cosh means hyperbolic cosine, cosh(x/50)=½(ex/50+e-x/50).
y=50cosh(x/50) where x is the distance between the posts, and y is the height in feet above ground of the cable when suspended. So at its lowest point the cable is 50 feet above the ground
The cable forms a catenary (curve) and we need the (arc) length of the curve between x=-33 and x=33.
dy/dx=y'=sinh(x/50) where sinh(x/50)=½(ex/50-e-x/50).
For an infinitesimal length of arc ds, we have ds2=dx2+dy2, so (ds/dx)2=1+(dy/dx)2.
We can write this s'2=1+y'2, so ds=√(1+y'2)dx and:
S=-33∫33√(1+y'2)dx=-33∫33√(1+sinh2(x/50))dx.
sinh(x/50)=½(ex/50-e-x/50), so sinh2(x/50)=¼(ex/25-2+e-x/25) and:
1+sinh2(x/50)=1+¼(ex/25-2+e-x/25)=¼(ex/25+2+e-x/25)=cosh2(x/50).
[From this we can see that the derivative of sinh(A))=cosh(A) and vice versa. Also cosh2(A)-sinh2(A)=1.]
Cable length S=-33∫33cosh(x/50)dx=50[sinh(x/50)]-3333. Note: sinh(-A)=-sinh(A).
S=100sinh(0.66)=70.897ft approx.