184. We may remark, that the equation \( x = C + \int \frac{dp}{P} \), and

\[ dy = C' + \int \frac{pdq}{P}, \]

severally satisfy the differential equation \( \frac{dp}{dx} = P \); and that, by supposing their second members integrated, they will be only of the first order. This agrees with what was observed, art. 181, viz., that there are two equations of the first order which satisfy a differential equation of the second order.

185. Let us now consider differentials of the form \( \frac{dy}{dx^2} = Y \),

\( Y \) representing any function of \( y \). If we make \( dy = pdx \),

we thence deduce \( dx = \frac{dy}{p} \), which gives \( \frac{d^2y}{dx^2} = \frac{dp}{dy} \).

Substituting this in the proposed equation, there results from it \( p^2 = Ydy \); by integrating, we find \( p^2 = 2fYdy \),

C, and hence

\[ p = \frac{dy}{dx} = \sqrt{C + 2fYdy}, \quad \text{and} \quad x = \int \frac{dy}{\sqrt{C + 2fYdy}} + C. \]

It is proper to observe that the above integration may be effected by multiplying the proposed equation by \( dy \),

for thence we find \( \frac{dy}{dx} \cdot \frac{d^2y}{dx^2} = Ydy \); and since \( \frac{dy}{dx} = \frac{d}{dx} \),

we have \( \frac{1}{2} \frac{d^2y}{dx^2} = \int Ydy + C, \quad \text{or} \quad \frac{dy}{dx} = \sqrt{2C + 2fYdy}. \)

Ex. Let the equation be \( \frac{d^2y}{dx^2} \cdot \sqrt{ay} = dx^2 \); we shall have

\[ \frac{d^2y}{dx^2} = \frac{1}{\sqrt{ay}}, \quad \frac{dy}{dx} \cdot \frac{d^2y}{dx^2} = \frac{dy}{\sqrt{ay}}, \]

and, by integrating, \( \frac{1}{2} \frac{d^2y}{dx^2} = \frac{2}{a} \sqrt{ay} + C; \) changing \( C \) into \( \frac{2e}{\sqrt{a}} \), we shall deduce from thence

\[ \frac{dy}{dx} = \frac{4}{\sqrt{a}}(\sqrt{y} + e), \quad \frac{2dx}{\sqrt{a}} = \frac{dy}{\sqrt{c} + \sqrt{y}}. \]

Now, making \( c + \sqrt{y} = z \), there will result

\[ \frac{dz}{\sqrt{a}} = \frac{(z - e)}{\sqrt{z}} dz = (z^{\frac{1}{2}} - ez^{-\frac{1}{2}})dz; \]

and finally,

\[ \frac{x}{\sqrt{a}} = \frac{3}{2} z^{\frac{3}{2}} - 2ez^{\frac{1}{2}} + e = \frac{3}{2}(\sqrt{y} - 2e)\sqrt{y} + c. \]

186. We have seen in the differential calculus, art. 93, that beyond the first order the form of the differential equations would be changed, according as \( x \) or \( y \), or even a function of these quantities, was assumed as the independent variable; which comes to the same thing as assuming that \( dx \) or \( dy \), or a given function of these variables, was constant. It is therefore necessary, in integrating equations which exceed the first degree, to know upon which of these hypotheses it has been calculated. The preceding examples all correspond to the case of \( y \) being a function of \( x \), and consequently of \( dx \) being constant; but it will be easy to discover, among relations deduced relatively to other hypotheses, to which of them they may be referred.

It is immediately evident that, if we represent by \( Q \) any function whatever of \( \frac{dx}{dy} \), every equation of the form

\[ \frac{d^2x}{dy^2} = Q, \]

and in which \( dy \) is constant, may be treated in the same way as that in art. 183, by making \( \frac{dx}{dy} = q \), and

\[ \frac{d^2x}{dy^2} = \frac{dq}{dy}. \]

It may also be reduced immediately to the form \( \frac{d^2x}{dx^2} = P \), by passing by the formula of art. 93, to the hypothesis of \( dx \) being constant, which will be effected by the substitution of \( \frac{dxdy}{dy^2} \) in place of \( \frac{d^2x}{dy^2} \).

If the proposed equation had been taken on the supposition of \( \sqrt{dx^2 + dy^2} \) being constant, and if it involved only \( dx, dy, d^2y, \) or \( dy, dx, dx^2 \), it might still be treated in the same manner as that of art. 183, after transforming it into one in which \( dx \) was constant.

