YOUR
CLEAR ROUTE TO A-LEVEL SUCCESS ae
HEINEMANN MODULAR MATHEMATICS
EDEXCEL AS AND A-LEVEL
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ECIFICATIONS
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- GeoffMannall
Kenwood
7
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|nspiring genenitions
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West
Kent College
LIBRARY
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_ Geoff Mannall
Michael Kenwood a
1 Hyperbolic functions
2 Differentiation
3 Integration
4 Coordinate systems
rp
Review exercise
113
Examination style paper FP2
id
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© Geoff Mannall and Michael Kenwood 2005
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First published 2005
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13-digit ISBN 978 0 435511 O1 2
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Typeset and illustrated by Tech-Set Limited, Gateshead, Tyne and Wear
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Acknowledgements:
The publisher’s and authors’ thanks are due to Edexcel for permission to
reproduce questions from past examination papers. These are marked with an [E].
The answers have been provided by the authors and are not the responsibility
of the examining board.
About this book
This book is designed to provide you with the best preparation
possible for your Edexcel FP2 exam. The series authors are senior
examiners and exam moderators themselves and have a good
understanding of Edexcel’s requirements.
Finding your way around
To help to find your way
revising use the:
around
when
you
are studying
and
e edge marks (shown on the front page) — these help you to get to
the right chapter quickly;
e contents list — this lists the headings that identify key syllabus
ideas covered in the book so you can turn straight to them;
e index — if you need to find a topic the bold number shows where
to find the main entry on a topic.
Remembering key ideas
We have provided clear explanations of the key ideas and
techniques you need throughout the book. Key ideas you need to
remember are listed in a summary of key points at the end of each
chapter and marked like this in the chapters:
d
— (sinh x) = cosh x
dx
®
Exercises and exam questions
In this book questions are carefully graded so they increase in
difficulty and gradually bring you up to exam standard.
@ past exam questions are marked with an [E];
e the review exercise on page 113 helps you practise answering
questions from several areas of mathematics at once, as in the real
exam;
e exam style practice paper — this is designed to help you prepare
for the exam itself;
e answers are included at the end of the book — use them to check
your work.
Contents
1 Hyperbolic functions
1.1
Definitions in terms of exponential functions
1.2
Graphs of the hyperbolic functions
1.3
Identities
Osborn’s rule
1.4
Inverse hyperbolic functions
is
The logarithmic form of inverse hyperbolic
functions
BN
aD
Summary of key points
2 Differentiation
p54
The derivatives of hyperbolic functions
2.2
The derivatives of inverse hyperbolic functions
2.3
The derivatives of inverse trigonometric functions
20
Summary of key points
3 Integration
sel
More standard forms
3.2
Integration using identities
3.3
Some other methods of integration
3.4
A strategy for systematic integration
6
Be
Reduction formulae
3.6
Using reduction formulae in applications of
integration
Shy
Length of a curve
3.8
Area of surface of revolution
Summary of key points
34
Contents
4 Coordinate systems
4.1
The parabola
The focus—directrix property of the parabola
Equations of tangents and normals to a
parabola, and other properties
Equation of the tangent in cartesian form
4.2
The ellipse
Equations of tangents and normals to an ellipse
The focus—directrix property of the ellipse
The focal distances property of the ellipse
4.3
The hyperbola
Asymptotes
Equations of tangents and normals to a hyperbola
The focus—directrix property of the hyperbola
4.4
The rectangular hyperbola
Parametric equations, tangents and normals
4.5
Intrinsic coordinates and radius of curvature
101
Summary of key points
109
Review exercise
is
Examination style paper FP2
E34
Answers
133
List of symbols and notation
141
Index
145
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Hyperbolic functions
1.1 Definitions in terms of
exponential functions
The exponential functions e“ and e * can be combined to form
functions that have strong similarities to the trigonometric (or
circular) functions. These functions are called hyperbolic functions.
The hyperbolic sine of x, written sinh x, and pronounced ‘shine x’,
is defined by:
8
sinh x = }(e*-—e™*),
x ER
The hyperbolic cosine of x, written cosh x, and pronounced ‘cosh x’,
is defined by:
8
These
cosh x =}(e* +e),
two
definitions
are
basic
and
x ER
from
them
four
other
hyperbolic functions are defined as follows. The hyperbolic tangent
of x, written tanh x, and pronounced ‘tansh x’ is defined by:
a
tanh x =
sinhx
e*—e*
=
coshx
e*+e*
e?*—1
=
eee
e*+1
ER
The hyperbolic secant of x, written sechx and pronounced ‘sech x’,
is defined by:
co]
sech x =
The hyperbolic cosecant
‘cosech x’, is defined by:
cosech x =
5
The hyperbolic cotangent
‘coth x’, is defined by:
a
cothx =
1
tanhx
1
2
=
—,xeER
coshx
e*+e™*
of x, written
(ee
and pronounced
eeape
sinhx
e* —
cothx
of x, written
_coshx
=—
cosechx
sinhx
=
e*+e*
e*—e*
=
e*+1
e?*—1
and
pronounced
xEeR,
x40
‘s
2
Hyperbolic functions
1.2 Graphs of the hyperbolic
functions
You can see that:
e| -x)
sinh
—(—x)
(—x) =
;
ES ys mi e*
=
;
=
—sinhx
So sinh x is an odd function. Similarly you have:
ei)
cosh
ih e
x)=
(—.x)
(>)
7
eX re e*
5
5
= cosh xX
So coshx is an even function.
Also:
cosh x =
Gi
x
St
5
aX
=
e
x
—e
any
;
= sinhx
nN
for all values of x.
The curves with equations y = e* and y =e
Notice
that e‘ > 0 and
~ look like this:
e * > 0 for all values
of x. The
with equations y = sinhx and y = coshx look like this:
y = cosh x
y = sinh x
curves
Hyperbolic functions
2x
Since tanh x
=,
e* + ]
you can see that at x = 0, tanhx = 0.
;
Pie
Also:
tanh
=
(—x) = :
eel
bene
22s
Se”
tee
tanh x
So tanh x is an odd function.
2x
Now
tanh x =
ee
—2x
— 1 4 l-e~
ext]
[te
ax
As xX > oo, e ” — 0 and tanhx —
1
Asx
260, &* = 0 and tanhx = —1
The curve with equation vy = tanhx lies completely in the interval
—1 < tanhx < 1 and looks like this:
YA
The lines y = +1 are asymptotes to the curve.
Example 1
Sketch the graph of the curve with equation y = sechx,
Since
1
sech x = ———
coshx
1
= ————~
cosh(—x)
= sech(—x),
sechx
x
is an
ER.
even
function and so it is symmetrical about the y-axis.
Since coshx >1 then sechx lies in the interval 0 < sechx <1.
The curve with equation y = sechx looks like this:
y
3
4
Hyperbolic functions
Example 2
Find the value of x for which tanh x = 5.
Ox
er — |
You know that tanh x = ~——., so:
i
a
Taking logarithms to the base e gives
2% —in3
So:
3
Sin 3
1.3 Identities
Many identities exist for hyperbolic functions and these have
similarities to the ones you have already met in trigonometry.
Here are some examples. In nearly every case you can prove
hyperbolic identities by using the definitions given in section 1.1.
Example 3
Prove that cosh*x — sinh’ x = 1.
By factorising:
left-hand side = (cosh x — sinh x) (cosh x + sinh x)
From the definitions of coshx and sinh x:
eu Be eo
cosh:
eu az en
;
5
= sinh hoax =
x =i}
2
cosh x + sinhx =< ae
=
K
cae
2
Hence:
a
am
4x
ae
2
*)(e*) = |
cosh’ x — sinh? x = (e
cosh?x — sinh’?x = 1
a
Two further identities can be easily deduced from the identity
cosh’ x — sinh? x = 1. If you divide by cosh? x, you obtain
i
sinh x ae
cosh? x
1
cosh? x
that is:
a
1 — tanh*x = sech*?x
Hyperbolic functions
If x 4 0, dividing by sinh’ x gives you
cosh? x
=
sinh? x
1
sinh? x
=
40
that is:
rT]
coth*x — 1 = cosech’x, x +0
Example 4
Prove that
cosh(x + y) = coshx coshy + sinhx sinhy
x
SESE Dect oneal (Dec2
—Xx
y
-y
ee
a)
AL Z ey WO) (pclae2 |W po2
=e"? + 2a)
ad beer? ae e
+¥))
= cosh(x + y) as required on the left-hand side.
Thus:
cosh(x + y) = cosh x cosh y + sinh x sinh y
By writing x = y = A, you get:
F
cosh 24 = cosh? A + sinh’ A
which is another well known hyperbolic identity.
Example 5
Find an identity for sinh2A in terms of coshA and sinh A. Hence
find an identity for tanh 2A in terms of tanh A.
You have
sinh2A = 1(eA =e
4)
=tie* +e
“*)\(e* —e“)
”
et e-4|
(eA ae
-
y
y,
= 2cosh A sinh A
=
So:
sinh2A = 2coshA
sinh A
©
6
Hyperbolic functions
#)
tanh2A
=
sinh 2.4
,
)
by cosh*
A gives:
2 sinh A cosh A
=
cosh* A
7)
cosh* A_
}
cosh* A
w
5
cosh‘ A + sinh‘ A
Dividing numerator and denominator
=
y)
2cosh A sinh A
=
==
cosh2A
tanh2A
.
that cosh2A = cosh’ A + sinh’ 4.
In example 4, you were shown
You can now write:
tanh2A
=
nyt)
sinh” A
“
cosh* A
2 tanh A
5
1 + tanh’ A
Osborn’s rule
You notice that there is much similarity between trigonometric
identities and hyperbolic identities. Many are of the same form,
often with signs changed but not always:
trigonometric
hyperbolic
cose A-tsin: A = 1
sin2A =2sin Acos A
cosh? A — sinh? A = 1
sinh 2A = 2sinh A cosh A
Osborn’s rule gives you a simple way to remember when to make
a sign change when moving from a trigonometric identity to its
hyperbolic counterpart.
m
The rule is to replace each trigonometric function by the
corresponding hyperbolic function and change the sign of every
product (or implied product) of two sines.
The justification of this rule will be given in Book FP3.
Example 6
Give the equivalent
identity:
hyperbolic
identity
for
the
trigonometric
(a) cos2x =1—2sin’ x.
b)
eV
tan(A—
nan
tan A — tan B
3)=
)
1+tanAtanB
In (a), you have a product of sines: sinx sin.x, so the equivalent
hyperbolic identity is
cosh 2x = 1 + 2sinh? x
Hyperbolic functions
In (b), you have an implied product of sines in the term:
tan A tan 2S]
sin A sin B
=
cos A cos B
so the equivalent hyperbolic identity is
tanh A — tanh B
tanh
(A = 4) =
___pak
)
1 — tanh A tanh B
1.4 Inverse hyperbolic functions
You know that only one-one functions can have an inverse
function. Since sinh x is a one-one function, its inverse function is
arsinh x and the graphs of these functions look like this:
)
si
.
y = sinhx
y = sinh x
Similarly, tanhx is a one-one function, its inverse function
artanh x and the graphs of these functions look like this:
is
7
8
Hyperbolic functions
For the function coshx, you need to take the domain x > 0, so
that it is a one-one function. Then the inverse function arcosh x
is defined for the domain x >1 and range arcosh x >0. The
graphs of coshx and arcosh x look like this:
JA
y = cosh x
y = arcosh x
vA
10
|
x
y = arsech x
Example 7
In the same diagram, sketch the curves with
equations
(0, 1)
y= sechx, x eR, x > 0
y= arsechx, x eR, 0< x < 1
The curves are shown in the diagram. One is the
reflection of the other in the line y = x.
3
70
1.5 The logarithmic form of inverse
hyperbolic functions
Since hyperbolic functions are defined in terms of the exponential
functions
e‘ and
e “, it is possible
to express
their
inverse
functions in terms of natural logarithms, like this:
Write arsinh x = u, then:
sii
that is:
— 5(e ee")
e'—2x-e
=e
“=0
ee
1
This is a quadratic equation in e”. By adding x? to both sides,
you can write it as:
ot = yee
(e” =
=
x)
ad
x? ot. 1
s =x
Hyperbolic functions
Hence:
e“—-x= +4/ (x? a1)
a
6 = xe
1)
But e“ > 0 always, and therefore the negative sign can be rejected
because \/(x* + 1) > x.
So:
=xt/e
w= ln [x + </* 41)
That is:
%
arsinhx = In[x + \/(x? + DJ
Similarly, you can show that:
«
arcosh x = In[x + (x? — 1)], x > 1
a
artanh x= } In GS=) |x| <1
1
1-x
Example 8
Express (a) arsinh} (b) arcosh3 (c) artanh (—3) in logarithmic
form.
(a) arsinh } = In[¢+ /(%+ 1)] =InG+3) =In2
(b) arcosh3 = In[3 + /(3* — 1)] =In(3 + 2v2)
a5
(c) artanh (—?) =4In [=] = 41In(4) = —41n7
Example 9
Find the values of x for which
sinh? x + 5 = 4coshx
Since cosh’ x — sinh” x = 1, you have
cosh?
x
that is:
—1+5-—4coshx
=0
cosh? x — 4coshx + 4 =0
Factorising:
(cosh x — ay =)
SO:
Cosh x= 2
From the definition of cosh x in terms of e* you have
ete?
=4
Multiplying by e* and rearranging:
e* =4e*+1=0
4 + ees
(16
-4)
Se
SO:
Se
Hence:
x = In(2+ 3) or In(2 — /3)
Notice
that
the
equation
arcosh2 = In(2 + ,/3) only.
2
coshx=2
has
two
roots,
but
9
10
Hyperbolic functions
1 Express in terms of e:
(a) sinh2
(b) cosh}
(c) tanh (—3)
(d) cosh (,/2)
(e) sinha
(f) tanh 1 — tanh(-—1)
Find, to 3 decimal places, the values of x for which:
(a) sinh x =3
(b) sinhx=—3
(c) coshx =5
(d)* cosh =4/5°
(6)) tani 3
(f) tanhx = —
Find the value of each of the following, giving each answer to
4 significant figures:
(a) cosh4
(b) sinh4
(c) tanh (—2)
(d) sinh(—})
(e) coshn
(f) tanh (e)
Given that coshx = 3, show that sinh x = +4. Hence find the
values of e* and x.
Sketch, in separate diagrams, the curves with equations
(a), y = cosech x, x eR, x= 0
(bi y=cothx, x eK, x0
Give the equations of the asymptotes to each curve.
Sketch, in the same diagram, the curves with equations
y=sinn2x and y= sinh 3x.
Find the x-coordinates of the points where the curves meet the
line y = 2, giving your answer to 2 decimal places.
In questions 7—20, prove the given identity and, where appropriate,
check the identity independently by using Osborn’s rule when you
know the comparable trigonometric identity.
sinh A = — sinh(—A)
sinh 2A = 2sinh A cosh A
cosh2A = 2cosh’? A — 1
sinh 34 = 3sinh A + 4sinh? A
cosh 3A = 4cosh? A — 3cosh A
tanh” 4 + sech*4 = 1
sinh (4 — B) = sinh A cosh B — cosh A sinh B
cosh (A — B) = cosh A cosh B — sinh A sinh B
A+B
A-—B
COS =e
5)
cosh A + cosh B = 2cosh
sinh A + sinh B = 2 sinh
A+ B
5
Ne 5
cosh —,—
2 sinh A sinh B = cosh(A + B) — cosh(A — B)
:
cosh x — 1
coshx + |
= ‘tanh +x
Hyperbolic functions
19
;
2tanh 1x
sinh x = ——_+—
1 — tanh* $x
cosh x + sinhx + 1
20 ———
=
= wothx
cosh x + sinhx = |
2
21
Pp
Given that sinh x = tan @, 0 < @ < §, express coshx and tanh x
in terms of 0.
Given that x > 0, show that sinh (In x) =
‘
x-—]
Express cosh (In x) in a similar form.
23
Find the value, or values, of x for which
4sinh x — 3coshx = 5
giving your answer, or answers, to 3 significant figures.
24
Given that tanh t= 4, find the value of e”’. Hence find the
exact value of f.
2-3
Y
ie
Given that artanh x + artanh y = }1n5, show that y = ;
26
Given that y = In{tan(%+ 4.x)], show that sinh y = tan xand
=
=
2%
cosh y = sec x.
27
Given that sinh y = x, show that
y=In[x+
+ x?¥]
By differentiating this result, show that
oy=|
a+) : (S
28
Solve the equation 2 cosh x + sinh x = 2.
29
Solve the equation 13 cosh 6 + 12sinh@ = %.
30
Prove that cosh (x + y)=coshxcoshy + sinh x sinh y. Given
that acosht+ bsinht = Rcosh(t+«a), a> b> 0, show that
ae
a& =sln
ea
Find R in terms of a and b.
31
Using the definitions of sinh x and cosh x, in terms of e*, show
that for {x)= 1,
artanh += 10
32
Solve for x the equation
3sech*x +4tanhx+1=0
giving the root as a natural logarithm.
11
12
Hyperbolic functions
33
Solve the equation
cosh’ t + sinh? ¢ = 3
giving the answers in terms of natural logarithms.
34
Solve the equation 4 tanh ¢ — secht = 1
giving the answer in terms of a natural logarithm.
35
Solve the equation cosh 2x = 3 sinh x, giving your answers to 3 s.f.
36
Given that p = Sin 2, find the value of tanh p. Find also the
values of sinh 2p, cosh 2p and tanh 2p.
ST
Prove that coth A + cosech A = coth}A.
38
Given that x = sin@cosht and y = cos @sinht, find a relation
between (a) x,
39
yand@
(b) x, yandt.
Prove that cosh® 4 — sinh® A = 1 + 3sinh? 2A.
Hence show that 8(cosh® A — sinh® A) = 3cosh4A4 + 5
Differentiation
In chapter 1 you learned the definitions of the six hyperbolic
functions in terms of exponentials. You also learned how to
define the inverse hyperbolic functions in terms of their
logarithmic equivalents. In this chapter you will learn how to
differentiate hyperbolic functions, and expressions involving them,
as well as how to differentiate inverse functions, including inverse
trigonometric and inverse hyperbolic functions.
2.1 The derivatives of hyperbolic
functions
You know that ae
dx
:
;
d
=e” and —(e*)=-—e ”.
dx
Since sinh x = (&* =e 7),
Oe
d
=X
x
Gy
(sinh ») = 24,
© x —e*)=j(e
+e =i*)
But cosh x =4(e* +e *). So:
‘Ti
a
— (sinh x) = cosh x
dx
Similarly you have:
d (Osh) = 24,
Ocean
Gy
© 4 € 2) = 5(coee
— 6")
zg
—(cosh x) = sinh x
dx
As with the trigonometric functions, these results form the basis
for finding the derivatives
sech x, cosech x and coth x.
of the
other
four
functions
tanhx,
14
Differentiation
Example
1
Find Bd(tanh x).
dx
sinhx
oe so you can use the formula for
cosh x
differentiating a quotient. Remember that:
You know that tanhx =
dy
d
(“) zs) de =
dx
\v
an
v2
where u and vy are functions of x. So let
;
=
;
=
Differentiating:
du
dy
=-
coshx
v
sinhy
y= o0sh x
du
—=coshx
dx
dv
—=sinhx
dx
dy
"Gy
dx
SH
y=tanhx=
“ay _ coshx cosh x — sinh x sinh x
ve
cosh? x
2
.
cosh? x — sinh? x
cosh? x
But
cosh? x — sinh? x = 1
d
|
—
(see example 3 in chapter 1). So:
1
(tanhx) =
dx (
cosh? x
= sech’
.
