Suppose the curve &Gamma. can be represented by the parametric vector function _ #~r (~s) _ where ~s represents the arc length in some region of the curve, and ~#r is differentiable on this interval.
The #~{tangent vector} at a point ~#r(~s) is defined as the unit vector in the direction of the curve, i.e.
#~t (~s) _ #:= _ fract{d#~r,d~s}
This is a unit vector as | d#~r ./ d~s | = 1 _ from the definition of arc length.
The #~{normal vector} at the point #~r(~s) is the unit vector defined by:
#~n (~s) _ #:= _ fract{d#~t ./ d~s, | d#~t ./ d~s| }
i.e.
d#~t ./ d~s_ #:= _ &kappa. (~s) #~n (~s) , _ _ where _ &kappa. (~s) _ #:= _ | d#~t ./ d~s|
&kappa. is called the #~{curvature} of &Gamma.. _ The reciprocal of &kappa., _ &rho. (~s) _ #:= _ 1 ./ &kappa. (~s) , _ is called the #~{radius of curvature} of &Gamma. (at ~s)
Consider the circle in the ~x~y-plane (~z = 0) with centre at the origin and of radius 5:
#~r (&theta.) _ = _ 5 cos &theta. #~i + 5 sin &theta. #~j
&theta. can be regarded as the angle in radians between the ~x-axis and the radius at the point ~#r (&theta.). The arc length at this point is _ ~s _ = _ 5 &theta. . _ So we could write the equation in terms of arc length:
#~r ( ~s ) _ = _ 5 cos ( ~s ./ 5 ) #~i + 5 sin ( ~s ./ 5 ) #~j
So in this case:
#~t (~s) _ = _ fract{d#~r,d~s} _ = _ - fract{1,5} # 5 sin ( ~s ./ 5 ) #~i + fract{1,5} # 5 cos ( ~s ./ 5 ) #~j
_ _ _ _ _ _ = _ - sin ( ~s ./ 5 ) #~i + cos ( ~s ./ 5 ) #~j
fract{d#~t,d~s} _ = _ - fract{1,5} cos ( ~s ./ 5 ) #~i - fract{1,5} sin ( ~s ./ 5 ) #~j
So the curvature
&kappa. _ = _ | d#~t ./ d~s | _ = _ fract{&sqrt. ${ cos&powtwo. ( ~s ./ 5 ) + sin&powtwo. ( ~s ./ 5 )},5} _ = _ fract{1,5}
and so the radius of curvature, _ &rho. _ = _ 5 , _ giving:
#~n _ = _ - cos ( ~s ./ 5 ) #~i - sin ( ~s ./ 5 ) #~j
#{Note:} in this fairly simple case the radius of curvature and consequently curvature are constant for all points on the circle. In general these will be functions of the arc length.
#{Exercise:} Find the tangent, normal, and curvature (or radius of curvature) of the circle of radius 5 but centred on the point (3,8) :
#~r (&theta.) _ = _ ( 5 cos &theta. + 3 ) #~i + ( 5 sin &theta. + 8 ) #~j
Just to convince yourself that the tangent is not always perpendicular to #~r , and that the normal is not always in the opposite direction to #~r.
In the above example the original parameter, &theta., is just a fractional multiple of the arc length. More generally we will have _ &theta. _ = _ &theta. ( ~s ), the function will be monotonic - in fact we can assume that it is strictly increasing as if it is strictly decreasing we can simply measure arc length in the other direction. In these circumstances we use the chain rule of differentiation:
#~t (~s) _ = _ fract{d#~r,d~s} _ = _ fract{d#~r,d&theta.}fract{d&theta.,d~s}
It is not necessary to know explicitly &theta. as a function of arc length, if &theta. is a strictly increasing function of arc length then d#~r ./ d&theta. will be in the same direction as d#~r ./ d~s and as | ~#t | == 1, so
#~t (~s) _ = _ fract{d#~r ./ d&theta., | d#~r ./ d&theta. | } , _ _ i.e. _ fract{d&theta.,d~s} _ = _ fract{1, | d#~r ./ d&theta. | }
Similarly
&kappa. ( ~s ) _ = _ mod{fract{d#~t,d~s}} _ = _ mod{fract{d#~t,d&theta.} fract{d&theta.,d~s}} _ = _ mod{fract{d#~t,d&theta.}} -: mod{fract{d#~r,d&theta.}}
and
#~n (~s) _ = _ fract{d#~t,d&theta.} # fract{d&theta.,d~s} # fract{1,&kappa. (~s)}
_ _ _ = _ fract{d#~t,d&theta.} # script{mod{fract{d#~r,d&theta.}},,,-1,} # script{mod{fract{d#~t,d&theta.}},,,-1,} # mod{fract{d#~r,d&theta.}} _ = _ fract{d#~t,d&theta.} -: mod{fract{d#~t,d&theta.}}
Which makes sense as ~#n is a unit vector.
