Probability



1.4 Hyperbolic Functions

The hyperbolic functions are combinations of exponential functions that have many similarities to the trigonometric functions. In fact, for almost every formula involving a trigonometric function there is an analogous formula involving a hyperbolic function. The hyperbolic functions are useful for doing integrals and they also arise is the solution of problems involving mechanical vibrations and electric circuits where one frequently encounters combinations of exponentials such as 2e3x – 5e-2x. For certain calculations it is convenient to express such combinations in terms of the two special combinations

and

These two special combinations are the first two hyperbolic functions. The remaining four are related to these in the same way as the remaining trigonometric functions are related to the sine and cosine.

Example 1.

sinh 2 = ( ( 3.62686

Here are their graphs. As we go along we will explain why these graphs have the shape they do.

We are particularly interested in the derivatives of the hyperbolic functions and integrals involving the hyperbolic functions. The following table contains the derivatives of the hyperbolic functions. The formulas are similar to the trigonometric functions, but the sign is opposite in (8) and (11).

In order to prove these derivative formulas we will need the following formulas which are similar to the corresponding formulas for the trigonometric functions.

Proposition 1. (13) – (15) hold.

Proof. To prove (13) one has

cosh2x – sinh2x = 2 - 2 = - = 1

To prove (14), divide (13) by cosh2x. To prove (15), divide (13) by sinh2x. //

Proposition 2. (7) – (12) hold.

Proof. To prove (7) and (8) one has

sinh x = = = cosh x

cosh x = = = sinh x

Formulas (9) – (12) follow from (7), (8), (14) and (15) and the rules for derivatives. //

If we combine the formulas for the derivatives of the hyperbolic functions with the chain rule we get the formulas at the right which are useful for the situation where what is inside a hyperbolic is an expression u(x) involving x instead of just x.

Example 2. Find if

a. y = cosh( ln x )

b. y = tan-1( sinh x )

c. y = tanh( 1 + e2x )

d. y = sinh( sin-1 x )

e. y = coth( sec-1 x )

For a, using (17) one has = (cosh( ln x )) = sinh( ln x ) ( ln x ) = . For b, using the chain rule one has = (tan-1( sinh x )) = ( sinh x ) = . For c, using (18) one has = (tanh( 1 + e2x )) = sech2( 1 + e2x ) ( 1 + e2x ) = 2e2x sech2( 1 + e2x ).

Each of the differentiation formulas (7) – (12) has a corresponding integration formula. These are given in the table at the right.

Example 3. Find dx

Let u = = x1/2. Then du = (1/2)x-1/2 dx = . So 2 du = . So the integral becomes dx = 2 cosh u du = 2 sinh u = 2 sinh().

Problem 1. Find a. sinh(1 + 4x) dx b. tanh x dx

c. s sech2(x2) dx d. (sinh x) ecosh x dx

Now let's consider why the graphs of the hyperbolic functions look the way they do. First we consider positive x.

First, note that sinh x > 0 for x > 0 since ex > e-x. Also cosh x > 0 for x > 0 since ex and e-x are positive for all x. It follows that the remaining four hyperbolic functions are positive for positive x.

Second, recall that a function with a positive derivative is increasing and a function with a negative derivative is decreasing. So, it follows from the derivative formulas that sinh x, cosh x and tanh x are increasing for x > 0 and coth x, sech x and csch x are decreasing for x > 0.

Third, note that as x increases e-x goes to zero. In fact e-4 ( 0.02. So by the time x ( 4 one has sinh x ( and coshh x ( . In particular, sinh x and cosh x approach ( as x approaches (. On the other hand, it follows from tanh x = that tanh x ( 1 as x ( (. From the reciprocal formulas it follows that coth x ( 1 and sech x ( 0 and csch x ( 0 as x ( (.

The graphs for negative x follow from the graphs for positive x and the fact that sinh x, tanh x, coth x and csch x are odd functions and cosh x and sech x are even functions.

Here are the addition formulas for sinh x and cosh x. Note their similarity to the corresponding formulas for sin x and cos x.

Proposition 3. For all x and y one has

(28) sinh(x + y) = sinh x cosh y + cosh x sinh y

(29) cosh(x + y) = cosh x cosh y + sinh x sinh y

To prove (28) one has

sinh x cosh y + cosh x sinh y = +

= + = = sinh(x + y)

To prove (29) one has

cosh x cosh y + sinh x sinh y = +

= + = = cosh(x + y)

The following double angle formulas are a corollary. They will be useful when we do integrals.

Proposition 4. For all x one has

(30) sinh 2x = 2 sinh x cosh x

(31) cosh 2x = cosh2 x + sinh2 x

(32) sinh2x =

(33) cosh2x =

Proof. (30) and (31) follow from (28) and (29) by replacing y by x. (32) and (33) follow from (31) and the identity cosh2x – sinh2x = 1. //

Example 4. A cable hangs from two poles that are 40 feet tall and separated by 100 feet. In the middle the cable is 38 feet off the ground. Find

a. A formula for the height of the cable above the ground for a general point between the two cables.

b. The angle the cable makes with the horizontal where it meets the pole.

