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The integral momentum equation applies momentum conservation to a chosen control volume. For a fixed region $V$ with outward unit normal $\mathbf{n}$, its vector form isThe first term is accumulation inside the region; the second is net outward momentum transport. External forces include pressure, viscous traction, gravity and any force exerted by a wall.

In steady flow through a pipe bend, accumulation vanishes but momentum flux changes direction, so the fluid requires a resultant force. This makes the equation useful even when the detailed internal velocity field is unknown. Pressure must be included on the inlet and outlet faces; using only the speed change can miss the dominant load. The displayed flux uses a fixed control volume; a moving boundary requires relative transport velocity.

The Bernoulli equation relates speed, pressure and height along a streamline of a steady inviscid flow. For constant mass density $\rho$ in uniform gravity,where $B$ is constant along that streamline. Each term has units of energy per unit mass. In a barotropic compressible fluid, the pressure term is instead $\int dp/\rho$.

In a horizontal narrowing passage, mass conservation increases speed and the ideal streamline balance predicts lower pressure. Pressure taps connected to a manometer can test this prediction. The argument compares points along the same streamline and accounts for the stated assumptions; pumps, viscous losses and unsteady motion require additional terms or a different relation. Equal values of $B$ on different streamlines require extra structure, such as a connected irrotational flow with a conservative body force.

For $\mathbf{u}=\nabla \varphi +\beta \nabla \alpha$, the densitygives incompressibility and material conservation of $\alpha$ and $\beta$. Differentiating the Clebsch representation then yields the Euler momentum equation with

For time-dependent irrotational flow with $u=\nabla \varphi$, constant mass density, and no body force, the Euler equations for an inviscid fluid integrate toA time-dependent change of gauge in $\varphi$ may set $C(t)=0$.

This is the unsteady potential-flow version of Bernoulli's principle, with a time-dependent potential gauge.

A lattice polygon has all vertices in the integer lattice; triangulation shows that its area is a half-integer.

A spherical cap is the part of a sphere cut off by a plane.

A regular polygon has equal side lengths and equal interior angles. Its vertices on a circle are scaled and rotated roots of unity.

A right angle is an angle of $\pi /2$ radians; two nonzero vectors meet at a right angle exactly when their inner product is zero.

The perpendicular bisector of a line segment is the line through its midpoint at a right angle to the segment.

An ellipse is the locus of points whose Euclidean distances from two fixed foci of a conic have constant sum $2a$. If the distance between the foci of a conic is $d<2a$, the semi-major and semi-minor axes are $a$ and $\sqrt{4a^2-d^2}/2$. Writing $c=d/2$ gives the conventional relation $b^2=a^2-c^2$.^[ref]

For an original calculation, choose $a=5$, $b=3$, and hence $c=4$. The ellipsehas foci of a conic $F_{-}=(-4,0)$ and $F_{+}=(4,0)$. The point $P=(3,12/5)$ lies on it since $9/25+16/25=1$. Its focal distances areIts eccentricity of a conic is $e=4/5$. The matching directrix of a conic for $F_{+}$ is $x=a/e=25/4$, and the perpendicular distance from $P$ to that straight line is $13/4$. Therefore the ratio of focal distance to directrix distance is $(13/5)/(13/4)=4/5$, agreeing with $e$.

The constant sum can be turned into a construction. Fix a pin at each focus of a conic and pull a closed string loop taut with a pencil. In this example the segment between the pins has length $8$, so a loop of length $18$ leaves $10$ for the two pencil-to-pin segments. Moving the pencil while keeping that loop taut traces the ellipse. This is a geometric prediction from the length constraint, not an assertion about a measured drawing.

The parametrization $x=5\cos t$, $y=3\sin t$ maps the unit circle onto this ellipse. It makes plotting easy, but is an affine map rather than an isometry: horizontal and vertical lengths have different scale factors. An ellipse is not a circle under the original Euclidean metric merely because this map exists. Its two semiaxes and eccentricity of a conic retain the metric information lost by that equivalence.

A useful proof outline begins with $|X-F_{-}|+|X-F_{+}|=2a$. Isolate one square root, square, and then repeat; the resulting quadratic equation is $x^2/a^2+y^2/(a^2-c^2)=1$. Since $a>c$, the quantities isolated before squaring have the required signs, so the reverse implication introduces no extra branch. The condition $a>c$ is essential: at equality the focal-distance locus becomes the line segment between the foci of a conic.

A cyclic quadrilateral has all four vertices on one circle. Among quadrilaterals with fixed cyclically ordered side lengths, a cyclic quadrilateral has maximal area.

This is the planar case of the isoperimetric inequality. Every bounded planar domain with a sufficiently regular boundary of length $L$ and area $A$ satisfiesEquality holds exactly for circular domains.

