Curvature, Arc Length and Parametric Representations of Curves

One of the more important areas of higher mathematics involves finding the arc length of a curve, as well as the curvature.   A number of instructive examples are given in this post for doing so.  A logical place to begin is with the arc length formula (derivations can be found in any good Calculus textbook.  For example,  an element ds (du) of arc on a curve is:

But in derivation terms - treating dx (du1) and dy (du2) as 'legs' of a right triangle:
we can treat ds as the differential of arc length such that.

ds² =  dx² + dy ²

Or:   ds   =   Ö (dx² + dy ² )  =    dx Ö [ 1  +  (dy/dx) ² ]

This can then be integrated (between appropriate limits, say x1 and x2) to give the total length of a curve..   Thence the arc length  is given by:

where the function f(x) defines the curve for which the arc length is evaluated between the points x1 and x2.   For which we have:  f(x)   = y =    [(x)]^½

Consider then finding the arc length between x1 = 2 and x2 = 8, i.e. for the segment of the parabola shown:

 

Then evaluating the integral as shown, one obtains: s =   8.886

A more difficult problem involves a polar curve, i.e. a curve in polar coordinates such as:

This curve has the function:  r(q)  =  q  - sin (q)

And we wish to find its length between q = -p   and q = p .  

The relevant integral for the arc length for this example is:


Performing the integration one then obtains: L  = 8.764

Which interested and energized readers are invited to check!

Parametric Representations:

In advanced treatments of curvature including determining the curvature of a complex curve, parametric representations are the norm. This usually entails taking the tangent T  to the curve at a point.  From this the curvature  k  of the arc can also be computed.  For example, one can give the parametric representation of a specific type of curve using:

x =  6 sin 2 t,  y = 6 cos 2 t, z = 5t

For which the tangent to the curve at a point is given by:

T = dR/ ds =  i(dx/ds) + j(dy/ds) + k(dz/ds)

=

i (dx/dt) (dt/ds) +   j (dy/dt)(dt/ds) + k (dz/dt) (dt/ds)

Then:

dx/dt=  12 cost 2t, dy/dt = -12 sin 2t,  dz/dt = 5

Therefore, the tangent unit vector is:

| T |  =  1 = (dt/ds)² [(12 cost 2t)² + (-12 sin 2t)² + 5²]

|T |  =  1 = (dt/ds)² [(144 + 25)] =   (dt/ds)²  (169)

So that: (dt/ds)   = 1/ 13 

And:  T   = 1/ 13   [12 cost 2t) i + (-12 sin 2t) j + 5 k ] dt/ds

The curvature can then be obtained from: k = | dT/ ds | 

Thus, any curve can be given by a parametric representation:   u1 = u1(t)  and u2=  u2(t)

For such a curve, consider now the distance between two infinitely near points on the surface, i.e. the distance or interval ds between two specified points.

Example problem:

A curve is given in spherical coordinates xⁱ by:

x¹ = t,   x² = arcsin 1/t,    x³ =  (t2 – 1) ^1/2

Compute the length of the arc between t = 1 and t = 2  

Solution:

(ds/ dt)² =    (dx¹/ dt) ²  +  (x¹ ) ²  (dx²/ dt) ²    +   (x¹  sin x  ²) ² (dx³/ dt) ² 

And:

(dx¹/ dt) ²  =   1

(dx²/ dt) ²   = [ -1/  t²   / Ö {1 – (  1/  t²  )} = 1/ t²  (t²  -   1)

  (dx³/ dt) ²    = 2t/ Ö2(t²  -   1) =  t² / (t²  -   1)

Whence:

(ds/ dt)² =     

1 +  t²   · 1/ t²  (t²  -   1) + (t ·  1/  t ) ²   ·  t² / (t²  -   1)

=   2 t² / (t²  -   1)

Then the length of the curve is:

L = ò ₁ ²   Ö2 t/ (t²  -   1)^1/2  dt = Ö2(t²  -   1) ] ₁²      =  Ö6

Problem for Math Mavens:

 For the parametric example with:

 T   = 1/ 13   [12 cost 2t) i + (-12 sin 2t) j + 5 k ] dt/ds,

 Find the curvature and the length of the curve from t = 0 to t = p.