# Elements of the Differential and Integral Calculus/Chapter XII

## CHAPTER XII

99. Curvature. The shape of a curve depends very largely upon the rate at which the direction of the tangent changes as the point of contact describes the curve. This rate of change of direction is called curvature and is denoted by K. We now proceed to find its analytical expression, first for the simple case of the circle, and then for curves in general.

100. Curvature of a circle. Consider a circle of radius R. Let

 $\tau$ = angle that the tangent at P makes with OX; and $\tau + \Delta \tau$ = angle made by the tangent at a neighboring point P'. Then we say $\Delta \tau$ = total curvature of arc PP'.

If the point P with its tangent be supposed to move along the curve to P', the total curvature (= $\Delta \tau$) would measure the total change in direction, or rotation, of the tangent; or, what is the same thing, the total change in direction of the arc itself. Denoting by s the length of the arc of the curve measured from some fixed point (as A) to P, and by $\Delta s$ the length of the arc P P', then the ratio

|$\frac{\Delta \tau}{\Delta s}$

measures the average change in direction per unit length of arc.[1] Since, from the figure,

$\Delta s = R \cdot \Delta \tau,$

or

$\frac{\Delta \tau}{\Delta s} = \frac{1}{R},$

it is evident that this ratio is constant everywhere on the circle. This ratio is, by definition, the curvature of the circle, and we have

 (38) $K = \frac{1}{R}$.

The curvature of a circle equals the reciprocal of its radius.

101. Curvature at a point. Consider any curve. As in section,

 $\Delta \tau$ = total curvature of the arc PP', and $\frac{\Delta \tau}{\Delta s}$ = average curvature of the arc PP'.

More important, however, than the notion of the average curvature of an arc is that of curvature at a point. This is obtained as follows. Imagine P to approach P along the curve; then the limiting value of the average curvature $\left( = \tfrac{\Delta \tau}{\Delta s} \right)$ as P approaches P along the curve is defined as the curvature at P, that is,

Curvature at a point = $\lim_{\Delta s \to 0} \left( \tfrac{\Delta \tau}{\Delta s} \right) = \tfrac{d\tau}{ds}$.

 (39) ∴ $K = \frac{d\tau}{ds}$ = curvature.

Since the angle $\Delta \tau$ is measured in radians and the length of arc $\Delta s$ in units of length, it follows that the unit of curvature at a point is one radian per unit of length.

102. Formulas for curvature. It is evident that if, in the last section, instead of measuring the angles which the tangents made with OX, 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 in:dependent of the system of coordinates used. However, since the equations of the curves we shall consider are all given in either rectangular or polar coördinates, it is necessary to deduce formulas for K in terms of both. We have

 $\ \tan \tau$ $= \frac{dy}{dx}$, § 32 or $\ \tau$ $\arctan \frac{dy}{dx}$.
 Differentiating with respect to x, using XX (A) $\frac{d\tau}{dx}$ $= \frac{ \frac{d^2 y}{dx^2} }{ 1 + \left( \frac{dy}{dx} \right)^2 }$. Also (B) $\frac{ds}{dx}$ $= \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{\frac{1}{2}}$ From (24), p. 143 [§90] Dividing (A) by (B) 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 (40) K $= \frac{ \frac{d^2 y}{dx^2} }{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{\frac{3}{2}} }$. If the equation of the curve be given in polar coördinates, K may be found as follows: From (B), §67, $\tau$ $= \theta + \psi$. Differentiating, (C) $\frac{d\tau}{d\theta}$ $= 1 + \frac{d\psi}{d\theta}$. But $\ \tan \psi$ $= \frac{\psi}{\frac{d\rho}{d\theta}}$. From (A), §67 ∴ $\ \psi$ $= \arctan \frac{\rho}{ \frac{d\rho}{d\theta} }$. Differentiating by XX with respect to $\theta$ and reducing, (D) $\frac{d\psi}{d\theta}$ $= \frac{ \left( \frac{d\rho}{d\theta} \right)^2 - \rho \frac{d^2 \rho}{d\theta^2} }{ \rho^2 \left( \frac{d\rho}{d\theta} \right)^2 }$
 Substituting (D) in (C), we get (E) $\frac{d\tau}{d\theta}$ $= \frac{\rho^2 - \rho \frac{d^2 \rho}{d\theta^2} + 2 \left( \frac{d\rho}{d\theta} \right)^2}{\rho^2 + \left( \frac{d\rho}{d\theta} \right)^2}$. Also (F) $\frac{ds}{d\theta}$ $= \left[ \rho^2 \left( \frac{d\rho}{d\theta} \right)^2 \right]^{frac{1}{2}}$. From (30), §91 Dividing (E) by (F) gives $\frac{ \frac{d\tau}{d\theta} }{ \frac{ds}{d\theta} }$ $= \frac{ \rho^2 - \rho \frac{d^2 \rho}{d\theta^2} + 2 \left( \frac{d\rho}{d\theta} \right)^2 }{ \left[ \rho^2 + \left( \frac{d\rho}{d\theta} \right)^2 \right]^{\frac{3}{2}} }$. But $\frac{ \frac{d\tau}{d\theta} }{ \frac{ds}{d\theta} }$ $= \frac{d\tau}{ds} = K$. Hence (41) $K$ $= \frac{ \rho^2 - \rho \frac{d^2 \rho}{d\theta^2} + 2 \left( \frac{d\rho}{d\theta} \right)^2 }{ \left[ \rho^2 + \left( \frac{d\rho}{d\theta} \right)^2 \right]^{\frac{3}{2}} }$.

