A rope with its two ends held in place forms a curve called a catenary (assuming that the stiffness of the rope is negligible). A catenary takes the shape of the function:
f(x) = a cosh(x/a),
where cosh is the hyperbolic cosine function.

If a 50-foot rope hangs by its ends from two flagpoles, one 50 feet tall and one 40 feet tall, and at its lowest point is 20 feet above the ground, how far apart are the flagpoles?

Let ground be x-axis.drop perpendicular from lowest point to ground & extend it as y-axis
The lowest point corresponds to a in eqn.y=acosh(x/a)
hence a=20
now cosh(x/a)=(e^(x/a)+e^(-x/a))/2
put z=e^(x/a)
for 1st pole,
5=z+1/z
solving we get z=(5+sqrt(21))/2 or (5-sqrt(21))/2
we have taken 1st pole to the left of origin & so
x should turn out to be negative.For this z<1
because only then log z will be -ve & x will be -ve as
a is +ve.so 2nd value of z is taken & x is calculated(say x1)
similar procedure is adopted for 2nd pole &
value corresponding to z>1 is taken & 2nd value of x(say x2)
is calculated.
now x2-x1 gives the distance between poles(in this x1 will be -ve)


is got.

Posted by ananth on 2002-12-24 01:32:58