7.7
1
7.7 Hyperbolic Functions
7.7
2
Definition.
Hyperbolic Functions
Even and Odd Parts of an Exponential Function
We recall that a function f is called even if f (−x) = f (x). f is called
odd if f (−x) = −f (x). The sine function is odd while the squaring
function (x2 ) is even. Please review the graphs of each. Also review
the graphs of odd and even functions in general.
Notice that every function defined on an interval centered at the origin
can be “decomposed” into the sum of an odd and an even function.
(1)
f (x) =
f (x) + f (−x) f (x) − f (−x)
+
2
2
{z
} |
{z
}
|
even part
odd part
(2)
sinh x =
ex − e−x
2
(Hyperbolic sine)
(3)
cosh x =
ex + e−x
2
(Hyperbolic cosine)
(4)
tanh x =
ex − e−x
sinh x
= x
cosh x e + e−x
(5)
sech x =
1
2
= x
cosh x e + e−x
(Hyperbolic tangent)
(Hyperbolic secant)
We sketch the graph of each of the four functions below (more details
appear in the text).
Now write the exponential function in this way.
ex =
ex + e−x
ex − e−x
+
2 } | {z
2 }
| {z
even function
odd function
The even part of ex is call the hyperbolic cosine of x and the odd part
is called hyperbolic sine of x. These functions are interesting on their
own and will be the subject of this section.
y = cosh x
3
1
−2
−1
1
2
It turns out that a hanging cable lies along the graph of y = α1 cosh αx.
Here α ∈ R is a constant that depends on the weight of the cable and
the tension at its lowest point (see exercise 7.7.83).
7.7
3
3
y = sinh x
1
−2
−1
−1
1
2
−3
7.7
4
It should come as no surprise that these functions satisfy some
interesting identities. For example, using (2) and (3) we obtain
2 t
2
t
e + e−t
e − e−t
cosh2 t − sinh2 t =
−
2
2
1 2t
e + 2 + e−2t − e2t + 2 − e−2t
=
4
1
= (4) = 1
4
In otherwords,
cosh2 t − sinh2 t = 1
(6)
y = tanh x
1
−2
And one can also show that
1
−1
tanh2 t + sech2 t = 1 and
2
2 cosh2 x = cosh 2x + 1
−1
A complete list of useful hyperbolic identities can be found in the text.
Why “hyperbolic”? Recall that the graph of the equation y = 1/x is
called a hyperbola. More generally, we know that the equations
It is worth noting that, for example,
ex − e−x
1 − e−2x
1−0
= lim
=
=1
x
−x
x→∞ e + e
x→∞ 1 + e−2x
1+0
(x/a)2 − (y/b)2 = ±1
lim tanh x = lim
x→∞
as suggested by the sketch above.
generate a family of hyperbolas that are symmetric to either the x or
y-axis. Now let
x = cosh t and y = sinh t
(7)
Then (6) implies
1
−2
−1
y = sech x
1
Similarly, one can show
lim sech x = 0
x→∞
2
(8)
x2 − y 2 = 1
which we recognize as the equation of the
familiar hyperbola shown. In this case, we
call (7) a parametrization of the hyperbola
given by (8).
3
b
2
(cosh t, sinh t)
1
−3 −2 −1
−1
1
2
3
−2
−3
For more on this, see exercise 7.7.86. See also the next example and
the remarks that follow.
7.7
5
Example 1.
Do you recognize the following equation?
(9)
x2 + 2xy − y 2 = 4
We try completing the square.
4 = x2 + 2xy − y 2
= x2 − (y 2 − 2xy + x2 ) + x2
(10)
=⇒ 4 = 2x2 − (x − y)2
Of course, we could now solve this for y as “a” function of x to obtain
(11)
y =x±
√
2x2 − 4
and graph both equations on a graphing utility. Instead, we try a
different approach. Rearranging (10) we obtain
(12)
Now let x =
x2 (x − y)2
−
=1
2
4
√
2 cosh t and x − y = 2 sinh t. Then
√
2
2 cosh t
x2 (x − y)2
(2 sinh t)2
−
=
−
2
4
2
4
2
2
= cosh t − sinh t = 1
We have discovered the parametric equations for the curve given by
the cartesian equation in (9).
