141

29. Problem, to find the centre ofgravity of the catenarian curve.

By the principles of fluxions we have thedistance of the centre of gravity of any

f*xk 1

curve from its vertex, =-— = — f‘xz.

Z Z**

Now since y — — = —and

J * <^2cx+x l

_ {c+x)x cx . xx

^ ^ ^ < ' x + a V' 2 CX + X* ✓ 2 CiT-fa* *V 2CX + X*

which, by substituting y for its equal, be-

comes 8 —y -\~

or

*Z‘2CX-\- X 1 •J 2cx-\-X*

- z-y

•</ 2C*-f X*cxx

2 cx-f-x*

or, by multiplying each side by c,

— c {x-y) and consequently P

= c ( z ~y)-

Again assume xz =p.

Then, xz-\-xz =p or by substitution

. ,-, (c + x)xx x{2cx+x , ‘)-\-ctx + x' , x

V VScx + x* d- ,—X=S— = —--

1 V 2cx + x 1

X

3 cxx

V 2cx -f X'

3 cxx

+

2cx 7 xs/ 2,cx + x‘

'■V or ;

\/ 2cx+x 32 cx^x

^/2cx+x 2

orP

V' 2cx-\-z

V , 2cx-{-x a J