﻿ Curvature in the Plane

### Part 3: Curvature in the Plane

In order to better understand curvature, let's explore it for 2-dimensional curves.  If r( t) is the parametrization of a 2-dimensional curve, then its velocity can be written in polar form as
 v = á vcos( q) ,vsin( q) ñ
Factoring out the v then leads to v = v  á cos(q) ,sin( q) ñ , which reveals that the unit vector is
 T = á cos[ q( t) ], sin[ q( t) ] ñ
As a result, the derivative of T is given by
dT
dt
=
d
dt
cos[ q(t) ],
d
dt
sin[ q( t) ]
=
-sin[ q( t) ]
dq
dt
, cos[ q( t) ]
dq
dt
=
dq
dt
á -sin( q), cos(q) ñ
Since á -sin( q) ,cos( q) ñ is also a unit vector, the magnitude of dT/dt is
dT
dt

dq
dt
(9)
which via the chain rule is equivalent to
dT
dt
=
dq
ds
ds
dt
(10)
where s(t) is the arclength function of r(t) .  That is, the curvature of a 2-dimensional curve is given by
k =
dq
ds
or equivalently, moving a short distance ds along the curve causes a change dq in the direction of the velocity vector, where dq = kds.

Moreover, because of the relationship between curvature and the rate of change of q, we often use
k =  1 v
dq dt
=  1 v
dT dt
to calculate the curvature (thus illustrating this important relationship).

EXAMPLE 4    Find the curvature of the circle of radius 3 parametrized by
 r( t) = á 3cos( t) ,3sin(t) ñ
Solution: Since the velocity is v( t) = á -3sin( t) ,3cos( t) ñ , the speed is
v
9sin2( t) +9cos2( t)
= 3
As a result, the unit tangent vector is
 T( t) = á -sin( t) ,cos(t) ñ
from which it follows that
dT
dt
= á -cos( t) ,-sin(t) ñ
As a result, the curvature is
k =
1
v
dT
dt
=
1
3
cos2( t) +sin2( t)
=
1
3

As example 4 illustrates, curvature is closely related to the radius of a circle. Indeed, the osculating circle of a curve is the circle with radius
R =
1
k
or equivalently,

ds
dq
= R
and with a center at
 Center = r + R N
Consequently, the osculating circle is practically the same as a small section of the curve,
This allows us to interpret ds = Rdq mean that a short distance ds along the curve is practically the same as the small arc with angle dq on the osculating circle.

EXAMPLE 5    Find the curvature k of the curve r( t) = á ln| cos(t) |,  t ñ

Solution: The velocity is
v
1
cos(t)

d
dt
cos(t)  ,  1 = á -cos( t), 1 ñ
-sin(t)
cos(t)
,  1
which reduces to v = á -tan(t), 1 ñ.  Thus, the speed is
v =
1+tan2( t)
=
sec2( t)
= sec( t)
and  the unit tangent vector is
T =
1
sec( t)
á-tan(t), 1 ñ = á -sin(t),  cos(t) ñ
and the derivative of the unit tangent is
dT
dt
= á -cos( t), -sin(t) ñ
As a result, (5) implies that the curvature is
k
1
v
dT
dt
=
1
sec(t)
cos2( t)  + sin2( t)
=  cos(t)
It follows that R = 1/k and that
 Center( t)
 =
 r( t)  +  R N( t)
 =
á ln| cos(t)| ,t ñ +  1 cos( t)
á -cos( t) ,-sin(t) ñ
 =
 á ln| cos(t)| -1,  t-tan( t) ñ
Indeed, the osculating circle itself at a fixed time t is given by
 Osc( q)
 =
 Center( t)+R á cos( q) ,sin( q) ñ
 =
 á ln| cos(t) | - 1,  t - tan(t) ñ +  sec(t) á cos(q),sin(q) ñ
This is illustrated in the applet below:

If r(t) parameterizes a straight line, then its unit tangent is constant, and correspondingly, its curvature is k = 0. To illustrate, the curve r( t) = á ln| cos(t) |,  t ñ  in example 5 has a horizontal asymptote as t approaches p/2, and likewise, the curvature  k( t) = cos( t) a approaches 0 as t approaches p/2.  Since the relationship of the curvature and the radius of the osculating circle is
k =
1