The open unit disk in the Euclidean plane is

The Borsuk conjecture asserted that every bounded subset of ${\mathbb{R}}^d$ with positive diameter can be partitioned into $d+1$ subsets of strictly smaller diameter. It is false in sufficiently high dimension.

A triangular prism is a prism whose two bases are congruent parallel triangles. One base is a translation of the other by a vector not parallel to its plane. Joining corresponding vertices of a polygon gives three parallelogram side faces, nine edge of a polyhedron segments and six vertices of a polyhedron.

The triangular prism is right when the translation vector is perpendicular to the base plane. Only in that case is it an orthogonal Cartesian product of a triangle with a line segment. An oblique triangular prism still has volume $Ah$, where $A$ is its base area and $h$ the perpendicular separation of the base planes; equal parallel cross sections give this formula by integration.

A triangle is the convex hull of three noncollinear points: the three vertices must not lie on one line. Its boundary consists of the three joining line segments; depending on context, "triangle" can mean this boundary or the filled region it encloses. Here a triangle's area refers to the filled region. The noncollinearity assumption excludes a collapsed shape with zero area, for which ordinary interior angles are not defined.

For example, the positions $(0,0)$, $(4,0)$ and $(0,3)$ form a right triangle with a right angle at the origin. Taking the side of length $4$ as its base gives perpendicular height $3$, so its area is $4\cdot 3/2=6$. The height is measured at a right angle to the chosen base; it is not generally the length of another side. In this example it happens to be a side because the triangle is a right triangle.

By contrast, side lengths $3,4,6$ define a valid triangle: each side is shorter than the sum of the other two, as required by the strict triangle inequality. But $3^2+4^2=25\ne 36=6^2$, so this is not a right triangle and the Pythagorean theorem cannot calculate its longest side from the other two. The law of cosines instead gives the opposite interior angle $C$ throughThis example separates the definition of a triangle from the additional assumption of a right angle.

Every triangle in the Euclidean plane has interior angles summing to $\pi$ radians, a result proved in Euclid's Elements I.32.^[ref] It lets two measured angles determine the third, and underlies the angle-angle criterion for similar triangles. The qualification "Euclidean plane" matters: the earlier spherical triangle demonstrates a different sum.

A sphere is the set of points at one fixed positive Euclidean distance from a center.

A linear Diophantine equation in two variables is $ax+by=c$ with integer coefficients and unknown integers $x,y$. Assume $a,b$ are not both zero, and put $d=\gcd (a,b)>0$.

The linear Diophantine equation solvability criterion gives a solution exactly when $d\mid c$: every integer linear combination is divisible by $d$, and Bézout's identity supplies a combination equal to $d$ which can be multiplied by $c/d$. The extended Euclidean algorithm constructs the coefficients. All solutions of a linear Diophantine equation then describes the complete one-parameter family when both coefficients are nonzero.

For example, $6x+9y=3$ has $(x,y)=(-1,1)$, and all its solutions are $x=-1+3t$, $y=1-2t$ with $t\in \mathbb{Z}$. There is no solution with both coordinates nonnegative, since those inequalities require $t\ge 1$ and $t\le 0$. Thus an integer-solvability test does not automatically settle a problem requiring nonnegative integers. If both coefficients are zero, the separate equation $0=c$ has every integer pair as a solution when $c=0$, and none otherwise.

When enzyme complexes relax much faster than substrate concentrations, the quasi-steady-state approximation sets the time derivatives of the fast complexes to zero and solves their algebraic balance equations.

It approximates the fast intermediate by a steady state balance without requiring the entire chemical system to be at equilibrium.

The Michaelis-Menten equation describes a saturating enzyme reaction rate:Here $s$ is free substrate concentration, $V_{\max }$ is the limiting rate at large $s$, and the Michaelis constant $K_M$ is the concentration giving $v=V_{\max }/2$.

For the elementary mechanism $E+S\rightleftharpoons ES\to E+P$, use binding rate constant $k_1$, dissociation rate constant $k_{-1}$ and turnover number $k_{\mathrm{cat}}$. The quasi-steady-state approximation sets the fast enzyme-substrate complex balance towhere $e_T$ is total enzyme concentration and $c=[ES]$. Solving givesThis modern steady-state derivation does not require substrate binding to be in chemical equilibrium. In particular, $K_M$ is not generally the binding dissociation constant.

At $s\ll K_M$, $v\simeq (V_{\max }/K_M)s$; at $s\gg K_M$, the enzyme approaches saturation. The simple model assumes an appropriate timescale separation and negligible reverse product reaction; additional binding sites or inhibition can change the rate law.

L. Michaelis and Maud L. Menten investigated invertase acting on sucrose. They compared early reaction rates at different substrate concentrations using optical rotation, controlled temperature and acidity, and accounted for the separate optical change caused by sugar mutarotation. These controls helped distinguish catalytic conversion from changes in the measurement signal. ^[ref]