p. 529

89.

a) Show that a satisfies the equation 16  a  =  sinh (15  a).

Problem 88 shows that s  =  (1/a) sinh (a x) where s is the length of the cable from its lowest point to a horizontal distance of x to the right. We are given that the total length of the cable is 32 feet and that -15  ≤  x  ≤  15. Therefore, s  =  16 when x  =  15. Plugging these numbers into the equation above gives 16  =  (1/a) sinh (a 15).

b) Solve for a graphically.

It is difficult to see where the graphs of  y  =  16 a and a  =  sinh (15 a) cross without narrowing the plot range. After a bit of experimenting we see that they cross between a  =   .02 and a  =   .06. Based on the graph, a good initial guess would be a  =  0.42.

Plot[ { 16 a,  Sinh[15 a] },  {a,   .02,  .06}]

c) Solve for a numerically.

FindRoot[ 16 a  ==  Sinh[15 a],  {a,  0.042}]

d) Estimate the tension at the cable's lowest point.

We are given that a  =  w/H, where w  =  2 lb/ft is the unit weight of the cable and H is the horizontal tension at the lowest point. From part c) we know a  =   .0417525, so the horizontal tension in pounds is

H  =  
      w/a    /. 
           {w -> 2,  a -> 0.0417524632548766838`}

e) Graph the catenary

a  =  0.0417524632548766838`;

y[x_]  :=  (1/a) Cosh[a x]

Plot[y[x],  {x, -15, 15},  PlotRange -> {0, 30}]

The height of the cable in feet at the ends is

y[15]

and the height of the cable at its lowest point is

y[0]

so the "sag" is

y[15]  -  y[0]