Using a similar method, you can show that
d
1
—(coth x) = ————
ic
sinh? x
= —cosech? x
dx (
)
sechx
|
=——— = (coshx)!,
cosh x
Example 2
d
Find —(sech x).
dx
You
know
that
so
dy
4
dy
Ae foror yy== (cosh fo)
x)
y using h the chain rule =
As
—
you
dy dt
ee
If you take t = coshx, then you have
y=t!
dy
=
a
=f
t= coshx
a5)
= —(cosh x)”
dt
XO
aE ee
can
cy
sinh x-
find
Differentiation
dy
a
dy
iar
dt
ae —(cosh x)
e
i
coms
“(sinh
x)
sinh x
cosh? x
ae
sinh x
~
coshx
coshx
= —tanh x sechx
=
d
aon x) = —tanhx sechx
So:
Similarly, you can show that:
d
a
az (cosech x) = —coth x cosech x
You need to apply the rules of differentiation which you learned
in your Cl, C2, C3 and C4 courses to this work, so that you can
build hyperbolic and inverse hyperbolic functions into your stock
of skills and knowledge. The next two examples illustrate how
these skills are used.
Example 3
d*y
Given that y = cosx cosh x, find ant
x
.
d
Using the product formula a)
du
dv
= Me + ua
you get
d
:
= = —sinx coshx
+ cosx sinhx
x
Using the product formula again you get
2
d“y
= = —cosx
&'3
:
:
:
;
coshx — sinx sinhx — sinx sinh x + cos x coshx
= —2sin x sinhx
Example 4
A curve is given by the equations x = cosht,
y = sinht, where f is
a parameter.
(a) Find a cartesian equation of the curve and sketch the curve.
(b) Find an equation of the tangent to the curve at the point
where ¢—InZ,
15
16
Differentiation
(a) Since cosh’ t — sinh? = 1, you have
a
From the curve with equation
that coshx21>0.
So here
equation of the curve 1s:
y? at
y =coshx (page 2) you can see
x=cosht21>0.
A cartesian
x?—y? =],
x > 0
The curve is symmetrical about the x-axis and it looks like this:
YA
O
(1,0)
é
It is in fact one branch of a curve called a hyperbola which is
where the name hyperbolic functions came from originally. You
will meet this curve again in chapter 4.
(Db) eSince (ne
= —(e ),
=} =4)
— 2 and en
At this point:
x =cosht =4(e’ +e")
and:
y—sipht—sie =e.
=}24+))=3
)
=}2-)=3
Ca.
Also:
SO
—=
rr
sinh f,
dy
—_—=
Al
cosht
dy_dy di_ dy (dx)
dx
S fe):
dy
—
A
and at f=In2,
cosht
ee
=
d
dt
dx
dt\dze
1
except where sinh
t == 0
5
==+=$
:
4
The tangent at (7, 3) to the curve has gradient 3 and its equation is
y-3=$e-9
which can be arranged as 3y — 5x +4=0.
Differentiation
2.2 The derivatives of inverse
hyperbolic functions
If y =arsinh x, then sinhy = x. Differentiating with respect to x,
you get
dy
cosh hy
y
el
dx
cosh y
But you know that cosh? y — sinh? y = 1 and therefore
dy
|
l
dx Jl + sinh*y)
Remember
4/(1 + x?)
that the function arcoshx has the domain
x > 0 as we
have defined it. If y = arcoshx, x > 0, then
cosh y=
eee
Differentiating with respect to x, you get
dy
sinh hyp—=4
} rE;
uo
SO:
dx
|
sinhy
/(x?-1)
So we have the results:
d
'
ay Corsi =
1
oat,
and
d
1
ay Carcosh 5 3) Mee ea)
Notice that at x = 1, the curve with equation y = arcoshx (page 8)
has a gradient that is parallel to the y-axis, since \/(x~ — 1) = 0,
d
l
ee
and hence —(arcoshx) = ——.—— is infinite.
ax
J (x* — 1)
Example 5
Find an equation of the tangent at the point where x = —4 to the
curve with equation y = artanhx.
dy
First you need to find a
x
Since y = artanhx,
xX gives
x = tanhy and differentiating with respect to
dy
1 = sech? y —
* Ax
That is:
dy
——
dx
l
sech” y
17
18
Differentiation
But
sec
y= 1— tanh? y = 1 — x°, since tanh yp = x
dy
l
So Eee
=. That is to say:
Axi ® (xe
—2 (artanh
(ar
x) x) =—
:
dx
dy
Atx=-},—=
ek:
Va
l
een
=4
ix?
es
i
The tangent at (—4, —41n3) has gradient 3 and its equation is
In questions 1-20, differentiate with respect to x:
1 cosh2x
2
4 sech2x
5 cosech
4x
6 e*coshx
sinh? 3x
8 tanh’x
9 coth(Inx)
In(sinhx)
11.
12
10
sinh 5x
3 tanh3x
xsinh2x
x?cosh3x
13. In(tanhx)
14. esinhx
ee
pele
x
17 cosh x
[Soe
x
sh x
h
coshx
3
Hav 25%
x2
19
cose Ge)
x
20
In(tanh
x — sech x)
21
Given that y = arsinh(x — 1), find the value of = atx = Zz.
be
22
Find an equation of the normal at the point where x = In2 on
d
)
the curve with equation y = sinhx + 3coshx.
23
The curve with equation y = S5sinh x — 4cosh x crosses the
x-axis at the point A. Determine the coordinates of 4 and an
equation of the tangent to the curve at A.
24
Find the minimum value of y, where
y = 13 cosh x + 12 sinhx
and the value of x where this occurs.
25
The tangent at the point P with x-coordinate 2c on the curve
with equation
x
;
vy= ccosh —,
5 meets the y-axis at the point Q.
Find the distance OQ in terms of c, where O is the origin.
Differentiation
26
Find to 2 decimal places the coordinates of the stationary
points on the curve with equation y = 8 sinh x — 27 tanh x and
determine the nature of these stationary points.
27
Given that y = Acosh3x + Bsinh 3x, where A and B are
d“y
constants, show that q x-9)——9y= (0.
28 Given that y = artanh (4e*), prove that
4 _
(
e
raeas ee
6 0 ia 2») x =
has
e
0
sg
29
Given that y = cosh 3x sin x, find dy and s “
dx
dx
30
ar
dy
Given that ye* = sinh tx , Show that = =) atx =In3.
31
Find an equation of the tangent and an equation of the
normal at the point where x = 3 on the curve with equation
y= nh x.
In questions 32-47, differentiate with respect to x:
32
arsinhx
33
arcosh
5x
34
artanhx?
35
arsechx
36
arcosech x
37
atcoinzx
38
arsechx?
39
xarcoshx
40
ee
arsinhx
41
(artanh x)°
42
(arsech x)?
43
e* arsinhx
44
ma
arcosh x
46
artanh (sinh x)
48
Find an equation of the tangent to the curve with equation
45
artanh (sin x)
;
sin x
arsinh x
y = arsinh x at
(i) the origin and
(ii) the point where x = 1.
49 Given that y = (arsinh x)’, show that
(T+ x7)
50
d*y
dy
dx? "dx o)
Find an equation of the normal at the point where x = ; on
the curve with equation y = artanh x.
51
Given that y = arsinh x, show that
(a)
y=In[x+ V0 +x’)]
(b) (1x
2,
(dy ;
(2) =|
d?y
(c) Cee ENS ag
ay
= i ae
7
meee)
0
19
20
Differentiation
52
Show that the curve with equation y = 3cosh x — x sinh x has
a minimum point 4 on the y-axis. Find the coordinates of A.
Show further that the curve has another stationary value
between x =
1.9 and x = 2. Sketch the curve.
53
Show that y = e""""~ satisfies the relation
(+
2) dy hs dy
0
x*)—
+x
y=
aba
dx
54
Given that y = sinh x + k cosh x, show that the least value of y
is /(A* — 1) and that this occurs at x = 5 In ag
where k is
a constant and |k| > 1.
mnNn
Show that (cosh x + sinh x)* + (cosh x — sinh x) = 2coshkx,
where k is real.
Hence solve the equation
.
5
;
5
(cosh x + sinh x)? + (cosh x — sinh x) = 5
giving your answers to 2 decimal places.
56
Find the coordinates of the minimum point on the curve with
equation y = Scoshx — 3sinhx.
57
dy
2
Given that tan y = e*, show that ae =1sech x, and find
2)
6
a
dy
dx2-
58
For the curve with equation y = arsinh (x 4+ 1), find
(a) the coordinates of its point of inflexion P
(b) an equation of the normal to the curve at P.
2.3 The derivatives of inverse
trigonometric functions
You first met inverse trigonometric functions in Book C3, chapter 6.
Here is a reminder of the shapes of the curves with equations
7 = sin x
(-%3 <x<
v= cosy
(Os X= 7)
=
(—§ <o< 4)
tane
4)
Differentiation
together with the shapes of the curves with equations
y =aresinx(—1
<x <1)
y=arccosx(—l
<= x < 1)
y = arctan x(x € R)
ye sine
|
y = arcsin x
a .
y =
arccos x
y = arctan x
'
'
Nla'
1
f
'
(
1
i}
1
'
'
'
!
'
'
'
|
i)
Consider y = arcsinx,
—1 <x < 1, and for which =F Sy
in y= arcun.x, then sin'y = x:
Using the chain rule, differentiate with respect to x to obtain
dy
cosy — = 1
x
dy
4
SO:
dx
cosy
21
22 =~ Differentiation
Since cos” y + sin’? y = 1, cosy =+/(1- sin’ y).
So:
cosy = 4/1 sin’ y)
(We take the positive square root only, because cosy > 0 in the
mterval —— = y= 5.)
9
.
.
Hence cosy = ,/(1 — x*) since x = siny.
w
d
}
— (arcsin x) =
So:
dx
1
J — x?)
Consider y = arccosx, —1
<x <1 and for which O <y <7.
You have cos y = x, where
0 = y <7.
Differentiating with respect to x:
ky
— sin y— =
x
dy
l
nce —
:
that is:
0
dx
3
But sin“ y+ cos’
2
siny
;
D
y = 1, so siny = ,/(1 — cos’ y) and:
siny = /(1 — x’)
because siny > 0 forO<y<xz.
gw
1
d
— (arccos x) =
dx
So:
yad==)
You should notice also that the gradient of the curve with
equation y = arcsinx, shown on page 21 is positive; the gradient
of the curve with equation y = arccosx, shown on page 21, is
dy
negative. These signs agree with the signs of a just derived for
x
y = arcsinx and y = arccos x.
Given that y= arctan x,
2
then:
xER
a
5
SV,
tany = x, where
é
—F<y<%.
Differentiating with respect to x, you get
sec’ y ee
Pe
that is:
But
dx
=,
l
dy
—_
dx
sec” y =i
sec*y
tan ie
lea x?
and this gives
‘
“ (arctan x)
—(arctan x) =
dx
|
+x?
Differentiation
Exampie 6
|
oa
;
(a) Let
t= x7, then y = arccost,
Find ee when (a) y = arccosx (b) y = arctan (e*).
dy fe
Sa
dt
=]
di
/@=P)
dy
dy
dt
de
diode, (41
dy
=|
Z
J —x4
By the chain rule:
(b) Let
=2x
3)
w=e**, then y= arctan v,
du
3,
=o
dx
dy
|
— =<
du
l+u*
l
1+e%
By the chain rule:
Example 7
Find an equation
dy
dy dz
3c"
dx
dudx
1+e*
of the
normal
to the curve
with
equation
y = arcsin 2x at the point where x = "
y =-atcsin 2x, so:
dy
l
a
=x)
dx (1 — (2x)*) dx
2
sae
Aix =7. yates t—2
and:
ay)
dy
2
dx
J(1-l)
aac
v
The gradient of the normal at (2) is a2) An equation of the
normal is
1
Differentiate with respect to x:
(a) arcsin 3x
2
(b) (arcsin xy
Differentiate with respect to x:
C
(a) arccos (=)
(b) arccos (3x7)
(c) arcsin (5)
(c) arccos (<3)
23
24
Differentiation
3 Differentiate with respect to x:
4
Differentiate with respect to x:
rar
=
(d) arctan
‘a
l =
v
x
(e) arcsec x
1+ /x
7
Given that y = arcsin x, show that:
(
6
e
(¢
Farr
(c)
arn
x
x
(Bb). oe XY arccos
(a) xarcsin x
5
(In x)
(c) arctan
(by etane
(ay arctan (2x)
]—x*
pad
:
dx?
ays
x—
dx
= 0
Given that y = x — arctanx, show that:
d*y
dy\
——2x{1—-—]=0
ae
7
8
( =]
d
(b) re (arccot x).
x
Find (a) z (arcsec x)
8
2
Find an equation at the point where x = V2)of the tangent to
the curve with equation y = arccosec x.
9 Given that k is a positive constant, differentiate with respect to x:
x
(a) arccos a
k
10
k
(b) arcsin —
x
(c)
3
arctan é
B
l
Given that y = arctan (:= -), prove that
x
Cy
dx
11
xt— x24 |
The tangent at the point (i, 4 on the curve with equation
y = arcsin 4x meets the y-axis at the point P. Prove that
OP =%- 3 where O is the origin.
SUMMARY OF KEY POINTS
l (sinh x) = cosa x,
|
Sheaatay=
at
ann x) = sech” tech? x
* (cosh x) = sinhx
q, oth x)| = —cosech* A x
d
—(sech x) = —tanh x sech x
dx
d
qx (Cosech x) = —coth x cosech x
Differentiation
ys & (arsinh x) =
dx
1
Jee +1
- (arcosh x) = Spee
d
me (artanh x) = ae
5
(arcsini x) = ae
d
—|
a: (arccos x) = Ai
a (arcian x) =
dx
eo
l
b+ x?
25
Integration
3.1
More standard forms
Books Cl—C4 described integration as the inverse or the reverse
process to differentiation. In chapter 2 of this book you learned a
number of important results in differentiation concerning inverse
trigonometric
functions,
hyperbolic
functions
and _ inverse
Bea |
hyperbolic functions. Generally, if qn (8) = f(x)
o
then:
|
f(x) dx = g(x) + C
where C is a constant.
Using this general result, you can add the following integrals to
your list of standard forms as a direct consequence of the
derivatives already found in chapter 2.
Derivative
Integral
*
d
4x oo x) = sinhx
x
=
© (sinhx) = coshx
x
E] © (tanhed
x
sech” x
a
© (aresin x) =
.
1
v(l — x?)
m=
d
—(arctanx) =
dx
cae
=
d
—(arsinh x) =
dx
=
d
—(arcoshx) =
dx
1
V(l + x?)
1
J (x? — 1)
|
|
sinh
x dx = coshx + C
cosh x dx= sinhx + C
sech? x dx = tanhx + C
a=
lr
Va cy
a
V(x?
l=
J(l —
dx = aresinx + C, |x| <1
dx = arctanx + C
:
= arsinh x + C
= arcoshx + C,x > 1
28
Integration
You should note that in your Edexcel formula book the following
are given (with the constants omitted);
1
lass dx = aresin (=), |x| <a,a>0
J (a? — x?)
i
;
loo
x
dx =
arsinh
=), (ime oAW)
laece
dx =
arcosh
(=), x Sa
]
l
3%
a+x-*
a
a
]
SO
low dx = -—arctan (=i a)
You can easily check
the chain rule.
these four results by differentiation
using
Example 1
Show that
]
|=
5
= dx =
“4+ x-
-
Zarctan (*)ae 8
a
a
l
a
x
Let y = — arctan (*) and t=-,
a
a
a
1
then:
also:
;
y =— arctan? and dy =
a
Ge
a =
,
dx
So
—
wl
ard
ews
a
aye
:
you get
5
ang
Farctan (*) =
d
(
a
;
dy
“dy dt
U sing the the chain
ch
rulejc, Ae ==...
a ae
Cele
aN
~
=|
a
nx
YG
@
=
5
aw +
x-
l+—ee
:
That is:
l
Xx
OF a= J
a
\o— dx =-7 arctann(*) ac
Example 2
Find (a) | == 5 tes Xe
Ves
b |
Os
ae dx
3
Integration
In (a), use the result
1
es
= 5 dx = areosh(*) +c
and take a = 3 to give
I
lFe =)
dx = arcoshix+C
In (b) you need to rewrite \/(25 — 4x) in the form 2y/[G)'-x7|
so that the result
lies ‘aii aresin(*
*)+C
J(@* -
a
can be used with a = 3
1
Hence:
lana! i
/(25 — 4x?)
|
:
l
v|(6P-x?|
=} arcsin
dx
; +C
2
= 5 arcesin?x+C
Example 3
:
3
Evaluate
l
|—_——
to 3 significant
Notice that /(1 + 4x7)
=2/(1+. x7), and so you can write the
ne
easdx,Ca giving ge your answer
figures.
integral as
2
]
1
x
js
i)G+)” vs (|
=! Jarsinh 2x],
= }(arsinh 4 — arsinh 2)
= 5(2.0947 — 1.4436)
= 0.326 (3 significant figures)
29
30
Integration
Example 4
The region R is bounded by the curve with equation y = coshx
and the line y = 3, as shown in the diagram. Find the area of R,
giving your answer to 3 significant figures.
First you need to find values of x at A and at B, that is, at cosh x = 3
=
e€t¢e*=6
or
e*-6e*+1=0
Use the quadratic formula to find e*:
a = Sue—4
= 3422
Then by taking natural logarithms you get
x= —1763 or 1.763
The points P and Q lie on the x-axis
x-coordinates as A and B respectively.
and
have
the
same
Area of R = area of rectangle 4POB — area under curve between
A and B and above x-axis
1.763
= AP
ero = |
cosh x dx
~1.763
=o
e0
[sinh x| ee
= 10.578 — sinh 1.763 + sinh( — 1.763)
= 4.92 (3 significant figures)
3.2 Integration using identities
You can use hyperbolic identities to replace an expression that
you cannot integrate directly by one that you can integrate. This
was done with trigonometric identities in Book C4 (chapter 6).
Example 5
Find
(a) |cosh* 2xce
(1b)
|tanh*seds.
Integration
(a) There are two possible methods.
