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Unformatted text preview: Chapter 2 Hyperbolic Functions 2 HYPERBOLIC
FUNCTIONS
Objectives
After studying this chapter you should
• understand what is meant by a hyperbolic function; • be able to find derivatives and integrals of hyperbolic
functions; • be able to find inverse hyperbolic functions and use them in
calculus applications; • recognise logarithmic equivalents of inverse hyperbolic
functions. 2.0 Introduction This chapter will introduce you to the hyperbolic functions which
you may have noticed on your calculator with the abbreviation
hyp. You will see some connections with trigonometric functions
and will be able to find various integrals which cannot be found
without the help of hyperbolic functions. The first systematic
consideration of hyperbolic functions was done by the Swiss
mathematician Johann Heinrich Lambert (17281777). 2.1 Definitions The hyperbolic cosine function, written cosh x, is defined for all
real values of x by the relation cosh x = ( 1 x
e + e− x
2 ) Similarly the hyperbolic sine function, sinh x, is defined by sinh x = ( 1 x
e − e− x
2 ) The names of these two hyperbolic functions suggest that they
have similar properties to the trigonometric functions and some of
these will be investigated. 33 Chapter 2 Hyperbolic Functions Activity 1
Show that cosh x + sinh x = e x and simplify cosh x − sinh x. (a) By multiplying the expressions for ( cosh x + sinh x ) and (cosh x − sinh x ) together, show that
cosh 2 x − sinh 2 x = 1 ( cosh x + sinh x )2 + (cosh x − sinh x )2 (b) By considering cosh 2 x + sinh 2 x = cosh 2x show that ( cosh x + sinh x )2 − (cosh x − sinh x )2 (c) By considering 2 sinh x cosh x = sinh 2x show that Activity 2
Use the definitions of sinh x and cosh x in terms of exponential
functions to prove that
(a) cosh 2x = 2 cosh 2 x − 1
(b) cosh 2x = 1 + 2 sinh 2 x Example
Prove that cosh ( x − y ) = cosh x cosh y − sinh x sinh y
Solution cosh x cosh y =
=
sinh x sinh y =
= ( ) ( 1 x
1
e + e− x × e y + e− y
2
2 ) ( 1 x+y
e
+ e x − y + e−( x − y) + e−( x + y)
4 ( ) ( 1 x
1
e − e− x × e y − e− y
2
2 ) ) ( 1 x+y
e
− e x − y − e−( x − y) + e−( x + y)
4 ) Subtracting gives cosh x cosh y − sinh x sinh y = 2 ×
=
34 ( ( 1 x−y
e
+ e−( x − y)
4 ) ) 1 x−y
e
+ e − ( x − y ) = cosh( x − y )
2 Chapter 2 Hyperbolic Functions Exercise 2A
3. cosh ( x + y ) = cosh x cosh y + sinh x sinh y Prove the following identities.
1. (a) sinh ( − x ) = − sinh x (b) cosh ( − x ) = cosh x 2. (a) sinh ( x + y ) = sinh x cosh y + cosh x sinh y
(b) sinh ( x − y ) = sinh x cosh y − cosh x sinh y 2.2 4. sinh A + sinh B = 2sinh A − B
A + B
cosh 2 2 5. cosh A − cosh B = 2sinh A − B
A + B
sinh 2 2 Osborn's rule You should have noticed from the previous exercise a similarity
between the corresponding identities for trigonometric functions.
In fact, trigonometric formulae can be converted into formulae for
hyperbolic functions using Osborn's rule, which states that cos
should be converted into cosh and sin into sinh, except when there
is a product of two sines, when a sign change must be effected. cos 2x = 1 − 2 sin 2 x For example, can be converted, remembering that sin 2 x = sin x.sin x,
into cosh 2x = 1 + 2 sinh 2 x . But sin 2 A = 2 sin A cos A simply converts to sinh 2 A = 2 sinh A cosh A because there is no
product of sines. Activity 3
Given the following trigonometric formulae, use Osborn's rule to
write down the corresponding hyperbolic function formulae.
