Lecture 1:
Hyperbolic Functions
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Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
REVIEW:
Inverse Trigonometric Functions
also denoted by
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Identities for Inverse Trigonometric Functions
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
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Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Integration Formulas
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
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Example
Evaluate
Solution
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Example
Evaluate
Solution
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Application of Inverse Trigonometric Functions
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Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
A soccer player kicks a ball with an initial speed v=14 m/s at an angle θ with the horizontal. The ball lands 18 m down the
field. If air resistance is neglected, then the ball will have a parabolic trajectory and the horizontal range R will be given by
where
is the acceleration due to gravity. Find the two values of θ,
at which the ball could have been kicked. Which angle results in the shorter time of flight? Why?
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
v=14 m/s
R=18 m
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
v=14 m/s
R=18 m
Rg
sin 2θ = 2
ν
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
v=14 m/s
R=18 m
18m × 9.8m/s2
Rg
sin 2θ = 2 =
(14m/s)2
ν
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
v=14 m/s
R=18 m
18m × 9.8m/s2
Rg
=
0.9
sin 2θ = 2 =
(14m/s)2
ν
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
v=14 m/s
R=18 m
18m × 9.8m/s2
Rg
=
0.9
sin 2θ = 2 =
(14m/s)2
ν
2θ = sin−1 0.9 = 64.15806723683288
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
v=14 m/s
R=18 m
18m × 9.8m/s2
Rg
=
0.9
sin 2θ = 2 =
(14m/s)2
ν
2θ = sin−1 0.9 = 64.15806723683288
or
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
v=14 m/s
R=18 m
18m × 9.8m/s2
Rg
=
0.9
sin 2θ = 2 =
(14m/s)2
ν
2θ = sin−1 0.9 = 64.15806723683288
or
2θ = 180 − sin−1 0.9 = 115.84193276316712
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
v=14 m/s
R=18 m
18m × 9.8m/s2
Rg
=
0.9
sin 2θ = 2 =
(14m/s)2
ν
2θ = sin−1 0.9 = 64.15806723683288
or
2θ = 180 − sin−1 0.9 = 115.84193276316712
θ = 32.1
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
v=14 m/s
R=18 m
18m × 9.8m/s2
Rg
=
0.9
sin 2θ = 2 =
(14m/s)2
ν
2θ = sin−1 0.9 = 64.15806723683288
or
2θ = 180 − sin−1 0.9 = 115.84193276316712
θ = 32.1
or
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
v=14 m/s
R=18 m
18m × 9.8m/s2
Rg
=
0.9
sin 2θ = 2 =
(14m/s)2
ν
2θ = sin−1 0.9 = 64.15806723683288
or
2θ = 180 − sin−1 0.9 = 115.84193276316712
θ = 32.1
or
θ = 57.9
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
An Earth-observing satellite has horizon sensors that can measure the angle θ.
R be the radius of the Earth (assumed spherical) and
h the distance between the satellite and the Earth's surface.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
A camera is positioned x feet from the base of a missile launching pad. A missile of length a feet is launched vertically.
What is the angle θ subtended at the lens by the missile when the base of the missile is b feet above the camera lens?
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
7.8 Hyperbolic Functions and Hanging Cables
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
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7.8 Hyperbolic Functions and Hanging Cables
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
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7.8 Hyperbolic Functions and Hanging Cables
hyperbolic sine of x
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
7.8 Hyperbolic Functions and Hanging Cables
hyperbolic cosine of x
hyperbolic sine of x
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
7.8 Hyperbolic Functions and Hanging Cables
hyperbolic cosine of x
hyperbolic sine of x
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Graphs of the Hyperbolic Functions
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
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Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Derivatives of Hyperbolic Functions
d e +e
d
cosh x =
dx
dx
2
x
−x
1
=
2
�
d −x
d x
e +
e
dx
dx
�
�
1� x
−x
e −e
=
= sinh x
2
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Derivatives of Hyperbolic Functions
d e +e
d
cosh x =
dx
dx
2
x
−x
1
=
2
�
d −x
d x
e +
e
dx
dx
�
�
1� x
−x
e −e
=
= sinh x
2
d
cosh x = sinh x
dx
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Derivatives of Hyperbolic Functions
d e +e
d
cosh x =
dx
dx
2
x
−x
1
=
2
�
d −x
d x
e +
e
dx
dx
�
�
1� x
−x
e −e
=
= sinh x
2
d
cosh x = sinh x
dx
x
d e −e
d
sinh x =
dx
dx
2
−x
1
=
2
�
d −x
d x
e −
e
dx
dx
�
�
1� x
−x
e +e
=
= cosh x
2
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Derivatives of Hyperbolic Functions
d e +e
d
cosh x =
dx
dx
2
x
−x
1
=
2
�
d −x
d x
e +
e
dx
dx
�
�
1� x
−x
e −e
=
= sinh x
2
d
cosh x = sinh x
dx
x
d e −e
d
sinh x =
dx
dx
2
−x
1
=
2
�
d −x
d x
e −
e
dx
dx
�
�
1� x
−x
e +e
=
= cosh x
2
d
sinh x = cosh x
dx
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Example
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Example
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Why They Are Called Hyperbolic Functions
parametric equations
represent the unit circle
Analogously,
the parametric equations
represent a portion of unit hyperbola
this is the reason why these functions are called
hyperbolic functions.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Example
A 100-ft wire is attached at its ends to the tops of two 50-ft poles that are positioned 90 ft apart.
How high above the ground is the middle of the wire?
Solution
the wire forms a catenary curve with equation
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
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gives
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gives
so
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
gives
so
Thus, the middle of the wire is
ft above the ground.
Calculus: Late Transcendentals, 8/E by Howard Anton, Irl Bivens, and Stephen Davis
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.