√
√
(13)
x = x(t) = 2 cosh t and y = y(t) = 2 cosh t − 2 sinh t
We will have more to say about parametric equations in chapter 11.
7.7
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Using the parametric plotting feature of any modern modern graphing
calculator (or any one of a number of computer algebra systems), we
obtain the graph shown below.
To convince yourself that (9) and
(13) are somehow equivalent, we
suggest that you try plotting a few
points with each. For example,
from (9) it is clear that the
hyperbola passes through the
point (2, 0). On the other hand, it
is not difficult
√to show that if
t0 = ln 1 + 2 , then
x(t0 ) =
√
4
2
b
−4
2 cosh t0 = 2 and y(t0 ) =
2
−2
4
−2
−4
√
2 cosh t0 − 2 sinh t0 = 0
Now try it on your own by setting y = 1 and solving for x in (9), etc.
Continuing with the last example, notice that the graph appears to have
tangent lines everywhere. Can we locate the vertical tangent lines
shown in the sketch below?
According to the Implicit Function theorem, we may differentiate both sides of
(9) with respect to x to obtain
4
2
2x + 2(y + xy ′ ) − 2yy ′ = 0
Rearranging yields
−4
y=x
dy
x+y
=
dx y − x
−2 b
b
2
√ √ 2, 2
−2
−4
Substituting y = x into (9) (or equivalently into (12)), we see that
√
x2 /2 − 0 = 1 or x = ± 2
as shown in the sketch. We will revisit this example in chapter 11.
4
7.7
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We can say more about the previous example. Recall the trigonometric
identity that resembles (6).
7.7
8
Derivatives and Integrals of the Hyperbolic Functions
sec2 t − tan2 t = 1
Notice that, for example, the hyperbolic cosine function is differentiable
(being the sum of two differentiable functions). In fact,
√
2 sec t and x − y = 2 tan t. Then
√
2
2 sec t
(2 tan t)2
x2 (x − y)2
−
=
−
2
4
2
4
= sec2 t − tan2 t = 1
d
1 d
cosh x =
ex + e−x
dx
2 dx
1 x
=
e − e−x
2
= sinh x
(14)
This time let x =
In other words, we have obtained another parametrization of (9) of the
hyperbola shown in Example 1.
√
√
x = 2 sec t and y = 2 sec t − 2 tan t
We used this parametrization to generate the last sketch.
The other derivative formulas can be derived in a similar manner. We
have
Remark. The alert reader may soon discover that the parametrization
x = cosh t and y = sinh t, say for t ∈ [−6, 6], generates only the right
branch of the hyperbola x2 − y 2 = 1 since cosh t > 0 for all real t. This is
indeed correct.
(15)
In fact, we actually need to parameterize each branch separately. It is
easy to verify that the parametrization (− cosh t, sinh t) generates the
left branch.
(17)
b
(− cosh t, sinh t)
3
b
2
(cosh t, sinh t)
1
−3 −2 −1
−1
1
−2
−3
What about the hyperbola in Example 1?
2
3
(16)
(18)
du
d
(sinh u) = cosh u
dx
dx
d
du
(cosh u) = sinh u
dx
dx
d
du
(tanh u) = sech2 u
dx
dx
d
du
(sech u) = − sech u tanh u
dx
dx
7.7
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This leads immediately to the corresponding integral formulas.
Z
(19)
cosh u du = sinh u + C
Z
(20)
Z
(21)
(22)
Z
sinh u du = cosh u + C
sech2 u du = tanh u + C
sech u tanh u du = − sech u + C