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The length of a curve is $L(\alpha )={\int }_a^b\Vert {\alpha }^{\prime }(t)\Vert dt$. Its arc length parameter is $s(t)={\int }_a^t\Vert {\alpha }^{\prime }(u)\Vert du$. Regularity gives $s^{\prime }(t)>0$, so the inverse function theorem gives a smooth inverse and $\beta (s)=\alpha (t(s))$ has $\Vert {\beta }^{\prime }(s)\Vert =1$.

Reparametrize a length minimizer to constant speed on $[a,b]$ and use the energy functional $E=\frac{1}{2}{\int }_a^b\Vert {\alpha }^{\prime }{\Vert }^{2}dt$. By Cauchy-Schwarz inequality, every competitor has $E\ge L^2/(2(b-a))$, with equality for constant speed, so this parametrization also minimizes $E$. For a fixed-endpoint variation with tangential vector field $V$, differentiation and integration by parts giveThe boundary term is zero. Arbitrary smooth tangential $V$ of compact interior support are realizable by the stated variation fact. Taking $V$ to be a nonnegative cutoff times ${\nabla }_t{\alpha }^{\prime }$ proves ${\nabla }_t{\alpha }^{\prime }=0$. Hence a length minimizer is a geodesic after constant-speed reparametrization. The original parameter need not be affine: arbitrary varying-speed parametrizations preserve length but have nonzero covariant acceleration. This qualification is necessary for the literal wording.

For distinct nonantipodal $p,q$ on the punctured unit sphere, there is a unique shorter great circle arc, of length $d=\arccos (p\cdot q)<\pi$. A minimizing curve exists exactly when that shorter arc avoids the removed north pole. If it avoids the pole it attains the spherical lower bound. If it passes through the pole, arbitrarily small smooth detours have lengths tending to $d$, but equality would force the unique shorter arc, which is unavailable. Thus the infimum is not attained; the longer great circle arc is not a substitute minimizer.

For antipodal $p,q$ in the punctured sphere, one can choose a great circle semicircle avoiding the north pole; it attains length $\pi$. Thus every admissible antipodal pair has a minimizer. If identical endpoints are included despite the earlier distinctness stipulation, the infimum is zero, but no smooth regular curve attains it; only a nonregular constant curve does.

Write the nonzero tangent vector as $U=aV+bW$. Since $g(V,V)=g(W,W)=0$ and $g(V,W)=-2$, nullness gives $g(U,U)=-4ab=0$. Thus exactly one coefficient can be nonzero, proving $\boxed{U\parallel V\text{ or }U\parallel W}$. This pointwise argument does not require the curve to be a geodesic.

Let the contact point have positive horizontal coordinate $x_0$. Equality of slopes between the tangent line and the logarithmic curve requires $k=1/x_0$. The point must also satisfy both equations, giving $\ln x_0=k{x}_0+1=2$. HenceThe contact point is $(e^2,2)$, and the resulting line $y=e^{-2}x+1$ has exactly the required derivative there.

The product rule and chain rule give $y^{\prime }=\sin (x^2)+2x^2\cos (x^2)$. At zero the slope is zero and the curve passes through $(0,1)$. Consequently the tangent line is

The curve starts at $y(0)=-\sqrt{3}$ and initially rises. It crosses zero at $x=\pi /3$ and $4\pi /3$, attains its maximum at $(5\pi /6,2)$ and minimum at $(11\pi /6,-2)$, then returns to $y(2\pi )=-\sqrt{3}$. These follow either from the shifted sine form or from $y^{\prime }=2\cos (x-\pi /3)$.

Away from zeros of $\sin x$, an integrating factor for a differential one-form is $\mu (x)=1/\sin x$. Multiplication converts the form toTherefore the implicit solution family is $xy+\ln |\sin x|=C$. On each graph interval between consecutive multiples of $\pi$, this becomesNo such open interval contains zero. At $x=k\pi$, the undivided graph equation would read $\cos (k\pi )=0$, which is impossible for a finite derivative; hence no differentiable graph extends across one of these points.

If solutions are instead understood as arbitrary parametrized integral curves of the differential form, the vertical lines $x=k\pi$ are additional solutions: along them $dx=0$ and $x\sin xdy=0$. They were lost on division by $\sin x$ and are not graphs $y(x)$ over an interval.

