The length of the normal at any point on the catenary y=c{(^x/c + e^-x/c)/2} varies as

Dear student,

First of all find the normal by differentiating, normal.dy/dx=-1

dear praveen ,

length of normal = y{1+(dy/dx)²}^1/2

y = c{(e^x/c+e^-x/c)/2}        ...................1

dy/dx = { (e^x/c-e^-x/c)/2}           ...................2

length = y{1+(dy/dx)²}^1/2 

         = y { 1+(e^x/c-e^-x/c)²}^1/2

        =y { (e^2x/c+e^-2x/c+2)/4}^1/2

        = y { ((e^x+e^-x)/2)²}^1/2

       = y { e^x+e^-x)/2          ..................3

from eq 1 putting (e^x+e^-x)/2 = y/c in eq 3 we get

length = yy/c = y²/c

so , length of normal varies as square of abcissa ....

approve if u like my ans