In the first you can use the identity
cosh 24 = 2cosh* A — I
and take A = 2x so that:
cosh? 2x = $(1 + cosh 4x)
Hence:
|cosh’ 2xdx = | (1 + cosh 4x) dx
= 5x - ;sinh 4x +C
In the second, use the exponential form of
cosh 2x = 1(e** ie)
and square the identity to obtain:
cosh? 2x = 1(e* eer)
Hence
|cosh? 2207 = 7 (er ae)
dx
=te*+4x-he*+C
which is equivalent to the first answer, since sinh 4x = 1(e* —e fay
(b) In the identity sech* 4 = 1 — tanh’ A, put A = x and then
tanh? x = 1 — sech? x
So:
|
tanh? x dx = |
(1 — sech?x) dx
But |sech’x dx = tanh x + C is a standard result, given on page 27.
So:
|tanh* xdx = x — tanha.+-C
Example 6
Find
(a) [tanhxd
(b)
|sechxa
inh;
(a) Use the identity tanhx = ee
cosh x
Then:
é
Sink x
|
(snubadxy = |
ax
} cosh %
Let coshx = ¢t; then, by differentiation, sinh x fr=]
sinh x
Bo:
sith x dx
= |——
Fee 2
|r
— dt= |-d
dt
\;
=Int+C
That is:
[tanhvds = In(coshx)+C
31
32
Integration
(b) Use the identity sechx =
|
sech xdx =
=)
A
. Pnen:
cosh x
l
J cosh x
dv=
COs x
: —— dx
J cosh x
2
But cosh” x = | + sinh* x, so you have
cosh x
|
sech xdx = |
——
¢
}
J 1+ sinh* x
;
Write
sinhx =f,
dx
coshx—-=1
then
written as
the
'
integral
Le!
|
can
be
;
given
on
dt
cosh
|sechwe =
But
and
: —
} 1 sinh-x
dx
dz
iam
’
| qe de = arctan + € is a
l=
|
~ df
jJi+f
standard
result
page 27.
So:
|
sech x dx = arctan t + C = arctan (sinh x) + C
Integrate each of the following with respect to x:
1 cosh 2x
2
sinh3x
3
—sinh 5X
4
e* coshx
5
e* sinhx
6
sinhxcoshx
7
cosh’ x
8
sinh? x
9
sech* 2x
10
tanh? 2x
11
coth? 5x
12
cosech 2x coth 2x
13.
tanh3xsech3x
14
i
—W——
15
l
———
16
l
———_
V(x? — 4)
17
|
—
x-+4
18
l
—
x-=—4
19
|
—__—___
20
l
————_—
2
22
nee
/ (64 + 9x?)
23
:
9x? + 64
24
!
J (16x? — 1)
|
a
Ta + 16x?)
26
s
2%2x
sech
ny
anh
2x
Sram
28 cosh?3x
29 sinh?
2x
/ (25 — x?)
J/(4 — x7)
/(25 — 9x?)
J/(4+ x?)
1
/ (16x? — 9)
30 tanh* 5x
Integration
Evaluate, using integration, giving your final answer to
3 significant figures:
»]
31
]
|cosh 2x dx
0
3
34
40
35
eaaeES 4
33
1
|tanh x dx
0
03
=
——
eee
1
Se
IAiea es
39
|e cosh 2x dx
»]
GX
————
38
iA+ x2 dx
l
———-
IV6
41
3
d
‘
2
|sinh 5.x dx
0
l
2
37
32
X
of
,
|e‘sinhxdx
x
.
dx
=x)
:
Find the area of the region bounded by the curve with
equation y\/(25 + x*) = 25, the x-axis and the lines x = +5.
42
Find the area of the finite region bounded by the curve with
equation x7 — y* = 4 and the line x = 5.
43
Find, giving your answer to 2 decimal places, the area of the
finite region bounded by the curve with equation y = sech x,
the x-axis and the lines x = 1 and x = 4.
44
Use the identity sinh x= 5(e* —e ~*) and the substitution
v =e’ to find |cosech a:
45
d
Show that ce [In (sinh x)] = coth x.
x
Hence find the area of the finite region bounded by the curve
with equation y = coth x, the x-axis and the lines x = 2 and
ae,
46
The region R is bounded by the curve with equation
|
y= Vor +9)’ the x-axis and the lines x = +2.
(a) Find the area of R.
The region R is rotated through 360° about the x-axis.
(b) Find the volume of the solid generated.
47
Acurve is given by the equations x = sinh? ¢, y = sinh, where
t is a real parameter. The finite region R is bounded by part of
the curve, the x-axis and the lines
x = 1 and x = 3. Find, to
2 decimal places, the area of R.
48
Use the substitution
1
x = sinhw to show that
|VL + x’)dx = 35/2 +5ln[l + V2]
0
33
34 = Integration
49
The finite region R is bounded by the curves with equations
y =coshx, y = sinh x, the y-axis and the line x = |. Find the
area of R.
50
A curve C is given by x = 2cost, y = sint, 0 <t < 27, where ¢
is a parameter. Find the area enclosed by C.
3.3 Some other methods of integration
Integrals of the types
l
————
dx,
px°- +qxt+r
l
dx,
/(px* + qx +1)
5
where p, g and r are constants can be integrated by using one of
the standard forms, where C is an arbitrary constant and a is a
positive constant.
1
!
1
dx = — arctan
x? + a
a
|
ca +C
a
x—a
|
+ C, where |x| >a
x+a
3
|
1
1
a* — x?
dx = 5 In|“
2a
+C, where |x| <a
la-—x
dx = arcsin (5)+ C
a
_
dx = arsinh (=)+C
1
x
ra
dx =
arcosh
(=)ae GC
Integrals 1, 4, 5, 6 have been discussed in section 3.1. Integrals 2 and 3
are obtained by first using partial fractions
It is important that you can complete
expression with confidence.
(Book C4, chapter 1).
the square
in a quadratic
The following examples illustrate the methods required.
Example 7
Find
(a) |=
x“
l
—2x —3
dx
caer Ee
(b) los
dx.
(a) Notice that x? -2x-—3=x?-2x+1-4=(x-1Y-4
So:
|
jaa
x“ —2x—3
,
=|
JOe—1)" =4
Integration
Peer
eg
Then:
1
ae
He
du
L
a
Oi
zo
3
iy = |
ae
Jaye
If you compare this with the standard result for | 5s,
have:
pa
]
:
dx == 41
eee
YOu
u—2
a nfpos|+c
x-—3
= 1n|~
qin iat
Se C
You could also get this result
using partial fractions.
by factorising
x7 —2x—3
and
(b) Notice that x? —2x+5=x7-2x4+14+4=(x-1) +4
-
a
ee
l
l
x-—2x+5
(x=1)" +4
:
dx
Again, as in (a), putx—1=wand— =.
u
1
as :
Wax
———
ee
os
-
|
lFemeces
:
|
1
ea
oe
,
= 5 arctan su + C (see standard result 1 on page 34)
al
=} arctan (>) +C
Example 8
]
Find |
Viegas
Notice that
Then:
Put
dx.
Ve
x?—xtl=x?-x+143
Sashes
|
ee dx
J(x* — x +1)
d
x —4=u, fie
du
35
36
;
0:
Integration
|
eereceay
x=
1
dx
ersecs du “
= arsinh
l
-
7, du
eerec ay
wm)
+c (see standard result 5, page 34)
2
= arsinh (=) +C
In examination questions, you will often be given a substitution to
use when evaluating an integral.
Example9
Use the substitution
4.5
|
x = 3(1+sin?‘)
to
evaluate
the
integral
| ey,
3
(6x = xe
dx
For x=
and:
3+ 3sint,
— = 3coss
6x
— x7 = 18+ 18sint — 9(1 + sin 1)
= 184 18sint—9— 18sint—9sin? 4
= 9 —9sin?t
= 9(1 — sin” t) = 9cos’t
So:
At
(6x — xp =Bicost
x =3, snt=0=>1r=0
Atx=45, sint=5>1=2
4.5
Hence:
I
| Sea
3 (6x
— x*)?
z
_ dx
l
:
0(6x
— x2 dt
ee f
5
a
=| son
3
seca
d
= | I dr
0
=(ib=3
You will often need to do integration by parts to find the integrals
of inverse trigonometric and inverse hyperbolic functions. Here are
some examples of this.
Integration
Example
10
Use integration by parts to find |arsinh x dx.
Book P3 (page 115) gave the integration by parts formula, which is
dy
du
[>e dx = uv — jus dx
where wu and vy are functions of x.
You know also that sil(arsinh x) = ee aan: (page 17) so in the
Vd + x?)
dx
integral |arsinh x dx, you can take
Then:
en
v = arsinhx
and
ae ==} |
dx
dy
1
——
a= SS Se: 2)
nd
Uf = ax
Using integration by parts:
arin
|
x
a
ea oe)
0 = Lain.
Now you need to find | aos dx which can be done by writing
Lx
a)
4.
= 1, 80 that:
VAG! +
x*)
dx
2xx —=2
a
—
Then:
So:
[ope
=
dx
x—=t
as aE
[peste ete
ca
vatsiec
|arsinh x dx = xarsinhx —/(1+x’)+C
Example 11
:
Giving your answer to 3 significant figures, find the area of the
region bounded by the curve with equation y = artanh x, the x-axis
and the line x = 5.
The diagram shows the region whose area is required. It is found
1
y
by evaluating the integral |artanh x dx.
0
d
You know that — (artanh x) =
dx
Praartanh x.
So:
o =
dx 1—x?
l
(page 18), so you take
oe
x2
and
d
pace l
dx
and
ex
37
38
Integration
Using integration by parts you obtain
artanhxdx = xartanh x — |
bg
a dx
= xartanhx
+ $1n |1 — x |*C
(The integral F
= dx consists of a function and its derivative,
the function being (1 — x’).)
1
2
le
Area of region = [xartanh x+ $In |1 —
=f artanh }+3In3— (0+3In1)
= 0.2747 — 0.1438
= 0.131 @ significant figures)
In questions 1—24 integrate the given expressions with respect to x:
1
l
l
1
2 ——
3
x = Dx
f(x? — 2x)
\/(2x — x?)
.
1
x*+x+1
o)
-
1
1
a
J/(x* +x + 1)
x(x?+ 1)
cosh x
7 ——
1 + sinh x
§
sinh x
———-—_
coshx — |
hee
hes
bx
gona
2x
+6
(ec
W
x—2
Beas
ae
12 Bik xcae
a+ Map a
14
arsinh
rsinh
x
15
x? +6x+1
B
Gx
Dx
x*—4x—2
(x2 +1)
(1 + x2)
(1 — x2)?
160
4x+7
————
17
i
~~~
18
19
2x — 3
——_—_
V(x? — 9)
20
4x —]
————_—~
/(x? + 16)
21
22
/ (2x? + 7x + 3)
l
5
J/(1 — x — x?)
Zo
2x? + 7x +3
|
13 —4x+ x?
arcsinin. i
24
x7 4+8x+17
l
—~—~
4x2+4x+2
l
/(5 — 4x — x?)
Use (a) differentiation (b) integration by parts to prove each of
the following where C is a constant:
25
|arcsinx dx = xarcsinx + (1 — x2) =k
26
|arccos x dx =x arccosx — (1 — xy +C
Integration
27 | arctanxdx
= xarctanx
—4In(1+x7)+C
28 | arcsec x dx = xarcsec x — In [x + V(x? — 1)] + C
29
|varetan.xdy = 4$(x* + Iarctanx —}x+C
30 |xarcosh
xdx = }(2x? — 1) arcosh x — $xV/(x? -1)+C
31
Use the substitution e* = u to find
|
sech x dx
32
Show that |sec x dx = In(sec x + tan x) + C, where you may
assume that secx + tanx > 0.
2
P
f
sec’ x + sec x tan x
eat
Hint: write sec x as ———————— and use the substitution
sec x + tan x
u = secx + tan i
33
Use the substitution cos x = u to find
sin x
34
Use the substitution ¢t = tanx to find
1
lire
35
Use the substitution u = e* to find
=
Scoshx+4sinhx
dx
~
In questions 36-50, evaluate the definite integrals using integration,
giving answers to 3 significant figures.
13
l
—_—_——_d
sta gees
13
.
»VO+ x?)dx
i
37
9
]
—
eae
ae th
[7 .
4
x
| ae
fo
41
42
_sec Dax
43
|
ear
[cosec x dx
1%
39
40
Integration
44
|arsinh 2x dx
45
iarcosh x dx
46
[arcsinx dx
47
| arctan x dx
V3
8
V3
48 |G16
a
a
2
49 | 3 dx
2)"
3.4 A strategy for systematic
integration
You
should make a list of all the standard integrals, often called
standard forms, that you have met so far in your Pure
Mathematics course. These standard forms include those given in
Books Cl1—C4 as well as those learned in this chapter. A list of
standard forms is given in the Summary of key points at the end
of this chapter. You should appreciate, however, that this list 1s
not exhaustive.
You should learn and memorise standard forms. Only a few
standard forms are given in the Edexcel formulae booklet
provided in your examinations.
At this stage of your work, you should take stock of all the
methods and results that you have studied. Any one of the
integrals you have learned may be required in the examination,
and so here is a clear strategy for you to use.
(a) Whenever it is possible, obtain an integral directly by
comparing it with and using the appropriate standard form.
(b) Consider the replacement of the integrand (that is, the
function to be integrated) by an identical function which you
know you can integrate. Partial fractions in algebra and
trigonometric and hyperbolic identities can be particularly
effective in replacement strategies.
(c) Consider simplifying the integration by changing the variable
using a substitution.
(d) Consider using integration by parts.
With regular practice you will find that you learn to choose
quickly the most favourable method in each case. Here are some
more examples.
Integration
Example
12
‘ |Cea
2,
ae
First you need to express
.
.
l
1
x
eal Cap
ac
: ls r+ |
¢ —
ds
‘i
s
j
j
a
this:
(x + 1)(x? +1) in partial fractions like
2
oe.
eee
(oe DiGe ET) 24d
ee
where A, B and C are constants.
That is:
2=A(x* +1) 4+(Bx + Ce
+1)
Put x= —1, then 47]
Put
= 0) then 2 = 4
= C=
Equating coefficients of x? gives; 0 = A+ BS B=]
Hence you get:
Zz
=
ee!
(xt DGDxe]
14x
2
0 |ee
< lemme
You should recognise at once that
|
l
dx =In|x+1|
xd
and
|
1
1+x
; dx = arctan x
The third integral Feet
should be recognised as a function
oy
:
mre
:
dx
of x? and its derivative 2x. So you write u = x and |= 2) —
du’
Then:
x
gp
jae
fy ee
omega
eae
le
=—1
l
ieee
=4|
1
l+u
hs
dx
d
a) i
du
=1In|1+u| =41n|1 + x?|
So you have:
lerner’
@+ D640
x =In|x+ 1] + arctan
x — 5In|1 ie PEG
41
42
|ntegration
Example
13
(A)
should
and you
| oa
aed
1 + sinhx
of sinhx and its derivative coshx.
and then cosh es ==
du
cosh x
iS:
—_—
a
|1 + sinh x 2
dx =
=
(b) [ovtax
(a) |»dx.
Given that y = Peceliee find
1+ sinhx
a function
recognise
So you substitute sinh x = wu
l
|—————_-
dx
hx—)d
|;+ sinhx (cos a) i
l
lm
d
=In|jl+ul/+C
= In|) sinh
inhx
(b) You have y! = I+s
cosh x
+ ©
|
sinh x
coshx
coshx
= sechx + tanh x
So:
[tas a} |sechxedx + |tanhvdx
both of which are standard forms.
[tax = 2arctan(e*) + In(coshx) + C
In examples
h
| Se
1+ sinhx
12 and
'
D)
13 you met the integrals 4, Te
ae>dx and
These are known
:
.
eee
as ‘a function and its derivative’.
After practice, you will find that you can write down the answers
+In {1 +x} and In|1+sinhx| without any working. When you
have sufficient confidence, you should do this and mentally check
by using differentiation of your answer.
Example 14
The diagram shows the curve C given by
x= COsny,
yp=sinht.
7-0
where ¢ is a parameter.
The finite region R is bounded by part of C, the x-axis
and
the line
x= cosh2.
At 4, ¢=0
and
at B.7=2.
Find the area of R and the volume generated when R is
rotated through 360° about the x-axis.
Integration
2
Area of
So:
The
Area of
R= [ras = |rae
dt
2
2
0
0
R= |(sinh )(sinh ¢) dt = |sinh? t dt
identity cosh2¢= 2sinh?¢+1
is used
to replace sinh?/ by
$(cosh 2t — 1).
Then:
area of
2
R= 1](cosh 2¢ — 1) dt
0
= [4sinh21 a,
= fsinh4 — | © 5.82 (3 significant figures)
2
2) dx
Volume generated = x |
y?dx = n| y —dt
dt
0
2
= n| sinh? ¢ sinh ¢ dt
0
The
identity
cosh? +—sinh?t=1
is used
to replace
sinh*t
cosh? ¢ — 1.
2
Then:
volume generated = x| (cosh? ¢ — 1) sinh td
0
2
s
= x| cosh? tsinh t dt — n| sinh
¢ dt
0
0
I a [[+cosh” 3417
t])—z[cosh ¢]52
I a + cosh? 2 —
4] — x[cosh2 — 1]
[cosh’ 2+2—3cosh 2]
wa
~ 46.0 (3 s.f.)
Example
15
Given that tan}x = t, show that
(a) SNR
PEE
aren
SE
Hence find (c) |cosee wd
1-0
ces ey
)
ee
(d) sae
dx
=?
by
43
44
Integration
(a)
2x
ae
ep
(2 a
cos’! bo x
=
a
(1+ tan’ $x) cos* 5x
2 sin 5x COS 5x
cos?
5.x + sin’*tx
= 2'sintxcostx = sinx
1-72
1-tan*4x
Cee
1 + tan?5x
cos’
(Notice
how
the
sin2A =2sinAcosA,
here.)
Dailiners
5x —
one
sin’
5x
= cos*}x
— sin*4x
= cos x
identities
cos’? A +sin? A =1,
cos2A = cos’ A — sin* A,
(b) Differentiate tan}x = t, with respect to x:
But sec? tx =1+tan’} il
ee ae eo
dx
oO
Saag
oySk
So:
‘
di
Tear
(c) Using the substitutions ¢ = tan}.x, sinx =
you have:
COSCO
|
1
dx
* dt
sin x dt
= In|tandx|+C
~
sin A
have
A
= tan A,
been
used
Integration
(d) Using
the
dx
2
i ee
you have:
substitutions
y=tanix,
cosx=
~
:
ee | ei
Lam
Su
ie
dx
l
|;
and
a
¢7
l+t
2
l
,)a
(
-\l——
]\l+¢r
54+ 5t°+3-3t°
1+?
||
lz
2
ee
~|dzt
a
8+ 2f J \1 +e
442°
II 1 arctan 41 +C
(see standard result 1 on page 34)
2
1 arctan (}tan$x) + C
2
exercise 0
In questions
re
1—SO integrate the given expressions
with respect
LOX
1 co
49 + x?
x
peas 2
44x?
I
——
(x +3)
l
ee
5
I
I
6.
“| ics
8
x(x— 3)
C———
11
tan? 2x
13
./xe"*
14
I
SS
7
/(5 — 2x?)
2x" +3
45
V(x = 2)
a
5x? + 13x —6
oe
J + 5x?)