Solution. Choose a coordinate system with the x axis lying along the ground, the y axis pointing upward and half way between the poles. Then be base of the poles are at (- 50, 0) and (50, 0) and the tops of the poles are at (- 50, 40) and (50, 40). The midpoint of the cable is at (0, 38). See picture at right.

In more advanced books it is shown that the equation of the cable is

y = c + a cosh

where c and a are numbers depending of heights of the poles and the distance between them. Plugging in x = 0 and y = 38 we get

38 = c + a

Plugging in x = 50 and y = 40 we get

40 = c + a cosh

Subtracting these two equations we get

a cosh - a = 2

If we let u = then this becomes

cosh u = 1 +

This is an equation that we can't solve algebraically, so we need to solve it numerically. Using some mathematical software (e.g. ), we get u ( 0.08000. So a ( ( 625 and c ( 38 – 625 = -587. Since ( 0.0016, the equation of the cable is approximately

y = 625 cosh(0.0016x) - 587

We need to find . One has

= (c + a cosh) = a sinh = sinh

= sinh = sinh u = sinh(0.08) = 0.08004

( = tan-1 = tan-1(0.08004) = 0.0798 radians = 4.58(

Inverse Hyperbolic Functions

The inverse hyperbolic functions are formed from the hyperbolic functions in the same way the inverse trigonometric functions are formed from the regular trigonometric functions. The only difference is the restrictions one places on the hyperbolic functions to make them one-to-one. To summarize

As with the inverse trigonometric functions, we get the graphs of the inverse hyperbolic functions by reflecting the graphs of the hyperbolic functions across the line y = x. Their graphs are the following.

As opposed to the inverse trigonometric functions, there are formulas for the inverse hyperbolic involving logarithms. At the right are these formulas.

Example 5. Show how to obtain the formula sinh-1 x = ln(x + ).

Solution. We use the fact that y = sinh-1x is that number y such that sinh y = x. Using the definition of sinh y this gives x = = . Now let u = ey. So x = and 2x = u - . This implies u2 - 2xu - 1 = 0. Using the quadratic formula, this gives u = = x ( . We choose the + sign since u > 0. So ey = u = x + . Taking ln of both sides gives y = ln(x + ).

In calculus one of the main applications of hyperbolic functions is to do certain integrals. In order to see how they are used, we first have to look at the derivative formulas for the inverse hyperbolic functions.

Example 6. Show that sinh-1 x =

Solution. Let y = sinh-1x so that x = sinh y. Use implicit differentiation. Take the derivative of both sides of this last formula with respect to x to get 1 = (cosh y) . Thus = . Using cosh2x – sinh2x = 1 gives = = .

-----------------------

(1) hyperbolic sine of x = sinh x =

(2) hyperbolic cosine of x = cosh x =

(3) hyperbolic tangent of x = tanh x = = =

(4) hyperbolic cotangent of x = coth x = = =

(5) hyperbolic secant of x = sech x = =

(6) hyperbolic cosecant of x = csch x = =

(7) sinh x = cosh x

(8) cosh x = sinh x

(9) tanh x = sech2 x

(10) coth x = - csch2 x

(11) sech x = - tanh x sech x

(12) csch x = - coth x csch x

sin x = cos x

cos x = - sin x

tan x = sec2 x

cot x = - csc2 x

sec x = tan x sec x

csc x = - cot x csc x

compare

with

cos2x + sin2x = 1

sec2x - tan2x = 1

cot2x – csc2x = 1

compare

with

(13) cosh2x – sinh2x = 1

(14) sech2x + tanh2x = 1

(15) coth2x – csch2x = 1

(16) sinh u(x) = cosh u(x)

(17) cosh u(x) = sinh u(x)

(18) tanh u(x) = sech2 u(x)

(19) coth u(x) = - csch2 u(x)

(20) sech u(x) = - tanh u(x) sech u(x)

(21) csch u(x) = - coth u(x) csch u(x)

(22) cosh x dx = sinh x

(23) sinh x dx = cosh x

(24) sech2 x dx = tanh x

(25) csch2 x dx = - coth x

(26) tanh x sech x dx = - sech x

(27) coth x csch x dx = - csch x

sin(x + y) = sin x cos y + cos x sin y

cos(x + y) = cos x cos y - sin x sin y

compare

with

sin 2x = 2 sin x cos y

cos 2x = cos2 x - sin2 x

sin2x =

cos2x =

compare

with

(34) y = sinh-1 x = that number y such that sinh y = x

(35) y = cosh-1 x = that number y such that y ( 0 and cosh y = x

(36) y = tanh-1 x = that number y such that tanh y = x

(37) y = coth-1 x = that number y such that coth y = x

(38) y = sech-1 x = that number y such that y ( 0 and sech y = x

(39) y = csch-1 x = that number y such that csch y = x

(40) sinh-1 x = ln(x + )

(41) cosh-1 x = ln(x + )

(42) tanh-1 x =

(43) coth-1 x =

(44) sech-1 x = ln

(45) csch-1 x = ln

sin-1 x =

cos-1 x =

tan-1 x =

cot-1 x =

sec-1 x =

csc-1 x =

(46) sinh-1 x =

(47) cosh-1 x =

(48) tanh-1 x =

(49) coth-1 x =

(50) sech-1 x =

(51) csch-1 x =

compare

with

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