187. When the equation contains \( \frac{d^2y}{dx^2} \) and the variable \( x \), it may, as before, be transformed into a differential equation of the first order, by substituting \( pdx \) for \( dy \),

and \( pdx^2 \) for \( d^2y \); then, if the primitive of that differential equation, and thence the value of \( p \) in terms of \( x \), can be found, we may have the value of \( y \) from the formula \( y = f(pdx) \); or else, if we have the value of \( x \) in terms of \( p \), then, because \( f(pdx) = px - fxdp \), we shall have \( y = px - fxdp \).

Ex. Let the equation be

\[ \frac{(dx^2 + dy^2)^{\frac{3}{2}}}{dx^2} = X \quad \text{or} \quad \frac{(1 + p^2)^{\frac{3}{2}}}{dp} = X, \]

where \( x \) denotes a given function of \( x \); then

\[ \frac{dx}{X} = \frac{-dp}{(1 + p^2)^{\frac{3}{2}}} \quad \text{and} \quad \int \frac{dx}{X} + C = \frac{-p}{\sqrt{1 + p^2}}. \]

Let \( V \) represent \( \int \frac{dx}{X} + C \), then \( p = \frac{V}{\sqrt{1 - V^2}} \),

and \( y = f(pdx) + C = f\left(\frac{Vdx}{\sqrt{1 - V^2}}\right) + C; \) \( y \) is now expressed by means of \( x \) only. This is the analytical solution of the geometrical problem to find the nature of a curve whose radius of curvature shall be a given function of \( x \), the abscissa.

If we suppose \( X = \frac{a^2}{2x} \), then \( \frac{-p}{\sqrt{1 + p^2}} = \int \frac{2xdx}{a^2} + C, \)

\[ p = \frac{dy}{dx} = \frac{x^2 + c}{a^2 - (x^2 + c^2)^{\frac{3}{2}}} \quad \text{and} \quad y = \int \frac{(x^2 + c)dx}{\sqrt{a^2 - (x^2 + c)^2}}; \]

here \( c \) is put generally for an indeterminate constant.

This is the equation of the elastic curve.

188. If the proposed differential equation is composed of \( \frac{d^2y}{dx^2} \) and \( y \), we may, as before, put \( p = \frac{dy}{dx} \),

which we get \( \frac{d^2y}{dx^2} = \frac{dp}{dx} = \frac{pdq}{dy} \); the equation will now involve \( dp, dq, p \) and \( y \) only. When the primitive equation can be found, and thence the value of \( p \) in terms of \( y \), we may find \( x \) by the formula \( x = \int \frac{dy}{p} \); but when \( y \) is expressed by \( p \), we may then employ the formula \( x = \frac{y}{p} + \int \frac{qdp}{p^2}. \)

189. Differentials of the second order, which have this form,

\[ \frac{d^2y}{dx^2} + P \frac{dy}{dx} + Qy = R, \] in which P, Q, R are any functions of x, are called linear, also equations of the first degree, because y is only of one dimension. It will easily be understood that the difficul- ty of finding the primitive equation will be greater than in the like equation of the first order; and, indeed, except in particular cases, there is no known method of reducing the integration to the finding of the differential of a sin- gle variable; that is, to the quadrature of a curve. If R = 0, in which case the equation is

\[ \frac{dy}{dx} + P \frac{dy}{dx} + Qy = 0, \]

it may be reduced to a differential equation of the first degree, by a very simple transformation. Let e, as usual, denote the base of Napier's logarithms; assume \( y = e^{fudx} \), then, regarding dx as constant, we have

\[ dy = udx e^{fudx}, \quad d^2y = e^{fudx} (du + u^2dx^2): \]

the value of dy and d^2y being substituted in the equation, and the common factors rejected, it becomes

\[ du + (u^2 + Pu + Q)dx = 0. \]

When P and Q are constants, which may be represented by A and B, the equation becomes

\[ du + (u^2 + Au + B)dx = 0; \]

in which the variables are separated, if we give it the form

\[ \frac{du}{u^2 + Au + B} + dx = 0. \]

As it is merely necessary to satisfy this equation, we may make \( u = m, m \) being any constant; then \( du = 0, \) and \( m^2 + Am + B = 0. \) This last equation gives, in general, two values of \( m; \) if we represent them by \( a \) and \( b, \) we shall have two values of \( fudx; \) viz. \( ax + c \) and \( bx + c', \) and hence we have two values of \( y, \) viz.

\[ y = e^{ax+c}, \quad y = e^{bx+c}; \]

or, putting C for \( e^c, \) and C' for \( e^{c'}, \)

\[ y = Ce^{ax}, \quad y = Ce^{bx}. \]

These, however, are only particular values of the func- tion \( y, \) because each contains only a single arbitrary con- stant; but by adding them, we get

\[ y = Ce^{ax} + C'e^{bx} \]

as the complete primitive equation.