1 — 4x2
ev*
\/(2x? — 5)
9 sin?2x
12
15
vx
I
J(2 — 8x?)
ie
/(4x? — 3)
;
ecos2x
Pe:
:
(3— x)
I
—————
1 + 5x?
21
l
x7 +4x+8
45
46
Integration
22
——_______-
l
23
V(35 + 4x — 4x7)
24 tanh? 5x
ie
es
(1 + cos x)*
;
tanh 5x
25 x? Inx
26 (1 +cosx)
reex-—4
29
l
—e+]
XxX
30
e™* cosx
31
sec?xIn(tanx)
32 Ta+x
33
2x + |
x29
34
x2
ox 21
nN
aeCOs 5XDk
36 x*/(1 — x?)
37 arsinh 4x
38 arctan 3x
39
40
41
42
e*cosx
cosh? x
cos*x
43 Eee rere,
caged
45 xcoshx
46 ane
(2x2 + 5)
47
coth2x
:
48
49
50
cosec* 2x
9x* + 12x + 20
/(9x? + 12x)
cosech 4x
tan*x
ss
In questions 51-60, evaluate the given definite integrals, giving
answers exactly or to 3 significant figures, as appropriate.
»/3
|
1
| ——
0
x
dx
52
=F 9
|cosh? x dx
0
4
l
53
|cos’ x dx
i
54
|arctan x dx
0
55
|—————
dx
56
| ———_——
57
|cin x + cos x)* dx
0
p)
58
|——~ dx
gal(x — 2)
59
[ee
oa,d x
60
;fee
61
Use the substitution uw = — to evaluate | —.——
62
By differentiating
08
|
1.5
1 (x2 —2x + 2)
4
6x8
ae
x22
x-—2
=
1
f Vee xe)
.
dx
xX
ANG? EON
ae d x
8)
l
g
1
x
2,XV/(x? —1)
4 x2
, find the value of|
3 x2(x —2\
dx.
Integration
47
V3
63
Evaluate |
ude
1
64
dx
1+ De
(a) Show that isin? x dx = x
0
7
(b) Evaluate i sin’ x dx, giving your answer in terms of z.
0
65
Using the substitution u* = 1 + x?, find [eva se yidy,
66
Use the substitution x = 2sint to show that
v3
x3
i Cee
ice ie
ea
ey fh
ral
|
ua
67
ES
Use the substitution tan x = ¢ to evaluate |cosec 2x dx.
4
4
68
Find emer dx in the cases when
(a) #= 9
69
(b) n=1
Oc). ee
Show that
Ja ~ x’ dx = +faresinx+ x/(1 — x?)}
+C
(a) by using differentiation
(b) by using the substitution x = sint¢.
70
Show that
4
5
©, haat, 0
once
matin: | i mee 1g)Dos ©
ioe
:
i
a
’
fa
Show that |xInxdx = In4-3
1
72
The curve with equation a*y* = x?(a? — x”) has two loops, as
shown.
Show, by integration, that the area enclosed by a loop is Sa’.
T
73
1 +cosx
Show that Isin x(
3
1 —cosx
)av=m4—4
In2
74
Show that | cosh* x dx = 21n2+ Ti:
aS aa
oe Nati
x
48
Integration
75
Show that |
Lt een:
dx =n +2 —4arctan 3.
3
3
76
Show that
77
Use the substitution x = sin? to show that
J0
arcsinhxdx = 3arcsinh3 + 1 — ,/10.
ae
Jox+ VC — x?)
78
dx =,
4
Use the substitution tan5x =? fOulitegtate
a)
l
b)
|
(a
3+ S5cosx
(
5 —3cosx
3.5 Reduction formulae
When using integration by parts, as in Book C4, you found it
necessary with some integrals to apply integration by parts more
than once.
|eetay
was
such
an
example.
icra
|vera the integral /5, and let
Then
du
Suppose
you
start
by calling
dv
w= x° and = =e,
x
;
x
The integration by parts formula is
dy
du
fueas = uy — |oShas
One application of this gives you
|x’e*dx = x*e* — Je@n dx
That is:
or
teh
|
ay
ae
aera
lL =x’e* — 21,
|setae,
By one application of integration by parts you have found
formula to reduce the integral J, to a less complicated one, J).
a
Suppose now that you had started with the integral J, = |weds,
where n is a positive integer.
Integration
3
Write
dy
w= x”,
du
Pam
x
— = ¢@
dx
iS
ie
and one application of integration by parts gives you
Lax.
= Je(n Sie dx
x"e* —nI,_;,
The relation
where/,_, = [x"terdx
/, = x"e‘ —nJ,,_;,n > 1, is called a reduction formula.
It has reduced
involving n — |.
an
integral
involving
n to a simpler
integral
By repeated application of the reduction formula you can often
reduce a complicated integral to one of those integrals that you
know how to evaluate already.
Example
16
Find [vera
If you take J, = [x"eras, then the integral you need is /3, and you
know also that
£23"
n
I; =
xe
—nlh n—|\
27
nS 1, So:
2= 31,
by using the reduction formula repeatedly for J;, 5 and J,.
Now
l= |
xe*dx = |e*dx =e”
and you can complete the solution by writing
l= xe — 3x’e* + 6xe* —6e* + C
=e*(x° — 3x* +6x-6) +C
Notice that it is customary when using reduction formulae to
complete all the stages of reduction and to add an arbitrary
constant to the final answer only at the end of the calculation.
Reduction formulae can be used to find indefinite integrals, as
shown in example 16, but, more often, they are used to evaluate
definite integrals.
49
50
Integration
Example
17
rl
Find a reduction formula for J, = | Si
0
UX,
2
Integration by parts is used on the integral written as
lA
| sin” !x sinx dx
0
ee
where
u-—=sin"
dy
'x,
—
dx
du
:
Since : fuae
a
a2
Bon
os = (7— L)sin
iG
|Sess
= - sin”!
0
n—2
~~“XCOS.x,
.
= sinx
v = —cos x
[>= dx you have
x
2
52
6
ye
x cos 1 -| (n — 1)sin” “x cos x(—cos x) dx
0
0
z
But
= sin”“x C08. = —0—(—0)=0
0
Hence
|sinsdx = (n— 1) |Sil’ ex cos Cd x
0
0
x
eerie
3)
Also:
yn
D
cos?x = 1 — sin’x
Sin d= (7
2
iS
Nila
1)|(sin” *x) (1 - sin’ x)dx
0
Nl
=(n—1) isin”*x dx —(n—
0
But
——
ws=
la
|Sint eda
0
.
vane)
IE =e (n Se li
That is:
L2G
ofa =| silty ech
0
-
A
=
9)
II,
eee
nl, = (n —
and
1) |sin”x dx
0
19) a
n—1
:
f= hg [,-2, 0 = 2, 1s the reduction formula.
Integration
Example 18
us
Use the result established in example 17 to evaluate (a) |sin’xdx
0
(b) isin®x dx.
0
(a) You have
|sin’ocds =
0
But
7s
(using i, = sl Reactor
= 7)
aed
=e
: ~)b(repeating the formula for n = 5)
= (8) (2) (=) T, (repeating the formula for n = 3)
= 9) @)G)n
i= [sin dx
But
0
cos.x| =-0-(-l)=1
=
0
Hene
r= 9OQM=%
us
(b) You have [sin’ dx = Ik
0
But
i=
—]
;
—]
3 ip, using J, = =
yoo 10r W= 8)
=@@(4-\s
2-1
By repeated application of the reduction formula you have:
Is = (3) (6) (@) G)oBut
2
2
0
0
I) = |sin’xdx = |f=
2
| =k
0
51
52
Integration
You do not always need to use integration by parts when
establishing a reduction formula.
Example 19
Use the identity sec?4 = 1 + tan’ to find a reduction formula for
Tt
4
[= [cans AN:
0
Writing the integral as
us
(ee Jans tan’x dx and putting tan’x = sec’x — 1 gives
0
at
.
4
3|
sy
iio |tan” “x(sec x — 1) dx
0
=
4
(an
n—2
0
d
But
—(tanx) =sec*x
2
n—2
“Vay
a
and
dx
4
“esecx Gai |) tan
0
tan”-?x sec*x dx =
n—1
tan” |x
since you have a function of tan x and its derivative sec? x.
4
|
tan” n—2~xdx = J,,_5
Also:
J0
[=
I
mali
tan”” x —
ge
0
I>
So. 1, = aries [,-> 18 the reduction formula for n > 2.
n—
Example 20
Use the result established in example 19 to evaluate (a) Jtan'xdx
0
ia
(b)
tan x dx.
Jo
4
(ayisince:
ae
te
I
/ 5 — |tan diy andy
0
n—1
you have:
and
Is =}4-]
eee 5 —f,
by applying the formula twice.
Integration
b= 4-1)
—
—i4+h
4
ri
I =| tan x dx = [in
sec 7
0
0
But
=In(V2) —-In1
= 5In2
So:
Is =5In2—4
a
(b) Similarly,
J, = [tan’¥Gy
and.
J; ee
0
fh
I¢ =1_-J,
y=, —h
16) =i-l
is=}=)
=;-f+1-h
zi
l= Jtan?oeds
0
a
= | ax
0
i
=
[> =f
0
Ig =+ —5+ ae
LMriveniduL.,
= |cos" dx, show that, for n > 2,
nl, = cos”"!xsinx + (n— 1)I,_2
a
TT
Hence evaluate /g for J, = |cos”x dx, and Jy for |COS idx,
0
JO
2
Aaityen that d, = Janx)"dx, show that, for n > 1,
=x)
2
Hence evaluate |(In x)? dx.
1
— nb
53
54
Integration
21
Given that J, = | sin”xdx, find a reduction relation between
0
I, and I,» for n = 2. Hence find I; and Je.
Evaluate (a) [sin Ocos*6d@
0
(b) |sin’ @.cos°@ dé.
Jo
Use the substitution x = sin¢ and an appropriate reduction
formula to evaluate
]
|l=
J0
2 dx
Use the substitution x = sin’ ¢ and an appropriate reduction
formula to evaluate
|x(1 — x): dx
0
Check your answer by using another method.
Using an appropriate reduction formula, evaluate, in terms of e,
1
the integral |xe* dx.
0
Given that J, = [cosh dx, show that, for n > 2,
nI, = cosh” !x sinh x+ (n — 1)Jn-2
Hence find |cosh dx,
1
Given that J, = |x 6dx, show that, forn = 1.
0
|
Hence evaluate |xe
0
e =
OS)
oa Ee!
“dx.
T
10
3
Given that
= |sec’ xdx
0
show that, for
gn-2
IE =
n—\
=
2:
9)
V3 + es
a
n—1|
Hence evaluate /,.
11
1
Given that J, = |ee
Gi
jn
Hence find /3.
laaa))
oe SOCAL
Dia?
On
TOR me
3)
Integration
12
Given that J, = |
2 sin(
—
1)x
I, =— oi
wks ey Aer
show that
n=
> sin 4x
Hence evaluate
:
x
Sin XxX
7
dx and check your answer by using
4
another method.
13
Given thatJmex" sin x dx, show that, for n > 2,
i = n(5)" eee “= 1)
Hence evaluate /3.
14
By setting up an appropriate reduction formula for
|x" sinhxdx
|x sinh
and applying the formula show that
xdx = cosh x(x + 20x° + 120x) — sinh x 6x" + 60x7 + 120)4+.C
where C is a constant.
15
Given that
J, = |(4 —x*)"dx, show that, for n > 1,
0
8n
i
2n+ 1
n—|
Evaluate Jy.
3.6 Using reduction formulae in
applications of integration
As you have seen, integration is used for finding areas of regions
and volumes of solids generated by revolving regions about the
axes. The integrals that arise when you use formulae such as
[vax and
x|y’dx
for these
applications
are
sometimes
best
evaluated using reduction formulae. Here is an illustration of when
they are useful and economical.
55S
56
Integration
Example 21
The diagram shows a sketch of the curve called an astroid which is
given parametrically by
Ve dcoc hh, vet
1,9
0S
re 27
where a 1s a positive constant.
aman
Find (a) the area of the region enclosed by the curve (b) the volume
generated when the part of the curve for which y > 0 is rotated
completely about the x-axis.
(a) The coordinates of the four vertices of the astroid are at (a, 0),
(0, a), (—a, 0) and (0, —a) and the values of the parameter f¢ at
these points are 0, 5, 7 and a respectively.
The area of the astroid is four times the area of the region under
the curve in the first quadrant, so you have:
*X=d
area of astroid = 4 | ydx
and, changing to variable f:
0
4 |asin’t (—3a cos’ fsin t) dt
nu
bck}
(0
= —12q° |sin? ¢ cos’ ¢ dt
nls
Change around the limits and write cos? ~ = | — sin‘ f¢ to obtain:
a
=)
5
area Ob-astroid =
a!
rls
.
nis
9d
é
12a7 |sin? tdt — 12¢ |sin® ¢ dt
JO
J()
a
Integration
us
From example’ 17,
i, = pes i,9
where
IJ, | Si
0
vy,
Using this reduction formula you have:
area of astroid = 12a (I, — I¢)
= 12a°| (2) (3)
le
= gia
(b)
Volume of solid formed = 2 |: ty" dx
Jx=0
ua
= —2n |a’ sin®t (—3acos*tsin t)dt
0
= 6na° |sin’va a sin?) dr
J0
=
6na’ [I —
Ty|
by using the same reduction formula as in (a).
So:
volume of solid = 78 2a? = na 3
:
:
1 The curve with equation
8
y=sin°x
between x = 0 and x = §,
the x-axis and the line with equation x = 5 are the boundaries
of a finite region R which is rotated completely about the
x-axis. Find the volume of the solid so formed.
2
The finite region R is bounded by the part of the y-axis
between
y = +4 and the curve with equation x = cos*y. The
region is rotated completely about the y-axis to form a solid S.
Find the volume of S.
57
58
Integration
The region R is bounded by the half-lines with equations 0 = 0
and 6 = 4 and the polar curve with equation r = a tan°0, where
a is a positive constant. Find the area of R.
Given that J, = [sectsdx, show that for n > 2,
(n —1)J, = tanxsec”-*x + (n — 2)Jn_2
The region R is bounded by the half-lines with equations
@ = +4 and an arc of the curve with equation r = sec?0.
Find the area of R.
.
.
.
9
Express the curve with cartesian equation (x? + y?)° = 4axy?
in polar form and show that the area of the loop in the
Ba,
9)
quadrant where both x and y are positive has area tna’.
]
1
Evaluate (a) |x'4/(1 —x?)dx
0
(b) |(1 — cos fy di.
0
us
Given that
J, = [xtcos x dx, show that forn > 1,
0
I, =
(2)"—n(n —
1)In-2
Hence find the area of the finite region bounded by the curve
4
with equation y = x"cos x and the x-axis between 0 and 5.
Show that the area of the region bounded by the curve with
equation y = 1 + tan®x, the x-axis and the lines with equations
x = 0 and x =7is 1B
15°
a
Given that 7, = Jseors dx, show that, for n > 2,
0
Dem
a
n—2
oe
n—-l1
ai In-2
n—-1
The finite region bounded by the curve with equation
y= sec?x and the lines with equations
x = 0, x =F and
y=0
is rotated completely about the x-axis. Find the volume of the
solid so formed.
10
1
Given that f, = |(1
0
— x) dx, show that, for n > 2,
n
TI,
|—~=
)Ih-?
1 = (5)
1—2
Hence find |ol
0
da,
Integration
3.7 Length of a curve
Suppose that a curve is continuous, with each point on the curve
having a unique tangent associated with it. Consider a part of the
curve from the point A to the point B. Next suppose that this part of
the curve is divided into very small bits of which PQ, as shown, is
typical.
YA
B
Q
P
A
oo
O
x
If P and Q are chosen to be sufficiently close together, then you
may assume that
length of are PO
length of chord PQ
Furthermore, if Q approaches P along the curve, which in effect
increases the number of very small bits indefinitely, then you
assume that
length of arc PQ
im <=
o—p length of chord PQ
At a more advanced stage in mathematics, it is necessary to
determine whether the assumption made about this limit is always
the same when the points P and Q are selected in a different way. For
your syllabus, however, you may take it that the assumption is valid.
Let the curve have cartesian equation y=f(x) and take the
coordinates of P and the neighbouring point Q to be (x,y) and
(x
+ dx, y+ dy) respectively. The length of the chord PQ is, by
Pythagoras’ theorem, i)|(6x)°+ (67) ].
-
of
VV
59
60
Integration
If the length of arc of the curve PQ is called ds, then you have
és
y/(6x)°+(8»)"|
That is:
(ds)? (5x)°+(dy)?
and dividing by (ox)° gives
Os
:
oy
Ox
:
Ox
As Q — P along the curve, 6x — 0 and
Os
odse
oy
dy
In the limit you have
Give the end points of the arc AB coordinates (x4, y4) and (gz, Vp).
ds
me
dx
dy\’
/|a (=) |
so integrating with respect to x:
dy\? 2
XR
a
Arc length AB = | 1
XA
GJ
—
d
The length of arc AB is found by evaluating this integral.
Notice that you can also obtain from the relation
(6s)° & (dx)? + (Oy)
the formula
ds\
mr
(=)
ene
dx\
(eee
eee |
(S) si
by dividing by dy and proceeding to the limit as dy > 0.
Integration
61
This leads to the formula
a
YB
Arc length 4B = | 1
YA
TPA, 2
ars
2 Ge i
In parametric form, you have on dividing by dr:
(dx th dy\
bt
\ot
ds)
6t)”
and proceeding to the limit as dr > 0,
ds\"_
(dx a dy\’
dt}
\dr
dr
tz
2
In this form, the formula is
&
Arc length 4B = | as
eae
ps
+ ty,
dt
|
|
:
where ¢, and tg are the values of the parameter ¢ at A and B.
3.8 Area of surface of revolution
YA
O
B
-
x
When the small arc PQ is rotated through 27 radians about the
x-axis a narrow ring, or circular band, of approximate surface area
2nyos is formed. The sum of the areas of all such rings from A to B
gives an approximate value of the area of the surface formed when
the arc AB is rotated through 27 radians about the x-axis.
Let O — P so that ds — 0 and the area of the surface is then
|
;
62
Integration
1
dy ak
1+ (32)
=
dsx
XB
wi
Surface area = | 2ny
Xy4
Similarly, if the arc AB is rotated through 27 radians about the
y-axis the area of the surface formed 1s given by
1
VB
r
Surface area =
2mx
1
(i) ‘
dx\7
—
|’
dy
YYA
In parametric coordinates the area of the surface formed when the
arc AB is rotated through 2z radians about the x-axis is
'B
ty
ds
dt
dt
| ny —
That is:
a
a
dx\?
1
/dy\’|’
urface area =| | 2zyTy | (i)
Surf:
|— +(Z)
-2.
dt
Example 22
Find the length of the arc 4B of the curve with equation y = 2x2,
where the x-coordinates of A and B are 3 and 8 respectively.
y
oe
=? 93 — 4)
=4027-8)
38
| ,
I|
—=
w NO
WwIrh
Integration
Example 23
The arc of the curve with equation y = cosh x from the point A
where x = 0 to the point B where x = In 2 is rotated through 22
radians about the x-axis to form a surface S.
