mat01a1: hyperbolic functions€¦ · the hyperbolic functions are made by combining ex and e x in...
TRANSCRIPT
MAT01A1: Hyperbolic Functions
Dr Craig
Week: 11 May 2020
This is a short “lecture” on an important
type of functions: the hyperbolic functions.
Hyperbolic functions have many applications
in physics, engineering, and applied
mathematics.
Please note that we only cover pages
159–161 of this section. We do not cover
inverse hyperbolic functions.
You can of course look over the inverse
functions and their derivatives on your own.
Hyperbolic functions
The hyperbolic functions are made by
combining ex and e−x in different ways.
They are related to hyperbolas similarly to
how trig functions are related to circles.
This is why their names are similar to trig
functions.
However, you need to be careful: many of
the identities and derivatives look like the
corresponding ones for trig functions, but
they are slightly different.
The main two hyperbolic functions:
sinh(x) =ex − e−x
2cosh(x) =
ex + e−x
2
The full name of the first function is
“hyperbolic sine” while the second one is
“hyperbolic cosine”.
sinhx is often read as “shine x”
coshx is often read as “cosh x”
The main two hyperbolic functions:
sinh(x) =ex − e−x
2cosh(x) =
ex + e−x
2
The full name of the first function is
“hyperbolic sine” while the second one is
“hyperbolic cosine”.
sinhx is often read as “shine x”
coshx is often read as “cosh x”
Definition of the Hyperbolic Functions:
sinh(x) =ex − e−x
2csch(x) =
1
sinhx
cosh(x) =ex + e−x
2sech(x) =
1
coshx
tanh(x) =sinhx
coshxcoth(x) =
coshx
sinhx
ran(sinh(x)) = (−∞,∞)
ran(cosh(x)) = [1,∞)
Remembering the definitions
y = c+ a cosh(x/a)
The cosh function is used to describe, for instance,a chain hanging between two fixed points. Youshould know the shape of the two graphs and thefunction definitions. If you think coshx↔ chainthen you will remember which one is which.
Hyperbolic Identities:
sinh(−x) = − sinh(x) cosh(−x) = cosh(x)
cosh2(x)− sinh2(x) = 1
1− tanh2(x) = sech2(x)
sinh(x + y) = sinhx cosh y + coshx sinh y
cosh(x + y) = coshx cosh y + sinhx sinh y
The proofs of all the identities follow directly
from the definition of cosh(x) and sinh(x).
Hyperbolic Identities:
sinh(−x) = − sinh(x) cosh(−x) = cosh(x)
cosh2(x)− sinh2(x) = 1
1− tanh2(x) = sech2(x)
sinh(x + y) = sinhx cosh y + coshx sinh y
cosh(x + y) = coshx cosh y + sinhx sinh y
The proofs of all the identities follow directly
from the definition of cosh(x) and sinh(x).
Proofs of identities:
On the next slide we will prove
cosh(x + y) = coshx cosh y + sinhx sinh y
Use the proof there as a guide to writing up
your own proof for
sinh(x + y) = sinhx cosh y + coshx sinh y
Substituting gives LHS = cosh(x+y) =ex+y + e−(x+y)
2
We have
RHS =
(ex + e−x
2
)(ey + e−y
2
)+
(ex − e−x
2
)(ey − e−y
2
)
=ex+y + ex−y + ey−x + e−x−y
4+
ex+y − ex−y − ey−x + e−x−y
4
=2ex+y + 2e−x−y
4=
ex+y + e−(x+y)
2= LHS
Substituting gives LHS = cosh(x+y) =ex+y + e−(x+y)
2We have
RHS =
(ex + e−x
2
)(ey + e−y
2
)+
(ex − e−x
2
)(ey − e−y
2
)
=ex+y + ex−y + ey−x + e−x−y
4+
ex+y − ex−y − ey−x + e−x−y
4
=2ex+y + 2e−x−y
4=
ex+y + e−(x+y)
2= LHS
Substituting gives LHS = cosh(x+y) =ex+y + e−(x+y)
2We have
RHS =
(ex + e−x
2
)(ey + e−y
2
)+
(ex − e−x
2
)(ey − e−y
2
)
=ex+y + ex−y + ey−x + e−x−y
4+
ex+y − ex−y − ey−x + e−x−y
4
=2ex+y + 2e−x−y
4=
ex+y + e−(x+y)
2= LHS
Substituting gives LHS = cosh(x+y) =ex+y + e−(x+y)
2We have
RHS =
(ex + e−x
2
)(ey + e−y
2
)+
(ex − e−x
2
)(ey − e−y
2
)
=ex+y + ex−y + ey−x + e−x−y
4+
ex+y − ex−y − ey−x + e−x−y
4
=2ex+y + 2e−x−y
4
=ex+y + e−(x+y)
2= LHS
Substituting gives LHS = cosh(x+y) =ex+y + e−(x+y)
2We have
RHS =
(ex + e−x
2
)(ey + e−y
2
)+
(ex − e−x
2
)(ey − e−y
2
)
=ex+y + ex−y + ey−x + e−x−y
4+
ex+y − ex−y − ey−x + e−x−y
4
=2ex+y + 2e−x−y
4=
ex+y + e−(x+y)
2= LHS
Derivatives of hyperbolic functions
In order to find the derivatives of both sinhx
and coshx, you need the following important
example of the chain rule at work.
d
dx
(e−x
)= e−x · d
dx(−x) = −e−x
Now, as an exercise, use this to calculate the
derivative of both sinhx and coshx. The
final answers for both are on the next slide.
