Formulas for curvature

It is evident that if, in the last section, instead of measuring the angles which the tangents made with the $x$-axis, we had denoted by $\tau$ and $\tau + \Delta \tau$ the angles made by the tangents with any arbitrarily fixed line, the different steps would in no wise have been changed, and consequently the results are entirely independent of the system of coordinates used. However, since the equations of the curves we shall consider are all given in either rectangular or polar coordinates, it is necessary to deduce formulas for $K$ in terms of both. We have $\ \tan \tau = \frac{dy}{dx}$ by §4.9, or $\tau = \arctan \frac{dy}{dx}$. Differentiating with respect to $x$, using XX in §5.1,

$$\frac{d\tau}{dx} = \frac{ \frac{d^2 y}{dx^2} }{ 1 + \left( \frac{dy}{dx} \right)^2 }.$$

Also

$$\frac{ds}{dx} = \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{\frac{1}{2}},$$

by (9.4). Dividing one equation into the other gives

$$\frac{ \frac{d\tau}{dx} }{ \frac{ds}{dx} } = \frac{ \frac{d^2 y}{dx^2} }{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{\frac{3}{2}} }.$$

But

$$\frac{ \frac{d\tau}{dx} }{ \frac{ds}{dx} } = \frac{d\tau}{ds} = K.$$

Hence

$$K = \frac{ \frac{d^2 y}{dx^2} }{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{\frac{3}{2}} }.$$ (12.3)

If the equation of the curve be given in polar coordinates, $K$ may be found as follows: From (6.13),

$$\tau = \theta + \psi.$$

Differentiating,

$$\frac{d\tau}{d\theta} = 1 + \frac{d\psi}{d\theta}.$$

But

$$\tan \psi = \frac{\rho}{\frac{d\rho}{d\theta}},$$

from (6.12). Therefore,

$$\psi = \arctan \frac{\rho}{ \frac{d\rho}{d\theta} }.$$

Differentiating with respect to $\theta$ using XX in §5.1 and reducing,

$$\frac{d\psi}{d\theta} = \frac{ \left( \frac{d\rho}{d\theta} \rig... ...ac{d^2 \rho}{d\theta^2} }{ \rho^2 + \left( \frac{d\rho}{d\theta} \right)^2 }.$$

Substituting, we get

$$\frac{d\tau}{d\theta} = \frac{\rho^2 - \rho \frac{d^2 \rho}{d\th... ...d\rho}{d\theta} \right)^2}{\rho^2 + \left( \frac{d\rho}{d\theta} \right)^2}.$$

Also

$$\frac{ds}{d\theta} = \left[ \rho^2 \left( \frac{d\rho}{d\theta} \right)^2 \right]^{\frac{1}{2}},$$

by (9.9). Dividing gives

$$\frac{ \frac{d\tau}{d\theta} }{ \frac{ds}{d\theta} } = \frac{ \r... ...eft[ \rho^2 + \left( \frac{d\rho}{d\theta} \right)^2 \right]^{\frac{3}{2}} }.$$

But

$$\frac{ \frac{d\tau}{d\theta} }{ \frac{ds}{d\theta} } = \frac{d\tau}{ds} = K.$$

Hence

$$K = \frac{ \rho^2 - \rho \frac{d^2 \rho}{d\theta^2} + 2 \left( \f... ...\left[ \rho^2 + \left( \frac{d\rho}{d\theta} \right)^2 \right]^{\frac{3}{2}} }.$$ (12.4)

Example 12.4.1   Find the curvature of the parabola $y^2 = 4px$ at the upper end of the latus rectum.

The latus rectum of a conic section is the chord parallel to the directrix and passing through the single focus, or one of the two foci. For more details, see for example http://en.wikipedia.org/wiki/Semi-latus_rectum.