To prove this, by differentiating, we get

\[ \frac{dy}{dx} = aCe^{ax} + bC'e^{bx}, \quad \frac{d^2y}{dx^2} = a^2Ce^{ax} + b^2C'e^{bx}. \]

From these and the above primitive equation, we have, after eliminating C and C',

\[ \frac{d^2y}{dx^2} + (a + b) \frac{dy}{dx} - aby = 0. \]

This will agree with the proposed equation, if we give \( a \) and \( b \) such values that \( a + b = P, ab = -Q. \)

If \( a \) and \( b \) come out impossible quantities, the exponents of \( a \) and \( b \) in the value of \( y \) will have the form

\[ a = a + \beta \sqrt{-1}, \quad b = a - \beta \sqrt{-1}; \]

we have then

\[ y = Ce^{ax + \beta x \sqrt{-1}} + C'e^{ax - \beta x \sqrt{-1}} \] \[ = e^{ax}(Ce^{\beta x \sqrt{-1}} + C'e^{-\beta x \sqrt{-1}}); \]

This result is rendered real by eliminating the imagi- nary quantities, by means of sines and cosines; thus we have (ALGEBRA, art. 269),

\[ e^{\beta x \sqrt{-1}} = \cos \beta x + \sqrt{-1} \sin \beta x, \] \[ e^{-\beta x \sqrt{-1}} = \cos \beta x - \sqrt{-1} \sin \beta x, \] \[ y = e^{ax} \{(C + C') \cos \beta x + (C - C') \sqrt{-1} \sin \beta x \}; \]

and making \( C + C' = c, (c - c') \sqrt{-1} = c', \)

\[ y = e^{ax} (c \cos \beta x + c' \sin \beta x); \]

or, making \( c = p \sin q, \quad c' = p \cos q, \)

\[ y = pe^{ax} \sin (\beta x + q). \]

When the roots \( a \) and \( b \) are equal, the value of \( y \) being reduced to

\[ Ce^{ax} + C'e^{ax} = (C + C')e^{ax}, \]

it becomes incomplete. In this case we may proceed, as in art. 58, by supposing that \( a \) and \( b \) differ by a very small quantity. Let us suppose that \( b = a + k, \) there thence results

\[ y = Ce^{ax} + C'e^{ax+kx} = e^{ax}(C + C'e^{kx}); \]

developing \( e^{kx} \) according to the powers of \( k, \) we have

\[ y = e^{ax}(C + C' + C'kx + C'\frac{k^2x^2}{2} + &c.) \]

or, putting \( C + C' = c \) and \( c'k = c' \)

\[ y = e^{ax}(c + cx + c'\frac{x^2}{2} + &c.). \]

This last expression, which satisfies the proposed equa- tion for all values of \( k, \) agrees with it likewise if \( k = 0, \) or \( b = a; \) and, in that case, it becomes

\[ y = e^{ax}(c + cx). \]

190. When P and Q are variable quantities, functions of \( x, \) then if \( v \) and \( v' \) are two values of \( y \) which each sat- isfy the equation

\[ \frac{dy}{dx} + P \frac{dy}{dx} + Qdy = 0, \]

we may take

\[ y = cv + c'v' \]

for the complete primitive equation. For then \( dy = cdv + cdv', \quad d^2y = cdv + cdv', \) and the differential equation becomes

\[ \frac{d^2v}{dx^2} + P \frac{dv}{dx} + Qv = 0; \quad \frac{d^2v'}{dx^2} + P \frac{dv'}{dx} + Qv' = 0. \]

Now, by hypothesis,

\[ \frac{d^2v}{dx^2} + P \frac{dv}{dx} + Qv = 0; \quad \frac{d^2v'}{dx^2} + P \frac{dv'}{dx} + Qv' = 0. \]

Therefore the equation is identical, and so the value of \( y \) is truly determined.

As an example of the integration of a differential equa- tion of the second degree, see the article ASCII (art. 57).

191. We shall now give an example of the integra- tion of a differential equation of the first degree and the se- cond order.

Let the equation be

\[ \frac{dy}{dx} + ax^2ydx^2 = 0. \]

If we suppose

\[ y = Ax^n + Bx^{n+1} + Cx^{n+2} + &c. \]

and that the series of exponents is an increasing one, or that \( n \) is positive, we may, when \( x \) is supposed very small, conceive \( y \) to reduce itself to its first term, since the others are too small to be compared with this first. On this supposition, we may confine ourselves to as- suming

\[ y = Ax^n; \] then \( \frac{dy}{dx} = a(a-1)Ax^{n-2}dx^2, \)

and the proposed equation will become

\[ a(a-1)Ax^{n-2} + aAx^{n-1} = 0. \]

It is not possible to determine \( a \) so as to make the two exponents \( n-2 \) and \( n-1 \) equal, except in the parti- cular case of \( n = 2; \) but the exponent of \( x \) being greater in the second term than the first, we may neglect one of these terms in comparison with the other; and the equation may then be verified in two ways (by approxi- Inverse mation), viz. by making \(a = 0\), and \(a = 1\), since upon either hypothesis the term \(a(a - 1)Ax^2\) (the greatest in the equation) vanishes; \(A\), therefore, remains indeterminate, and we have two series, one beginning with \(A\), and the other with \(Ax\).

