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The binormal is $B=T\times N$, the cross product of the unit tangent vector and principal normal vector. It is a unit vector perpendicular to both factors and completes the oriented Frenet frame.

Along the boundary of an oriented surface, the outward unit conormal lies in the tangent plane of the surface and is perpendicular to the boundary unit tangent vector. If $\mathbf{n}$ is the chosen unit normal vector, the positively oriented unit tangent vector is $\mathbf{n}\times \mathbf{\nu }$.

For a regular curve with positive curvature of a space curve, the Frenet-Serret formulas give $dT/ds=\kappa N$. If the unit tangent vector has constant component along a fixed vector $a$, differentiation of $T\cdot a$ gives $\kappa N\cdot a=0$. Hence every principal normal vector is perpendicular to that direction. Nonzero curvature is needed to define the principal normal.

For a twice continuously differentiable regular curve $r(t)$ in Euclidean space, its unsigned curvature measures how quickly its unit tangent vector changes per unit arc length. WriteThe curvature iswhere the norm is the Euclidean norm. Differentiating with respect to arc length removes the effect of traversal speed; differentiating $T$ with respect to $t$ alone generally does not. See Ghomi's university lecture notes, section 1.7.

For a circle of radius $R>0$, use $r(t)=(R\cos t,R\sin t)$. Then $v=R$, $T=(-\sin t,\cos t)$ and $\Vert dT/dt\Vert =1$, giving $\kappa =1/R$. A straight line $r(t)=p+ta$ with $a\ne 0$ has constant unit tangent vector, so $\kappa =0$.

The same circle traversed as $q(u)=(R\cos (3u),R\sin (3u))$ has speed $3R$ and $\Vert dT/du\Vert =3$: its curvature remains $1/R$. By contrast, enlarging the circle to radius $2R$ changes its geometry and gives curvature $1/(2R)$. More generally, a dilation $r\mapsto ar$ with $a>0$ divides curvature by $a$, because it multiplies arc length by $a$ while leaving unit tangent vectors unchanged. Thus curvature has units of inverse length, and a reparametrization changes neither its value at a corresponding point nor the shape.

Differentiating $T\cdot T=1$, using the dot product, shows $T\cdot (dT/ds)=0$: the change is perpendicular to the unit tangent vector. Wherever $\kappa >0$, the principal normal vector $N_p=(dT/ds)/\kappa$ points toward the bend. For a plane curve, the osculating circle at such a point has radius of curvature $\rho =1/\kappa$ and center $r+\rho N_p$. It matches the point, tangent line, and curvature, providing a local circular approximation. At $\kappa =0$ there is no finite osculating circle defined by this formula. See Ghomi's university lecture notes, section 1.13.

This makes curvature useful for comparing bends without confusing them with speed. If $t$ denotes time, differentiating $r^{\prime }=vT$ gives acceleration $r^{\prime \prime }=v^{\prime }T+v^2(dT/ds)$, whose component perpendicular to the unit tangent vector has magnitude $v^2\kappa$. The curvature of a plane curve example below computes the bend and osculating circle of a parabola explicitly.

For smooth surfaces, bending has more than one direction. The principal curvatures $k_1,k_2$ instead define Gaussian curvature $K=k_1{k}_2$ and mean curvature $H=(k_1+k_2)/2$. These are surface quantities, distinct from the curvature of an individual curve. In particular, the circle value $1/R$ must not be confused with the Gaussian curvature $1/R^2$ of a sphere of radius $R$. See Ritter's Oxford lecture notes, section 12.2.

For a continuously differentiable nonzero vector field $F$, the curvature of a space curve following its direction isTake the unit tangent vector $t=\pm F/|F|$ and differentiate with respect to arc length. The derivative of $|F{|}^{-1}$ is parallel to $F$ and disappears from $t\times t^{\prime }$. Since $t\cdot t^{\prime }=0$, we have $|t\times t^{\prime }|=|t^{\prime }|=\kappa$, proving the formula. It remains valid with zero curvature, even where the usual Frenet frame is undefined.

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.

An inextensible filament keeps the length of every material segment fixed. With material arc length $s$, its centerline obeys $|{\mathbf{r}}_s|=1$ and has unit tangent vector $\mathbf{t}={\mathbf{r}}_s$. A filament tension acts as a Lagrange multiplier for this constraint.

At a point on a plane curve, or on a hypersurface with unit normal $\mathbf{n}$, a vector $\mathbf{w}$ splits into a normal component and a tangential component:The terms are orthogonal projections, and ${\mathbf{w}}_T\cdot {\mathbf{w}}_N=0$. Changing $\mathbf{n}$ to $-\mathbf{n}$ reverses the scalar coefficient $\mathbf{w}\cdot \mathbf{n}$, but leaves the normal component vector unchanged.