In the example _ #~r (&theta.) _ = _ 5 cos &theta. #~i + 5 sin &theta. #~j _ :
fract{d#~r,d&theta.} _ = _ - 5 sin &theta. #~i + 5 cos &theta. #~j , _ _ _ _ mod{fract{d#~r,d&theta.}} _ = _ 5
#~t _ = _ - sin &theta. #~i + cos &theta. #~j
fract{d#~t,d&theta.} _ = _ - cos &theta. #~i - sin &theta. #~j , _ _ _ _ mod{fract{d#~t,d&theta.}} _ = _ 1
&kappa. _ = _ mod{fract{d#~t,d&theta.}} -: mod{fract{d#~r,d&theta.}} _ = _ fract{1,5}
~#n ( &theta. )_ = _ fract{d#~t,d&theta.} -: mod{fract{d#~t,d&theta.}} _ = _ - cos &theta. #~i - sin &theta. #~j
#{Exercise:} Use the above method to find the tangent, normal, and curvature (or radius of curvature) of the elipse:
#~r (&theta.) _ = _ 2 cos &theta. #~i + 3 sin &theta. #~j
The unit #~{binormal vector} at the point #~r(~s) is the unit vector defined by:
#~b (~s) _ #:= _ #~t (~s) # #~n (~s)
Obviously, #~b is normal to the plane of ~#t and #~n, so if the curve lies in a plane then #~b is constant, otherwise the rate of change of #~b, _ d#~b ./ d~s , _ indicates how quickly the plane through ~#t and #~n is changing, i.e. how twisted the curve is.
fract{d#~b,d~s} _ = _ fract{d( #~t # #~n ),d~s} _ = _ fract{d#~t,d~s} # #~n _ + _ #~t # fract{d#~n,d~s} _ = _ #~t # fract{d#~n,d~s}
as
fract{d#~t,d~s} # #~n _ = _ &kappa. #~n # ~#n _ = _ #0
so
fract{d#~b,d~s} # #~n _ = _ rndb{#~t # fract{d#~n,d~s}} # ~#n _ = _ rndb{#~t &dot. ~#n } fract{d#~n,d~s} _ - _ rndb{fract{d#~n,d~s} &dot. ~#n} #~t
(see vector triple product )
but _ #~t &dot. ~#n _ = _ 0, _ and
fract{d#~n,d~s} &dot. ~#n _ = _ fract{1,2} fract{d ( #~n &dot. ~#n ),d~s} _ = _ fract{1,2} fract{d ( 1 ),d~s} _ = _ 0
since
fract{d ( #~n &dot. ~#n ),d~s} _ = _ ~#n &dot. fract{d#~n,d~s} _ + _ fract{d#~n,d~s} &dot. ~#n _ = _ 2 rndb{fract{d#~n,d~s} &dot. ~#n}
So
fract{d#~b,d~s} # #~n _ = _ 0 , _ i.e. _ fract{d#~b,d~s} _ .// _ #~n , write _ _ fract{d#~b,d~s} _ = _ - &tau. #~n
&tau. = &tau.(~s) _ is called the #~{torsion} of the curve (at the point #~r(~s) ).
Note again that
fract{d#~b,d~s} _ = _ fract{d#~b,d&theta.} fract{d&theta.,d~s} _ = _ fract{d#~b,d&theta.} -: mod{fract{d#~r,d&theta.}}