Find the area of S.
-
y = cosh x
,
So:
But
y =coshx
dy
> — =sinhx
dx
1+ (2) = 1 +sinh* x
dx
cosh? x = 1 + sinh’ He 80;
In2
2
Area of S = | 2ny|1 + (2)
0
In2
| 2n cosh’ x dx
Ni
dx
63
64
Integration
Now
So:
x
x=In2>e*=2 and e~ =)
area of S=4[4(4)+2In2—4(4)
-(§+0-}5)
= £(15
+16In2)
Example 24
A curve has equations
where f is a parameter. The arc of the curve from the point A where
t= 7 to the origin O is rotated completely about the y-axis. Find
(a) the length of the arc OA
(b) the area of the surface generated.
O
a8
(a) If you differentiate the equations with respect to ¢ you get
dx,
> dy _
ao
Then:
en
dx\-
ee
/dpy- Se i
mae
pas
(=)
+(3)
5
2t- oy + Or 4 4
2
=24 217 +924 = (1+ 372)?
Bl
(dx\?
dy\°
Length
—
ate |dr
gth of of arc OA = I ()
+(Z)
Length of arc OA =3 3.
Integration
tO
(b) Area of surface generated =
we)4 *
SCSk
SSEEaEeeerd
|
=f.
a.|.2
&
2.|
|es
omSe eSWe
DN.
Sh
N=
rel
~
no)
as
|
Ww
~
+
WI)
oe
———
eS)~
ww
7
QO.
jak~
ee
\|
i)4
el &
bs walBo
||
alal
Nw—
4
Area of surface generated =
In questions 1—6, find the length of the arc of the curve given
between the points specified.
1 The curve with equation y = 2x7 between the points with
x-coordinates 7 and 11.
2
The curve with equation y = 5 cosh 2x between the points
with x-coordinates 0 and arsinh 3.
3
The curve with equation y = 2 + In sin x between the points
where x = 3 and x = 5.
4
The curve with equation y = xi — 1x3 between the points with
x-coordinates
| and 4.
5 The curve with equation y = Incosx between the points where
6 The curve with equation y =+x*° +4x | between the points
whete «=
1 and x=,
In questions 7-12, find the length of the arc of the curve given in
parametric form between the points where ¢ = 4 and t = fy.
T
Kap
sin iy
HL Cost
between the points where ¢; = 0 and fz = 5.
65
66
Integration
x =sin tcos t,
y= 22 sin t
between the points where ¢; = 0 and f = 3.
Keer vast
between the points where f; = | and tf = 2.
10
Meare sin t,y =e cost
between the points where ¢; = 0 and fn = 5.
11
=A
pf
—2 ny
between the points where f; = 1 and fp = 4.
12
x = sinh? 1, y = 2 sinhs.
between the points where t; = 0 and f = 2.
In questions 13-18 find the area of the surface generated.
13
The curve with equation
y=coshx
from
x =—ltox=1
is
rotated completely about the x-axis.
14
The line with equation y = 2x from x = | to x = 3 is rotated
completely about the x-axis.
15
The curve with equation y = 2x? from x =3
to x=8is
rotated completely about the x-axis.
16
The curve with equation
y=cosx
from
x=0
to x =§ is
rotated completely about the x-axis.
17
The curve with parametric equations x = tf — sint,
from t=0
18
t=7
is rotated completely about the x-axis.
The curve with parametric equations x = 2t,
to
19
to
y= 1 —cost
y= ¢* from
r=0
t=2 is rotated completely about the y-axis.
Show that the perimeter of the curve with equation xi + yi a
is 6.
Find, in terms of z, the surface area generated when the
quadrant of the curve for which x > 0 and y > 0 is rotated
completely about the y-axis.
20
For the curve with equation
y = (1 — »/(). show that
= (2) - (Gx 41)
dx
Px
(a) Find the length of the curve from the origin O to the
point A (1, 0).
The arc of the curve OA is rotated through 360° about the x-axis.
(b) Find the area of the surface generated.
Integration
21
The curve C has equation y = sinh? x. The arc of C from the
origin O to the point A where x = 2 is rotated completely
about the x-axis to form a surface S.
(a) Show that the arc of the curve has length 5 sinh 4.
(b) Find, in terms of z, the area of the surface S.
22
The equations
x = t+sint,
y=1—cost,0<t<
2n, definea
curve C parametrically from the origin O to A (27, 0), which is
rotated completely about
Ox
to form a surface S.
(a) Find the length of the curve from O to A.
(b) Find the area of S.
23
The arc of the curve given parametrically by
x= tanh st, p = sech7
between the points where ft = 0 and ¢ = | is rotated completely
about the y-axis. Show that the area of the curved surface
formed is
2n(e — 1)?
e?+1
24
A sphere has radius a. Find the curved surface area of a zone
of the sphere cut off by two parallel planes that are a distance
b apart.
25
The arc of the parabola with equation y? = 4ax, where ais a
positive constant, from the origin O to A (a, 2a) is rotated
through 27 radians about the x-axis. Show that the area of the
surface generated is
1 { sinhdx == COS X +C
/ 2 {cosh xdx = sinhx +
a .|
67
68
Integration
ae
Vl +x”)
a
xX+a
l
aS.
ae,o
:
=-Inje*
II oe
J+ Cal <a
a
14 To integrate expressions of the form
|
De ane
dx
|
lana
GS
tax en
where p, g, r are constants, you complete the square in
the denominator and then use one of the standard forms
in 8-13 above.
15 Some integrals involving powers of the variable or powers
of a function of the variable can be related to an integral
involving a lower power of the same form. For example,
Ly = [innsdx.
i3 = [sin"x dx
If a relation is established between J, and J/,_, (or J,_>),
then this relation is called a reduction formula and it can be
applied as many times as required for a positive integer n.
16
Typical reduction formulae are:
ug
it
2
n—-|
0
a
L= |sinteds, i=
1
n= |xe dx,
0
1,=¢-nh
pon ee
4.221
Such formulae are often proved using integration by parts.
©
Integration
17
69
Some reduction formulae such as
4
Le |tan"ds,
I, SS ie nS
0
n—-|
are
proved
by
sec’ x = 1 + tan~x.
using
an
identity:
2
here
it
is
y
O
18
r
|
Arc length AB is
d
1
Area
fdyV\
=
f;(3) (3) :
OR
19
Sade
—-
of
surface
formed
when
arc
AB
is
rotated
completely about Ox is
on
20
mf's (4) (2)
a
dx\?
/dy\
Integrals arising from the evaluation of lengths of arcs,
areas and volumes sometimes require a _ reduction
formula in the calculation process. In examination
questions you will often be asked to establish a reduction
formula and then to use the formula to evaluate a
definite integral.
Coordinate systems
In this chapter the work on curve sketching and the properties of
curves is extended from the work covered in Books Cl to FPI. In
particular, some specific properties of the parabola, ellipse,
hyperbola and rectangular hyperbola are covered, including the
use of both cartesian and parametric coordinates. Each of these
curves can be described as a locus. A locus is defined as a set of
points which satisfy given conditions.
Another coordinate system is introduced later in section 4.5:
intrinsic coordinates. This leads on to the radius of curvature.
4.1 The parabola
The curve given by the parametric equations
x=at*,
y=2at,
teR
where a is a positive constant is called a parabola.
y
Now y = 2at,so
t= = and
2a
So:
a
ig=
yp?
(z
That is, yr = 4ax is a cartesian equation of
y
os
y? = dax
the parabola.
Since y> >0 for all real y and a>0, you
know that x > 0 and for every positive value
of x there are two values of y, one positive
and the other negative. A sketch of the curve
looks like this:
The curve is symmetrical about the x-axis and
the y-axis is a tangent to the curve at O.
0
a
72
Coordinate systems
The focus-directrix property of the parabola
Take
a general
point
P with
coordinates
(at, 2at) on
the
parabola. The point S(a,0) is called the focus and the line with
equation x +a=0
is called the directrix. Notice that the focus
lies on the axis of symmetry of a parabola and that the directrix
is perpendicular to the axis of symmetry.
P (at*, 2at)
In the diagram, the line through P parallel to the x-axis meets the
directrix at the point Q.
Now
PO =a
ta=all+ 77)
SP* = (at — a) +(2at)?
=
lat hat - 4a
=¢ 120
=a
400
(1+20? +1")
=a(l+i)
So:
SP= ail +1”)
That is, PQ = SP for all points P on the parabola or, in words,
every point on the parabola is equidistant from the focus (a, 0)
and the directrix
x+a=0. You can easily prove, by reversing
this argument, that if a point is equidistant from (a, 0) and the
ae x +a=0, then the point lies on the parabola with equation
= 4ax. This property is often called the focus—directrix property
a the parabola.
Coordinate systems
Equations of tangents and normals to a
parabola, and other properties
Take a general point P(ar*,2at) on the parabola with equation
y = 4ax.
P (at?, 2at)
If you differentiate the equation of the parabola with respect to x
you get
dy
2y—=4
ae
dy
and at P:
So:
dy
1
peer
x
2(2at)—
= 4a
dx
gradient of the tangent to the parabola at P
An equation of the tangent to the parabola at P is
y — 2at =—l (x — at2 )
=>
a
That is:
y=
lar =x-at?
ty =x + at?
is an equation of the tangent to the parabola at P(at’, 2ar).
The gradient of the normal at P is
73
74
Coordinate systems
The equation of the normal to the parabola at P is
y —2at = -t(x — at”)
a
That is:
y+tx = 2at+ at?
is an equation of the normal to the parabola at P(at’, 2at).
Suppose now that the tangent at P meets the x-axis at T and that
the normal at P meets the x-axis at N, as in the diagram.
Putting y = 0 in the tangent equation gives x = —at? and OT = at’.
Putting
y = 0 in the normal equation gives x = 2a+ at”, that is
ON = 2a + at’.
So the focus S(a,0) is the mid-point of TN and you know that
SP = a(1 + fal from the focus—directrix property (page 72).
Thus:
Sl=SN
=SP
Equation of the tangent in cartesian form
Many properties of the parabola are best proved by using
parametric coordinates but you sometimes need to find an
equation of the tangent to a parabola at a point P whose
coordinates are (h,k). For the parabola with equation y* = 4ax
you have
2y
— =4
, ax
=
and at (h,k)
22
ess
S
An equation of the tangent is then
2
y-k=— (x—h)
That is:
yk —k* =2ax —2ah
But (4,k) lies on the parabola, so k* = 4ah and an equation of the
tangent can be written as
yk — 4ah = 2ax — 2ah
a
That is:
ky = 2a(h+ x)
Note: This equation is easily remembered from y? = 4ax when
you write y? as yy and 4ax as 2a(x+ x), replacing one y by k
and one x by h.
Coordinate systems
Example 1
Find an equation of the chord joining the points P(ap’,2a p) and
Q(aq’, 2aq) on the parabola with equation y~ = 4ax. ae that if
the chord PQ passes through S(a,0) then pg = —1.
a)
The gradient of the chord PQ = ae
|
ap — aqv
___2a(p~4)
alp — q)(p + 4)
As P and Q are distinct points,
p # q and the gradient of PQ is
2
p+q
An equation of PQ is
2
y — 2ap =——
cn
(x-ap
p+-Gg)—2ap — 2apq = 2x —2ap”
That is, v(p + g) — 2x = 2apgq is an equation of PQ.
Put
x =a,
y = 0 for the chord to pass through S, then you have
—2a4=20p9
=>
pa=—!
as required.
Example 2
Find an equation of the normal at P(47,4¢7) on the parabola
with equation 4y = x°. The normal at P meets the y-axis at the
point Q. Find, in cartesian form, an equation of the locus of the
point M, the mid-point of PQ, as ¢ varies.
Ay = x2
The diagram shows the parabola, which passes through O and is
symmetrical about the y-axis.
75
76
Coordinate systems
Differentiating 4y = x° with respect to x gives
dy
A ===):
dx
:
and at P(4t,40°),
)
)
ax
4
That is, the gradient of the tangent to the parabola at P is 2/ and the
;
l
gradient of the normal at P is therefore Say
An equation of the normal at P is
l
y—4t? = =a
At)
The normal meets the y-axis at Q where x = 0.
>
SO:
At
y-4t° = ¥
=
PS 2+ 41? at O
The mid-point M of the line joining P(41,4¢7) and Q(0,2
+ 41°)
is given by
_4t
that is:
Now
4° 42440
ae
y=
<=we
=
5
Ae
|
as f¢ varies the parametric equations of the locus of M are
x=2t,y=4074+1.
By eliminating f, you get, since x? = 417,
y= al
which is a cartesian equation of the locus of M.
Example 3
The line with equation
y= mx-+c is a tangent to the parabola
with equation y* = 4ax. Prove that mc =a.
You know that the line with equation ty = x + at isa tangent to
the parabola with equation y* = 4ax for all values of 1.
:
l
Notice that the equation of the tangent can be written y =-—x-+ at
and comparing with y = mx + c you have
;
m=
l
:
Aid
Ge — a
Eliminating ¢ from these equations gives the required condition
mc =a for the line
y= mx-+c to be a tangent to the parabola
*
.
9)
with equation y* = 4ax.
Coordinate systems
Exercise 4A
1 Find, in cartesian form, an equation of the parabola whose
focus and directrix are respectively:
2
Cade 2.0). see DQ)
(b3.00)s xe
= 10
(c) (0,4), ¥y+4=0
(ad) Cyl) oy
0
Find the coordinates of the focus and an equation of the
directrix of the parabola with equation
(ay p=
(c)
3
1x
(be
—x? = l6oy
= 4s
(yay Se
16
Find an equation of the tangent and of the normal at the
point given, to the parabola with equation:
(a) y?=xat
(1, 1)
(c) x? = 16y at (16,16)
4
(b)i oy? = 32% at CQ, 8)
(d) y=4(x+1) at (-3, 16)
Find a cartesian equation for the parabola given in parametric
form by the equations
(Cs ae
ae
(ce) x=—44,
(oy) Xf.
5
ety
(Qc) Meee = ay
p= 329°
(Gd) cu
10
dip Sar? ap
Vv S(T = 2)0 =3)
Find an equation of the tangent at the point (x, y) to each
parabola in question 4.
6
Prove that the point (at?, 2at) always lies on the curve with
equation y~ = 4ax. State the value of 1 at the point:
(a)
7
(a, 2a)
(b) (a, —2a)
(c) (25a, 10a)
(d) (4a, —a)
Find an equation of the normal at the point (1, 2) on the
curve with equation y* = 4x. This normal meets the curve
again at the point C. Find the coordinates of C.
8
The tangent with equation fy = x + t to the parabola with
equation y* = 4x passes through the point (5, —6). Find the
possible values of t, and the coordinates of the point of
contact of the tangent with the parabola for each value of ¢.
9 A line with gradient 4 touches the parabola with equation
y* = 4x. Find the coordinates of the point of contact of this
tangent with the parabola and an equation of the normal to
the parabola at that point.
77
78
Coordinate systems
10
Show that the tangents to the parabola with equation Vom 4ax
at the points P(at;, 2at,) and Q(ar5, 2at)) meet at the point
T[at)to,a(t; + t)]. If PO passes through the focus (a, 0) show
that
Ath
11
41
(b) T lies on the directrix x +a =0.
The line with equation y = m(x + a), where m can vary but a
is constant, meets the parabola with equation y* = 4ax in two
points P and Q. Find, in terms of a and m, the coordinates of
the mid-point R of PQ. As m varies show that R lies on the
curve with equation y* = 2a(x 4+ a).
12
The points P(ap*, 2ap) and Q(aq*, 2aq) lie on the parabola
with equation y* = 4ax. Show that the tangents to the
parabola at P and Q meet at Riapq, a(p + q)|. Show further
that the area of APQR is $.a’|p — ql’.
13
Show that the normals at P(att, 2at,) and Q(at3, 2at2) on the
parabola with equation y? = 4ax meet at the point R where the
coordinates ofR are [a(t; + tyto+ i + 2); — at) to(t) + ty) ].
Given that the line PQ passes through the focus S(a, 0), show
that as ft; and ft) vary, R lies on the curve with equation
Vv =a — 3a).
14
The parabolas with equations ay = x* and y* = 8ax meet at
the origin O and at a second point P. Calculate the
coordinates of P in terms of a.
The tangent at P to the curve with equation ay = x* meets the
x-axis at X and the tangent at P to the curve with equation
y* = 8ax meets the y-axis at Y. Find an equation of the circle
whose diameter is XY.
15
The normal to the parabola with equation y* = 4ax at the point
P(at*, 2at) meets the curve again at the point Q(as*, 2as).
Show that s = —t——
Show that, as ¢ varies, the locus of the mid-point of PQ has
the equation
y* + 2a(2a — x) y? + 8a4 =0
16
The variable chord PQ on the parabola with equation y? = 4x
subtends a right angle at the origin O. By taking P as (7, 21))
and Q as (t3,2t), find a relation between ¢; and f) and hence
show that PQ passes through a fixed point on the x-axis.
Coordinate systems
17
The coordinates of A and B which lie on the parabola with
equation y* = 4x are (t?,2r) and (417,
47) respectively. The
tangents to the parabola at A and B meet at the point C and
M is the mid-point of the chord AB. Show that the line CM is
parallel to the x-axis for all values of t, where t + 0.
18 Show that the normal at (1°, 27) to the parabola with equation
y? = 4x has equation y + tx = 2r+ 4°. At the points A and Bon
the parabola, t = 2 and t = — 5respectively. The normals to the
parabola at A and B meet at C. Find the coordinates of C and
find an equation for the third normal which passes through C.
19
Show that the line with equation 4y — 3x = 20 1s a tangent to
both the circle with equation x* + y? = 16 and the parabola
with equation 7 = 15x.
20
Find an equation of the line which is a tangent to both the
parabola with equation y* = 4ax and the parabola with
equation x? = 4ay.
4.2 The ellipse
The curve given by the parametric equations
Pa
t=
cost,
yovosnt,
Vat.27
where a and > are positive constants, is called an ellipse.
x
cosy = —,
a
Now
So:
int =F
and
ye nhs
i|
—
—_
a
h2
=
which is a cartesian equation of the ellipse.
2
x* =a (5)
Since
that is:
Also
2
y* = b? (- =
a
that is:
for real x,
|y| <b
for real y, so:
2
:
cos’t+sin?t=1
79
80
Coordinate systems
Differentiating the equation of the ellipse with respect to x, you get
ax
ec
dy
Va es
pr
dy
dy
bx
ary
So at (acost, bsin ft):
dy
ao
dx
Since cot
bcost
b
—= =— cot?
asin t
a
= cot 3!=0, the tangents to the ellipse at 1=5 us and
t=3x+, that is, at (0,5) and at (0, — 5), are parallel to the x-axis
and since cot0 = cotz = oo the tangents at t= 0 and t=7,
is, at (a,0) and (—a, 0) are parallel to the y-axis.
that
The curve looks like this, for a > Db:
)
Notice that for a = b, the equation becomes x” + y* = a’, which is
an equation of a circle. So a circle is a special case of an ellipse.
