Hyperbolic Functions Homework
Using the definitions of hyperbolic functions, prove the following:
1. cosh2x – sinh2 = 1
2. cosh 2x = cosh2x + sinh2x
3. sinh 2x = 2 cosh x sinh x
4.
sinh2 x = ½ (cosh 2x – 1)
5. cosh2 x = ½ (cosh 2x + 1)
6.
cosh (x + y) = cosh x cosh y + sinh x sinh y
7.
cosh ln x  sinh ln x
 x2
cosh ln x  sinh ln x
Evaluate (correct to three decimal places) the indicated hyperbolic or inverse hyperbolic functions.
8. sinh 2
9. tanh–1 0
10. cosh 0.5
11. sech–1 0.2
12. coth 3
13. csch 0
Find dy/dx for the following:
14. y = sinh x
15. y = y = tanh (3 x)
17. y = sinh(ln x) cosh(ln x)
18.
16. y = sech (cosh x)
y = ln [coth (3x2 + 1)]
Evaluate the following integrals:
19.

sinh(ln x)
dx
x
22.  [ex sech (ex )tanh (ex) ] dx
20.
 x cosh(1  x
23.
 x2 cosh x
2
)dx
dx
ln 2
21.

24.
 sinh2 x
0
tanh xdx
dx
Graphing Calculator Exploration:
25. Graph the parametric equations x1 = cosh t and y1 = sinh t on your graphing calculator in
PARAMETRIC graphing mode using a standard window. The hyperbolic functions are found on the HYP
sub-menu of the MATH menu on the TI-86. Can you guess what the resulting graph is?
Complete the graph by graphing these sets of parametric equations
x2 = – cosh t and y2 = sinh t
x3 = cosh t and y3 = – sinh t
x4 = – cosh t and y4 = – sinh t
If you could not identify the graph from equations x1 and y1, you should be able to identify it now.
Do you remember what happens with the trigonometric functions when you graph the parametric equations
x1 = cos t and y1 = sin t ? If you don't remember, graph them to see.
St. Louis Arch
 x
a
The curve y  a cosh   is called a catenary.
The catenary models the shape of a hanging flexible rope, chain, cable, or wire that is suspended between
two points at the same height. The Latin word catena means “chain”.
The Saint Louis Arch was constructed using a catenary curve, using a function of the form
 x
y  c  b cosh 
a
The height of the arch is approximately 625 feet and the width of the arch at the bottom is approximately
599.5 feet. ( Based on information from website: http://www.nps.gov/jeff/equation.htm)
(The arch height of 625 feet is approximately the height of a building with 50 or more stories. It's about half
the height of the Empire State Building, about 3/4 of the height of the TransAmerica Pyramid in San
Francisco, and slightly taller than the Washington Monument in Washington D.C.)
1. Imagine that the arch is on a coordinate system with
the origin (0,0) located in the center of the arch on the
ground. Using the top of the arch and the two points at
the bases of the arch, show that the general arch
shape approximately satisfies the equation:
 x 
y  693.8598  68.7673 cosh 

 99 .671 
for –299.224 x  299.224.
2. A recent visitor to the arch described the shape to me as a parabola. From the perspective of
constructing the arch, the unique structural engineering properties of the catenary do not apply to a
parabola. Use your calculator graph the function for the arch. Then graph the parabola
y = 625.0925 – 0.006982x2 on the same screen to see how the shape of the catenary and the
parabola differ.
3. Find the slope of the arch at
a. The bottom ends of each arch (x = -299.2 and x = 299.2)
b. At the points on the arch that are 150 feet away from the center of the arch.
4. Find the area enclosed in the space between the arch and the ground.
5. Find the arclength along the arch between the two bases.