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?

Before I start, let me say that the heights of the two poles are just to throw people off.

Either with calculus or recognizing that cosh(x) looks somewhat like a parabolic, the mininum occurs at x=0.

f(0)=a*cosh(0/a)=a=20

f'(x)=sinh(x/a) (derivative of f(x))
Let R=distance between two poles
Length of rope = integral(sqrt(1+(f'(x))^2),x,0,R)
=integral(sqrt(1+(sinh(x/a))^2),x,0,R)
=integral(sqrt((cosh(x/a))^2),x,0,R)
=integral(cosh(x/a),x,0,R)
=sinh(x/a),0,R
=sinh(R/a)=50

NOTE: There should be an absolute value which is omitted since cosh(x/a) is nonnegative.

Therefore R=20*asinh(50).

I know it's a lot of wordy math. It'd be much better if I could use the integral sign. Also, the length formula could be found in a calculus book but it's proved by first writing the length as sqrt((dx)^2+(dy)^2) and factoring out the dx...

Posted by np_rt on 2002-12-24 23:41:13