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:
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)
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:
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.
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