Thursday, September 20, 2018

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 - with dx and dy also shown as 'legs' of a right triangle:


Where we treat ds as the differential of arc length.

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.   Consider then finding the arc length between x1 = 0 and x2 = 8 for the parabola section below:

 For which we have:  f(x)   =   [(x)]½

Then evaluating the integral as shown, one obtains: s =   8.732

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) +   (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,   x= 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/  t)} = 1/ t(t-   1)

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

Whence:

(ds/ dt)2 =     

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

=   2 t2 / (t-   1)

Then the length of the curve is:

L = ò 1 2   Ö2 t/ (t-   1)1/2  dt = Ö2(t-   1) ] 1    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

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