(a) sin A − sin B = 2 cos A + B A − B
sin
2 2 (b) sin 3A = 3sin A − 4 sin 3 A
(c) cos 2 θ + sin 2 θ = 1 2.3 Further functions Corresponding to the trigonometric functions tan x, cot x, sec x
and cosecx we define
tanh x = sinh x
,
cosh x coth x = 1
cosh x
=
,
tanh x sinh x 35 Chapter 2 Hyperbolic Functions sech x = 1
cosh x and cosech x = 1
sinh x By implication when using Osborn's rule, where the function
tanh x occurs, it must be regarded as involving sinh x.
Therefore, to convert the formula sec2 x = 1 + tan2 x
we must write sech 2 x = 1 − tanh 2 x . Activity 4
(a) Prove that tanh x = ex − e− x
2
and sech x = x − x ,
x
−x
e +e
e +e and hence verify that
sech 2 x = 1 − tanh 2 x . (b) Apply Osborn's rule to obtain a formula which corresponds to
cosec 2 y = 1 + cot 2 y .
Prove the result by converting cosech y and coth y into
exponential functions. 2.4 Graphs of hyperbolic
functions You could plot the graphs of cosh x and sinh x quite easily on a
graphics calculator and obtain graphs as shown opposite. y cosh x 1 0 y The shape of the graph of y = cosh x is that of a particular chain
supported at each end and hanging freely. It is often called a
catenary (from the Latin word catena for chain or thread). 36 x sinh x x Chapter 2 Hyperbolic Functions Activity 5
(a) Superimpose the graphs of y = cosh x and y = sinh x on the
screen of a graphics calculator. Do the curves ever
intersect?
(b) Use a graphics calculator to sketch the function
f : x a tanh x with domain x ∈ R. What is the range of the
function?
(c) Try to predict what the graphs of
y = sechx, y = cosech x and y = coth x
will look like. Check your ideas by plotting the graphs on a
graphics calculator. 2.5 Solving equations 3
and we wish to find the exact value of x.
4
Recall that cosh 2 x = 1 + sinh 2 x and cosh x is always positive, so
3
5
when sinh x = , cosh x = .
4
4
Suppose sinh x = From Activity 1, we have sinh x + cosh x = e x 3 5
+ =2
4 4
x = ln 2 . ex = so
and hence Alternatively, we can write sinh x =
so sinh x = ( 1 x
e − e− x
2 ) 3
means
4 ( ) 1 x
3
e − e− x =
2
4 ⇒ 2e x − 3 − 2e − x = 0 and multiplying by e x 2e 2 x − 3e x − 2 = 0 (e x )( ) − 2 2e x + 1 = 0
e x = 2 or e x = − 1
2 But e x is always positive so e x = 2 ⇒ x = ln 2.
37 Chapter 2 Hyperbolic Functions Activity 6
Find the values of x for which cosh x = 13
5 expressing your answers as natural logarithms. Example
Solve the equation 2 cosh 2x + 10 sinh 2x = 5
giving your answer in terms of a natural logarithm.
Solution cosh 2x =
So ( ) ( 1 2x
1
e + e −2 x ; sinh 2x = e 2 x − e −2 x
2
2 ) e 2 x + e −2 x + 5e 2 x − 5e −2 x = 5
6e 2 x − 5 − 4e −2 x = 0
6e 4 x − 5e 2 x − 4 = 0 (3e 2x )( ) − 4 2e 2 x + 1 = 0 e2 x = 4
1
or e 2 x = −
3
2 The only real solution occurs when e 2 x > 0 2x = ln So 4
1 4
⇒ x = ln
3
2 3 Exercise 2B
1. Given that sinh x = 5
, find the values of
12 2. Given that cosh x = 5
, determine the values of
4 (a) cosh x (b) tanh x (c) sech x (a) sinh x (d) coth x (e) sinh 2 x (f) cosh 2 x Use the formula for cosh ( 2 x + x ) to determine the
value of cosh 3x. Determine the value of x as a natural logarithm. 38 (b) cosh 2 x (c) sinh 2 x Chapter 2 Hyperbolic Functions 3. In the case when tanh x = 1
1
, show that x = ln 3.
2
2 Hence write down the minimum value of
25cosh x − 24sinh x and find the value of x at which
this occurs, giving your answer in terms of a
natural logarithm. 4. Solve the following equations giving your answers
in terms of natural logarithms.
