A History of Mathematics From Mesopotamia to Modernity

188 A History ofMathematics

6. The simplest method of solution is to divide bywand multiply byx(so all like terms are
   together):
   dw
   w

=a·dx.

(We have had to assume that we can do this, but this is quite allowable on the assumption

that differentials are ‘quantities’ of some kind. We now suppose that we know thatdw/w=

d(ln(w)), that is, effectively we integrate both sides: ln(w)=ax+k, or equivalentlyw=Aeax.

7. (Not easy.) Let the arc length from A to B bes, then the weightWisρgs(ρ=density/unit
   length). The horizontal components giveT 0 =Tcosψ, and the vertical giveTsinψ=ρgs.
   Hence,

ρgs=T 0 tanψ

This is called the ‘intrinsic equation’. Note that tanψ=dy/dx. On the other hand,ds^2 =

dx^2 +dy^2 (from an infinitesimal triangle), sods/dx=

√

1 +(dy/dx)^2. Writeufordy/dx, and

differentiate the intrinsic equation. You get:

k·

ds

dx

=k

√

1 +u^2 =

du

dx

wherek=ρg/T 0. Now, if you know that the integral of 1/(

√

1 +u^2 )is sinh−^1 (u)(look it up),

the rest follows:

kx=sinh−^1 (u); u=sinh(kx); y=

1

k

cosh(kx)

usingu=dy/dx, and noting thatu=0 whenx=0.

8. You have:θ =0,r = 2 a, which has cartesian coordinates(2, 0);θ =±π/2,r=a, with
   coordinates(0,± 1 ); andθ=−π,r=0, giving(0, 0). These provide a basis, and you can add
   others.