If we take successively

\[y = A + Bx^3 + cx^2 + \ldots\]

and substitute for \(y\) these values, and for \(dy\) its corresponding values, we shall then find, by properly arranging the terms, that \(b\) must be \(= 2\); and, in either case, determining the values of the co-efficients \(A, B, C, \ldots\) we arrive at these two series,

\[ \begin{align*} A &= \frac{aAx^n}{(n+1)(n+2)} + \frac{a^2Ax^{n+4}}{(n+1)(n+2)(2n+3)(2n+4)} \\ &\quad + \frac{a^3Ax^{n+6}}{(n+1)(n+2)(2n+3)(2n+4)(3n+5)(3n+6)} \\ &\quad + \ldots \end{align*} \]

These developments are only particular cases, since they contain each but one arbitrary constant \(A\); but, on account of the particular form of the proposed example (189), we shall obtain a general expression for \(y\), by writing in the latter of them \(A_1\) for \(A\), and taking their sum.

Having now given as full a view of the principles of the fluxional calculus as we conceive to be compatible with the nature of our work, we shall conclude with a few more examples of its application to geometry.

**Problem 1.**

To find the length of the enlarged meridian in Mercator's, or rather Wright's projection of the sphere.

In this projection, the meridians and parallels of latitude are straight lines, which intersect each other at right angles; and the projection of any small arc of a meridian (as one minute), reckoned from any parallel, is to the projection of a like arc of longitude in that parallel, as radius to the cosine of the latitude of the parallel, that is, in the same ratio as the arcs themselves in the sphere. Supposing the radius of the sphere to be unity, let \(v\) be any arc of latitude reckoned from the equator, \(z\) its projection on the chart, or the enlarged meridian. Let \(v'\) and \(z'\) be any small increments of \(v\) and \(z\); then, by the principles of the projection,

\[ \frac{z'}{v'} = \frac{\text{rod}}{\cos v}; \]

therefore, putting the ratio of the differentials for the limit of the ratio of the increments,

\[ \frac{dz}{dv} = \frac{1}{\cos v}, \quad \text{and} \quad dz = \frac{dv}{\cos v}. \]

Therefore, taking the integral (art. 155),

\[ z = 1 \left\{ \tan (45^\circ + \frac{1}{2}v) \right\}. \]

Here no correction is wanted, because when \(v = 0\), then \(z = 1 \cdot (\tan 45^\circ) = 1 \cdot (1) = 0\), as it should be.

Henry Bond, in the year 1650, discovered, by chance, that the enlarged meridian might be expressed by the logarithmic tangents of half the complements of the latitudes, a rule easily found from the preceding solution; but the difficulty of proving this was then considered so great, that Mercator offered to wager a sum of money against any person that should undertake to prove it either true or false. James Gregory, however, proved it in his *Exercitationes Geometricae*, published in 1668; also Barrow, in his Geometrical Lectures: their demonstrations, however, were intricate. Afterwards Dr Wallis and Dr Halley gave demonstrations which were sufficiently simple and elegant.

**Problem 2.**

If any number of straight lines are drawn according to some determinate law, it is required to find the nature of a curve to which these are tangents.

For example, let AE be a straight line given by position, and K a given point without it; let any number of lines KD, KD', &c. be drawn to meet AE in D, D', &c.; and let perpendiculars DC, D'C', &c. be drawn to these lines; it is required to find the nature of the curve ACC', to which these perpendiculars are tangents.

Without attending to the particular case, we shall resolve the general problem, and suppose AE to be the axis of the curve, A being the origin of the co-ordinates, and CD any position of the tangent, which meets the axis in D. From the point of contact C draw the perpendicular CB; and, considering C as a point in the curve, put \(AB = x\), \(BC = y\); but again considering C as any point whatever in the tangent, put \(AB = x'\), \(BC = y'\). Then, whatever be the conditions that determine the position of the tangent, the relation of \(x'\) and \(y'\), the co-ordinates of any point in it, may be expressed by the equation \(y' = Px' + Q\), where P and Q are put to denote generally certain functions of constant quantities, and some quantity \(p\), which has the same value for any given position of the tangent, but which changes its value if the tangent changes its position. For example, \(p\) may express the angle which the tangent makes with the axis, or it may represent the subtangent BD, &c.