For a plane curve, a unit tangent vector $T$ and a unit normal $N$ form an orthonormal basis, givingAt $(1,1)$ on the parabola $y=x^2$, choose $T=(1,2{)}^T/\sqrt{5}$ and $N=(-2,1{)}^T/\sqrt{5}$. For $\mathbf{w}=(3,1{)}^T$,Their sum is $(3,1{)}^T$ and their inner product is zero. A negative normal coefficient means that the normal component points opposite the chosen $N$, rather than giving it a negative Euclidean norm.

These are local directions: being tangent at a point does not mean that a straight displacement stays on the curve. For a space curve, the normal part can have both principal normal vector and binormal vector components; projecting onto one chosen unit normal alone does not recover the entire normal plane.

A normal vector at a point of an embedded smooth manifold is orthogonal to every tangent vector in its tangent space, using the surrounding dot product. Its direction describes departure from the local tangent geometry. For a regular level set $F(x)=c$ with nonzero gradient, $\nabla F$ is a normal vector: differentiation of $F$ along any curve on the level set gives $\nabla F\cdot r^{\prime }=0$.

For a plane curve, the normal direction is a straight line perpendicular to its unit tangent vector; the two unit normals differ by sign. For a curve in three-dimensional Euclidean space, the orthogonal directions form a two-dimensional plane. A principal normal vector selects one of them using curvature, so it must be distinguished from an arbitrary normal vector.

For a light ray parametrized by arc length $s$, with unit tangent vector $t$ in a smooth positive refractive index $\mu$, the optical momentum is $p=\mu t$. This is the momentum for optical path-length normalization. The Euler-Lagrange equation for Fermat principle, $\int \mu (r)|r^{\prime }(\tau )|d\tau$, gives $d(\mu t)/ds=\nabla \mu$. Therefore an index independent of $z$ conserves $p_z=\mu dz/ds$. In coordinates $r(z),\theta (z)$ this is $\mu /\sqrt{1+r^{\prime 2}+r^2{\theta }^{\prime 2}}$, the first integral supplied by the Beltrami identity for several dependent variables.

At a point of a twice continuously differentiable regular curve with positive curvature $\kappa$, its osculating circle has radius $1/\kappa$ and center $c=r+N_p/\kappa$, where $N_p$ is the principal normal vector. It matches the point, tangent line and curvature. For a space curve, it lies in the plane spanned by the unit tangent vector and principal normal vector. The parabola example under curvature of a plane curve gives center $(0,1/2)$ and radius $1/2$ at the origin.

Let $s$ be arc length, so inextensibility makes $r^{\prime }=d_3$ a unit tangent vector. The curvature is $\kappa ={\theta }^{\prime }$, hence $m^{\prime }=EI{\theta }^{\prime \prime }{e}_z$ for uniform bending stiffness. Balance of linear momentum in static equilibrium gives a constant resultant $n=F{e}_x+G{e}_y$.

Taking the cross product directly,The scalar component of conservation of angular momentum therefore becomesTogether with $x^{\prime }=\cos \theta$ and $y^{\prime }=\sin \theta$, this Euler elastica equation determines the centerline once the appropriate boundary conditions are supplied.

Let $R$ be reflection in the plane. Choose a point of the regular intersection curve and a unit tangent vector $w$ to it. Both the point and $w$ are fixed by $R$. The unique local geodesic with these initial data is carried by the surface isometry $R$ to another geodesic with the same initial data. Uniqueness therefore gives $R\gamma =\gamma$. Its image lies in the fixed-point set of the reflection, namely the plane, hence locally in the intersection curve. Thus the intersection curve is a geodesic, when parametrized by arclength.

Use the unit normal from part (c), and orient ${\gamma }_a$ by increasing $v$. Write $T={\gamma }_a^{\prime }/|{\gamma }_a^{\prime }|$ for its unit tangent vector. The signed geodesic curvature convention used below iswhere $s$ is arc length. The second equality follows by differentiating $T$: the term parallel to ${\gamma }_a^{\prime }$ disappears in the scalar triple product. It makes the correction for a parameter which is not arc-length parametrization explicit.

Put $C=\cos a$, $S=\sin a$, $D=1+S^2$, and $G=3+2C$. The speed is the constant $\sqrt{G}$. At $v=0$,soTaking its dot product with $N=(C,S,-S)/\sqrt{D}$ yields $-S(3+2C)/\sqrt{D}=-SG/\sqrt{D}$. ThereforeThe screw motion of part (b) makes this independent of $v$. Thus ${\gamma }_a$ is a geodesic precisely when $a$ is an integer multiple of $\pi$. Reversing the direction of traversal or the unit normal changes the sign of signed geodesic curvature; its magnitude is the absolute value of the displayed expression. The signed convention is needed in part (e).

