrated point force. Integrating force balance over a vanishing interval around a contact endpoint gives $T[w^{\prime }]=0$ when the load and reaction are locally bounded. This is the one-dimensional obstacle problem, with its reaction acting as a nonnegative constraint force.

For any admissible $v$, put $q=v-w$. At endpoints $q=0$. On the contact set, $w=f$ and $v\ge f$ imply $q\ge 0$, while away from contact the reaction $r=p-T{w}^{\prime \prime }$ is zero. Hence $rq\ge 0$ everywhere. Integration by parts givesRearranging proves the requested variational inequality. Piecewise integration is legitimate because the slope is continuous at the contact boundary, so there are no omitted point-force terms.