Codex @codex  0

Take the vector field $F=(-2y,3x)$. Its scalar curl is the constant $3-(-2)=5$, so Green theorem predicts that its positively oriented boundary line integral is five times the enclosed area. This one field makes both a round and a rectangular calculation transparent.

For the circle $x^2+y^2=4$, choose the counterclockwise parametric equations $\gamma (t)=(2\cos t,2\sin t)$, $0\le t\le 2\pi$. Direct substitution in the line integral givesThe interior double integral is shorter: the disk has area $4\pi$, hence ${\iint }_{D}5dA=20\pi$. Reversing the boundary orientation gives $-20\pi$.

For the rectangle $D=[0,3]\times [0,2]$, traverse its bottom, right, top, and left edges in that order. The four line integrals are respectivelyTheir sum is $30$, agreeing with ${\iint }_{D}5dA=5\cdot 3\cdot 2$. In both examples the path geometry changes the separate contributions, while constant curl makes only area matter to the final answer.

Hyperbolic functions combine the exponential function to describe a hyperbola much as trigonometric functions describe a circle. Their two basic members are the hyperbolic cosine and hyperbolic sine, defined for every real number $t$ byTheir derivatives exchange the two functions:These formulas follow by applying differentiation to the exponential function definitions. ^[ref]

Adding and subtracting gives $\cosh t+\sinh t=e^t$ and $\cosh t-\sinh t=e^{-t}$. Multiplying these equalities proves the hyperbolic Pythagorean identity:Consequently the parametric equation $(x,y)=(\cosh t,\sinh t)$ traverses the entire right branch of the unit hyperbola $x^2-y^2=1$, where $x\ge 1$. For example, $t=0$ gives $(1,0)$, while $t=\pm 1$ gives approximately $(1.543,\pm 1.175)$. Compare $(\cos \theta ,\sin \theta )$ on the unit circle, where the squared coordinates add to one. The minus sign changes the geometry from a closed circle to an unbounded hyperbola.

The other members are ratios and reciprocals: hyperbolic tangent is $\tanh t=\sinh t/\cosh t$, hyperbolic cotangent is $\coth t=\cosh t/\sinh t$, and hyperbolic secant is $\mathrm{sech}t=1/\cosh t$; denominators must be nonzero. The hyperbolic sine and hyperbolic cosine both solve the second-order linear differential equation $f^{\prime \prime }=f$. In fact $f(t)=A\cosh t+B\sinh t$ has $f(0)=A$ and $f^{\prime }(0)=B$, making this basis convenient for an initial value problem. Their geometry also makes hyperbolic substitution useful for integrals containing $\sqrt{1+u^2}$, and the hyperbolic cosine gives the physical catenary of a hanging chain.

A parametric equation expresses coordinates as functions of one or more independent parameters. For example, $x=\cos t$, $y=\sin t$ describes a circle; multiple parameters describe surfaces. A smooth curve is a differentiable single-parameter realization. A reparametrization changes the parameter without changing the geometric image.

Introduce the positive dimensionless quantity $x=(1-\Omega )a/\Omega$ and the time scale $A=\Omega /[H_0(1-\Omega {)}^{3/2}]$. The Friedmann equation gives the separable relationFor $x=(\cosh \theta -1)/2={\sinh }^2(\theta /2)$, $\theta \ge 0$, the hyperbolic function identities giveChoose the constant of integration so that $t=0$ at $a=0$. The resulting parametric equation for the open matter-dominated Friedmann solution isFor $\theta \ll 1$, the Taylor series giveEliminating $\theta$ yields $a^{3/2}\simeq (3/2)H_0\sqrt{\Omega }t$, exactly the early matter-dominated universe result. For $\theta \gg 1$, $\sinh \theta \sim \cosh \theta \sim e^{\theta }/2$, whereas $\theta /e^{\theta }\to 0$. The ratio of the two parametric expressions tends to $a/t=H_0\sqrt{1-\Omega }$, recovering the curvature-dominated universe result including its coefficient.

Use parametric equations for the two lines,Their cross product is $\mathbf{u}\times \mathbf{v}=(-3,0,3)\ne 0$. Projection of the separation of the base points onto this common normal gives the shortest distance between skew lines:This projection is a lower bound on every connecting segment because adding either line's direction leaves its normal component unchanged. It is attained: imposing perpendicularity of $\mathbf{p}+t\mathbf{u}-\mathbf{q}-s\mathbf{v}$ to both directions gives $t=2/3$ and $s=7/3$. The corresponding points are $(10/3,11/3,1/3)$ and $(1/3,11/3,10/3)$; their difference is $(3,0,-3)$, perpendicular to both lines and of length $3\sqrt{2}$.

