Monday, October 4, 2021

Revisiting Parametric Curves in 3-space

The helix - a curve in 3-space to be analyzed.

Curves in 3-space play a major role in many areas of physics and are worthwhile exploring.  Adopting a positive direction for measuring distance along such a curve, say from a point P 0 at t =0, then the arc length can then be measured as P moves from  P 0.   Then the  position of P becomes a function if the arc length  s  from P 0  to P.  Then the vector:

R =  ix  + y + z

is also a function of s.   Of particular interest will be the derivative:

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

By taking the limit:

dR/ ds =    lim  s   -> 0   (Rs)

It can be shown one obtains a unit vector tangent to the curve at P, say, and pointing in the direction in which arc length s increases along the curve. One thereby arrives at the vector T:

T = dR/ ds

Consider for reference the helix shown which is described by:

x = a cos wt,   y =  a sin wt,  z = bt

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

We get:

i( - sin wt  dt/ds)   +   j(cos wt  dt/ds) + k( b dt/ds)

Which is the unit vector to the helix at any point P.  It is also of interest here to determine the magnitude of the scalar dt/ds.  This is accomplished by using the fact that:

| T |  =  1   or  T·T   =   1

Factoring out (dt/ds) from the earlier eqn.  and by using some algebra:

( a 2 2  +  2)  (dt/ds) 2  =   1

Whence we take the square roots, and make the scalar factor the subject, so:

dt/ ds  =   +   1/   Ö ( a 2 2  +  2)

Since we've agreed s (arc length) is an increasing function of t we can take dt/ds as a positive constant and  write for the tangent vector:

T  =    (- i sin   wt  +  j  cos  wt )  + b k /  Ö ( a 2 2  +  2)

The length (or arc length) of a given space curve can be found by computing ds from:

ds   =   + Ö (dx2 + dy 2  + dz 2 )

And integrating between appropriate limits. In this sense the procedure is much like that for obtaining the arc length of a curve in 2 dimensions.   In addition, the curvature  k  of a space curve (i.e. in 3 dimensions) is defined by the same vector equation as for a plane curve.  This means taking the derivative dT/ ds which will be either 0 (straight line or  k = 0) or  will be normal to T.  Defining a unit vector N, i.e. a principal normal to the curve at P, we may write:

| N |  =  1     and:  dT/ ds =       k N

Then the curvature for the helix is obtained:

|  dT/ ds |    =  [(dT/ dt)/  ds/ dt]  =

a (- i sin   wt  +  j  cos  wt )  + b k /  Ö ( a 2 2  +  2)  {Ö ( a 2 2  +  2) }

=  (- i sin   wt  +  j  cos  wt )  + b k /  ( a 2 2  +  2)

Factor out the non-trig term:

dT/ ds =    /  ( a 2 2  +  2)    [  i sin   wt  +  j  cos  wt   + b ]

Then by inspection (i.e. comparing with  dT/ ds =       k N):

k  =  /  ( a 2 2  +  2

Example problem:

Given a curve defined by:

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

Find an expression for  | T |   and T in terms of dtds

Solution:

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

Therefore:

| 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:

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

Problems for the Math whiz:

1) For the example problem:

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

2)  For a space curve defined by:

. x = exp(t) cost t,   y = exp(t) sin t,  z =  exp (t)

Find  the unit vector T =  dR/ ds