Illustrative Example 1. Find the curvature of the parabola $y^2 = 4 px$ at the upper end of the 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 (40), $K$ $= -\frac{4p^2}{( y^2 + 4p^2 )^{\frac{3}{2}}}$, giving the curvature at any point. At the upper end of the latus rectum (p, 2p) $K = -\frac{4p^2}{ (4p^2 + 4p^2)^{\frac{3}{2}} }$ $= - \frac{4p^2}{16 \sqrt{2} p^3} = - \frac{1}{4 \sqrt{2} p}$.[2] Ans.

Illustrative Example 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 (41), $K = \tfrac{1}{\rho \sqrt{1 + a^2}}$ Ans.

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.

Illustrative Example 3. The transition curve on a railway track has the shape of an arc of the cubical parabola $y = \tfrac{1}{3}x^3$. 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 $(2, \tfrac{8}{3})$? (c) the point $(1, \tfrac{1}{3})$?

 Solution. $\frac{dy}{dx} = x^2, \frac{d^2 y}{dx^2} = 2x$. Substituting in (40), $K = \frac{2x}{(1 + x^4)^{\frac{3}{2}}}$. (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° 16′ per mile. Ans. (c) At $(1, \frac{1}{3})$, $K = \frac{2}{(2)^{\frac{3}{2}}} = \frac{1}{\sqrt{2}}$ radians per mile = 40° 30′ per mile. Ans.

103. Radius of curvature. By analogy with the circle (see (38), § 99), the radius of curvature of a curve at a point is defined as the reciprocal of the curvature of the curve at that point. Denoting the radius of curvature by R, we have

 $R = \frac{1}{K}$;[3] Or, substituting the values of X from (40) and (41), (42) $R = \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{\frac{3}{2}} }{ \frac{d^2 y}{dx^2} }$; (43) $R = \frac{ \left[ \rho^2 + \left( \frac{d\rho}{d\theta} \right)^2 \right]^{\frac{3}{2}} }{ \rho^2 - \rho \frac{d^2 \rho}{d\theta^2} + 2 \left( \frac{d\rho}{d\theta} \right)^2 }$[4]

Illustrative Example 1. Find the radius of curvature at any point of the catenary $y = \tfrac{a}{2} (e^{\tfrac{x}{a}} + e^{-\tfrac{x}{a}})$.