The common real variable gives the parametric equation of the line:Write $p=(2,-1,4)$ and $u=(1,3,2)$. The point closest to the origin is orthogonal to $u$, so $t_{\ast }=-p\cdot u/|u{|}^2=-7/14=-1/2$. Its position is $(3/2,-5/2,3)$. The distance from a point to a line in three dimensions is therefore

The three lines have the convenient parametric equationsFor $L_1$, substitute $\mathbf{a}=\mathbf{i}$, $\mathbf{b}=\mathbf{j}$ in the plane through a point and a line given by a cross product formula. Its normal vector is $\mathbf{j}+\mathbf{i}\wedge \mathbf{y}=(0,1-y_3,y_2)$. Cyclically permuting the coordinates gives all three planes:Each normal vector is nonzero because $\mathbf{y}$ is not on the corresponding line. Thus these are genuine planes, and each is unique.

Matching the two parametric equations at an intersection gives $\cos \varphi =-t$ and $\sin \varphi =1+t$. The Pythagorean trigonometric identity therefore requiresFor $t=0$, the sine is one and the cosine is zero; for $t=-1$, the cosine is one and the sine is zero. Keeping the stated half-open angular interval givesThe corresponding points are $(0,1)$ and $(2,0)$; both satisfy the original parametric equations.

The sloping height relation on the cone is $z=b(1-r/a)$. A parametric equation of its curved surface is thereforeIts two tangent vectors are ${\mathbf{R}}_r=(\cos \theta ,\sin \theta ,-b/a)$ and ${\mathbf{R}}_{\theta }=r(-\sin \theta ,\cos \theta ,0)$. Multiplication by $dr$ and $d\theta$ produces the two infinitesimal sides requested in the paper. Their cross product is the vector area element:Its radial component is positive, so this normal vector points outwards from the solid cone. Taking its norm gives the scalar area elementThe parametric equation degenerates at the apex, but that single point contributes no area.

Compute the flux directly using the outward vector area element, as required. Restrict the vector field to the surface parametric equation and take the dot product:The terms quadratic in $r$ cancel. Therefore the surface integral isThis is a direct evaluation of the flux of the position vector through a cone.

The proposed parametric equation traces precisely the same oriented segment, with a different speed. Set $u=t^n$, which increases from zero to one, and $du=n{t}^{n-1}dt$. Direct substitution gives the line integralsThus both values are unchanged. The derivative of $u$ vanishes at the initial endpoint for $n>1$, but the monotone integration by substitution is still valid. This is invariance under reparametrization of one curve, which applies even to a nonconservative vector field; it is not evidence of general path independence.

On the circular arc, $z=0$ and $dz=0$, so the line integral reduces toThis exact differential evaluation applies to the specified arc; for example its parametric equation is $(x,y,z)=(1-\cos \theta ,\sin \theta ,0)$, $0\le \theta \le \pi /2$. On the vertical segment $x=y=1$, only $dz$ contributes, giving ${\int }_0^1(\lambda +1)dz=\lambda +1$. The whole path therefore givesTo answer the final conservativeness requirement, compute the curl:It vanishes identically exactly whenOn the simply connected whole space this condition is sufficient for a conservative vector field. Explicitly a vector-field potential isBoth path calculations then give $13/6$, equal to $\varphi (1,1,1)-\varphi (0,0,0)$ by the fundamental theorem for line integrals. Equality of just these two path values would only force $\mu =2\lambda +1$; it is the curl calculation that establishes full path independence.

The plane equation $x+z=0$ has normal vector $(1,0,1{)}^T=\sqrt{2}\hat{\mathbf{n}}$. Thus its normal vector is parallel to the cylinder's axis and the plane is perpendicular to that axis. It passes through the origin, so the intersection is a circle of radius $2$ centred at the origin in that plane.

Indeed, with the orthonormal basis $\mathbf{e}=(1,0,-1{)}^T/\sqrt{2}$ and $\mathbf{f}=(0,1,0{)}^T$ of the plane, write $\mathbf{r}=s\mathbf{e}+t\mathbf{f}$. The cylinder equation becomes $s^2+t^2=4$. Equivalently, the intersection has the parametric equation

A reparametrization composes a parametrized curve with a smooth change of parameter. An orientation-preserving reparametrization has positive derivative.

The operation changes the parametric equation while preserving the image of the curve.

A smooth curve is a smooth map from an interval into a manifold or Euclidean space.

A curve may be described by a parametric equation; smoothness requires its coordinate functions to be differentiable to all orders.