Revisiting Differential Geometry (3): Arc Length & Curvature

 

In the usual treatment of differential geometry,  arc length  is given by:

L =Öå ³ _i =1 ( y _i  -  x _i ) ²

Where ( x _i , y _i) define the end points of a curve segment. Hence, when defining the length of an arc C of a curve we can technically approximate it using a sequential series of broken lines or chords, as depicted below:

The length of such broken lines can easily be determined if the end points of each chord are known. Let ℓ_i  then be a series of broken lines or chords whose end points lie on C.  Then if ℓ_i   tends to some limit s,  C is said to be rectifiable and is called the length of the arc C. 

Let x(t) then (a <  t  <  b) be an allowable representation of an arc of a curve of class r > 1.  Then the arc C has length: 

s = ò ^b _a    Ö {å ³ _i =1  (dx _i /dt) ² } dt  

And s is independent of the choice of allowable representation. Note that if we replace the fixed value b in the above with the variable t then s becomes a function of t. Note also that a can be replaced with any other fixed value. Thereby we obtain the integral:

s(t) =  ò ^b _a     Ö(x' · x') dt 

The function s(t) is then called the arc length of C. If  t _o  > t then s(t) is positive and equal to the arc a(t _o) b (t)  of C. If t< t _o  then s(t) is negative and the length of a(t _o) b (t)  is given by - s(t). Instead of:

s² =  å ³ _i =1 x _i ² =  x' · x'

We can write: 

 ds² =å ³ _i =1 dx _i ²   =  dx · dx

where  ds is called the element of arc.  For example, looking at the circular helix we find:  

x'  =  (- r sin t, r cos t, c),  x' · x'  =  r² + c ²

So that:  s(t)  =  tÖr² + c ²

Then if we also wish to represent the circular helix by a parametric representation with arc length s as parameter, we can write:  

x(s) =  (r cos (s/w), r sin (s/w), cs/w ) 

And:  w =   Ör² + c ²

 What about?: 

x ₁ = t,   

x ₂ = t sin 1/t   {t ≠  0  

                            0  {t = 0

x ₃ = 0   (0 <  t   <  t ₁

 We note the function is continuous even at t = 0, but the corresponding point set - which doesn't form an arc according to the definition, has no length.

Example (3): If the arc length in polar coordinates can be obtained from, e.g.:  

a)Find the arc length between  0  and  p/2, 

b) then the arc length between  -p/2  and  p/2 for the Archimedean spiral:  r  =  q - sin  q  and sketch the resulting  curve.  

(a) Solution.  From the polar equation we will have:

r(q) ²   = (q - sin  q) ²  =

q ² - 2 q sin  q + sin ²  q

And:   d r(q)  / d q   =  1 -  cos  q

(d r(q)  / d q ) ²   = 

(1 - cos  q) (1 - cos  q) = 1 - 2 cos  q + cos ²  q

Whence:

r(q) ²  +   (d r(q)  / d q ) ²   = 

q ² - 2 q sin  q +sin ²  q + 1 - 2 cos  q + cos ²  q

Leaving the integral:

ò^{p/ 2} ₀Ö{(q² - 2 q sin  q + sin ²  q) +

 ( 1 -2 cos  q + cos ²  q)}d q

=>ò^{p/ 2} ₀   Ö{q² - 2cos  q - 2 q sin  q +2 } d q

(Rem:  sin ² q  +  cos ²  q = 1)

The end result for the integral computation:

(b)The second integral is analogous - simply changing the limits - whereupon we find (using Mathcad):

The resulting curve is sketched below :

Suggested Problem:

Obtain the arc length s of the curve between x=0, x=2 for:

f(x) =  x³ / 2   -   x² / 3