Solution. $\frac{dy}{dx} =\ \frac{2p}{y}$; $\frac{d^2 y}{dx^2} = - \frac{2p}{y^2} \frac{dy}{dx} = - \frac{4p^2}{y^3}$. Substituting in (12.3), $K = -\frac{40-p^2}{( y^2 + 4p^2 )^{\frac{3}{2}}}$, giving the curvature at any point. At the upper end of the latus rectum $(p, 2p)$,

$$K = -\frac{4p^2}{ (4p^2 + 4p^2)^{\frac{3}{2}} } = - \frac{4p^2}{16 \sqrt{2} p^3} = - \frac{1}{4 \sqrt{2} p}.$$

While in our work it is generally only the numerical value of $K$ that is of importance, yet we can give a geometric meaning to its sign. Throughout our work we have taken the positive sign of the radical $\sqrt{1 + \left( \frac{dy}{dx} \right)^2}$. Therefore $K$ will be positive or negative at the same time that $\frac{d^2 y}{dx^2}$ is, i.e., (by §8.8), according as the curve is concave upwards or concave downwards.

We shall solve this using SAGE.

[fontsize=\small,fontfamily=courier,fontshape=tt,frame=single,label=\sage]

sage: x = var("x")
sage: p = var("p")
sage: y = sqrt(4*p*x)
sage: K = diff(y,x,2)/(1+diff(y,x)^2)^(3/2)
sage: K
-p^2/(2*(p/x + 1)^(3/2)*(p*x)^(3/2))

Taking $x=p$ and simplifying gives the result above.

[fontsize=\small,fontfamily=courier,fontshape=tt,frame=single,label=\sage]

sage: K.variables()
(p, x)
sage: K(p,p)
-p^2/(4*sqrt(2)*(p^2)^(3/2))
sage: K(p,p).simplify_rational()
-1/(4*sqrt(2)*sqrt(p^2))

Example 12.4.2   Find the curvature of the logarithmic spiral $\rho = e^{a\theta}$ at any point.

Solution. $\frac{d\rho}{d\theta} = a e^{a\theta} = a \rho$; $\frac{d^2 \rho}{d\theta^2} = a^2 e^{a\theta} = a^2 \rho$.

Substituting in (12.4), $K = \frac{1}{\rho \sqrt{1 + a^2}}$.

In laying out the curves on a railroad it will not do, on account of the high speed of trains, to pass abruptly from a straight stretch of track to a circular curve. In order to make the change of direction gradual, engineers make use of transition curves to connect the straight part of a track with a circular curve. Arcs of cubical parabolas are generally employed as transition curves.

Now we do this in SAGE:

[fontsize=\small,fontfamily=courier,fontshape=tt,frame=single,label=\sage]

sage: rho = var("rho")
sage: t = var("t")
sage: r = var("r")
sage: a = var("a")
sage: r = exp(a*t)
sage: K = (r^2-r*diff(r,t,2)+2*diff(r,t)^2)/(r^2+diff(r,t)^2)^(3/2)
sage: K
1/sqrt(a^2*e^(2*a*t) + e^(2*a*t))
sage: K.simplify_rational()
e^(-(a*t))/sqrt(a^2 + 1)

Example 12.4.3   The transition curve on a railway track has the shape of an arc of the cubical parabola $y = \frac{1}{3}x^3$. At what rate is a car on this track changing its direction ($1$ mi. = unit of length) when it is passing through (a) the point $(3, 9)$? (b) the point $(2, \frac{8}{3})$? (c) the point $(1, \frac{1}{3})$?

Solution. $\frac{dy}{dx} = x^2$, $\frac{d^2 y}{dx^2} = 2x$. Substituting in (12.3), $K = \frac{2x}{(1 + x^4)^{\frac{3}{2}}}$. (a) At $(3, 9)$, $K = \frac{6}{(82)^{\frac{3}{2}}}$ radians per mile = $28'$ per mile. (b) At $(2, \frac{8}{3})$, $K = \frac{4}{(17)^{\frac{3}{2}}}$ radians per mile = $3^o 16'$ per mile. (c) At $(1, \frac{1}{3})$, $K = \frac{2}{(2)^{\frac{3}{2}}} = \frac{1}{\sqrt{2}}$ radians per mile = $40^o 30'$ per mile.