For a> b, tae lines AB and CD are called the major axis and the
minor axis, respectively of the ellipse.
Equations of tangents and normals to an ellipse
The tangent at P (acost,bsin t) to the ellipse has gradient
and its equation is
y — bsint
=
—
bcost
—
sic
bcost
—(x —acos?)
asint
ay sint — absin? t = —bx cost + abcos’ t
bxcost+aysint = ab(sin° t+ cos’ t)
and since sin? t+ cos*f= 1, you have on dividing by ab:
a
a
Vx
— cost += sint = 1
a
b
which is an equation of the tangent to the ellipse at P(acost, bsint).
Coordinate systems
Notice further that if h = acost and k = bsint then the equation
of the tangent at the point (h,k) is
Xx
(h
a\a
a
That is
fe
iy
bby
ua AY = 1
a
b’
Remember this equation by writing the equation of the ellipse as
AX
py
= =f = =1 and replacing one x by / and one y by k.
The normal at P(acost,bsint) has gradient
nt
a
eres
which is the
negative reciprocal of the tangent’s gradient, and its equation is
y= bsnt=
a
That is:
asint
bcost
(x — acos ft)
ax sint — by cost = (a” — b?) sintcost
is an equation of the normal at P(acost, bsinf).
The focus-directrix property of the ellipse
An ellipse is often defined as the locus of a point P which moves
in such a way that its distance from a fixed point S (the focus) is
in a constant ratio e, where 0 < e < 1, to its distance from a fixed
line (the directrix). The constant ratio e is called the eccentricity
of the ellipse. For a parabola e = | and for a hyperbola e > 1, as
you will see in section 4.3.
Take the points A(a,0) and B(—a,0) on the x-axis with origin O.
The point S(ae,0), where
0 < e <1
lies between A and O, and is
taken as the focus and the line x = = which meets the x-axis at
e
D(<,0) is taken as the directrix.
e
”*
i
B (a, 0)
O
81
82
Coordinate systems
a(l-—e)
a-—ae
=e
==
AD
rig
aie =e)
SA
:;
Notice that
so A lies on the ellipse.
BS
ade
_ a(1+e)
BD
ats
“(1 +e)
Also:
so B also lies on the ellipse.
Suppose that P(x,y) is on the ellipse, then
PS_e
PE —
;
where E lies on the line with equation
Then:
d
and
e
/PED=90°.
;
PS =e PE
D)
=)
=
(*«—de) oy
2
7/4
-
=e (-— x)
e
x? —2aex+are* + we = a — 2aex + ex?
=
x
(l-e)+
y= a (1 — e”)
x
y
++ art fare BSG
That is:
If you
x =-
a
ane")
b° Sor — e”) in this equation you get the
ama:
usual equation of the ellipse, = He =1. This should help you
a
to remember that
al
substitute
b? = a(1 — e*)
The focal distances property of the ellipse
The symmetry of the ellipse about the x-axis and the y-axis gives
you the position of a second focus at (—ae,0) and a second
mee
;
ad
directrix with equation x = —-.
e
Coordinate systems
The ellipse shown has equation
a
2 +
f
is
Be =
1, the foci S and 7 are
;
at (tae,0) and the directrices have equations
a
x = +-.
e
The points
A(a,0) and B(—a,0) are at the ends of the major axis (the longer
axis). The foci lie on the major axis and the directrices are
perpendicular to the major axis.
For any point P on the ellipse, by definition for the focus S and
;
5
directrix
a
x =-:
e
PS =ePe
(1)
and for the focus 7 and directrix x = =
PY Serr
(2)
Adding (1) and (2) gives:
PS
But
So:
=
PILSH=APe aT)
PE+
ps+
2a
PF = EF=—
2a
pT =e(=) eat |
That is, at any point on an ellipse the sum of the focal distances is
constant.
PS + PT =2a
83
84
Coordinate systems
Example 4
For the ellipse with equation
)
2
> a 5 =,
find
(a) the eccentricity
(b) the coordinates of the foci
(c) the equations of the directrices.
(a) Here, you have a’ = 4, b’ = 3 by direct comparison with the
:
2
p aaia
ellipse a + Shae IIe
Using 6? = a*(1 —e’):
3=4(1 —e*)
That is:
SA
6244
Ses
(b) For the standard ellipse, the foci are at (tae,0) and so for
the given ellipse the foci are at (+1,0), since
a= 2 and e = I
;
;
(c) For the standard ellipse, the directrices are
x
a
= +e
so in this
case where a = 2 and e= ; the directrices have equations x = +4.
Example 5
The point P(acos@,asin@) les on the circle with equation
x° +) =a’, and N is the foot of the perpendicular from P to the
x-axis, as shown. The point Q divides the line PN so that a = us
a
0<b<a.
As @ varies, find the locus of the point Q.
4
P (acos @, a sin 8)
The angle PON = 0 and OP =a, the radius of the circle with
equation x? + y? =a’, which has centre O.
Coordinate systerns
PN=asin@
and
Oe
rN
8
a
b
ON = —(asin@)
= bsin@
a
So:
.. The point Q has coordinates (acos 6, b sin 0).
As @ varies, the relations x=acos@,
y=bsin@
are the
parametric equations of the locus of Q and by using the identity
rye
cos’ 0+ sin? @=1 you have
as the equation of the locus of Q. So the locus of Q is an ellipse.
For this ellipse, the angle 6 is often called the eccentric angle.
This example illustrates also the relation between the ellipse with
Ps
equation
4
x
y
te
ee:
2
and
;
its
bounding
,
circle
with
equation
x + y? =a’, because respective y-coordinates on each are in the
ratio
—.
This bounding circle is often called the auxiliary circle
a
of the ellipse.
Example 6
P (acos 0, b sin 9)
The perpendicular from the origin to the tangent at P(acos 0, b sin @)
to the ellipse
a
tg
a
2
=]
meets the tangent in the point Q, as
shown. As 6 varies, find, in parametric form, the locus of Q.
From previous work (page 80) you know that the tangent at P to
the ellipse has equation
86
Coordinate systems
Ngai sia
a
b
_
The gradient of the line O@ 1s
‘
That 1s:
»—0)=
asing
;
sesh
and its equation 1s
bcos@
asin 0
0
bcos 60 2
ax sin 0 — bycos@ = 0
You can find the coordinates of Q by solving the equations of
PQ (the tangent) and OQ simultaneously.
On simplifying this gives for Q:
5)
ab- cos @
es
29
a sin’ 0 + b? cos’ 0
a
3}
.
bsin@
SS
a’ sin’ 0 + b? cos? 0
and these are also the parametric equations of the locus of Q as 0
varies.
1. The foci and equations of the directrices of an ellipse are
given. Find a cartesian equation of the ellipse in each case:
(a) Foci at (+4, 0), directrices
x = +9
(b) Foci at (4,0), directrices x = +2
(c)) Poerat (O}222), cdiréctriceswy = a6
(d) Foci at (0,410), directrices
2
y = +40
For the ellipse given, find the coordinates of the foci and
equations for the directrices:
x
3
ye
9
(b) x*
x* +4) a
,
;
Cyr
2
ey
(c) 2x°+y eal
oie
d)
—+—=1
(d) rac
=
Find the eccentricity of the ellipse with equation:
Dn
og
a)
ae
i
(a)
eG—$ =
—
]
(6) 3x75? = 1
4
ry
(a) <a
45 + ||
9
—
(b) —13 +3
(Oca
=
l
all
Prove that the tangent at (x;, 1) to the ellipse with equation
ax’ + by” = | has equation axx; + byy; = 1.
Coordinate systems
Find an equation of the tangent and the normal to the ellipse
at the point stated in each case:
(a) x* + 2y? =3 at (1, 1)
(b) 2x? + 3y? = 35 at (2, —3)
a
(c) jy
(d)rat
oe
eal (g, 2)
P
;
a Oy? =
13-4t' (2541)
The eccentric angle corresponding to the point (2, 1) on the
ellipse with equation x? + 9)? = 13 is @. Find tan 0.
Express in terms of a single parameter ¢ the coordinates of any
point on the ellipse with equation:
5
ag
==]
(bay? 465y7 = 1
(c) x? +25y? = 100
The tangents at the points (acos«, bsin«) and (acos f, bsinB)
d 2, y52
aR
ye”
he =1
on the ellipse with equation
are perpendicular.
Prove that
tana tan B = ——Op
The point P(acos 6, bsin@) is a general point on the ellipse with
2D
equation
\
=
ae +i=
i:
ihe
Toci@ares:S
«and» 7, and
@ is the
eccentricity of the ellipse. Prove that
|SP* — TP?| = 4a’ecos0
10
Find equations of the tangent and the normal to the ellipse
with equation 4x* + 25y? = 100 at the point P(5cos f, 2 sin).
11
Find cartesian equations for the curves given parametrically as:
Cae
COS Laas
(bhe¥ Sosins,
(c) Y= cost+1,.
12
OSU
y= 4cosi
y=2sn¢—1
An ellipse has focus 5(/5, 0) and equation
sna hy?
974
as
ES
|
The variable point 7(3 cos ¢, 2 sin f) is joined to S. The line STis
produced to P so that = = ;. Find the locus of P as ¢ varies.
88
Coordinate systerns
13
The ellipse with equation Po “7 Be = 1 passes through the
é
;
F
x
D)
y
D
points (2,4) and (2,3). Find the eccentricity of the ellipse
and the equations of its directrices.
14
Show that the line with equation y = 2x + 5 is a tangent to the
ellipse with equation 9x* + 4y* = 36. Find an equation of the
corresponding normal to the ellipse.
15
Find the coordinates of the foci and equations of the directrices
for the ellipse with equation 16x* + 25y* = 100. Find
also an equation of the tangent to the ellipse at (v5, 3).
16
Show that $(/3,0) is a focus of the ellipse with equation
3x° + 4y* = 36. The origin is O and P is any point on the
ellipse. A line is drawn from O perpendicular to the tangent to
the ellipse at P and this line meets the line SP, produced if
necessary, in the point Q. Show that the locus of Q is a circle.
17
The point P(7 cos t, 5sin ¢) is on the ellipse with equation
Ds
2
. The line through P parallel to the y-axis meets
49
25
the x-axis at XY. The point Q is on the line YP produced so
that YQ = 2XP. Find, in cartesian form, an equation of the
locus of Q, as ¢ varies.
18
The point P(acos 6, bsin 8) on the ellipse with equation
2
ie
Bes = 1 is joined to the point A(a, 0) and M is the mid-point
@'R
of AP. As @ varies, find a cartesian equation of the locus of M.
19
2
The tangents to the ellipse with equation
Spe
ei
e + Be
2 = 1) at the
points P(acos 0, bsin @) and QO(—asin 6, bcos @) intersect at the
point R. As @ varies, show that R lies on the curve with equation
20
The foci of an ellipse are at S(4,0) and 7(—4,0) and for any
point P on the ellipse SP + TP = 10. Find a cartesian
equation of the ellipse.
Zi
Prove that the line with equation
y = mx + c is a tangent to
2
the ellipse with equation ee
S
=
= lutvandoniyil
BP
am
+b? = c?.
Coordinate systems
4.3 The hyperbola
The curve given by the parametric equations
w
Xx =asect,
y=bdtant,
0<t<27
where a and + are positive constants and t 4 3,tF# 3" is called a
hyperbola.
*
Since
xe
sect =-—,
a
y
tant= ; and
sec-
A
.
f= tan“? = 1,
ake:
a
ES
a
LI,
fp
which is a cartesian equation of the hyperbola.
Now
y=b
so for real y:
([
)
~
—
e — 120, that is |x|
>a
No part of the curve lies in the region for which |x| < a.
If you differentiate the equation of the hyperbola with respect to
x, you get
Se 2yayy
ap
ar
be
>
dy
dx
5b2
= .@7y
So at the point (asec ¢, b tan ft):
dy
beasect
b
’
— = ——— = — cosec
dx
a@btant
and, as there is no value of ¢ for which cosect is zero (Book C3,
page 76), there are no turning points on the hyperbola.
Since cosec0 = cosecz = oo, the curve is parallel to the y-axis at
b= O-and
Aly = 2.
As
x — +00,
y > +00.
The curve consists of two branches, symmetrical about the x-axis
and the y-axis, as shown overleaf.
The line AB is called the transverse axis of the hyperbola. The
line CD, where C is (0,5) and D is (0, —A) is called the conjugate
axis of the hyperbola. The point O is known as the centre of the
curve.
89
90
Coordinate systems
Asymptotes
If you consider the equation of the hyperbola in the form
oe
pad (=
2 = )
2
you can see that
and for both
VD
Qos
x > 0 and y>0, y< =
The part of the hyperbola for which x > 0 and y > 0, then always
lies below
hyperbola
the line
y— a
y=
as
and it can
x
increases
be shown
indefinitely.
that for the
Under
these
circumstances the line y = = is called an asymptote to the curve.
A similar argument can be used in each quadrant and the lines
bx
y=—
a
bx
y=-—— are therefore equations of the asymptotes
a
2
of the hyperbola a - = =
ao
ab
and
= A detailed study of asymptotes which are not parallel to the
coordinate axes, such as for the hyperbola discussed here, falls
outside the syllabus, but you should be able to state equations of
the asymptotes for the hyperbola
Coordinate systerns
Here is a sketch of the hyperbola and its asymptotes:
as
y
Se ;
)
-
4
4
4
4
Equations of tangents and normals to a
hyperbola
The tangent at P(asect,btant) to the hyperbola
b
: cosec t (see page 89) and its equation is
has gradient
b
y — btan t= — cosect(x
— asec?)
a
y
=>
xX:
—+-— tant =—cosect — cosecfsect
a
Multiply by tan sr:
5]
y
2
x
9)
— tant
— tan” tf = — sect
— sec’ f
a
Use sec” t — tan? ¢= 1 and rearrange:
x
~ sect
a
a
—~ tant =1
b
This is an equation of the tangent to the hyperbola at (asec f, b tant).
Notice further that if h = asect and k = btant, then the equation
k) is
of the tangent at the point (A,
gw That is:
a
Cages
91
92
Coordinate systems
Remember this equation by writing the equation of the hyperbola as
S — S =1
and replacing one x by / and one y by k.
The normal at P(asect,btanf) has gradient
—
asin t
ae
ee
which is the
negative reciprocal of the tangent’s gradient, and its equation is
asint
y—btant=-—
a
That is,
(x — asec t)
ax sint + by = (a* + 5’) tant
is an equation of the normal to the hyperbola at P(asec t, b tan ft).
The focus-directrix property of the hyperbola
A hyperbola is often defined as the locus of a point P which
moves in such a way that its distance from a fixed point S (the
focus) is in a constant ratio e, where e > 1, to its distance from a
fixed line (the directrix). The constant ratio e is called the
eccentricity of the hyperbola.
Take the points A(a, 0) and B(—a, 0) on the x-axis with origin O. The
point S(ae,0) lies on OA produced, where e > 1, and is taken as
:
a
a
the focus and the line x =—, which meets the x-axis at p(= .0)
e
€
between O and A, is taken as the directrix.
B(—a, 0)
Notice that
O
—
D
=
So A lies on the hyperbola.
A (a, 0)
eee
S (ae, 0)
ep
x
Coordinate systems
Also:
ee eee Bae lav
BD
a+ 2
ee
ee)
so B also lies on the hyperbola.
Suppose that P(x, y) is on the hyperbola, then:
oa
PE
where E£ lies on the line
x = : and (PED = 90".
e
So:
PS* =e PE’
ay
(Gex)"5 yr o) Ste 2 (x-=)
SS
=>
e
2
i
2=e'x? —2aex+a’
x°—2Jaext+are’+y*
=>
That is:
xe = lh=y* =a (e* - 1)
5 See
eae
If you substitute b> = a°(e* — 1) in this equation you get the usual
equation of the hyperbola,
2
i
= — = =1.
a
This should help you to
remember that
b* = a*(e? — 1)
The symmetry of the hyperbola about the x-axis and the y-axis gives
you the position of a second focus at (—ae, 0) and a second directrix
a
with equation x = —-.
e
Example 7
For the hyperbola with equation
ey?
So a
1, find the eccentricity,
the coordinates of the foci and equations for the directrices.
Since a =9, b?> = 16 and
6=96-9
be =a’ (e7 - 1) you have
=
9e=05
=
223
Foci are at (ae, 0) that is, at (+ 5,0)
Direcitices ate,
eae)
x=+%.
X= =
so the equations of the directrices are
93
94
Coordinate systems
Example 8
.
The normal at P(asect,btant) on the hyperbola with equation
ZB
p
aco!
meets the x-axis at Q and the y-axis at S. The
a
quadrilateral
OQORS,
where O is the origin, is a rectangle. As f¢
varies find, in cartesian form, an equation of the locus of R.
My
The normal at P has equation
ax sint + by = (a + b’) tant
(see page 92).
The point Q is where
That is, Q is at
y=0 and
Phe
(:a
a
sin¢
axsint= (a + b°) ve
sec f, |
2
2,
Similarly, Sis at («.
a _ tan a
Since OQRS is a rectangle, the point R(x,y) is given by
a+b
ye
x=
a
secf,
y=
b
tant
and as ¢ varies these are parametric equations of the locus of R.
For R:
SCO =
aX
yee
Gb
t
i
antl
by
So
be
a
Substitute these in the identity sec?¢= 1+ tan?
-)2
(ax),
_(oy2
(a? + b’)
(a? +b?)
That is, the locus of R is the hyperbola with equation
ax? — py? = (@ pe
Coordinate systems
4.4 The rectangular hyperbola
You will recall that for a = b the ellipse with equation
s
ie =
=
is a circle centre O and radius a with equation x7 + yp=a.
aed
ee
a
ae
e
2 =1
Similarly, when a= 5 the hyperbola
has equation
=
= 7, and, in this special case, the curve is called a
rectangular hyperbola. The asymptotes have equations y = +x and
are at right angles to each other — hence the name ‘rectangular’
hyperbola.
Since
a = b and b° = a’(e*
— 1), you have
a =a (e* —1)
=
legaf
2
feat
=
2a
The eccentricity of a rectangular hyperbola is V2.
The graph of the rectangular hyperbola with equation x7 — y* = a°
looks like this:
Remember that the eccentricity is /2 and the asymptotes
equations y = x and y = —x.
have
The curve with equation xy= c’ is another form of rectangular
hyperbola; it is, in effect, similar to the first one but the coordinate
axes are also the asymptotes. The curve looks like this:
z
a
The eccentricity is /2 and the asymptotes
and: y= 0;
have equations x = 0
95
96
Coordinate systems
Parametric equations, tangents and normals
The
rectangular
hyperbola
x? —y*=a’
can
the
have
parametric
equivalent
parametric
(asect,atanrf)
coordinates
equations of the curve are
and
a
y=atant, O<t<27
x =asect,
where ¢ 42, t 4 22 and a is a positive constant.
