Page:Somerville Mechanism of the heavens.djvu/120

Chap II.] and integrating, $$log\frac{dx}{dt} = log C + log \frac{dy}{dt}$$. 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 $$\frac{d^2y}{dt^2} = -A\frac{dy}{dt}$$ 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;

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

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

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

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

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

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