Let us now suppose that the variable quantity \(p\) changes its value, and becomes \(p + h\), and that \(C'D'\) is the new position of the tangent corresponding to \(p + h\); then, considering P and Q as functions of \(p\), by Taylor's theorem (art. 31),

\[ P \text{ becomes } P + \frac{dP}{dp}h + \frac{d^2P}{dp^2}h^2 + \ldots \]

\[ Q \text{ becomes } Q + \frac{dQ}{dp}h + \frac{d^2Q}{dp^2}h^2 + \ldots \]

The relation of \(x'\) to \(y'\) in the new position of the tangent will now be expressed by the equation

\[ y' = Px' + Q + \left( \frac{dP}{dp}x' + \frac{dQ}{dp} \right)h + Kh^2 + \ldots \]

Where \(Kh^2 + \ldots\) is put for all the remaining terms of the series.

Now, as this equation holds true of every point in the Inverse tangent \( C'D' \), and the equation \( y' = Px + Q \) holds true of every point in the tangent \( CD \), it follows that at \( e \), the intersection of the two tangents, both equations must be true at the same time; therefore at \( e \) we have

\[ \left( \frac{dP}{dp} x' + \frac{dQ}{dp} \right) h + Kh^2 + \text{&c.} = 0; \]

and, dividing by \( h \),

\[ \frac{dP}{dp} x' + \frac{dQ}{dp} + Kh + \text{&c.} = 0. \]

Conceive now the two tangents to approach to coincidence; when \( C' \) comes to \( C \), then \( e \) will also fall at \( C \); and \( h \), and all the terms into which it enters, vanish; also \( x' \) and \( y' \) become \( x \) and \( y \); and, to determine the nature of the curve, we have these two equations:

\[ y = Px + Q \quad \text{(1)} \]

\[ 0 = \frac{dP}{dp} x + \frac{dQ}{dp} \quad \text{(2)} \]

By eliminating \( p \) from these, the resulting equation will express the nature of the curve.

Ex. 1. Let us now recur to the particular case of \( AE \), a straight line given by position, \( K \) a given point, and \( KDC \) a right angle: Draw \( KA \) perpendicular to \( AE \); put \( AB = x, BC = y, KA = a \), and let \( AD \) be the variable quantity \( p \). The triangles \( KAD, DBC \) are manifestly similar; therefore \( KA : AD = DB : BC \), that is, \( a : p = x : p : y \); hence \( ay = px - p^2 \), and \( y = \frac{p}{a} x - \frac{p^2}{a} \).

Compare this with equation (1), and it will appear that

\[ P = \frac{p}{a}, \quad Q = -\frac{p^2}{a}; \]

therefore \( \frac{dP}{dp} = \frac{1}{a}, \quad \frac{dQ}{dp} = -\frac{2p}{a} \); hence the nature of the curve is expressed by the two equations

\[ y = \frac{p}{a} x - \frac{p^2}{a}; \quad 0 = \frac{1}{a} x - \frac{2p}{a}. \]

The second of these equations gives \( p = \frac{1}{2}x \); and hence the first becomes \( y = \frac{x^3}{2a} - \frac{x^2}{4a} \); therefore \( 4ay = x^2 \) is the equation of the curve, which is evidently a parabola, of which \( AK \) is the axis, \( K \) the focus, and \( A \) the vertex.

Ex. 2. Suppose a ray of light \( RD \), coming from the sun, to fall upon \( FEG \), the concave surface of a sphere, at \( D \), and to be thence reflected in the direction \( DH \); it is proposed to find the nature of the curve to which this, and all rays reflected in the same manner, are tangents.

Draw \( AD \) the radius of the sphere, and \( AE \) parallel to the incident ray \( RD \); let \( C \) be the point in which the reflected ray touches the curve; let \( DC \) meet \( AE \) in \( H \), and draw \( CB \) perpendicular to \( AE \). Put \( AD = a, AB = x, BC = y \), and let \( p \) be the variable angle \( DAE \).

By the principles of optics, \( AD \) bisects the angle \( RDH \), which is equal to \( DHE \), that is, to the sum of the angles \( DAH, ADH \); therefore the angles \( ADH, DAH \) are equal, and angle \( DHE = 2p \). Now, by trigonometry,

\[ AH = \frac{\sin ADH}{\sin AHD} \times AD = \frac{\sin p}{\sin 2p} a, \]

\[ BH = \frac{\cos CHB}{\sin CHB} \times CB = \frac{\cos 2p}{\sin 2p} y; \]

Hence \( x = (AH + BH) = \frac{\sin p}{\sin 2p} a + \frac{\cos 2p}{\sin 2p} y \),

and \( y = \frac{\sin 2p}{\cos 2p} x - \frac{\sin p}{\cos 2p} a \).