(a) 4 cosh x + sinh x = 4 8. Determine a condition on A and B for which the
equation (b) 3sinh x − cosh x = 1 Acosh x + Bsinh x = 1 (c) 4 tanh x = 1 + sech x has at least one real solution. 5. Find the possible values of sinh x for which
12 cosh 2 x + 7sinh x = 24 9. Given that a, b, c are all positive, show that when
a>b then a cosh x + bsinh x can be written in the
form Rcosh ( x + α ). (You may find the identity cosh 2 x − sinh 2 x = 1
useful.) Hence determine a further condition for which the
equation Hence find the possible values of x, leaving your
answers as natural logarithms. a cosh x + bsinh x = c
has real solutions. 6. Solve the equations
(a) 3cosh 2 x + 5cosh x = 22 10. Use an appropriate iterative method to find the
solution of the equation (b) 4 cosh 2 x − 2sinh x = 7 cosh x = 3x 7. Express 25cosh x − 24sinh x in the form
Rcosh ( x − α ) giving the values of R and tanh α . 2.6 giving your answer correct to three significant
figures. Calculus of hyperbolic
functions Activity 7
(a) By writing cosh x = ( ) 1 x
e + e − x , prove that
2 d
( cosh x ) = sinh x.
dx (b) Use a similar method to find d
(sinh x ).
dx (c) Assuming the derivatives of sinh x and cosh x, use the
sinh x
quotient rule to prove that if y = tanh x =
cosh x
then dy
= sech 2 x.
dx Note: care must be taken that Osborn's rule is not used to
obtain corresponding results from trigonometry in calculus.
39 Chapter 2 Hyperbolic Functions Activity 8
Use the quotient rule, or otherwise, to prove that
(a) d
(sech x ) = −sech x tanh x
dx (b) d
(cosech x ) = −cosech x coth x
dx (c) d
(coth x ) = −cosech 2 x
dx Example
Integrate each of the following with respect to x.
(a) cosh 3x (b) sinh 2 x (c) x sinh x (d) e x cosh x Solution
(a) ∫ cosh 3x d x = 3 sinh 3x + constant
1 (b) sinh 2 x d x can be found by using cosh 2x = 1 + 2 sinh 2 x
giving 1
2 ∫ (cosh 2x − 1) d x
= 1
1
sinh 2 x − x + constant
4
2 Alternatively, you could change to exponentials, giving
sinh 2 x = ∫ sinh 2 xdx = ( 1 2x
e − 2 + e −2 x
4 ) 1 2x 1
1
e − x − e −2 x + constant
8
2
8 Can you show this answer is identical to the one found earlier?
(c) Using integration by parts, ∫ x sinh x d x = x cosh x − ∫ cosh x d x = x cosh x − sinh x + constant 40 Chapter 2 Hyperbolic Functions (d) Certainly this is found most easily by converting to
exponentials, giving
e x cosh x = ∫e x 1 2x 1
e +
2
2 cosh x d x = 1 2x 1
e + x + constant
4
2 Exercise 2C
1. Differentiate with respect to x 4. A curve has equation
y = λ cosh x + sinh x (a) tanh 4 x (b) sech2 x (c) cosech ( 5x + 3) (d) sinh e x where λ is a constant. (e) cosh 3 2 x (f ) tanh (sin x ) (a) Sketch the curve for the cases λ = 0 and λ = 1. (g) cosh 5x sinh 3x (h) ( ) ( coth 4 x ) 2. Integrate each of the following with respect to x.
(b) cosh 2 3x (a) sinh 4 x
2 2 (c) x cosh 2 x
(e) cosech2 x coth 3x (d) sech 7x
(f ) tanh x (g) tanh 2 x (h) e 2 sinh 3x ( (i) x cosh x + 4
2 3 ) (k) cosh 2 x sinh 3x 4 ( j) sinh x
(l) sech x (b) Determine the coordinates of the turning
point of the curve in the case when λ = 4
. Is
3 this a maximum or minimum point?
(c) Determine the range of values of λ for which
the curve has no real turning points.