Derivatives of hyperbolic functions
In order to find the derivatives of both sinhx
and coshx, you need the following important
example of the chain rule at work.
d
dx
(e−x
)= e−x · d
dx(−x) = −e−x
Now, as an exercise, use this to calculate the
derivative of both sinhx and coshx. The
final answers for both are on the next slide.
Derivatives of hyperbolic functions
In order to find the derivatives of both sinhx
and coshx, you need the following important
example of the chain rule at work.
d
dx
(e−x
)= e−x · d
dx(−x) = −e−x
Now, as an exercise, use this to calculate the
derivative of both sinhx and coshx. The
final answers for both are on the next slide.
The derivative of both sinh(x) and cosh(x) can becalculated using just their definition and the chainrule. The rest of the derivatives can be calculatedusing the quotient rule and/or chain rule.
The derivative of both sinh(x) and cosh(x) can becalculated using just their definition and the chainrule. The rest of the derivatives can be calculatedusing the quotient rule and/or chain rule.
Combining the derivatives from the previous
slide with our existing differentiation
techniques we can attempt the more
complicated examples on the next slide.
Examples:
1. Find y′ if y = ln(coshx).
2. Find g′(x) if g(x) = x sinhx− coshx.
3. Find f ′(t) if f (t) = arctan(sinh t).
Attempt these on your own. The solutions
are on the slides that follow.
You should aim to simplify them as far as
possible.
Solutions:
(1.) We apply the chain rule to get:
y′ =1
coshx· ddx
(cosh) =sinhx
coshx= tanhx
(2.) We use the product rule:
g′(x) = (d
dx(x) sinhx + x
d
dx(sinhx))− d
dx(coshx)
= sinhx + x coshx− sinhx
= x coshx
Solutions:
(1.) We apply the chain rule to get:
y′ =1
coshx· ddx
(cosh) =sinhx
coshx= tanhx
(2.) We use the product rule:
g′(x) = (d
dx(x) sinhx + x
d
dx(sinhx))− d
dx(coshx)
= sinhx + x coshx− sinhx
= x coshx
Solutions continued...
(3.) We have f (t) = arctan(sinh t). Recall
thatd
dx(arctanx) =
1
1 + x2.
Therefore
f ′(t) =1
1 + (sinh t)2· ddt(sinh t) =
cosh t
1 + sinh2 t
=cosh t
cosh2 t=
1
cosh t
= sech t
Solutions continued...
(3.) We have f (t) = arctan(sinh t). Recall
thatd
dx(arctanx) =
1
1 + x2. Therefore
f ′(t) =1
1 + (sinh t)2· ddt(sinh t)
=cosh t
1 + sinh2 t
=cosh t
cosh2 t=
1
cosh t
= sech t
Solutions continued...
(3.) We have f (t) = arctan(sinh t). Recall
thatd
dx(arctanx) =
1
1 + x2. Therefore
f ′(t) =1
1 + (sinh t)2· ddt(sinh t) =
cosh t
1 + sinh2 t
=cosh t
cosh2 t=
1
cosh t
= sech t
Solutions continued...
(3.) We have f (t) = arctan(sinh t). Recall
thatd
dx(arctanx) =
1
1 + x2. Therefore
f ′(t) =1
1 + (sinh t)2· ddt(sinh t) =
cosh t
1 + sinh2 t
=cosh t
cosh2 t
=1
cosh t
= sech t
Solutions continued...
(3.) We have f (t) = arctan(sinh t). Recall
thatd
dx(arctanx) =
1
1 + x2. Therefore
f ′(t) =1
1 + (sinh t)2· ddt(sinh t) =
cosh t
1 + sinh2 t
=cosh t
cosh2 t=
1
cosh t
= sech t
Solutions continued...
(3.) We have f (t) = arctan(sinh t). Recall
thatd
dx(arctanx) =
1
1 + x2. Therefore
f ′(t) =1
1 + (sinh t)2· ddt(sinh t) =
cosh t
1 + sinh2 t
=cosh t
cosh2 t=
1
cosh t
= sech t
Prescribed tut problems:
Complete the following exercises from the
8th edition (Ch 3.11):
I 2, 10, 11, 17, 23, 34, 40, 41, 47, 52
I If you are using the 7th edition then do
the numbers above except for the
following changes:
34 → 33, 40 → 38, 41 → 39
Do these exercises before you start on the
slides for Ch 4.4.