Page 224 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 18: Theory of Space Curves

Notes
L
=  x  x ds  Lr.
2
2
2
1
0
Since the geometric mean of two numbers is  their arithmetic mean, it follows that
1 1
2
2
A  r  (A   r )  Lr.
2 2
This gives, after squaring and cancellation of r, the inequality (14).
Suppose now that the equality sign in (14) holds. A and r have then the same geometric and
2
arithmetic mean, so that A = r and L = 2r. The direction of the lines g, g’ being arbitrary, this
2
means that C has the same “width” in all directions. Moreover, we must have the equality sign
everywhere in (15). It follows in particular that
2
2
2
'
'2
'
'2
(x ,x  x x ) = (x  x )(x  x ),
2
2
1
1
2
1
2
1
which gives
2
x 1   x 2  x  x 2 2   r.
1
x ' 2 x ' 1 x  x '2 2
'2
1
From the first equality in (15), the factor of proportionality is seen to be r, i.e.,
'
'
x  rx , x   rx .
2
1
2
1
This remains true when we interchange x and x , so that
2
1
'
x  rx .
1
2
Therefore, we have
2
2
2
x  x  r ,
2
1
which means that C is a circle.
18.10 Summary
A parametrized curve in Euclidean three-space e is given by a vector function
3

x(t) = (x (t), x (t), x (t))
2
3
1
that assigns a vector to every value of a parameter t in a domain interval [a, b]. The
coordinate functions of the curve are the functions x (t). In order to apply the methods of
i
calculus, we suppose the functions x (t) to have as many continuous derivatives as needed
i
in the following treatment.
One of the most useful ways to parametrize a curve is by the arc length s itself. If we let

s = s(t), then we have
s’(t) = |x’(t)| = |x’(s)|s’(t),
from which it follows that |x’(s)| = 1 for all s. So the derivative of x with respect to arc
length is always a unit vector.
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