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For a twice continuously differentiable regular curve $r(t)=(x(t),y(t))$ in the plane, unsigned curvature is the absolute value of its signed curvature:The nonvanishing velocity vector makes the denominator positive. For a graph of a function $(t,f(t))$, this reduces to $|f^{\prime \prime }(t)|/(1+f^{\prime }(t{)}^2{)}^{3/2}$; omitting the absolute value gives the signed curvature of a plane graph for increasing $t$.

For the parabola $r(t)=(t,t^2)$, the derivatives are $r^{\prime }=(1,2t)$ and $r^{\prime \prime }=(0,2)$. ThereforeThe curvature is greatest at the vertex $t=0$, where $\kappa =2$ and the radius of curvature is $1/2$. At $t=\pm 1/2$ the curvature is $1/\sqrt{2}$, so the parabola bends less sharply despite its larger speed $\sqrt{2}$.

Its unit tangent vector and left unit normal areSince $k>0$, $N_L$ is also its principal normal vector. At $t=0$ the osculating circle consequently has center $(0,0)+(1/2)(0,1)=(0,1/2)$ and equationThe lower branch $y=1/2-\sqrt{1/4-x^2}$ has value $0$, derivative $0$, and second derivative $2$ at $x=0$, matching the parabola $y=x^2$ through second order. Away from the vertex the two curves separate: an osculating circle describes a local bend, not the whole curve.

Vanishing curvature at one point does not imply a straight segment. The graph of a function $y=x^3$ has signed curvature $6x/(1+9x^4{)}^{3/2}$, which vanishes at its inflection point $x=0$ and changes sign across it. By contrast, if curvature vanishes throughout a connected regular interval, $dT/ds=0$ throughout that interval: the constant unit tangent vector forces the image to lie on a straight line.

The intrinsic signed curvature is the curvature of a stress-free filament, measured along material arc length. An elastic filament with actual curvature $\kappa$ stores bending energy $\frac{B}{2}\int (\kappa -{\kappa }_0{)}^{2}ds$, with filament bending modulus $B$. In a planar small-slope Monge representation, $\kappa \simeq {\zeta }^{\prime \prime }$ and the leading energy is quadratic in ${\zeta }^{\prime \prime }-{\kappa }_0$.

Introduce a constant filament bending modulus $B$, which sets the mechanical force scale. The filament has intrinsic curvature of an elastic filament ${\kappa }_0(s)$, and its actual signed curvature in a small-slope Monge representation is $\kappa \simeq {\zeta }_{xx}$. To quadratic order one may identify material arc length $s$ with $x$ in ${\kappa }_0$, whileAt fixed positive extension force, the mechanical potential is bending energy minus $FL$. Replacing $L$ by $\mathcal{L}-(\mathcal{L}-L)$ leaves, up to a force-dependent constant,Its first variation isThe higher-order Euler-Lagrange equation is thereforeFor a Fourier transform relation, take a periodic bulk segment, or a long-filament bulk approximation with negligible endpoint contributions; these are necessary boundary conditions because the question specifies no filament end conditions. Using $f(x)=\int \hat{f}(q)e^{iqx}dq/(2\pi )$, every nonzero wavenumber obeysA zero displacement mode is an arbitrary translation. In a periodic representation a constant intrinsic-curvature mode contributes a constant bending cost but no bulk slope, so it does not contribute to the following deficit. The Parseval identity givesFor clarity about finite normalization, on a periodic interval of length $L$ define ${\kappa }_n=L^{-1}{\int }_0^L{\kappa }_0(x)e^{-i{q}_{n}x}dx$, $q_n=2\pi n/L$. Then the equivalent exact quadratic-order formula isThis is force-extension of an intrinsically curved filament, determined by its particular fixed shape rather than by a thermal orientation entropy.

For an ensemble with quenched intrinsic curvature, first minimize each realization and then average the last expression, replacing $|{\kappa }_n{|}^2$ by its ensemble mean. For stationary bulk disorder, let $S_{\kappa }(q)$ be the power spectrum, normalized so that its correlation is $⟨{\kappa }_0(x){\kappa }_0(x+r)⟩=\int S_{\kappa }(q)e^{iqr}dq/(2\pi )$. The measurable mean deficit per unit projected length isIf the curvature is smooth enough that the second spectral moment is finite, one can take the large-force limit under this integral, obtaining the smooth-curvature high-force extension lawFor a stationary ensemble the integral is $L⟨({\kappa }_{0,x}{)}^2⟩$. Equivalently, the bulk slope is ${\zeta }_x\simeq -(B/F){\kappa }_{0,x}$; a plot of the mean deficit against $F^{-2}$ measures the mean squared intrinsic-curvature gradient, if $B$ is known. The $F^{-2}$ law is a smooth-curvature bulk result, not a universal law for every random filament. For example, ideal white intrinsic-curvature disorder with $S_{\kappa }(q)=S_0$ has an infinite second spectral moment; retaining the full integral instead gives $⟨\mathcal{L}-L⟩/L=S_0\sqrt{B}/(8\sqrt{F})$ in that ideal continuum model.