5. Find the area of the region bounded by the
coordinate axes, the line x = ln 3 and the curve
with equation y = cosh 2 x + sech2 x. 3. Find the equation of the tangent to the curve
with equation y = 3cosh 2 x − sinh x
at the point where x = ln 2. 2.7 Inverse hyperbolic functions The function f : x a sinh x ( x ∈ R ) is oneone, as can be seen
from the graph in Section 2.4. This means that the inverse
function f −1 exists. The inverse hyperbolic sine function is
denoted by sinh −1 x. Its graph is obtained by reflecting the
graph of sinh x in the line y = x.
Recall that y
sinh–1 x x d
(sinh x ) = cosh x , so the gradient of the graph of
dx y = sinh x is equal to 1 at the origin. Similarly, the graph of
sinh −1 x has gradient 1 at the origin. 41 Chapter 2 Hyperbolic Functions Similarly, the function g: x a tanh x ( x ∈ R ) is oneone. y
1 You should have obtained its graph in Activity 5. The range of
g is { y:−1 < y < 1} or the open interval ( −1, 1). y = tanh x 0 x –1 Activity 9
Sketch the graph of the inverse tanh function, tanh −1 x. y Its range is now R. What is its domain? y = cosh x The function h: x a cosh x ( x ∈ R ) is not a oneone function
and so we cannot define an inverse function. However, if we
change the domain to give the function
f : x a cosh x with domain 1
0 x { x: x ∈ R, x ≥ 0}
y then we do have a oneone function, as illustrated.
So, provided we consider cosh x for x ≥ 0, we can define the
inverse function f −1 : x a cosh −1 x with domain y = cosh–1 x { x: x ∈ R, x ≥ 1}.
1 This is called the principal value of cosh −1 x. 2.8 0
Logarithmic equivalents Activity 10
Let y = sinh −1 x so sinh y = x.
Since cosh y is always positive, show that cosh y = (1 + x )
2 By considering sinh y + cosh y, find an expression for e y in terms
of x.
Hence show that sinh −1 x = ln x + 42 (1 + x 2 ) 1 x Chapter 2 Hyperbolic Functions Activity 11
Let y = tanh −1 x so tanh y = x.
Express tanh y in the terms of ey and hence show that
e2 y = 1+ x
.
1− x Deduce that tanh −1 x = 1 1+ x
ln
.
2 1− x (Do not forget that tanh −1 x is only defined for x < 1. ) Activity 12
Let y = cosh −1 x, where x ≥ 1, so cosh y = x.
Use the graph in Section 2.7 to explain why y is positive and
hence why sinh y is positive. Show that sinh y = ( x 2 − 1) . Hence show that
cosh −1 x = ln x + ( x 2 − 1) . The full results are summarised below. sinh −1 x = ln x + tanh −1 x = ( x 2 + 1) 1 1+ x
ln
2 1− x cosh −1 x = ln x + ( x 2 − 1) (all values of x) ( x < 1) ( x ≥ 1) 43 Chapter 2 Hyperbolic Functions Exercise 2D
3. Express sech −1x in logarithmic form for 0 < x < 1. 1. Express each of the following in logarithmic
form.
3
(a) sinh −1 4 4. Find the value of x for which 1
(c) tanh −1 2 (b) cosh −1 2 sinh−1 x + cosh−1( x + 2 ) = 0. 2. Given that y = sinh −1 x, find a quadratic equation 5. Solve the equation satisfied by e y . Hence obtain the logarithmic
form of sinh −1 x. Explain why you discard one of
the solutions. 2.9 Derivatives of inverse
hyperbolic functions Let y = sinh−1 x so that x = sinh y.
dx
= cosh y but cosh2 y = sinh2 y + 1 = x 2 + 1
dy
and cosh y is always positive.
So dx
=
dy ( x2 + 1) In other words, dy
=
dx and therefore d
(sinh−1 x ) =
dx (x 1 2 (x 1 2 + 1) + 1) Activity 13
(a) Show that d
(cosh−1 x ) =
dx (b) Show that d
1
tanh−1 x =
dx
1 − x2 ( 1 ( x2 − 1) ) An alternative way of showing that
to use the logarithmic equivalents. 44 d
(sinh−1 x ) =
dx (x 1 2 + 1) is x −2
= ln 2.
