Consider a fixed point **f**(*u*) and two moving points ** P**
and

f(u) +pf'(u) +qf''(u)

The *binormal* vector **b**(*u*) is
the unit-length vector of the cross-product of
**f**'(*u*) and **f**''(*u*):

Thus, the binormal vectorb(u) = (f'(u) ×f''(u)) / | (f'(u) ×f''(u)) |

The *normal* vector is the vector
perpendicular to both tangent and binormal vectors with its
direction determined by the right-handed system. That is, the unit-length
normal vector **n**(*u*) is defined to be

The linen(u) = (b(u) ×f'(u) ) / |b(u) ×f'(u) |

The first and second derivatives are as follows:f(u) = (acos(u),asin(u),bu)

The non-unit-length binormal vector is the cross-product off'(u) = ( -asin(u),acos(u),b)

f''(u) = ( -acos(u), -asin(u), 0 )

The non-unit-length normal vector is the cross-product of the binormal vector and the tangent vector, in this order:b(u) =f'(u) ×f''(u) = (absin(u), -abcos(u),a^{2})

If you comparen(u) =b(u) ×f'(u) = ( -a(a^{2}+b^{2})cos(u), -a(a^{2}+b^{2})sin(u), 0 )

From the definition of osculating plane, we know that this circle of curvature must be on the osculating plane. Since the circle of curvature is tangent to the curve and hence the tangent line, the center of curvature lies on the normal line.

The curvature at *u*, ** k**(

(ku) = |f'(u) ×f''(u) | / |f'(u) |^{3}

Let us continue with the previous example of the circular helix curve.
From the computations of **f**'(*u*) and **f**''(*u*),
we have

Therefore, the curvature at any point on the curve is a constantf'(u) = ( -asin(u),acos(u),b)

f'(u) ×f''(u) = (absin(u), -abcos(u),a^{2})

|f'(u) | =SQRT(a^{2}+b^{2})

|f'(u) ×f''(u) | =aSQRT(a^{2}+b^{2})

(ku) =a/ (a^{2}+b^{2})

Since the radius of the circle of curvature is 1/** k**, we see
that the center of circle of curvature is located at a distance of
(

- Consider a straight line:
**f**(*u*) = (*a*+*up*,*b*+*uq*,*c*+*ur*)**f**'(*u*) = (*p*,*q*,*r*)

|**f**'(*u*) | =**SQRT**(*p*^{2}+*q*^{2}+*r*^{2})

**f**''(*u*) = ( 0, 0, 0 )

**f**'(*u*) ×**f**''(*u*) = ( 0, 0, 0 )

(*k**u*) = 0**f**''(*u*) is a zero vector. - Consider a circle on the
*xy*-plane:**f**(*u*) = (*r*cos(*u*) +*p*,*r*sin(*u*) +*q*, 0 )*xy*-plane, the third coordinate function is always 0. From the given circle's equation we have the following:**f**'(*u*) = ( -*r*sin(*u*),*r*cos(*u*), 0 )

**f**''(*u*) = ( -*r*cos(*u*), -*r*sin(*u*), 0 )

**f**'(*u*) ×**f**''(*u*) = ( 0, 0,*r*^{2})

|**f**'(*u*) | =*r*

|**f**'(*u*) ×**f**''(*u*) | =*r*^{2}

**b**(*u*) = (**f**'(*u*) ×**f**''(*u*)) / |**f**'(*u*) ×**f**''(*u*) | = ( 0, 0, 1 )

**n**(*u*) = (**b**(*u*) ×**f**'(*u*)) / |**b**(*u*) ×**f**'(*u*) | = ( -cos(*u*), -sin(*u*), 0 )

(*k**u*) = 1/*r**u*), cos(*u*), 0), the binormal vector is ( 0, 0, 1), and the normal vector is ( -cos(*u*), sin(*u*), 0). The binormal vector is always perpendicular to the*xy*-plane while both the tangent and normal vectors lie on the*xy*-plane. The curvature of a circle is a constant 1/*r*. As a result, the radius of the circle of curvature is*r*and the circle of curvature is the given circle itself. - Consider the space cubic defines as follows:
**f**(*u*) = (*u*,*u*^{2},*u*^{3})(*k**u*):**f**'(*u*) = ( 1, 2*u*, 3*u*^{2})

|**f**'(*u*) | =**SQRT**(1 + 4*u*^{2}+ 9*u*^{4})

**f**''(*u*) = ( 0, 2, 6*u*)

**f**'(*u*) ×**f**''(*u*) = ( 6*u*^{2}, -6*u*, 2 )

|**f**'(*u*) ×**f**''(*u*) | = 2**SQRT**(1 + 9*u*^{2}+ 9*u*^{4})

(*k**u*) = 2**SQRT**(1 + 9*u*^{2}+ 9*u*^{4}) / (**SQRT**(1 + 4*u*^{2}+ 9*u*^{4}))^{3}

There is one more parameter, the *torsion*, that describing the way of
twisting of a space curve. A planar curve does not twist and hence its
torsion is zero. Since we will not use torsion in this course, it will not
be introduced here. See your calculus or differential geometry books for
the details.