Endpoint conditions also matter. If each end is moment-free and has no transverse applied force, the natural conditions from the variation areat each end. These can produce a free-end curvature boundary layer of width $\ell =\sqrt{B/F}$. For a constant intrinsic filament curvature $\kappa$, integration of the higher-order Euler-Lagrange equation with zero transverse force gives $B({\zeta }_x{)}_{xx}-F{\zeta }_x=0$ and henceThe last formula follows by integrating ${\sinh }^2[(x-L/2)/\ell ]$, and tends to ${\kappa }^2{\ell }^3/2\propto F^{-3/2}$, despite the vanishing bulk curvature gradient. Thus a finite free-ended measurement must include or suppress endpoint bending before interpreting a bulk $F^{-2}$ fit.

For the freely jointed chain, rigid links have independent orientations, so the restoring force is entropic. Let $T$ now denote absolute temperature, and $\beta =1/(k_{B}T)$ with Boltzmann constant $k_B$. In the usual three-dimensional chain, a segment at polar angle $\theta$ has force energy $-Fb\cos \theta$. Its angular canonical partition function and the full chain partition function areDifferentiating $\log Z_N$ with respect to $\beta F$ yields the exact mean projected displacementHere $\mathcal{L}$ is the Langevin function; this is force-extension of a three-dimensional freely jointed chain, in the fixed-force canonical ensemble. At high extension, $\coth \xi =1+2e^{-2\xi }+\cdots$, soThus the smooth quenched-curvature bulk deficit falls as $F^{-2}$, whereas the three-dimensional freely jointed chain deficit falls as $F^{-1}$ and depends explicitly on temperature. The distinction is mechanical straightening of a fixed intrinsic shape versus orientational entropy; thermal bending fluctuations of a stiff filament have not been included in the first calculation.

If the comparison chain is also restricted to two dimensions, it is a planar freely jointed chain rather than the three-dimensional model. In this case,The last identity follows directly by differentiating the integral defining $I_0$; $I_0$ and $I_1$ are Modified Bessel functions of the first kind. At large $\xi$ the angular weight is proportional to $e^{-\xi {\theta }^2/2}$, giving $⟨\cos \theta ⟩\simeq 1-⟨{\theta }^2⟩/2=1-1/(2\xi )$. Consequently $Nb-⟨L⟩\sim N{k}_{B}T/(2F)$: the same inverse-force exponent, with half the three-dimensional coefficient. Both versions assume inextensible links and forces below the regime where the links themselves stretch.

For the anticlockwise boundary of the region, the inward normal is the left normal, so the signed curvature obeys the turning tangent theorem ${\int }_0^{L}k(s)ds=2\pi$. Therefore some $s_0$ has $k(s_0)\le 2\pi /L$. The enclosed area satisfies $A\ge d_1{d}_2$, and the planar isoperimetric inequality gives $L\ge 2\sqrt{\pi A}$. ConsequentlyThis uses the interpretation of the curve as a simple boundary traversed once; the inward normal and enclosed region in the question have that meaning.

Fix the standard orientation of a vector space on the plane. For an oriented regular curve in that plane, let $T=(T_x,T_y)$ be its unit tangent vector and let $N_L=(-T_y,T_x)$ be its left unit normal, obtained by an anticlockwise rotation through $\pi /2$. Its signed curvature $k$, also commonly written ${\kappa }_s$, is defined byHere $s$ increases with the chosen traversal and $\det$ is the determinant. Positive signed curvature means turning left; negative signed curvature means turning right. The left unit normal exists even when $k=0$; when $k<0$ it points opposite to the principal normal vector. See Ghomi's university lecture notes, section 1.10.

For an arbitrary regular parametric curve $r(t)=(x(t),y(t))$ traversed with increasing $t$, the signed curvature isFor the anticlockwise circle $r(t)=(R\cos t,R\sin t)$, the numerator is $R^2$ and the denominator is $R^3$, so $k=1/R$. For the clockwise traversal $q(t)=(R\cos t,-R\sin t)$, the numerator is $-R^2$, so $k=-1/R$, while unsigned curvature is $1/R$ in both cases. On this circle $N_L$ points inward for anticlockwise traversal and outward for clockwise traversal. For any simple closed regular curve traversed anticlockwise, $N_L$ is the inward unit normal, but signed curvature can still be negative along nonconvex portions.

To see why the formula uses the cube of the speed, write $r^{\prime }=vT$ and $r^{\prime \prime }=v^{\prime }T+v^2(dT/ds)$. The term parallel to $T$ contributes zero to the determinant, giving $\det (r^{\prime },r^{\prime \prime })=v^3\det (T,dT/ds)=v^{3}k$. Under a reparametrization $q(u)=r(\varphi (u))$ with ${\varphi }^{\prime }\ne 0$, the numerator gains a factor $({\varphi }^{\prime }{)}^3$ and the denominator gains $|{\varphi }^{\prime }{|}^3$. Thus orientation-preserving reparametrization preserves signed curvature, while reversing traversal changes its sign at corresponding points. Unsigned curvature is preserved in either case.

Locally write the unit tangent vector as $T=(\cos \theta ,\sin \theta )$. Differentiating with respect to arc length gives $k=d\theta /ds$: signed curvature records the direction as well as the amount of turning. Its integral therefore measures net turning, as in the turning tangent theorem; the integral of $|k|$ is total curvature and counts both left and right turns positively.

The unit tangent of a simple regular closed plane curve traversed anticlockwise turns once. Consequently its integrated signed curvature is $2\pi$; clockwise orientation gives $-2\pi$. Simplicity and one traversal are essential to this value.