An equation of the
P(asec t, atanf) is
a
tangent
to
the
rectangular
hyperbola
at
xsect—ytant=—a
and an equation of the normal at P is
a
xsint
+ y = 2atant
You can derive these equations at once by putting b =a
standard results for the hyperbola
BX
2
in the
2
Ve
ie
1. ‘(pages Ot 592).
The parametric coordinates most widely used for the rectangular
hyperbola xy = c* are
C
ae
(cr,=), where c is a positive constant.
It is clear that, apart from ¢ = 0, all values of ¢ give you
é
ae
The equivalent parametric equations of the curve are
Coordinate systems
Differentiate xy = c” with tespect 10 x:
J
dy
-x—=0
~ aa
S|
dy
y
—=-—=
ax
x
e
At
P(ct,“),
Sea
t
dx
ee
ct
t?
and you can see from the sketch of the curve that
: <0
for all
Xx
non-zero
f.
The tangent at P has equation
ny
m=
That is:
} »>—--=——(x-c
2
t)
ie
x+t7y = 2et
The normal at P has equation
c
ye —
a
That is:
2
t“(x — ct)
tix —ty=c(t*-1)
Example 9
A line is drawn from the origin O at right angles to the tangent at
C
P(ct, =) to the hyperbola with equation xy = Cc meeting the
tangent at the point
cartesian form.
Q. As
¢ varies,
find
the locus
of QO in
The normal at P to the hyperbola has gradient ¢*. Since the line
OQ is parallel to the normal to the curve at P, it has equation
y =t’x. The tangent at P has equation
wan ty ess
The coordinates of Q are obtained
simultaneously to give
Zen
by solving these equations
2ct?
Tage
Hie
As t varies, these are the parametric equations of the locus of f.
You can get the cartesian form of the equation of the locus by
eliminating ¢ from these equations. This is equivalent to
eliminating ¢ from the equations
x+t?y=2ct
and
y= ix
97
98
Coordinate systems
and algebraically this is easier because
re
a
x
x+
(=
)y = 20
y
|=
b
which on multiplying by x and then squaring gives
ie =e y?)= 4c?xy
This is then a cartesian equation of the locus of Q.
1 The foci and equations of the directrices of a hyperbola are
given. Find a cartesian equation of the hyperbola in each case.
(a) Foci at (+13, 0), directrices x = ag
(b) Poct at ((2:4, 0), directrices. x= +4
(c) Foci at (0, +,/ 5), directrices y = ++a
(d)* Foci at ((0,2), directrices
py= +1
Show that equations of the asymptotes of each hyperbola
given in question | are respectively
(Ayie2
ee aya 0
(Cis
(b) /7x+3y=0
(d) y= =x
For the hyperbola whose equation is given, find (i) the value
of e (i1) the coordinates of the foci (iii) the equations of the
directrices (iv) the equations of the asymptotes.
(a) 3x7-y?=3
(b) 8y?-x? =16
wen
(c) apo
Prove that the tangent at (x), y;) to the hyperbola with
equation
ax* — by” = 1 has equation
ax ,x — byy = 1.
B
z
P is any point on the hyperbola with equation
9
Pps
ae
Ee - = =),
Sis the focus (ae,0) and S’ is the focus (—ae, 0). Show that
|SP — S'P| = 2a
Coordinate systems
6 Find an equation of the tangent and an equation of the
normal at the point given to the hyperbola with equation:
(a) x7 = 2)? = 1 at G, 2)
(Cy 32 = 5x?'= 7 at (1, =—2)
(by
= 27" SP are, 1)
Find a cartesian equation of the curve given parametrically as:
(ay x= 4seci y= tan.
f= ont se 3) BS
(D) 2 = Stan fay = 3007,
2 0
(c=
t+i,yat—2t40
(d) x= acosht,
y =Osinht, a> 0
(¢) 2=44,7=
~™
|Ww
Prove that the triangle formed by the asymptotes of the curve
with equation x* — 2y* = 4 and any tangent to the curve is of
constant area.
The foci S and S’ of a hyperbola are at (5, 0) and (—5,0)
respectively. Given that the eccentricity is 3 find
(a) a cartesian equation of the hyperbola
(b) an equation of the tangent to the hyperbola at (2, 6/2).
10
Find an equation of the tangent to the curve with equation
xy = 12 at the point P where x = 6. The tangent meets the
coordinate axes at Q and R. Show that PO = PR.
11
The normal at P(6,2) to the curve with equation xy = 12
meets the curve again at Q. Find the coordinates of Q.
12
The tangents at (or. “) and
QO (ca<) to the hyperbola
with equation xy = c * meet at the point R.
14
2c
Show that at R, oe See and
yi=
: :
p+4q
pr+4d
Find the locus of R when p and q vary and
(a) pqg=2
(b) ptg=2
(c) p’+q°=2
giving your answer, in each case, in cartesian form.
13
Show that the line with equation
y + x + 1 = 01s a tangent to
both the hyperbola with equation xy = 4 and the parabola
with equation y? = 4x.
Find also the distance between the points of contact.
99
100
14
Coordinate systems
Fae
6
c
Show that the chord joining P(ct -) and Q (cr,| where
l
2
equation
has
c?
=
xy
equation
with
curve
the
on
P and Q are
x+
thhy =c(t) + t>)
Given that ¢; and f> vary but that PQ always passes through
the point (2c, 2c), show that the locus of the mid-point of PQ
is the curve with equation xy = c(x + y).
15
Given that S and S’ are the foci of the hyperbola with equation
ae ee
= 1, show that SP and S’P are equally inclined to the
a)
tangent at any point P on the hyperbola.
16
The variable point P(asec t, b tan f) is on the hyperbola with
re
equation en
a
= 1 and Nis the point (3a, 3b). The point Q
lies on the line PN so that PO = 2QON. As ¢ varies find, in
.
a
ae
=
.
.
cartesian form, the locus of Q.
2
’
,
Ly, The normal at the point (2 "| to the curve with equation
xy = 4 meets the lines with equations y = x and y = —x at the
points QO and R respectively. Prove that PO = PR.
18
j
;
5
The normal at P(ct, _) to the curve with equation xy = c~
meets the curve again in the point Q. Show that the length of PQ
4\3
isfag 5© (hee.
19
Explain why the parametric coordinates (acosh ¢, asinh ¢) can
represent only one branch of the rectangular hyperbola with
equation x* — y* = a’. Find similar parametric coordinates for
the second branch.
The normal at P(acosht,asinh?), where ¢ > 0, passes through
the point (4a, 0). Find the non-zero value of ¢, giving your
answer as a natural logarithm.
20
C
C
The tangents at P(on, -) and ofcn<) to the rectangular
2
hyperbola with equation xy = c” meet on the rectangular
hyperbola with equation xy =
—. Prove that PQ is a tangent to
&|
the curve with equation xy = 4c?.
Coordinate systems
4.5 Intrinsic coordinates and radius
of curvature
A curve does not have a constant gradient. The gradient of a
curve is continuously changing. In fact the gradient of a
continuous curve with equation
y=f(x) at a general point P is
defined as the gradient of the tangent to the curve at P.
a
O
Here is the curve with equation y = f(x) and with the tangent drawn
at the point P(x,y). From some fixed point 4 of the curve the length
of the arc AP is s and the angle w, often called the gradient angle, is
the angle made by the tangent to the curve at P with the positive
direction of the x-axis. It is clear that for every point P on the curve,
x and y continuously change and so do s and w.
;
:
my
,
At P, you know that the gradient of the curve is esand the gradient
Be
of the tangent to the curve is tanw. So:
dy
=
dx
a
an 17
From your work in section 3.7, you know that at P
ds\"_
4
dxy
But
SO:
ti
or reciprocating,
since
2
(=) =1+tan’y
dx
see
=
tan? Wy
ds
7
ee
7" = &3
sec W
y ou have
dx
a
:
réls
d
That 1s:
dy
d
meee
1
101
102
Coordinate systems
Similarly, you can show that
dy
.
ia sin os
&
As you already know, (x,y) are the cartesian coordinates of a point
P. In the same way (s, ) are called the intrinsic coordinates of P.
The differential relations
x
— =tany, <= cosy)
and
dy
:
=.= sin
make it possible to move from an equation of a curve in cartesian
form, y = f(x), to the equivalent equation of the curve, s = g(wW),
in intrinsic form. In many cases, equivalent forms can be very
complicated and unhelpful, but there are a few curves with
important properties which are best illustrated and investigated
by using intrinsic coordinates.
At the point P on the curve with equation y = f(x) where s and w
are defined as before, the rate of change of s with respect to ys, that
ocean
:
f
iS dy’ is called the radius of curvature of the curve at P. With the
.
conventions
used
:
here, s increases
;
with
x and
therefore
ds.
dy 1S
positive if yw increases with x and is negative if w decreases with x.
If you are given the cartesian equation of a curve, the question
arises about how to find the radius of curvature.
;
d
The radius of curvature is p = ay and you have
d’y
Now:
= a
d
(tan) = sec” y a
d*y
ty
dy
eee
dx
dx
sec*y
a
de
sec”
wey
dx?
N
=
$
dx
ds
=
— =
ds
COD ss dx
cosy
—_
dy = tan.
dx
Coordinate systems
Se
_
ds
ds”
“dx
But
1 + tan’ wy = sec” W
so:
sec = ,/(1 + tan? p)
Ss
3
5
dy
p=
So
pale
dx
1
a
dy
Tats (2)
y)P=
(1 4+ tan
sec Y=
and:
3
:
HAS
2) |
ay
dx
For a curve given in parametric coordinates, say x = f(t), y = g(t)
dy
Gy.
‘di
eee — ———
7»
Patt
where
ax
Xx =——
7
and d
xX
Pay
ee
jy ——— F
dt
ee \x
dx? dx
d
ae
x2
_
80)
dt
~ Cords
at dx? di dx
x2
=
eeSs vx 2
*3
xy—yx
x where
2
(3fe
are
2
d*y
and Youn
103
104
Coordinate systems
iw
eo
Be
a
x
:
5
x? “+ y
5
f
=
ese
ee
[x? + y?|
ras)
C=
3
SS
xy-—yx
In your examination you may be given the equation of a curve in
cartesian, parametric or intrinsic form and asked to find the
radius of curvature at some point on the curve.
Example 10
The point A(0, 1) lies on the curve with equation y = coshx.
Measuring the arc length s from 4A, find the intrinsic equation of
this curve.
dy
y= coshy =
dx
—=—sinhx
dy
xX
;
w=
and
0
is
dx
Die
dx
Thatisaees— |(1 + sinh? x)’dx = |cosh
x dx = sinh x
0
0
(because 1 + sinh? x = cosh’ x)
Bu
d
= tab
dx
and
9s = sinh y= gy
dx
So: s=tanwy, which is the intrinsic equation of the curve. This
curve is called a catenary.
Coordinate systems
105
Example 11
A cycloid has parametric equations
x=tf—snty
= l—cost
Show
that
Hence find the intrinsic equation, where the arc length s is measured
from the point where t = 0.
dx
—=x=1-cost
dt
dy
—=y=sint
dt
J
and
dy
yp
sin ¢
dx
x
1—cost
Using the double angle formulae (Book C3, page 77)
sint = 2sin 5tcos 51
cost =1-—2sin° 7 }t
and
y
dy
2sinitcosit
ee
=
ou have:
dx
2 sin? ly
cosit
cont
sin it
The arc length from O (0, 0) to P (t—sint, 1 — cos f) is
t
1
sl pe +r far
0
t
1
|(1 —2cost+ cos’ t+ sin? t)’ dt
0
:
1
= |(2 — 2cost)?dt
0
t
= |(4 sin? Ly) idt
0
t
=| 2sin +td¢
0
Hence
s = [—4cos ap
as
= —4cos $1 + 4cos0
=4-4cos}t
106
Thus
Coordinate systems
s=4-—4cos(}1—-w) =4-—4sinp
The intrinsic equation of the cycloid is
s=4(1 —siny)
Example 12
Find the radius of curvature at the point where
y=4
and at
the point where
y=
on the curve with intrinsic equation
s=4(1 —siny).
;
For the curve s = 4 — 4siny,
d
p= Gy= ~4cos
At W =%, p =|—4cos
#| =2V2
At Wy=%, p=|-4cos
#| =2
Example 13
Find the radius of curvature at
hyperbola with equation xy = 4.
P(1,4)
on
wats yaa
z
dy
and
5)
eed.
at 1.4)
the
rectangular
Coordinate systems
Example 14
The point P lies in the first quadrant at t¢=arctan+,
on the
ellipse with equations x = 3cost,
y= 2sint, 0 <¢t < 2m. Find the
radius of curvature at P.
x=
3C0sf,
§==SSint,
X=
—3 cost
y=2sint,
y=2cost,
y=-—2sint
t= arctan
5=>tant=}
sinini
¢
=
alld
COs at
t= Fe
V5
Hencee at
t= arctan 5 (first quadrant):
= be iu
aad
5"
a A
y= BS
6
2
(4? + i)?
Radius of curvature = ~————~
xy—ypx
) i Jee
aa
7 ce
cans
1
3
aes
Bee
Given that the arc length is measured from the point (0, »),
find the intrinsic equation of the curve with cartesian equation
4y = cosh 4x.
2
Measuring arc length from the origin O, find the intrinsic
equation of the curve with cartesian equation y = Insec x.
3
Measuring arc length from the origin O, find the intrinsic
equation of the curve given parametrically by
xX=21+2sint,
4
y=2—2cost
Show that the curve with intrinsic equation s = wis a circle,
radius |.
;
107
108
Coordinate systems
5 The intrinsic equation of a curve is 5 = >. The curve passes
through the point (2, 0). Show that x = 2(cosw + wsiny) and
find y in terms of w.
For the curve with equation sin y= e*, where s is measured
from the point (0, 5), show that e* = tan y
Find the radius of curvature at the point (\/2, 2) on the curve
with equation y = x.
Find the radius of curvature at the point (3, 3) on the curve
with equation 3y* = x°.
Find the radius of curvature at the point where x = 4 on the
curve with equation
10
y = x*-—x +1.
Find the radius of curvature at the point t = 7 on the curve
with equations
x=3cost—cos3?,
y=3sint—sin 3?
where ¢ is a parameter.
11
Find the radius of curvature at the point where ¢ = In2 on the
curve with parametric equations
w= Cosh
12
—7,
y=cosnt
+7
A curve has intrinsic equation s = atan? w, where a is a
positive constant. Find the radius of curvature of the curve at
the point where (a) W = | (b) w =.
13
For the curve given by the parametric equations
x = 2cos*t,
=
2 sin? PomOizaneel
find the radius of curvature at the point where f¢ is
(a)
14
Fb) &
The intrinsic equation of a curve is s = 12 sin? w, where s is
measured from the point (—8, 0).
(a) Find the radius of curvature at the point where y = 7.
(b) Show that x3 + y3 = 41s a cartesian equation of the curve.
15
The parametric equations of a curve are
KS
Cosi
tsink
pesmi
7cost
0-7 <<
Find the radius of curvature at any point on the curve in
terms of ¢.
16
Show that the curves with equations x? — y* = 3 and xy =2
intersect at right angles at the points P(2, 1) and O(—2, —1).
Show that the radii of curvature of the curves at P are in the
ratio 4:3.
Coordinate systems
17
A curve has intrinsic equation s = asec* w, where a is a
positive constant. Find the radius of curvature of the curve at
the points where y = Fand wy =%
18
For the curve with equation y = sin? 2x, find the radius of
curvature at the origin O and at the point P where x = 5.
19
Find the radius of curvature of the curve with equation
x3 + y? =4xy
20
at the point (2, 2).
Find, in terms of a and d, the radius of curvature of the ellipse
with equation
x? n y? =
a
at t th the point A
SUMMARY
ob
:
ah We
OF KEY POINTS
1 The parabola with equation y? = 4ax has focus at (a, 0)
and directrix with equation x +a = 0.
Parametrically the parabola with equation ye = vi is
given by
x = at’, y= ial
3 The tangent at (4, k) to the parabola has the equation
ky Stole
hy
and the tangent at (at’, 2at) has the equation
fy =x + at
The corresponding normal has the equation
yt+tix= at + at
4 The
ellipse
with
equation
saus =1,
where
b? = a (1-— e*), has foci at co ae,0)and apes with
d
equations x = + x
x
> Parametrically the ellipse with equation ie os
aye by
x= acost,
y —dsint,
051<20
109
110
Coordinate systems
6 The tangent
2
xy
at (4,k)
to
the
ellipse
with
equation
—--+-<~= 1 has equation
oy
4
hx
hy
tbe
and the tangent at (acos¢, bsin ¢) has equation
x
v..
—cosi+ > sint= |
a
b
The corresponding normal has equation
ax sint — bycost = (a — b°) costsint
7 The
hyperbola
with
equation
~-—-+.=1,
where
b’ =a’(e’ — 1), has foci at (+.e,0) and directrices with
equations x = ee
2
8 Parametrically the hyperbola with equation
is givenby
x% = asecl,
with
t #41
y = biani,
2
=a
eo
0< 1 < 2x
A 3,
:
b
9 The hyperbola has asymptotes with equations y = =x.
10 The tangent at (4,k) to the hyperbola has equation
Ax
ky
ec
and the tangent at (asec f, b tan f) has equation
x
= sect
tant = l
a
b
The corresponding normal has equation
axsint + by = (a +b°) tant
11
The equations x? — 8 =d
and xy =’ each represent a
rectangular hyperbola with e = /2 and asymptotes which
are perpendicular, and with equations
y= +x for the
first and y = 0 and x = 0 for the second.
id
:
c
FOrx) = oF parametric equations are x = ct, y = ; iz 0.
2
For x? -—y = a, xX = asecl,
representation often used.
y= atant
*
.
is a parametric
Coordinate systems
13 The equations of the tangent and the normal at
,
(cr,-)
to the rectangular hyperbola with equation xy = c* are
| respectively
xt Py =2ct
=ty=e(t*—1)
14
The equations of the tangent and the normal at
(asect,atant)
to the rectangular hyperbola
with
equation x7 — y?=a’ are respectively
xsect—ytant=a
xsint+y=2atant
15 The parabola, ellipse and hyperbola are each loci of a
point P which moves so that its distance from a fixed point
(the focus) is in a constant ratio (e, the eccentricity) to its
distance from a fixed line (the directrix). This focus—
directrix property is such that e=1 for the parabola,
0 < e < 1 for the ellipse and e > 1 for the hyperbola. For a
rectangular hyperbola e = V2.
16 Intrinsic coordinates —
O
x
The length of the arc AP on the curve shown is s, the
tangent to the curve at P makes an angle w with the
positive
x-axis
and
(s, w)
are
galled the
coordinates of the point Pf.
_ In particular,
:
oi7
dy
oo ny,
ay
oe Met
dy.
Pee yak
{ne radius of curvature, p, at P(x, y) is
d
GY)
1
ee
dx?
2
3
eat
:
213
ip — %
intrinsic
111
Review exercise
1 Find the equation of the tangent to the parabola with
equation )* = 4ax at the point T(ar’, 2at). If S is the focus,
find the equation of the chord OSR which is parallel to the
tangent at 7. Prove that OR = 4TS.