By comparing this with the general formula (1), it appears that

\[ P = \frac{\sin 2p}{\cos 2p} Q = -\frac{\sin p}{\cos 2p} a; \quad \text{hence}, \]

\[ \frac{dP}{dp} = \frac{2}{\cos^2 2p}, \quad \frac{dQ}{dp} = -\frac{\cos p}{\cos 2p} a - \frac{2 \sin 2p \sin p}{\cos^2 2p} a. \]

Hence, by formula (2),

\[ \frac{2x}{\cos^2 2p} = \left\{ \frac{\cos p}{\cos 2p} + \frac{2 \sin 2p \sin p}{\cos^2 2p} \right\} a. \]

From equations (a) and (b) we readily find

\[ x = \left( \frac{1}{2} \cos 2p \cos p + \sin 2p \sin p \right) a, \]

\[ y = \left( \frac{1}{2} \sin 2p \cos p - \cos 2p \sin p \right) a; \]

and hence, by observing that \( \sin 2p = 2 \sin p \cos p \), and that \( \cos 2p = 2 \cos^2 p - 1 = 1 - 2 \sin^2 p \) (Arithmetic of Sines, Algebra), we have also

\[ = \frac{1}{2} \cos p (1 + 2 \sin^2 p) a; \quad y = (\sin^2 p) a. \]

From these equations it is easy to eliminate the trigonometrical quantities \( \cos p \) and \( \sin p \), and the result will be the equation of the curve, which is an epicycloid. The curve in question is the catenary curve to a circle.

It is easy to see that the general problem, of which we have now given two examples, is very comprehensive. We may evidently find, by the formulas (1) and (2), catenary and diastatic curves in all cases whatever. These, and an infinite number of other geometrical problems, are contained in the following more general problem.

Problem 3.

Determine the nature of a curve which touches an infinite number of lines of a given kind, described upon a plane according to some determinate law. For example, suppose the lines to be parabolas described by a projectile thrown from an engine with a given velocity, at every possible elevation in a vertical plane.

Let \( HCD \) be any one of the lines of a given kind (a parabola, for example), and suppose it referred to an axis \( AB \), by the rectangular co-ordinates \( AB = x' \) and \( BC = y' \). Let \( p \) denote some quantity belonging to the line or curve \( HCD \), which has always the same value in the same curve, but which has different values in different curves. Thus, if the curve be a circle, \( p \) may be its radius; or if the curve be a parabola, \( p \) may be its parameter, &c. This quantity \( p \) may also by analogy be called the parameter of the curve \( HCD \). Let us suppose the nature of this curve to be expressed by the equation

\[ f(x', y', p) = 0 \quad \text{(1)} \]

that is, let some function of \( x', y' \), and \( p \), be supposed \( = 0 \). If \( p \) now be supposed to change its value, and become \( p + h \), then the curve \( HCD \) will change its figure, and have some other position \( H'C'D' \). Let \( AB = x' \), and \( BC' = y' \). Inverse \( y' \) be any co-ordinates of this other curve, and as, by Method hypothesis, the two curves are expressed by equations of the same form, we must have

\[ f(x', y', p + h) = 0; \]

and this expression, by Taylor's theorem, is equivalent to

\[ f(x', y', p) + \frac{d}{dp} \left\{ f(x', y', p) \right\} h + Kh^2 + &c., \]

the differential being taken upon the hypothesis that \( p \) alone is variable, and \( Kh^2 + &c. \) being put for all terms of the series following the second, each of which is multiplied by a power of \( h \).

Let the two curves intersect each other in \( c \), and let \( Ab = x \), and \( cb = y \), be the common co-ordinates; then, as equation (1) holds true of every point in the curve HCD, and equation (2) holds true of every point in H'CD', both must hold true at once, if we substitute in them \( x \) and \( y \), the co-ordinates belonging to their common point \( c \); that is, we must have

\[ f(x, y, p) = 0, \]

\[ f(x, y, p) + \frac{d}{dp} \left\{ f(x, y, p) \right\} h + Kh^2 + &c. = 0; \]

and hence we must also have

\[ \frac{d}{dp} \left\{ f(x, y, p) \right\} + Kh + &c. = 0. \]

Let C and C' be now supposed the points in which the curves HCD, H'CD' touch the curve PCQ, whose nature is required; then, if we suppose \( h \) to decrease continually, and at last to vanish, the points C' and c will approach to C, and at last will coincide with it, so that \( x \) and \( y \), which are co-ordinates of \( c \), the intersection of the two curves HCD, H'CD', will then become the co-ordinates of the curve PCQ. As all the terms which contain \( h \) will then vanish, we have evidently this rule.

