]> 15.4 Computation of Curvature

15.4 Computation of Curvature

Computation of curvature and the various directions of interest with respect to the curve is quite straightforward, given a parametric representation of the curve, so that we can instruct a spreadsheet to compute everything here along our curve with a handful of instructions.

The vector d T ^ d s , by the chain rule is d T ^ d t d t d s and d t d s is the reciprocal of d s d t which is | v | and also can be written as | d r d t | .

Since T ^ is the unit vector in the direction of d r d t , it is d r / d t | d r / d t | .

The derivative of this latter expression with respect to t is, by the quotient rule

d T ^ d t = d ( d r / d t | d r / d t | ) / d t = d ( d r / d t | v | ) / d t = d 2 r d t 2 | v | v d | v | d t | v | 2

We can identify d 2 r d t 2 with the acceleration of the motion which we denote by a ( t ) .

We need here differentiate | v | with respect to t , which, given that | v | is ( v · v ) 1 / 2 obeys d | v | d t = a · v | v | .

Putting all this together we find

d T ^ d s = 1 | v | d T ^ d t = a | v | v ( a · v ) | v | | v | 3 = a ( v · v ) v ( a · v ) ( v · v ) 2

This result looks somewhat messy but it actually not so bad. Recall that a ( v · v ) v ( a · v ) ( v · v ) is the projection of a normal to v . Therefore we have here that d T ^ d s is the projection of a normal to v divided by the square of the magnitude of v .

Thus the curvature κ , which is the magnitude of this vector, is the component of a normal to v divided by the square of the magnitude of v .

κ = | a ( v · v ) v ( a · v ) v · v | v · v

Consider the example of the helix: x = cos t , y = sin t , z = t .

We can compute: v = d r d t = ( sin t , cos t , 1 ) and a = d v d t = ( cos t , sin t , 0 ) .

Here a and v are perpendicular, so that we get κ = | a | v · v = 1 2 for all values of t .

The center of curvature is at the reflection of the point on the curve at which we compute it, through the z axis, that is, at the point with coordinates ( cos t , sin t , t ) , a distance 2 (or 1 κ ) from ( cos t , sin t , t ) in the direction of the projection of a normal to v .

Exercises:

15.1 Show that the curvature of a circle is 1 r . (This proves that the radius of curvature is 1 κ .)

15.2 Find the curvature κ , position r , and center of curvature at t = j 100 for j = 0 to 700 for the following curve (using a spreadsheet)

x = cos t , y = cos 2 t , z = sin 3 t .

Chart it as best you can.

15.3 Set this curve up on the applet. Where is the curvature greatest.