Curvature of Plane Curves

[Pages:3]Curvature of Plane Curves

What is arc length parametrization? Let : [a, b] R R2, t (s) = (x(s), y(s))

be a nice curve. We define its arc length from t = a to t = b to be

b

|| (t)||dt.

a

We say that (s) is an arc length parametrization provided || (s)|| 1. We also call such a parametrized curve a unit speed curve.

Let c : R R2, t c(t) be a curve We can always (assuming that ||c (t)|| never equals 0) construct an arc length parametrization for a curve. Define a function

=t

s : R R, t s, t

||c( ) ||d.

=a

Observe that

ds

= ||c (t)||. dt

()

Define

: R R2, s (s)

by s = c.

We have the diagram

R

s

AAAAAcAA

R / R2

Using the chain rule we have

ds dc (s) ? = .

dt dt

From

equation

we

have

that

||

dc dt

||

=

ds dt

.

Hence

||

||

1.

We say that a curve (s) that is an arclength parametrization is a unit speed curve.

We have that the length of the curve from (a) to (b) is b - a for any a, b > a.

Definition of Curvature

We start with an arclength parametrized curve , so that || (s)|| 1. Hence we can write (s) = (cos((s)), sin((s))) = t(s). The curvature is how fast the direction of the tangent vector changes, that is, the curvature is

(s) = (s).

Notice that

(s) = (s)(- sin((s), cos((s)) = (s)n(s).

1

and that n(s) = (- sin((s), cos((s)) is a normal to the curve so that the two unit vectors

(n(s), t(s))

form an orietnted basis of R2. From this we can see the difference between curvature that

is positive and curvature that is negative. See the figure.

Observe that the curvature is the area of the rectangle spanned by (t) and (t).

Hence

(s) = det

(s) gamma (s)

.

A formula for the curvature of an arbitrarily (not arclength) parmetrized curve.

Given c(t) we can theoretically construct a corresponding arc length parametrized curve

(s). But only theoretically. How do we explicitly compute the curvature of c(t) in terms

of

c (t), c

(t),

ds dt

?

We

use

the

diagram

RO

s

R

AAAAAcAA

/ R2

The chain rule tells how to compute derivatives of c in terms of the derivatives of s and

d ds

.

We

solve

this

relation

to

obtain

d ds

in

terms

of

ds dt

and

dc dt

.

We

apply

this

idea

a

second

time

to

obtain

d2 ds2

in

terms

of

, d2c

dt2

dc dt

,

ds dt

.

Applying the chain rule to

s=c

we obtain

d ds dc

=.

()

ds dt dt

Differentiating this with respect to t, we obtain

ds 2

d2s d2c

(s) + (s) = .

dt

dt2 dt2

Using the relation replace (s) with a multiple of c (t) we find we have

ds 2 dc d2c

(s) + = .

dt

dt dt2

Solve for (s) gives for an appropriate constant .

d2c dc

-

(s) =

dt2 dt

(

ds dt

)2

2

We use the determinant formula for curvature. This gives

(t) =

(s) (s)

c (t)

ds

=

det

dt

=

c

(t)

-

c

(t)

( ds )2

dt

ds dt

-3

det

c (t) c (t)

.

If c(t) = (x(t), y(t), this gives

xy -x y

=

.

(x 2 + y 2)3/2

Specializing more to the case of a curve y = f (x) which we parametrized by (x(t) = t, y(t))

we get

y

=

,

(1 + y 2)3/2

. but y

=

. d2y

dx2

3

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