Let the equation of the given curves be

\[ f(x, y, p) = 0. \]

\( x \) and \( y \) being the co-ordinates, and \( p \) a variable parameter. From this equation, by taking the differential, supposing \( p \) to be variable, and all the other quantities constant, we deduce this other equation,

\[ \frac{d}{dp} \left\{ f(x, y, p) \right\} = 0. \]

By these eliminate \( p \), and the result will be an equation, which expresses the nature of the curve that touches all the given curves.

Ex. Let ACD, ACD', &c. be parabolas described by a projectile thrown from a given point A, with a given velocity, in a given vertical plane. It is proposed to find the curve PCQ which touches them all. Let EF be the axis of any one of the curves, AD an ordinate to the axis, \( AP = a \) the height due to the velocity of projection, \( AB = x \), \( BC = y \), the co-ordinates of C, any point in the curve. Put the parameter of the axis \( = p \), and considering AD as a function of \( p \), which is to be regarded as variable, put \( AD = q \).

By the theory of projectiles, \( EF = a - \frac{1}{2} p \), and by the nature of the parabola, \( AF^2 = p \times EF \), and \( AB \times BD = p \times BC \); hence we have these two equations,

\[ qx - x^2 = py, \quad (1) \]

\[ q^2 = 4ap - p^2, \quad (2) \]

From the first of these,

\[ f(x, y, p) = x^2 + py - qx = 0; \]

and, taking the differentials, considering \( x \) and \( y \) as constant, and \( q \) as a function of the variable quantity \( p \),

\[ \frac{d}{dp} \left\{ f(x, y, p) \right\} = y - \frac{dq}{dp} x = 0, \]

therefore \( \frac{y}{x} = \frac{dq}{dp} \); but from equation (2), taking the inverse differentials, \( dq = 2adp - pdp \), and hence \( \frac{dq}{dp} = \frac{2a - p}{q} \),

and by the first equation \( \frac{y}{x} = \frac{q - x}{p} \), therefore \( \frac{q - x}{p} = \frac{2a - p}{q} \), and hence \( q^2 - qx = 2ap - p^2 \); and substituting for \( q^2 \) its value given by equation (2), we get \( qx = 2ap \);

hence, and by equation (1),

\[ P = \frac{x}{2a - y}, \quad q = \frac{2ax}{2a - y}. \]

These values of \( p \) and \( q \) being substituted in the second equation, and the common denominator rejected, it becomes

\[ 4a^2x^2 = (8a^2 - 4ay - x^2)x^2. \]

Hence \( 4ay = 4a^2 - x^2 \), and this is the equation of the curve PCQ, which is evidently a parabola having its focus at \( A \), the common intersection of all the parabolas, its axis perpendicular to the horizon, and its parameter \( = 4a \).

The geometrical theory comprehended in the second and third problems has a corresponding analytical theory relating to the integrals of certain differential equations. This is the theory of singular primitive equations, which are not included in the complete primitive equation. Thus, the differential equation

\[ dy \sqrt{(x^2 + y^2 - b)} - ydx - xdx = 0, \]

has for its complete primitive equation

\[ x^2 - 2ay - a^2 - b = 0, \]

where \( a \) is the arbitrary constant quantity; but, besides this, it has a singular primitive equation,

\[ x^2 + y^2 - b = 0, \]

which does not admit of an arbitrary constant, although it equally satisfies the differential equation, as is easily proved by differentiating. The bounds within which it was proper to confine this treatise have not allowed us to enter into this branch of the subject, which, although interesting, is yet not elementary.

Investigation of the Properties of the Catenary.

Suppose a chain or thread of uniform thickness, and perfectly flexible, but inextensible, to be fastened by its extremities at two points A, B in a vertical plane; by the action of gravity it will take the form of a curve ACPB, called the catenary.

The curve will evidently be all in one plane, and a horizontal line CD, in that plane, at C its lowest point, will manifestly touch the curve, which is retained at rest by a mutual balancing of the forces produced by the weight of the particles of the chain.

If we suppose its lowest point to be fixed at C, it is easy to understand that the part BC will not on that account change in the least its figure, because the tension of the chain at C, when it was free, is now replaced by the reaction of the force exerted in the horizontal direction CD. The forces exerted on the points B, C will evidently be the same in quantity and in direction, whether the chain be considered as flexible, or rigid like a solid wire; hence it may be regarded as kept at rest by three forces, viz. its gravity acting vertically, and the re-actions of the tensions at B and C; the former exerted in the direction of a tangent to the curve at B, and the latter in the direction CD, the tangent at the lowest point C.