2 tanh −1 x +1 Chapter 2 Hyperbolic Functions Since [ d
x+
dx ( x2 + 1) ] = 1 + 1 .2x ( x2 + 1)−
2
x = 1+ ( x2 + 1) = [ we can now find the derivative of ln x + { ( x2 + 1) + x
( x2 + 1) ( x2 + 1) ] ( x2 + 1) + x .
1
2
( x + 1) x + ( x2 + 1) ( x2 + 1) } = d ln x + dx 1
2 which cancels down to (x 1 2 + 1) d
(sinh−1 x ) =
dx So (x 1 2 + 1) You can use a similar approach to find the derivatives of cosh−1 x and tanh−1 x but the algebra is a little messy. Example
Differentiate
(a) cosh−1( 2x + 1) 1
(b) sinh−1 with respect to x ( x > 0 ) . x Solution
(a) Use the function of a function or chain rule. [ 1 d
cosh−1( 2x + 1) = 2.
dx (b) {(2x + 1) − 1}
2 = ( 4x −1
1
1
d −1 1 = − 2.
sinh x = x2 .
x
dx 1 + 1 2 x = 2 2 + 4x ) = (x 1 2 + x) x (1 + x2 ) −1 x (1 + x2 )
45 Chapter 2 Hyperbolic Functions Exercise 2E
7. Differentiate sech −1x with respect to x, by first
writing x = sech y. Differentiate each of the expressions in Questions
1 to 6 with respect to x.
1. cosh −1 ( 4 + 3x )
2. sinh −1 8. Find an expression for the derivative of cosech−1x
in terms of x. ( x) 3. tanh −1 ( 3x + 1) 9. Prove that
d
( coth−1 x ) = ( x2−1 ) .
−1
dx 4. x 2 sinh −1 ( 2 x )
1
5. cosh −1 x ( x > 0) 6. sinh −1 ( cosh 2 x ) 2.10 Use of hyperbolic functions
in integration
Activity 14
Use the results from Section 2.9 to write down the values of
⌠
(a) ⌡ (x 1 2 + 1) dx and ⌠
(b) ⌡ (x 1 2 Activity 15
x
Differentiate sinh−1 with respect to x. 3
⌠
Hence find ⌡ (x 1 2 + 9) ⌠
What do you think ⌡ d x. (x 2 1 + 49) d x is equal to? Activity 16
Use the substitution x = 2 cosh u to show that
⌠ ⌡ 46 (x 1 2 − 4) x
d x = cosh−1 + constant 2 − 1) dx Chapter 2 Hyperbolic Functions Activity 17
Prove, by using suitable substitutions that, where a is a constant,
⌠
(a) ⌡ x
d x = sinh−1 + constant a
(x + a ) ⌠
(b) ⌡ x
d x = cosh−1 + constant a
(x − a ) 1 2 2 1 2 2 Integrals of this type are found by means of a substitution
involving hyperbolic functions. They may be a little more
complicated than the ones above and it is sometimes necessary
to complete the square. Activity 18
Express 4x 2 − 8x − 5 in the form A( x − B) + C, where A, B and C
are constants.
2 Example
Evaluate in terms of natural logarithms
7 ⌠ ⌡
4 ( 4x 2 1 − 8x − 5) dx Solution
From Activity 18, the integral can be written as
7 ⌠ ⌡
4 1 {4( x − 1) − 9} dx 2 You need to make use of the identity cosh2 A − 1 = sinh2 A
because of the appearance of the denominator.
Substitute 4( x − 1) = 9 cosh2 u in order to accomplish this. So 2( x − 1) = 3cosh u and 2 2 dx
= 3sinh x
du
47 Chapter 2 Hyperbolic Functions The denominator then becomes {9 cosh2 u − 9} = (9sinh2 u) = 3sinh u
In order to deal with the limits, note that when [) and when
(
(so u = ln[4 + 15 ]) x = 4, cosh u = 2 so u = ln 2 + 3
x = 7, cosh u = 4 The integral then becomes
cosh −1 4 3
2 ∫ cosh −1 2 sinh u du
=
3sinh u = cosh −1 4 ∫ 1
du
2 cosh −1 2 [ { ( ( ) 1
1
cosh−1 4 − cosh−1 2 =
ln 2 + 3 − ln 4 + 15
2
2 Example
Evaluate ∫ (x
1 2 −3 ) + 6x + 13 d x leaving your answer in terms of natural logarithms.
