]> 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.