In like manner any portion PC between P and its lowest point may be considered as kept at rest by the joint action of the gravity of the mass PC, the tension of the chain at C, which is constant, and the tension at P, which is variable; and by the nature of an equilibrium (Mechanics), these forces will be to each other as the sides of any triangle which are parallel to their directions. Draw a horizontal line OX in the plane of the curve as an axis, meeting the vertical through C in O, which may be taken as an origin of co-ordinates; draw PQ perpendicular to OX, and PT touching the curve and meeting the axis in T. From what has been explained, the tension of the curve at C will be to the weight of the matter in PC as TQ to PQ, that is, as radius to the tangent of the angle which PT makes with any horizontal line.

Put OQ = x, PQ = y, arc CP = z, angle PTQ = ϕ; and let a be the tension of the curve at the vertex C: we have then a : z = 1 : tan ϕ; hence

\[ \tan \phi = \frac{z}{a} \]

This is the distinguishing property of the curve, from which we are to deduce its other properties.

From equation (1), by differentiation,

\[ \frac{d\phi}{\cos \phi} = \frac{dz}{a}. \]

Now in all curves,

\[ dz = -\frac{dx}{\cos \phi} = \frac{dy}{\sin \phi}; \]

hence we obtain

\[ dx = a \cdot \frac{d\phi}{\cos \phi}, \quad dy = a \cdot \frac{d\phi \sin \phi}{\cos \phi}. \]

From these, by integration,

\[ x = a l \tan (45° + \frac{1}{2} \phi) + \text{const. (art. 155)} \]

\[ y = \frac{a}{\cos \phi} + \text{const.} \]

If we assume that OC measures the tension at C, which is a constant quantity, then because when ϕ = 0, x = 0, and y = a, we have the relation of the quantities x, y, ϕ, expressed by these equations,

\[ x = a l \tan (45° + \frac{1}{2} \phi), \quad y = \frac{a}{\cos \phi} = a \sec \phi. \]

The constant line a is called the parameter of the curve, also its modulus. From the first of these equations, putting ϕ for 45° + \(\frac{1}{2}\) ϕ, we have

\[ \tan \phi = e^{\frac{x}{a}}, \quad \cot \phi = e^{-\frac{x}{a}}; \]

\[ \tan \phi + \cot \phi = e^{\frac{x}{a}} + e^{-\frac{x}{a}}. \]

Now, \( \tan \phi + \cot \phi = \frac{\sin \phi}{\cos \phi} + \frac{\cos \phi}{\sin \phi} = \frac{2}{\sin 2\phi} = \frac{2}{\cos \phi} = \frac{2y}{a}; \)

therefore \( y = \frac{a}{2} \left( e^{\frac{x}{a}} + e^{-\frac{x}{a}} \right) \)

\[ \sec^2 \phi = \frac{1}{4} \left( e^{\frac{2x}{a}} + e^{-\frac{2x}{a}} + 2 \right), \]

\[ \sec^2 \phi - 1 = \frac{1}{4} \left( e^{\frac{2x}{a}} + e^{-\frac{2x}{a}} - 2 \right), \]

\[ \tan \phi = \sqrt{\sec^2 \phi - 1} = \frac{1}{2} \left( e^{\frac{x}{a}} - e^{-\frac{x}{a}} \right). \]

But it was found that \( z = a \tan \phi, \)

therefore \( z = \frac{a}{2} \left( e^{\frac{x}{a}} - e^{-\frac{x}{a}} \right) \)

From formulae (3) and (4) we have

\[ y + z = ae^{\frac{x}{a}}, \quad y - z = ae^{-\frac{x}{a}} \]

hence again,

\[ y = \sqrt{a^2 + z^2}, \quad z = \sqrt{y^2 - a^2}, \]

and since from formula (5)

\[ \frac{x}{a} = 1, \quad \frac{y + z}{a} = 1, \quad \frac{a}{y - z} \]

therefore also,

\[ \frac{x}{a} = \frac{1}{2} \left( \frac{y + \sqrt{y^2 - a^2}}{a} \right), \]

\[ \frac{z}{a} = \frac{1}{2} \left( \frac{y - \sqrt{y^2 - a^2}}{a} \right). \]

Let s denote the area bounded by the arc PC and the straight lines CO, OQ, PQ; then, since \( dx = \frac{ad\phi}{\cos \phi}, \) and \( y = \frac{a}{\cos \phi}, \) therefore

\[ ds = ydx = \frac{a^2 d\phi}{\cos^2 \phi} = adz. \]

and \( s = az. \)

Thus it appears that the arc CP is always proportional to the conterminous area CPQO. It is evident that the catenary, if inverted, would form an equilibrated arch. From this property it appears that if a wall, whose height above the crown of the arch is equal to the line CO, be horizontal at the top, the structure thus formed will constitute an equilibrated arch with a straight roadway. (See on this subject our article Arcu.)