For the unit-speed curve, the unit tangent vector, curvature of a space curve, principal normal vector and binormal vector areDifferentiating $t\cdot t=1$ gives $t\cdot {\gamma }^{\prime \prime }=0$, so $t,p,b$ form a positively oriented orthonormal basis. The hypothesis ${\gamma }^{\prime \prime }\ne 0$ makes the principal normal vector well-defined.

To determine the sign of the torsion of a space curve without relying on a pictorial handedness convention, take $R>0$, $h\ne 0$, $L=\sqrt{R^2+h^2}$ and use the unit-speed circular helixWriting $u=s/L$, direct differentiation gives its Frenet frame:ConsequentlyAt $s=0$ the marked point is $(R,0,0)$, with $t=(0,R/L,h/L)$, $p=(-1,0,0)$, $b=(0,-h/L,R/L)$ and $b^{\prime }=(h/L^2,0,0)$. For $h>0$, $b^{\prime }$ points opposite to $p$ and the torsion of a space curve is positive. Replacing $h$ by $-h$ reflects the circular helix across the horizontal plane and makes the torsion of a space curve negative; then $b^{\prime }$ points in the same direction as $p$. The direction of increasing $s$ is fixed in both panels.

For a regular differentiable curve $\mathbf{x}(t)$, its speed is $ds/dt=|{\mathbf{x}}^{\prime }(t)|>0$, where $s$ is arc length. The oriented unit tangent vector is $\hat{\mathbf{T}}={\mathbf{x}}^{\prime }/|{\mathbf{x}}^{\prime }|$. The curvature of a space curve is $\kappa =|d\hat{\mathbf{T}}/ds|$. Equivalently, for a twice differentiable regular curve,For the helix, differentiation gives ${\mathbf{x}}^{\prime }=(-a\sin t,a\cos t,b)$ and speed $v=\sqrt{a^2+b^2}$. ConsequentlyIndeed $d\hat{\mathbf{T}}/dt=(-a\cos t,-a\sin t,0)/v$ has magnitude $|a|/v$, and division by $ds/dt=v$ gives the result. This requires $(a,b)\ne (0,0)$; the excluded constant curve has no well-defined unit tangent vector. For $a=0,b\ne 0$ the curve is a straight line with zero curvature of a space curve.

Divide the velocity by the speed calculated above. The unit tangent vector isIts derivative isBecause $ds/dt=\sqrt{2}(1+t^2)>0$, the principal normal vector has the direction of $T^{\prime }(t)$, givingThis is a unit vector, and $N\cdot e_z=0$ for every real $t$. The curvature of a space curve is $|T^{\prime }|/(ds/dt)=1/(1+t^2{)}^2>0$, so the principal normal vector is defined everywhere on this curve. The same perpendicularity follows from constant tangent component forces perpendicular principal normals: $T\cdot e_z=1/\sqrt{2}$ is constant, and differentiating this identity forces the normal component along $e_z$ to vanish.

For a regular curve parametrized by $u$, put $v(u)=|{\mathbf{x}}^{\prime }(u)|>0$ and orient arc length in the direction of increasing $u$. Since $ds=|d\mathbf{x}|=vdu$, the unit tangent vector isRegularity is the condition that permits division by $v$.

Differentiate the unit tangent vector with $d/ds=v^{-1}d/du$ and $v^{\prime }=({\mathbf{x}}^{\prime }\cdot {\mathbf{x}}^{\prime \prime })/v$. This givesThe curvature of a space curve is $\kappa =|d\mathbf{t}/ds|$. Squaring the displayed expression and simplifying givesThe last equality is the cross product norm identity. Equivalently, the hinted vector triple product rewrites $d\mathbf{t}/ds=-{\mathbf{x}}^{\prime }\times ({\mathbf{x}}^{\prime }\times {\mathbf{x}}^{\prime \prime })/v^4$.

For an arc-length parametrization, the unit tangent vector is $t=r^{\prime }(s)$, with $|t|=1$. Its derivative is perpendicular to $t$. The curvature of a space curve is $\kappa =|t^{\prime }|$; where $\kappa >0$, its principal normal vector is $n=t^{\prime }/\kappa$, so $t^{\prime }=\kappa n$. The binormal vector $b=t\times n$ completes the positively oriented Frenet frame.

Differentiate $b\cdot b=1$ to get $b^{\prime }\cdot b=0$. AlsoThus $b^{\prime }$ has only an $n$ component. Define the torsion of a space curve by $\tau =-b^{\prime }\cdot n$, giving $b^{\prime }=-\tau n$.

Next, $n^{\prime }\cdot n=0$, and differentiating the other inner products gives $n^{\prime }\cdot t=-\kappa$ and $n^{\prime }\cdot b=\tau$. Hence the Frenet-Serret formulas areThese statements require enough smoothness, for example a $C^3$ curve with positive curvature on the interval under discussion. If $\kappa =0$, the unit tangent vector is still defined, but the standard principal normal vector, binormal vector and torsion of a space curve need not be; their existence cannot be inferred just from smoothness of the curve.