Solution
Completing the square, x2 + 6x + 13 = ( x + 3) + 4
2 sinh 2 A + 1 = cosh 2 A this time
You will need the identity
because of the + sign after completing the square. Now make the substitution x + 3 = 2 sinh θ dx
= 2 cosh θ
dθ
When x = −3, sinh θ = 0 ⇒ θ = 0
giving ( When x = 1, sinh θ = 2 ⇒ θ = sinh−1 2 = ln 2 + 5
The integral transforms to
sinh −1 2 ∫ (4 sinh 2 ) θ + 4 .2 cosh θ dθ 0 sinh−1 2 = ∫ 4 cosh θ dθ
2 0 48 ) )} Chapter 2 Hyperbolic Functions You can either convert this into exponentials or you can use the
identity cosh 2 A = 2 cosh2 A − 1
sinh −1 2 ∫ (2 + 2 cosh 2θ )d θ giving 0 sinh −1 2 = [2θ + sinh 2θ ]0 sinh −1 2 = 2[θ + sinh θ cosh θ ]0 ( = 2 sinh −1 2 + 2 1 + 2 2 ( ) ) = 2 ln 2 + 5 + 4 5 Activity 19
Show that
d
(sec φ tan φ ) = 2 sec3 φ − sec φ
dφ
and deduce that ∫ sec 3φ dφ = 1 sec φ tan φ + ln (sec φ + tan φ )
{
}
2 Hence use the substitution
x + 3 = 2 tan φ in the integral of the previous example and verify that you
obtain the same answer. 49 Chapter 2 Hyperbolic Functions Exercise 2F
Evaluate the integrals in Questions 1 to 12. 15. Evaluate each of the following integrals.
3 2 ⌠
1. ⌡0 1 ( x2 + 1) 1 ⌠
3. ⌡0 ⌠
2. ⌡3 dx 1 ( x2 + 2 x + 5) dx 6. ∫1 ( x2 + 2 x + 2 ) d x ⌠
8. ⌡1 5 ⌠
9. ⌡4 0 ⌠
11. ⌡−1 ( x2 − 4 ) 2 1 ( x2 + x ) ∫ (3x
5 1 ( 2 x2 + 4 x + 7) dx 12. 2 4 + 6 x − 7) ∫ (x
3 dx (b) 2 ) + 6x − 7 d x 2 2 dx 2 ( x2 + 2 x ) ⌠
10. ⌡1 dx ( x2 − 9) 1 ( x2 + 6 x + 8)
1 (x 2 1 16. Transform the integral 3 2 x +1 dx ∫ ( x − 9) d x
4 2 0 2 7. 1 1 ⌠
4. ⌡−1 ∫ (4 + x ) d x
2 5. 4 ⌠
(a) ⌡2 dx dx 1
⌠
dx ⌡1 x 4 + x2 1
. Hence, by
u
means of a further substitution, or otherwise,
evaluate the integral.
by means of the substitution x = 17. The point P has coordinates ( a cosh t, bsinh t ).
Show that P lies on the hyperbola with equation
x 2 y2
−
= 1.
a2 b2 −12 x + 8) d x Which branch does it lie on when a>0? 13. Use the substitution u = e to evaluate
x 1 ∫0 sech x d x . Given that O is the origin, and A is the point
(a, 0), prove that the region bounded by the lines
OA and OP and the arc AP of the hyperbola has
area −1 14. (a) Differentiate sinh x with respect to x. 1
abt.
2 By writing sinh −1 x as 1 × sinh −1 x, use
2 integration by parts to find ∫ sinh −1 d x.
1
(b) Use integration by parts to evaluate
−1
∫1 cosh x d x .
2 2.11 Miscellaneous Exercises
.11 Miscellaneous Exercises 1. The sketch below shows the curve with equation
y = 3cosh x − x sinh x, which cuts the yaxis at the
point A. Prove that, at A, y takes a minimum
value and state this value. dy
= 0 at B, show that the
dx
xcoordinate of B is the positive root of the
equation Given that x cosh x − 2sinh x = 0. y Show that this root lies between 1.8 and 2.