Where $\mathbf{A}=|\mathbf{v}{|}^2\mathbf{w}$ is twice continuously differentiable, the identity divergence of a curl is zero gives $\nabla \cdot (\nabla \times \mathbf{A})=0$. All regular portions of the parameter interval lie in this region, henceas an improper line integral. Besides the $\mathrm{arctanh}(x^2)$ endpoint problem from (i), the curve touches $y=1$ at $\theta =\pi /2$, where the printed real $\arcsin y$ expression has no smooth open neighbourhood. The identity applies on the intervening open segments, and its zero value has a unique continuous extension along the curve. A literal classical integrand at every parameter is not supplied by the printed field.

The gradient part contributes zero around a closed path by the fundamental theorem for line integrals:Along the parametrization, $x^2+z^2=1$, so the first two components of $\mathbf{w}$ vanish wherever the printed expression is defined. The remaining component gives $\mathbf{w}\cdot d\mathbf{x}=xdz={\cos }^2\theta d\theta$. ThusThe printed field needs a qualification: $\mathrm{arctanh}(x^2)$ is undefined at the isolated parameters where $x^2=1$, so $0\mathrm{arctanh}(1)$ cannot literally be evaluated. The result above is the improper line integral, or equivalently the integral of the continuous extension of the restricted field along the curve. That restricted field tends to $(0,0,\cos \theta )$ at each omitted parameter. No extension of the full field to a neighbourhood is being assumed.

The derivative is $dy/dx=3x^2$, giving slope twelve at $x=2$. The point on the curve is $(2,9)$. The point-slope equation of the tangent line is consequently

Convert the cardioid from polar coordinates to Cartesian coordinates:The curve is symmetric about the horizontal axis. It has a cusp at the origin at $\theta =\pi$, passes through $(0,\pm 1)$, and reaches $(2,0)$ on the right. Its leftmost horizontal coordinate is $-1/4$, since $x=(\cos \theta +1/2{)}^2-1/4$.

For the largest $y$, differentiate the parametrization:The candidates have $\cos \theta =1/2$ or $-1$. The positive maximum occurs at $\theta =\pi /3$, where $r=3/2$. Comparing with the negative extremum and the zero cusp value proves

The cobweb plot uses the curve $v=\sqrt{2+u}$ and the diagonal $v=u$. From a point on the diagonal, a vertical step to the curve computes the next iterate; a horizontal step returns its value to the diagonal.

For $0\le u<2$, $(2+u)-u^2=(2-u)(1+u)>0$, so $u<\sqrt{2+u}<2$. Thus a start below $2$ gives a strictly increasing sequence bounded above by $2$. For $u>2$ the inequalities reverse to $2<\sqrt{2+u}<u$, giving a strictly decreasing sequence bounded below by $2$. Both types of staircase approach the same intersection.

The bounded monotone sequence theorem proves existence of a limit, and continuity then makes it a nonnegative fixed point. Part (i) givesThis proves the global convergence of a square-root iteration; the graph below illustrates both cases without relying on a finite numerical sample as proof.

The curve has $x\ne 0$ and can be written $y=1/x-x$. It is a hyperbola with vertical asymptote $x=0$ and oblique asymptote $y=-x$, crossing the horizontal axis at $(\pm 1,0)$. It is symmetric under $(x,y)\mapsto (-x,-y)$.

On each branch $dy/dx=-1/x^2-1<0$. For $x>0$, the branch descends from $+\infty$ as $x\downarrow 0$, passes through $(1,0)$, and approaches $y=-x$ from above as $x\to +\infty$. The negative branch follows by the central symmetry.

The endpoint constraints are $y(-a)=y(a)=0$. The constants in the three costs contribute only $2aK$, $2a{K}^{\prime }$ and $2a{K}^{\prime \prime }$ respectively, so they do not affect the minimizing shapes. For companies B and C, the fixed arc length isThe original PDF has $(y^{\prime }{)}^2$ as company A's first variable term. The local TeX transcription has lost that derivative dot and incorrectly repeats $y^2$; the calculation below follows the PDF.

