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

\frac{d \hat{T}}{d s}

, by the chain rule is

\frac{d \hat{T}}{d t} \frac{d t}{d s}

and

\frac{d t}{d s}

is the reciprocal of

\frac{d s}{d t}

which is

| \overset{⟶}{v} |

and also can be written as

| \frac{d \overset{⟶}{r}}{d t} |

Since

\hat{T}

is the unit vector in the direction of

\frac{d \overset{⟶}{r}}{d t}

, it is

\frac{d \overset{⟶}{r} / d t}{| d \overset{⟶}{r} / d t |}

The derivative of this latter expression with respect to

t

is, by the quotient rule

\frac{d \hat{T}}{d t} = d (\frac{d \overset{⟶}{r} / d t}{| d \overset{⟶}{r} / d t |}) / d t = d (\frac{d \overset{⟶}{r} / d t}{| \overset{⟶}{v} |}) / d t = \frac{\frac{d^{2} \overset{⟶}{r}}{d t^{2}} | \overset{⟶}{v} | - \overset{⟶}{v} \frac{d | \overset{⟶}{v} |}{d t}}{{| \overset{⟶}{v} |}^{2}}

We can identify

\frac{d^{2} \overset{⟶}{r}}{d t^{2}}

with the acceleration of the motion which we denote by

\overset{⟶}{a} (t)

We need here differentiate

| \overset{⟶}{v} |

with respect to

t

, which, given that

| \overset{⟶}{v} |

{(\overset{⟶}{v} \cdot \overset{⟶}{v})}^{1 / 2}

obeys

\frac{d | \overset{⟶}{v} |}{d t} = \frac{\overset{⟶}{a} \cdot \overset{⟶}{v}}{| \overset{⟶}{v} |}

\frac{d \hat{T}}{d s} = \frac{1}{| \overset{⟶}{v} |} \frac{d \hat{T}}{d t} = \frac{\overset{⟶}{a} | \overset{⟶}{v} | - \frac{\overset{⟶}{v} (\overset{⟶}{a} \cdot \overset{⟶}{v})}{| \overset{⟶}{v} |}}{{| \overset{⟶}{v} |}^{3}} = \frac{\overset{⟶}{a} (\overset{⟶}{v} \cdot \overset{⟶}{v}) - \overset{⟶}{v} (\overset{⟶}{a} \cdot \overset{⟶}{v})}{{(\overset{⟶}{v} \cdot \overset{⟶}{v})}^{2}}

This result looks somewhat messy but it actually not so bad. Recall that

\frac{\overset{⟶}{a} (\overset{⟶}{v} \cdot \overset{⟶}{v}) - \overset{⟶}{v} (\overset{⟶}{a} \cdot \overset{⟶}{v})}{(\overset{⟶}{v} \cdot \overset{⟶}{v})}

is the projection of

\overset{⟶}{a}

normal to

\overset{⟶}{v}

. Therefore we have here that

\frac{d \hat{T}}{d s}

is the projection of $\overset{⟶}{a}$ normal to $\overset{⟶}{v}$ divided by the square of the magnitude of

\overset{⟶}{v}

Thus the curvature $κ$ , which is the magnitude of this vector, is the component of $\overset{⟶}{a}$ normal to $\overset{⟶}{v}$ divided by the square of the magnitude of

\overset{⟶}{v}

κ = \frac{| \frac{\overset{⟶}{a} (\overset{⟶}{v} \cdot \overset{⟶}{v}) - \overset{⟶}{v} (\overset{⟶}{a} \cdot \overset{⟶}{v})}{\overset{⟶}{v} \cdot \overset{⟶}{v}} |}{\overset{⟶}{v} \cdot \overset{⟶}{v}}

We can compute:

\overset{⟶}{v} = \frac{d \overset{⟶}{r}}{d t} = (- \sin t, \cos t, 1)

and

\overset{⟶}{a} = \frac{d \overset{⟶}{v}}{d t} = (- \cos t, - \sin t, 0)

Here

\overset{⟶}{a}

and

\overset{⟶}{v}

are perpendicular, so that we get

κ = \frac{| \overset{⟶}{a} |}{\overset{⟶}{v} \cdot \overset{⟶}{v}} = \frac{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

\frac{1}{κ}

) from

(\cos t, \sin t, t)

in the direction of the projection of

\overset{⟶}{a}

normal to

\overset{⟶}{v}

15.1 Show that the curvature of a circle is $\frac{1}{r}$ . (This proves that the radius of curvature is $\frac{1}{κ}$ .)

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

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