B A 0 50 y = 3cosh x – x sinh x
y = 3 cosh x − x sinh x x Find, by integration, the area of the finite region
bounded by the coordinate axes, the curve with
equation y = 3cosh x − x sinh x and the line x = 2,
giving your answer in terms of e.
(AEB) Chapter 2 Hyperbolic Functions 2. Starting from the definition
1
1
cosh θ = ( eθ + e−θ ) and sinh θ = ( eθ − e−θ )
2
2
show that 7. Define cosech x and coth x in terms of exponential
functions and from your definitions prove that
coth 2 x ≡ 1 + cosech 2 x. Solve the equation sinh( A + B) = sinh Acosh B + cosh Asinh B. 3coth 2 x + 4cosech x = 23 There exist real numbers r and α such that 8. Solve the equation 5cosh x + 13sinh x ≡ r sinh ( x + α ). 3sech2 x + 4 tanh x + 1 = 0 3
Find r and show that α = ln .
2
Hence, or otherwise,
(a) solve the equation 5cosh x + 13sinh x = 12sinh 2
(b) show that 9. Find the area of the region R bounded by the
curve with equation y = cosh x, the line x = ln 2
and the coordinate axes. Find also the volume
obtained when R is rotated completely about the
xaxis.
(AEB)
10. Prove that 1
15e −10 dx
1
⌠
= ln ⌡1 5cosh x + 13sinh x 12 3e + 2 3. Given that x < 1, prove that ( ) 1
d
tanh −1 x =
.
1 − x2
dx Show by integrating the result above that
1 1+ x ln
2 1− x ∫ tanh −1 Given that
2 cosh y − 7sinh x = 3 and cosh y − 3sinh2 x = 2 ,
find the real values of x and y in logarithmic
form.
(AEB) 3 x dx. (a) 1+ t
(b) cosh x =
1− t2 2 ∫ (x
3 (b) 1 2 1 2 ) + 6x + 5 d x 12. Use the definitions in terms of exponential
functions to prove that 1
x
2 5. Solve the equations
1
3
(a) cosh ( ln x ) − sinh ln x = 1 2 4 (a) 1 − tanh 2 x
= sech2 x
1 + tanh 2 x (b) 2sinh x − cosh x = 2 tanh d
( tanh x ) = 1 − tanh2 x
dx Hence, use the substitution t = tanh x to find (b) 4sinh x + 3e x + 3 = 0
cosh x + sinh x = e ∫ ( x + 6 x + 5) d x
1 Hence, or otherwise, solve the equation 6. Show that and write down a similar expression for cosh −1 x. 1
x, prove the identities
2 2t
(a) sinh x =
1 − t2 2 11. Evaluate the following integrals Use integration by parts to find
4. Given that t ≡ tanh (1 + x ) sinh −1 x = ln x + (AEB) tanh −1 x = (AEB) ∫ sech2 x d x x n n
n−k
k
Deduce that cosh n x + sinh n x = cosh x sinh x k
k=0 ∑ Obtain a similar expression for cosh n x − sinh n x.
Hence prove that cosh 7x = 64 cosh 7 x −112 cosh5 x + 56 cosh3 x − 7cosh x (AEB) 13. Sketch the graph of the curve with equation
y = tanh x and state the equations of its
asymptotes.
Use your sketch to show that the equation
tanh x = 10 − 3x has just one root α . Show that α
lies between 3 and 3 1 .
3
Taking 3 as a first approximation for α , use the
NewtonRaphson method once to obtain a second
approximation, giving your answer to four
decimal places. 51 Chapter 2 Hyperbolic Functions (1 + x2 ) . 14. Prove that sinh −1 x = ln x + 19. Solve the equation 3sech 2 x + 4 tanh x + 1 = 0 .
(AEB) (a) Given that exp( z ) ≡ ez , show that ( ) y = exp sinh −1 x satisfies the differential
equation cosh y + sinh y 2 y = ln cosh y − sinh y dy
d
(1 + x2 ) d xy + x d x − y = 0
2
2 1 (b) Find the value of ∫ sinh −1 x d x, leaving your 0 answer in terms of a natural logarithm.