 Solution. $\frac{dy}{dx}$ $= \frac{1}{2} (e^{\frac{x}{a}} - e^{-\frac{x}{a}}); \frac{d^2 y}{dx^2} = \frac{1}{2a} (e^{\frac{x}{a}} - e^{-\frac{x}{a}})$. Substituting in (42), $R = \frac{\left[ 1 + \left( \frac{e^{\frac{x}{a}} - e^{-\frac{x}{a}}}{2} \right)^2 \right]^{\frac{3}{2}} }{\frac{ e^{\frac{x}{a}} - e^{-\frac{x}{a}} }{2a}}$ $= \frac{\left( \frac{ e^{\frac{x}{a}} - e^{-\frac{x}{a}} }{2} \right)^3}{ \frac{ e^{\frac{x}{a}} - e^{-\frac{x}{a}} }{2a}} = \frac{a (e^{\frac{x}{a}} - e^{-\frac{x}{a}})^2}{4} = \frac{y^2}{a}$. Ans. If the equation of the curve is given in parametric form, find the first and second derivatives of y with respect to x from (A) and (B), §97, namely: (G) $\frac{dy}{dx}$ $= \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }$, and (H) $\frac{d^2 y}{dx^2}$ $= \frac{ \frac{dx}{dt} \frac{d^2 y}{dt^2} - \frac{dy}{dt} \frac{d^2 x}{dt^2} }{ \left( \frac{dx}{dt} \right)^3 }$;

and then substitute the results in (42).[5]

Illustrative Example 2. Find the radius of curvature of the cycloid

$x = a(t - \sin t)$
$y = a(t - \cos t)$
 Solution. $\frac{dx}{dt} = a(1 - \cos t)$, $\frac{dy}{dt} = a \sin t$; $\frac{d^2 x}{dt^2} = a \sin t$, $\frac{d^2 y}{dt^2} = a \cos t$,
Substituting in (G) and (H), and then in (42), we get
 $\frac{dy}{dx} = \frac{\sin t}{1 - \cos t}, \frac{d^2 y}{dx^2} = \frac{ a(1 - \cos t) a \cos t - a \sin t a \sin t }{ a^3 (1 - \cos t)^3 } = \frac{1}{a(1 - \cos t)^2}$. $R = \frac{ \left[ 1 + \left( \frac{\sin t}{1 - \cos t} \right)^2 \right]^{\frac{3}{2}} }{ -\frac{1}{a(1 - \cos t)^2} } = -2a \sqrt{2 - 2 \cos t}.$ Ans.

104. Circle of curvature. Consider any point P on the curve C. The tangent drawn to the curve at P has the same slope as the curve itself at P (§ 64). In an analogous manner we may construct for each point of the curve a circle whose curvature is the same as the curvature of the curve itself at that point. To do this, proceed as follows. Draw the normal to the curve at P on the concave side of the curve. Lay off on this normal the distance PC = radius of curvature (= R) at P. With C as a center draw the circle passing through P. The curvature of this circle is then

$K = \frac{1}{R},$

which also equals the curvature of the curve itself at P. The circle so constructed is called the circle of curvature for the point P on the curve.

In general, the circle of curvature of a curve at a point will cross the curve at that point. This is illustrated in the above figure.

Just as the tangent at P shows the direction of the curve at P, so the circle of curvature at P aids us very materially in forming a geometric concept of the curvature of the curve at P, the rate of change of direction of the curve and of the circle being the same at P.

In a subsequent section (§ 116) the circle of curvature will be defined as the limiting position of a secant circle, a definition analogous to that of the tangent given in § 32.

Illustrative Example 1. Find the radius of curvature at the point (3, 4) on the equilateral hyperbola $xy = 12$, and draw the corresponding circle of curvature.

 Solution. $\frac{dy}{dx} = -\frac{y}{x}, \frac{d^2 y}{dx^2} = \frac{2y}{x^2},$ For (3, 4), $\frac{dy}{dx} = -\frac{4}{3}, \frac{d^2 y}{dx^2} = \frac{8}{9}$ ∴ R $= \frac{ [1 + \frac{16}{9}]^{\frac{3}{2}} }{ \frac{8}{9} } = \frac{125}{24} = 5\frac{5}{24}$.

The circle of curvature crosses the curve at two points.