Company A. The relevant functional is $J_A=\int (y^{\prime 2}+y^2-2xy)dx$. The Euler-Lagrange equation givesIts general solution is $y=x+c_1\cosh x+c_2\sinh x$. Adding and subtracting the endpoint equations gives $c_1=0$ and $c_2=-a/\sinh a$. ThusThis curve is odd, negative on the left half and positive on the right half; company A has no lower-half-plane restriction. It is the unique global minimum. Indeed, for any endpoint-zero variation $\eta$, integration by parts and the differential equation cancel all linear terms, leavingunless $\eta =0$.

Company B. Minimize $\int ydx$ under the length constraint. Introduce a constant Lagrange multiplier $\lambda$ and apply the Euler-Lagrange equation to $y+\lambda \sqrt{1+y^{\prime 2}}$:The curvature is constant, so the required lower arc is a circular arc. Writing its radius as $R\ge a$ and imposing the endpoints givesSince $\sqrt{1+y_B^{\prime 2}}=R/\sqrt{R^2-x^2}$, its length determines the radius:Equivalently, introduce $0<\theta \le \pi /2$ with $a=R\sin \theta$. Then $L/(2a)=\theta /\sin \theta$, which increases strictly from $1$ to $\pi /2$. The stipulated range of $L$ therefore selects one minor arc. At $L=\pi a$ it becomes the lower semicircle, with vertical endpoint tangents; it is still a single-valued finite-length graph.

Global minimality follows from the convexity of the augmented slope cost with $\lambda =R$. For a competitor $y=y_B+\eta$,Consequently the augmented cost difference is at least $\int (\eta +x{\eta }^{\prime })dx=[x\eta {]}_{-a}^a=0$. Competitors have the same length, so the multiplier contribution cancels and the original cost is minimized. This proves that a circular arc maximizes area below a fixed chord: minimizing negative signed height maximizes the area between the fence and the chord.

Company C. The variable cost is $J_C=\int yds=\int y\sqrt{1+y^{\prime 2}}dx$. The augmented integrand is $(y+\lambda )\sqrt{1+y^{\prime 2}}$. Since it has no explicit $x$ dependence, the Beltrami identity givesThe hanging branch with $b>0$ integrates to $y+\lambda =b\cosh [(x-x_0)/b]$. Equal endpoint heights force $x_0=0$, and the endpoint value fixes $\lambda =b\cosh (a/b)$. ThusIts length isWriting $q=a/b$, the equation $L/(2a)=\sinh q/q$ has one positive solution for every $L>2a$: its derivative has numerator $q\cosh q-\sinh q>0$. The curve is a catenary, is even, and lies strictly below the chord except at the endpoints.

A direct argument proves this is a global minimum, rather than merely an extremum of an augmented functional. Parametrize any competitor by arc length $s\in [-L/2,L/2]$, from its left endpoint to its right endpoint. Then $x^{\prime 2}+y^{\prime 2}=1$, $\int x^{\prime }ds=2a$, and integration by parts givesThe Cauchy-Schwarz inequality gives the last step. Equality requires $x^{\prime }=b/\sqrt{b^2+s^2}$ and $y^{\prime }=s/\sqrt{b^2+s^2}$, whose integrals are precisely the displayed catenary with the stated length. Thus it attains a lower bound valid for every competitor. This proves that a catenary minimizes the integral of height along a fixed-length curve.

The sketch uses $a=1$ and $L=2.7$ for the two fixed-length cases. The vertical scale in the company A panel is enlarged to show its small odd deflection. No comparison between the three companies' absolute prices is required or possible from the shapes alone, since their additive cost constants are independent.

For a piecewise differentiable curve given in polar coordinates by $r=r(\theta )$, differentiate $x=r\cos \theta$ and $y=r\sin \theta$. The cross terms cancel in $d{x}^2+d{y}^2$, leaving $d{s}^2=d{r}^2+r^{2}d{\theta }^2$. Therefore the arc length on an increasing parameter interval isThis formula remains valid through an isolated zero of the speed when the parametrization is piecewise smooth. A square root of a square must retain its absolute value.

For a unit-speed curve with positive curvature of a space curve $\kappa$, the principal normal is $N=T^{\prime }/\kappa$, with unit tangent vector $T$. It is perpendicular to $T$ and points in the direction in which the tangent is turning. At zero curvature this formula does not define $N$.