Fall 2014 Math 151 Supplements
Hand Examples
4. Hyperbolic Functions
On your calculator, find regular and inverse hyperbolic
c
Sun, 02/Nov
2014
Art Belmonte functions in the Catalog under Category 2. For limits
and derivatives, you may use calculator templates. To
approximate, press Ctrl-Enter. Note cosh x > 0 for all x.
Summary
Definitions of Hyperbolic Functions
ex − e−x
sinh x =
2
cosh x =
ex + e−x
2
sinh x
tanh x =
cosh x
287/2
1
csch x =
sinh x
sech x =
Find the value of (a) tanh 0 and (b) tanh 1.
1
cosh x
Solution
cosh x
coth x =
sinh x
0
Hyperbolic Identities
sinh (−x) = − sinh x
cosh (−x) = cosh x
cosh2 x − sinh2 x = 1
1 − tanh2 x = sech2 x
Find the value of (a) cosh 3 and (b) cosh (ln 3).
d
sinh x = cosh x
dx
d
csch x = −csch x coth x
dx
d
cosh x = sinh x
dx
d
sech x = −sech x tanh x
dx
d
tanh x = sech2 x
dx
d
coth x = −csch2 x
dx
Solution
(a) We have cosh 3 =
e3 +e−3
2
(b) Similarly, cosh (ln 3) =
≈ 10.07.
eln 3 +e− ln 3
2
=
3+ 13
2
=
10
6
= 35 .
287/6
Definitions of Inverse Hyperbolic Functions
p
y = sinh−1 x = ln x + x2 + 1 , x ∈ R
⇔ sinh y = x
Find the value of (a) sinh 1 and (b) sinh−1 1.
Solution
⇔ cosh y = x
1
−1
(a) We have sinh 1 = e −e
≈ 1.18.
2
√
√ (b) sinh−1 1 = ln 1 + 12 + 1 = ln 1 + 2 ≈ 0.88.
and y ≥ 0
y = tanh
≈ 0.76.
287/4
Derivatives of Hyperbolic Functions
−1
e1 −e−1
e1 +e−1
(b) Similarly, tanh 1 =
p
y = cosh−1 x = ln x + x2 − 1 , x ≥ 1
−0
= 1−1
(a) The definitions give sinh 0 = e −e
2
2 = 0 and
0
−0
1+1
sinh 0
0
e +e
cosh 0 = 2 = 2 = 1. Hence tanh 0 = cosh
0 = 1 = 1.
1
1+x
x = ln
, −1 < x < 1 ⇔ tanh y = x
2
1−x
Derivatives of Inverse Hyperbolic Functions
287/8
d
1
sinh−1 x = √
dx
1 + x2
d
1
csch−1 x = − √
Prove the identity cosh (−x) = cosh x. That is, cosh is an
dx
|x| x2 + 1
even function.
d
1
cosh−1 x = √
2
dx
x −1
d
1
sech−1 x = − √
dx
x 1 − x2
Proof
d
1
tanh−1 x =
dx
1 − x2
d
1
coth−1 x =
dx
1 − x2
We have cosh (−x) =
1
e(−x) +e−(−x)
2
=
ex +e−x
2
= cosh x.
288/16
288/48
Prove the identity cosh 2x = cosh2 x + sinh2 x.
Eliminate t from the parametric equations x = sinh t,
y = cosh t. Describe the curve as an equation in x and y.
Proof
=
e2x +2+e−2x +e2x −2+e−2x
4
=
Solution
2 x −x 2
ex +e−x
+ e −e
2
2
2e2x +2e−2x
e2x +e−2x
=
= cosh 2x.
4
2
We have cosh2 x + sinh2 x =
The parmetric equations are defined for all t ∈ R. Recall
that the range of sinh is R, whereas that of cosh is [1, ∞).
If sinh x = 43 , find the values of the other hyperbolic
functions at x.
We have 1 = cosh2 t − sinh2 t = y2 − x2 with y = cosht ≥ 1
and x = sinht ∈ R. This is the upper half of the hyperbola
y2 − x2 = 1. Verify with parametric and Cartesian graphs
on your calculator.
Solution
288/55
Start with the M.O.A.H.T.I. (Mother Of All Hyperbolic
Trig Identities): cosh2 x − sinh2 x = 1 and recall that
cosh x > 0 for all x.
s
2 r
25 5
3
1+
=
cosh x =
=
4
16 4
sinh x
3/4 3
tanh x =
=
=
cosh x 5/4 5
1
1
4
cschx =
=
=
sinh x 3/4 3
1
4
1
=
=
sechx =
cosh x 5/4 5
cosh x 5/4 5
coth x =
=
= .
sinh x
3/4 3
At what point on the curve y = cosh x does the tangent
have slope 1?
288/20
Solution
√ Solve y0 = sinh x = 1 to obtain x = sinh−1 1 = ln 1 + 2
from 287/6b above. Recall cosh x > 0. Then the MOAHTI
(cosh2 x − sinh2 x = 1) yields
p
p
√
y = cosh x = 1 + sinh2 x = 1 + 12 = 2.
√ √ Our point is (x, y) = ln 1 + 2 , 2 ≈ (0.88, 1.41).
Here is an illustrtive graph.
Stewart 288/55: The party’s gone out of bounds.
288/36
4
Find the derivative of f (t) = ln (sinht).
3
Solution
2
y
cosht
We have f 0 (t) =
= cotht.
sinht
1
288/44
0
√
Find the derivative of y = x tanh−1 x + ln 1 − x2 .
−1
−2
Solution
Write y = x tanh−1 x + 12 ln 1 − x2 . We have
y0 = (1) tanh−1 x + x ·
1
1 −2x
+ ·
= tanh−1 x.
2
1−x
2 1 − x2
2
−1
0
1
x
2
3
4