EXAMPLES

1. Find the radius of curvature for each of the following curves, at the point indicated; draw the curve and the corresponding circle of curvature:

 (a) $b^2 x^2 + a^2 y^2 = a^2 b^2, (a, 0).$ Ans. $R = \frac{b^2}{a}.$ (b) $b^2 y^2 + a^2 y^2 = a^2 b^2, (0, b).$ $R = \frac{a^2}{b}.$ (c) $y = x^4 - 4x^3 - 18x^2, (0, 0).$ $R = \frac{1}{36}.$ (d) $16y^2 = 4x^4 - x^6, (2, 0).$ $R = 2.$ (e) $y = x^3, (x_1, y_1).$ $R = \frac{(1 + 9{x_1}^4)^{\frac{3}{2}}}{6x_1}.$ (f) $y^2 = x^3, (4, 8).$ $R = \frac{1}{3}(40)^{\frac{3}{2}}.$ (g) $y^2 = 8x, (\frac{9}{8}, 3).$ $R = 7\frac{13}{16}.$ (h) $\left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^{\frac{2}{3}} = 1, (0, b).$ $R = \frac{a^2}{3b}.$ (i) $x^2 = 4ay, (0, 0).$ $R = 2a$ (j) $(y - x^2)^2 = x^5, (0, 0).$ $R = \frac{1}{2}$ (k) $(b^2 x^2 - a^2 y^2 = a^2 b^2, (x_1, y_1).$ $R = \frac{(b^4 {x_1}^2 + a^4 {y_1}^2)^{\frac{3}{2}}}{a^4 b^4}$ (l) $e^x = \sin y, (x_1, y_1).$ (p) $9y = x^3, x = 3.$ (m) $y = \sin x, \left( \frac{\pi}{2}, 1 \right).$ (q) $4y^2 = x^3, x = 4.$ (n) $y = \cos x, \left( \frac{\pi}{4}, \sqrt{2} \right).$ (r) $x^2 - y^2 = a^2, y = 0.$ (o) $y = \log x, x = e.$ (s) $x^2 + 2y^2 = 9, (1, -2).$

2. Determine the radius of curvature of the curve $a^2 y = bx^2 + cx^2 y$ at the origin.

Ans. $R = \frac{a^2}{2b}$.

3. Show that the radius of curvature of the witch $y^2 = \tfrac{a^2 (a - x)}{x}$ at the vertex is $\tfrac{a}{2}$.

4. Find the radius of curvature of the curve $y = \log \sec x$ at the point $(x_1, y_1)$.

Ans. $R = \sec x_1$.

5. Find K at any point on the parabola $x^{\tfrac{1}{2}} + y^{\tfrac{1}{2}} = a^{\tfrac{1}{2}}$. Ans. $K = \tfrac{ a^{\tfrac{1}{2}} }{ 2(x + y)^{\tfrac{3}{2}} }$.

6. Find R at any point on the hypocycloid $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$. Ans. $R = 3 (axy)^{\frac{1}{3}}$.

7. Find R at any point on the cycloid $x = r \operatorname{arcvers} \tfrac{y}{r} - \sqrt{2ry - y^2}$. Ans. $R = 2 \sqrt{2ry}$.

Find the radius of curvature of the following curves at any point:

 8. The circle $\rho = a \sin \theta$. Ans. $R = \frac{a}{2}$. 9. The spiral of Archimedes $\rho = a\theta$. $R = \frac{ (\rho^2 = a^2)^{\frac{3}{2}} }{ \rho^2 + 2a^2 }$. 10. The cardioid $\rho = a (1 - \cos \theta)$. $R = \frac{2}{3} \sqrt{2a\rho}$. 11. The lemniscate $\rho^2 = a^2 \cos 2 \theta$. $R = \frac{a^2}{3\rho}$. 12. The parabola $\rho = a \sec^2 \frac{\theta}{2}$. $R = 2 a \sec^3 \frac{\theta}{2}$. 13. The curve $\rho = a sin^3 \frac{\theta}{3}$.
 14. The trisectrix $\rho = 2 a \cos \theta - a$. Ans. $R = \frac{ a(5 - 4 \cos \theta)^{\frac{3}{2}} }{ 9 - 6 \cos \theta }$ 15. The equilateral hyperbola $\rho^2 \cos 2 \theta = a^2$. Ans. $R = \frac{\rho^3}{a^2}$. 16. The conic $\rho = \frac{a(1 - e^2)}{1 - e \cos \theta}$. Ans. $R = \frac{a(1 - e^2)(1 - 2e \cos \theta + e^2)^{\frac{3}{2}}}{(1 - e \cos \theta)^3}$. 17. The curve $\begin{cases}x = 3 t^2, \\ y = 3 t - t^3.\ t = 1.\end{cases}$ Ans. $R = 6.$ 18. The hypocycloid $\begin{cases}x = a \cos^3 t, \\ y = a \sin^3 t.\ t = t_1.\end{cases}$ Ans. $R = 3 a \sin t_1 \cos t_1.$ 19. The curve $\begin{cases}x = a(\cos t + t \sin t), \\ y = a(\sin t - t \cos t).\ t = \frac{\pi}{2}.\end{cases}$ Ans. $R = \frac{\pi a}{2}.$ 20. The curve $\begin{cases}x = a(m \cos t + \cos mt), \\ y = a(m \sin t - \sin mt).\ t = t_0.\end{cases}$ Ans. $R = \frac{4 ma}{m - 1} \sin \left( \frac{m + 1}{2} \right) t_0.$

