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 + j y + k 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)
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( - a w sin wt dt/ds) + j(a w 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 w 2 + b 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 w 2 + b 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 = a w (- i sin wt + j cos wt ) + b k / Ö ( a 2 w 2 + b 2)
The length (or arc length) of a given space curve can be found by computing ds from:
ds = + Ö (dx^{2} + 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 w (- i sin wt + j cos wt ) + b k / Ö ( a 2 w 2 + b 2) {Ö ( a 2 w 2 + b 2) }
= a w (- i sin wt + j cos wt ) + b k / ( a 2 w 2 + b 2)
Factor out the non-trig term:
dT/ ds = a w / ( a 2 w 2 + b 2) [ i sin wt + j cos wt + b k ]
Then by inspection (i.e. comparing with dT/ ds = k N):
k = a w / ( a 2 w 2 + b 2)
Example problem:
Given a curve defined by:
R = ix + j y + k 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_{ }D _{s} _{ -> 0} (D R/ D s)
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( - a w sin wt dt/ds) + j(a w 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 w 2 + b 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 w 2 + b 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 = a w (- i sin wt + j cos wt ) + b k / Ö ( a 2 w 2 + b 2)
The length (or arc length) of a given space curve can be found by computing ds from:
ds = + Ö (dx^{2} + 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 w (- i sin wt + j cos wt ) + b k / Ö ( a 2 w 2 + b 2) {Ö ( a 2 w 2 + b 2) }
= a w (- i sin wt + j cos wt ) + b k / ( a 2 w 2 + b 2)
Factor out the non-trig term:
dT/ ds = a w / ( a 2 w 2 + b 2) [ i sin wt + j cos wt + b k ]
Then by inspection (i.e. comparing with dT/ ds = k N):
k = a w / ( a 2 w 2 + b 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 dt/ ds
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} + 5^{2}]
| 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
Problems for the Math whiz:
1) For the example problem:
Find the curvature k and the length of the curve from t = 0 to t = p
Find the unit vector T = dR/ dsFind the curvature k 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)