Page:Somerville Mechanism of the heavens.djvu/120

and integrating, $log{\frac {dx}{dt}}=logC+log{\frac {dy}{dt}}$. Whence ${\frac {dx}{dt}}=C{\frac {dy}{dt}}$, or $dx=Cdy$, and if we integrate a second time,

$x=Cy+D$,

in which C and D are the constant quantities introduced by double integration. As this is the equation to a straight line, it follows that the projection of the curve in which the body moves on the plane $xoy$ is a straight line, consequently the curve $MN$ is in the plane $zox$, that is at right angles to $xoy$; thus $MN$ is a plane curve, and the motion of the projectile is in a plane at right angles to the horizon. Since the projection of $MN$ on $xoy$ is the straight line $ED$, therefore $y=0$, and the equation ${\frac {d^{2}y}{dt^{2}}}=-A{\frac {dy}{dt}}$ is of no use in the solution of the problem, there being no motion in the direction $oy$. Theoretical reasons, confirmed to a certain extent by experience, show that the resistance of the air supposed of uniform density is proportional to the square of the velocity;

hence,

$A=hv^{2}=h{\frac {ds^{2}}{dt^{2}}}$,

$h$ being a quantity that varies with the density, and is constant when it is uniform; thus the general equations become

(a)

${\frac {d^{2}x}{dt^{2}}}=-h.{\frac {ds}{dt}}.{\frac {dx}{dt}};{\frac {d^{2}z}{dt^{2}}}=g-h.{\frac {ds}{dt}}.{\frac {dz}{dt}}$;

the integral of the first is

${\frac {dx}{dt}}=C.c^{-hs}$

$C$ being an arbitrary constant quantity, and $c$ the number whose hyperbolic logarithm is unity.

In order to integrate the second, let $dz=udx,u$ being a function of $z$; then the differential according to $t$ gives

${\frac {d^{2}z}{dt^{2}}}={\frac {du}{dt}}.{\frac {dx}{dt}}+u.{\frac {d^{2}x}{dt^{2}}}$.

If this be put in the second of equations (a), it becomes, in consequence of the first,

${\frac {du}{dt}}.{\frac {dx}{dt}}=-g$;