15. Sketch the graph of y = tanh −1 x.
Determine the value of x, in terms of e, for
1
which tanh −1 x = .
2
The point P is on the curve y = tanh −1 x where
1
. Find the equation of the tangent to the
2
curve at P. Determine where the tangent to the
curve crosses the yaxis.
y= 16. Evaluate the following integrals, giving your
answers as multiples of π or in logarithmic
form.
2 ⌠
(a) ⌡0 2 dx (3x2 − 6 x + 4 ) ⌠
(b) ⌡0 20. Define sinh y and cosh y in terms of exponential
functions and show that dx (1 + 6 x − 3x2 ) 17. Find the value of x for which
3
4
+ sinh −1 x = sinh −1
sinh
4
3
−1 18. Starting from the definitions of hyperbolic
functions in terms of exponential functions,
show that 1
By putting tanh y = , deduce that
3
1
1
tanh −1 = ln 2 3 2 (AEB) 21. Show that the function f ( x ) = (1 + x ) sinh ( 3x − 2 )
has a stationary value which occurs at the
intersection of the curve y = tanh ( 3x − 2 ) and the
straight line y + 3x + 3 = 0. Show that the
stationary value occurs between –1 and 0. Use
NewtonRaphson's method once, using an initial
value of –1 to obtain an improved estimate of the
xvalue of the stationary point.
sinh x
, express the value
cosh x
of tanh x in terms of e x and e−x . 22. (a) Given that tanh x = (b) Given that tanh y = t, show that y =
for − 1 < t < 1. (c) Given that y = tanh −1 (sin x ) show that dy
= sec x and hence show that
dx
π
6 ∫ sec x tanh
0
2 (sin x ) d x = 1 3 − 2 + 1 ln 3 3
2 −1 cosh ( x − y ) = cosh x cosh y − sinh x sinh y and that (AEB)
23. Solve the simultaneous equations tanh −1 = cosh x − 3sinh y = 0 1 1+ x where −1 < x < 1.
ln
2 1− x (a) Find the values of R and α such that
5cosh x − 4sinh x ≡ Rcosh ( x − α ) Hence write down the coordinates of the
minimum point on the curve with equation y = 5cosh x − 4sinh x
(b) Solve the equation
9sech y − 3tanh y = 7 ,
2 leaving your answer in terms of natural
logarithms.
(AEB) 52 1 1+ t ln
2 1− t 2sinh x + 6 cosh y = 5
giving your answers in logarithmic form. (AEB)
24. Given that x = cosh y, show that the value of x is
either
ln x + ( x2 − 1) or ln x − Solve the equations
(a) sinh2 θ − 5cosh θ + 7 = 0
(b) cosh ( z + ln 3) = 2 ( x2 − 1) Chapter 2 Hyperbolic Functions 25. Sketch the curve with equation y = sech x and
determine the coordinates of its point of
inflection. The region bounded by the curve, the
coordinate axes and the line x = ln 3 is R. 28. Prove that
16sinh2 x cosh3 x ≡ cosh 5x + cosh 3x − 2 cosh x
Hence, or otherwise, evaluate
1 ∫ 16sinh x cosh x d x, Calculate
(a) the area of R
(b) the volume generated when R is rotated
through 2 π radians about the xaxis.
26. Solve the equation tanh x + 4sech x = 4.
27. Prove the identity
1
cosh x cos x − sinh x sin x ≡ (1 + cosh 2 x cos2 x )
2
2 2 2 3 0 2 2 giving your answer in terms of e.
29. Show that the minimum value of sinh x + n cosh x
is ( n2 −1)
x= and that this occurs when 1 n − 1
ln
2 n + 1 Show also, by obtaining a quadratic equation in
e x , that if k > ( n2 −1) then the equation sinh x + n cosh x = k has two real roots, giving your
answers in terms of natural logarithms.
(AEB) 53 Chapter 2 Hyperbolic Functions 54 ...
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This note was uploaded on 01/31/2012 for the course PHYS 418 taught by Professor Somalwar during the Spring '11 term at Rutgers.
 Spring '11
 somalwar

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