21. Find the radius of curvature for each of the following curves at the point indicated; draw the curve and the corresponding circle of curvature:

 (a) $x = t^2, 2 y = t; t = 1.$ (e) $x = t, y = 6 t^{-1}; t = 2$. (b) $x = t^2, y = t^3; t=1.$ (f) $x = 2 e^t,y = e^{-t}; t = 0.$ (c) $x = \sin t, y = \cos 2t; t= \frac{\pi}{6}.$ (g) $x = \sin t, y = 2 \cos t; t = \frac{\pi}{4}.$ (d) $x = 1 - t, y = t^3; t = 3.$ (h) $x = t^3, y = t^2 + 2t; t = 1.$

22. An automobile race track has the form of the ellipse $x^2 + 16y^2 = 16$, the unit being one mile. At what rate is a car on this track changing its direction

(a) when passing through one end of the major axis?

(b) when passing through one end of the minor axis?

(c) when two miles from the minor axis?

(d) when equidistant from the minor and major axes?

Ans. (a) 4 radians per mile; (b) $\frac{1}{16}$ radian per mile.

23. On leaving her dock a steamship moves on an arc of the semi cubical parabola $4 y^2 = x^3$. If the shore line coincides with the axis of y, and the unit of length is one mile, how fast is the ship changing its direction when one mile from the shore?

Ans. $\frac{24}{125}$ radians per mile.

24. A battleship 400 ft. long has changed its direction 30° while moving through a distance equal to its own length. What is the radius of the circle in which it is moving?

Ans. 764 ft.

25. At what rate is a bicycle rider on a circular track of half a mile diameter changing his direction?

Ans. 4 rad. per mile = 43' per rod.

26. The origin being directly above the starting point, an aëroplane follows approximately the spiral $\rho = \theta$, the unit of length being one mile. How rapidly is the aëroplane turning at the instant it has circled the starting point once?

27. A railway track has curves of approximately the form of arcs from the following curves. At what rate will an engine change its direction when passing through the points indicated (1 mi. = unit of length):

 (a) $y = x^3, (2, 8)$? (d) $y = e^x, x = 0$? (b) $y = x^2, (3, 9)$? (e) $y = \cos x, x = \frac{\pi}{4}$? (c) $x^2 - y^2 = 8, (3, 1)$? (f) $\rho \theta = 4, \theta = 1$?

1. Thus, if $\Delta \tau = \frac{\pi}{6}$ radians (= 30°), and $\Delta s$ = 3 centimeters, then $\tfrac{\Delta \tau}{\Delta s} = \tfrac{\pi}{18}$ radians per centimeter = 10° per centimeter = average rate of change of direction.
2. 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 $\sqrt{1 + \left( \tfrac{dy}{dx} \right)^2}$. Therefore K will be positive or negative at the same time as $\tfrac{d^2 y}{dx^2}$ that is (§85), according as the curve is concave upwards or concave dx downwards.
3. Hence the radius of curvature will have the same sign as the curvature, that is, + or -, according as the curve is concave upwards or concave downwards.
4. In § 98, p. 152, (43) is derived from (42) by transforming from rectangular to polar coordinates.
5. Substituting (G) and (H), and then in (42) gives $R = \tfrac{ \left[ \left( \tfrac{dx}{dt} \right)^2 + \left( \tfrac{dy}{dt} \right)^2 \right] }{ \tfrac{dx}{dt} \tfrac{d^2 y}{dt^2} - \tfrac{dy}{dt} \tfrac{d^2 x}{dt^2} }$.