2
[iE]
Starting from the definition of cosh x in terms of e*, find, in
terms of natural logarithms, the values of x for which
3
3coshx
[E]
3
Find the minimum value of (Scosh x + 3 sinh x).
4
Given that /,(t) = |sin” x dx, where n > —1, use integration
0
by parts to show that, for n > 1,
[E]
t
nl,(t) = —sin”~'t cos t+ (n — 1l)In_2(f)
Hence show that
x
Find (a) |‘sin*xdx
(b) |sin’xdx
0
[E]
0
5
f(x) =
|
(x + 2)(x? + 4)
Express f(x) in partial fractions.
Show that
;
Ge+ 2102)
|iiyjax = =
6
er
[E]
Using the substitution ¢ = tan x, evaluate
us
4
]
|
a
x
0 3cos?
x + sin* x
giving your answer in terms of 7.
[E]
114
Review exercise
Write down an expression for tanh x in terms of exponentials.
Hence, or otherwise, find the exact value of x for which
[E]
5:
tania
arsinh |
sinh” x dx, prove that, for n>2,
Given that J, = |
0
ni,
=
J/2—(n-
1 )In—2
[E]
Given that y = arcsin a where a is a constant, show that
a
d y
a
dy
ay
|)
dx? (=
10
3
SO
E
(FI
dx
(a) Find |
sinhx+2coshx
(b) Show that
4
3x-1
ie
aay:
|
-dx = 9(./2 — 1) + 2arsinh 1
L,= |Povo dia
11
(E]
0
0
By writing x”\/(1 — x?) as x""!{x,./(1 — x)}, show that
(n+2)f,=(n-l)h,_-2,
neB2
Hence evaluate Js.
[E]
12
The diagram shows a sketch of the parabola C with equation
y? = 8x and the line L with equation x = k, where k is a
positive constant.
(a) On a copy of this sketch, shade the regions for which
(C= kK)
ox)
0
Given that L passes through the focus of C,
(b) write down the value of k.
[E]
Review exercise
13
x
Write down the value
val o f len
|———~dx.
=
x
Hence, or otherwise, find |aresin x dx
Arh
eS aoneeemnoeereame
fa1SE,d
14
Sh
15
Solve the simultaneous equations:
Seen
|3sin2x+4cos2x
Pree |:
5 ae
[E]
(E]
cosh x + cosh y = 4
sinh x + sinh y = 2
giving your answers in terms of natural logarithms.
16
[E]
The foci of the ellipse with equation
x
2
y
2
are S, S’. Show that the normal at a point P on the ellipse is
equally inclined to the lines SP, S’P. The perpendicular from
the centre O of the ellipse to the tangent at P meets SP,
produced if necessary, in G. Show that the locus of G is a
circle of radius a whose centre is S.
17
(a) Show that
2
1
[E]
dx = 41n3.
ee
1
(b) Use the substitution x = — to find the value of
u
9
ie
|¢ Se
a
|
ae
2)
18
E
LE]
Use the substitution t = tanx to evaluate
eerpreety
9 sin 2x%++.3
T
giving your answer in the form 2 where p and g are integers.
q
19
fox -aresinx,
—lL< x < 1
(a) Evaluate f(—+)
(b) Find an equation of the tangent to the curve with
equation y = f(x) at the point where x = mre
[E]
115
116
20
Review exercise
x
Given that y = arctan —, where a is a constant, show that
a
dy
tany
“ax 1 + tan? y
V3
eaE3
21 Show that | eeey
1 V(4— x?)
22
é
Solve the equation
4tanh x — sechx = |
[E]
giving your solution in logarithmic form.
23
Evaluate
3
l
;
oe 5) x
(a) |Tae
1
(b) |arsinh
x dx
0
24
[E]
Prove that cosh? x — sinh” x = 1.
Solve the equation
3 sinh?x— 2coshx -2=0
[E]
Give the values of x as natural logarithms.
25
1EGe\— |tan"0d0, n>0, |x| <5
0
By writing tan 0 as tan”~* @ tan? 6, or otherwise, show that
Lxy)= 7
l
an
£
OD
Teoh) ah
Xtc
T
3
Hence evaluate |tan’ @d0, leaving your answer in terms of z.
0
[E]
26
The curve C has equation y = 2 x3, The arc of C from (0,0) to
(1, 2) is rotated through 27 about the x-axis. The surface
generated has area S.
Show that
!
S= “| x1
+ x)dx
Use the trapezium rule with 5 equally spaced ordinates to
find an approximate value for S, giving your answer to
2 decimal places.
[E]
Review exercise
27
A hyperbola of the form
has asymptotes with equation y? = m*x* and passes through
the point (a,0). Find an equation of the hyperbola in terms of
Xv y,@ and m.
A point P on this hyperbola is equidistant from one of its
asymptotes and the x-axis. Prove that, for all values of m,
P lies on the curve with equation
[oe = y)= Axe ey — a’)
28
[E]
Express tanh x in terms of (a) e* (b) e*.
Hence, or otherwise, prove that
Ce.
He
;
tanh x + tanhy
1 + tanh x tanh y
Determine whether tanh x is an even function, an odd
function or neither.
Starting from the expression for tanh x in terms of e* show that
d
— (tanh x) = sech? x
dx
Show that the volume of the solid generated when the finite
region enclosed by the curve with equation y = tanh x, the
line x = +1 and the x-axis is rotated through 272 radians
about the x-axis is pais)
Ls
29
Sketch the curve whose parametric equations are
x =acos*t,
y=asin't,
Oa t= +t, a0
The points A and B on the curve correspond to the values
t = 0 and t= 4n respectively. Calculate
(a) the length of the arc AB of the curve
(b) the area of the curved surface generated when the arc AB
is rotated through 2z radians about the x-axis.
30
[E]
(a) Given that sinh y = 1, find the value of cosh y.
arsinh |
(b) Given that J, = |
sinh” x.dxy 7.2.0,
0
show that nl, =4/2 —(n— 1) I,-2, nS 2.
Hence evaluate J4, leaving your answer in terms of surds and
inverse hyperbolic functions.
[E]
117
118
Review exercise
dx
31
(a)
Find
ec
acon
ax
(b) Find eee,
a0
(c) By using the substitution x = sin @, show that
[ x4dx Be Cire 7/3)
gay (1 = 27)
32
E]
64
Evaluate:
|
© |Tae
|a
E
© |yaaa”
33
a
(a) Evaluate
(b) Show that
3 3(x + 1)
| rere
34
ses
a
hind
43
[E]
If x and y satisfy the equations
cosh x cosh y = 2
sinh x sinh y = —1
show that x = —y = +In(1 + v2).
35
[E]
YA
O
:
:
x
The diagram shows the path of a stone embedded in the surface
of a tyre. This path is a curve with parametric equations
x=a(0-sin@),
y=a(l —cos@)
where a is a positive constant and @ is a parameter.
Find the length of the curve between x = 0 and x = za.
[E]
Review exercise
1
Given that J, = |Xe
0
36
Ux hv,
(a) show that na) =Hiw+ 1)=e7
(b) evaluate J3, leaving your answer in terms of e.
37
[E]
Prove that the gradient of the chord joining the point (or. ‘)
P
and the point o(ca, <) on the rectangular hyperbola with
q
;
‘
l
equation xy =c is =—.
Pq
The points P, Q and R lie on a rectangular hyperbola, the
angle OPR being a right angle. Prove that the angle between
OR and the tangent at P is also a right angle.
38
[E]
Given that t = tan @, show that
De
in20 = rae
(a) sin
dé
l
(6) dt
142
Hence, or otherwise, evaluate
5
—
real
}dé
giving your answer in the form pInq, where p and q are
[E]
numbers to be found.
39
Differentiate with respect to x:
(b)> ar arctan
(a) arccos 2x
pat
Pee?
[E]
Simplify your answers.
40
2
Given that J, = |x(In x)" dx, where n is a non-negative integer,
prove that, for n>1,
Votan leo
Evaluate
2
|x(In x)? dx
1
Alin)
[E]
119
120
41
Review exercise
A curve has cartesian equation 4ay = x? + 4a’, where a is a
positive constant. Find the radius of curvature of the curve at
the point (4a, 5a).
42
Find an equation of the tangent at the point P(4a, 4a) to the
parabola with equation y* = 4ax. Show that the coordinates
of the point R, where this tangent meets the x-axis, are
(—4a, 0).
The second tangent to the parabola with equation y? = 4ax
from R meets the parabola at Q. Obtain the coordinates of Q,
and calculate the area of the finite region enclosed between
the tangents RP, RO and the parabola with equation
y =Aax,
43
[E]
A circle with centre at the point P(h,k) touches the y-axis,
and passes through the point $(2,0). Show that P lies on the
curve with equation y* = 4x — 4, and sketch this curve.
44
[E]
A curve is given parametrically by the equations
es a(2 + ),
v= Zar
Find the values of the parameter ¢ at the points P and Q in
which this curve is cut by the circle with centre (3a,0) and
radius 5a.
Show that the tangents to the curve at P and Q meet on the
circle, and that the normals to the curve at P and Q also meet
on
45
the circle.
[E]
(a) Sketch the graphs of (i) y = sinhx (ii) y = tanhx.
Starting from the definitions of sinh x and cosh x in terms of
e*, show that cosh? x — sinh? x = 1.
Given that cosech x = — ran find the exact values of (ili) cosh x
(iv) tanh 2x.
(b) Find
l
(1) fe + 2x4 10)
¥
(11) |tanh 3x-sech3d
[E]
46
Use the substitution ¢t = sinhx to prove that
[a +e ar=far +02! +hinfa+ (1 + 02)
(E]
Review exercise
47
Using the definitions of sinh and cosh, prove that
artanh x = Sin
tor —1 <x
l+x
l-—-x
< 1,
j
d
(a) Find —(artanh
x).
dx
(Dy Sketen the sraph of » =artanh x for —1 = x =< 1.
(c) Calculate, to 3 decimal places, the area of the finite
region bounded by the curve with equation y = artanh x, the
x-axis, and the line x = Bi}
48
[E]
Are
a
Use the substitution x = sinh’ where a is a constant, to
sin
show that, for x
> 0,a>0,
|
|
ess dx = —~-arsinh (=) + constant
x/(x* + a’)
a
<
|
49
:
[E]
d-
Show that the curve with equation y = sechx has eats =) at
points where x = +Inp and state a value for p.
Sketch the curve with equation y = sech x.
[E]
T
50
Given that J, = |sin” x dx, where n EN,
Jo
(a) show that (n+ 2)J,42 =(n+
Dk.
(b) Evaluate |sin® 3¢ dt.
J0
(c)
(2n)!7
Show that 4, = ———_,
22n+1(n!)
(d) Show further that
1b, =—
4n
[E]
us
51
Given that /, = le cos xdx, prove that for n>2
0
I, = (§'-n@- Dhp—-2
Hence find J.
[E]
121
122
52
Review exercise
Show that an equation of the normal to the parabola with
equation y? = 4ax at the point P(ar’, 2ar) is
yttx = 2at+ at’
This normal meets the parabola again at the point QO(as”, 2as).
Show that
(a) stit==0,
) pg
53
T=
SE
Dae
#0
3
(F]
Find (a) Java - sys
0) |vid 2°)dx
54
[E]
Given that, for n>0,
1
F
L,= |eh
prove that, for n>1,
2(n ote 2)Ln =
(2n + n=
Show that /; = 7, and hence find J, and J;.
35)
[E]
Show that an equation of the tangent to the rectangular
hyperbola with equation xy = c* (with c > 0) at the
Cy
point (cr,=) 1S
a ie
a
Tangents are drawn from the point (—3, 3) to the rectangular
hyperbola with equation xy = 16. Find the coordinates of the
points of contact of these tangents with the hyperbola.
56
[E]
P is any point on the ellipse with equation 9x? + 25y? = 225,
and A and B are the points (—4, 0) and (4, 0) respectively.
Prove that PA + PB = 10.
Prove also that the normal at P bisects the angle APB.
[E]
Review exercise
57
(a) Express sinht and coshf in terms of e’ and hence show
that, if x = sinhr, then
(1) = = Cosh
Gi) f= In [x5
= 1),
(b) Using the substitution x = sinh¢, or otherwise, evaluate
the integral
1
{= |JCe* + 1) dx
0
1
IE}
(c) Evaluate |arsinh x dx.
0
58
Evaluate
=
59
.
0
l
e
ieWGetiw
L (3 = 2x = x7) Cee
az) dx
[E]
Find the values of x for which
Atanh- x—1=0
giving your answers in the form alnb, where a € Q and
beZ.
60
[EF]
ee
’
C
A curve is given parametrically by x = ct, y = -. Show that
Ce
an equation of the tangent to the curve at the point (ct, | 1S
HL
= ct
If P is the foot of the perpendicular from the origin to this
tangent, show that P lies on the curve whose equation is
(x? + y*)’= Ac? xy
a]
61
Given
7
that 1 a
[E]
1
| C2 +a)"
dx, where a and n are constants,
0,
(a) show that, for n 4 0,
2nd 2 Ina, == Ch
i
li, ear
Hence, or otherwise, evaluate
1
22
l
ee ee
ay
(x2 + 4) me
(P) |,
©)
]
|G2 22% 42)
dx
:
E
UE]
123
124
62
Review exercise
Find the gradient of the parabola with equation y? = 4ax at
the point P(at?,2at) and hence show that the equation of the
tangent at this point is x — ty + af = 0.
The tangent meets the y-axis at 7, and O is the origin. Show
that the coordinates of the centre of the circle through O, P
a
t
;
and 7 are (5 +a, 5) and deduce that, as ¢ varies, the
locus of the centre of this circle is another parabola.
63
[E]
An equation of the curve C is
n=
x40 — x)
where a is a positive constant.
(a) Sketch the graph of C, and find, in terms of a, the
coordinates of the stationary points and the coordinates of
any intersections with the coordinate axes.
(b) Show that the arc length of C in the first quadrant from
the point at which x = 4a to the point at which x = 9a 1s
[>)
fo
LlO+@)Je
and evaluate this integral.
64
[E]
(a) Using the same axes, sketch the graphs given by the
equations
y= sinh.
y=coshioe
xER
xeER
(b) Show that the area of the finite region bounded by these
curves and the lines
x = a,x =b,0<a<b,ise“-e”.
(c) Find the limiting value of this area as both a — 0 and
b — oo.
65
[E]
Given that y = e* sec x, show that
d
(a) ~ = yl +tanx)
dx
(b) ee
dy
66
de
an x) + ‘Vpsec?
SEC x:
Way
E
[ ]
Find the value of x for which
4tanh x — sechx = 1
giving your solution in terms of natural logarithms.
[E]
Review exercise
67
The curve C has parametric equations
x=a(0—cos#)
y=a(1 —sin@)
Show that the length of the arc of C from 0 = 0 to @ = Z is
given by
ce av2
J/(1 + sin 6) dé
0
By using an identity for | + sin @ in terms of half angles, or
otherwise, show that
s = 2a(/2— 1)
68
[E]
Show that an equation of the chord joining the points
P(acos@, b sing) and Q(a cos 6, b sin @) on the ellipse with
equation b?x? + ay? = a?b* is
bx cos +(6+
) +ay sin $(8 + $) = ab cos 4(6 — ¢)
Prove that, if the chord PQ subtends a right angle at the
point (a,0), then PQ passes through a fixed point on the
X-axis.
[E]
69
The diagram shows a sketch of the curve C with equation
y =coshx. The points P and Q on C have x-coordinates —1
and +1 respectively. The arc of C between P and Q is rotated
through 2z about the x-axis. Determine the area of the
curved surface formed.
[E]
eln 3
70
Show that | x sinh xdx =2In3 —%
0
[E]
71
(a) Express x? + 6x — 7 in the form (x + a)’ —b, where a and
b are integers.
(b) Using a suitable hyperbolic substitution, or otherwise, obtain
2
1
IVien
d
i we
giving your answer in terms of natural logarithms.
[E]
125
126
72
Review exercise
l
|x" e-* dx, where n>0, find the relation between J/,
0
and J, where = 1
I
i=
[E]
Express |x*e dx in terms of e.
0
gh, Given that J, = |seora dd, show by differentiating
0
74
sec”~?@tan 0, or otherwise, that for
n>2,
G=)G= Qe Dier
2?
[E]
Use the substitution f = tan5x and the formula
to evaluate
2
|
lames”
9 (1 + 3cosx)
giving your answer in terms of natural logarithms.
qs
[E]
Differentiate with respect to x:
(a) Incoshx
(b) sinh (x7)
(c) arcsin (2x)
76
‘
(a) Show that |Ine dia
I
4
(=).
(b) By using the substitution x = tan 9, or otherwise, show that
i
Ce
tell
0 (1+ x2)
q5)
ae
8
[E]
Starting from the definitions of cosh and sinh in terms of
exponentials, show that
(a) cosh? x — sinh? x = 1
x+1
(b) arcothx = 4im(3
Den
Nee asi)lt
(c) Solve the equation
cosech?x — 2cothx = 2
giving your answer in terms of natural logarithms.
3
(d) Find |arcoth
x dx, giving your answer in terms of natural
2
logarithms.
[E]
Review exercise
78
An arc of a parabola is given parametrically by the equations
PSH tat:
See
Meta 2
This arc is rotated completely about the x-axis. Show that the
area of the surface of revolution is
8na* (Sv/5 — 1)
3
Show also that the length L of the arc is given by
fis 2a | J/(1+
2) de
0
Hence, using the substitution t = sinh 0, or otherwise, find L.
[E]
fie) Given that J, = fer reoss sin” x dx, n>0, show that
0
i, =¢ =f
tel
Hence evaluate /3, giving your answer in terms of e.
80
[E|
Evaluate
i
ia
0
Ae ide
On a sketch, indicate a region R which has area J.
81
[E]
Starting from the definitions of cosh x and sinh x in terms of
e*, prove that
(a) cosh2x = Zeosh’ x=
I
(b) cosh? 2x + sinh? 2x = cosh 4x
82
[E]
By use of the substitution u* = 1 + x, or otherwise, find the
exact value of
che
|
=F
J3 x/(1 + x)
83
(E]
(a) Starting from the definition of tanh x in terms of e*,
prove that
l+x
aveanee — {1n( a 4
Z
l—x
Sketch the curve with equation y = artanhx, x ER.
(b) Solve the equation x = tanh[In ,/(6x)] forO<x<
1.
[E]
127
128
84
Review exercise
An ellipse has parametric equations
Sa wcos8,
v= hsind.
Ws 0 =e
ees
where b? = a?(1 — e’).
(a) Show that the length, s, of the arc of the ellipse from
A= 0 tod = 1 18 civen by
c= a| \/(1 — e? cos?)dé
0
(b) Write down the first two non-zero terms in the expansion
of /(1 — e* cos” 0), 0 < e < 1, as a power series in ascending
powers of e. Given that e is sufficiently small so that powers
of e higher than e” can be neglected,
(c) show that an approximate value of s is
na(1 — 4e7)
85
[E]
Using the substitution w* = x? — 1, or otherwise, evaluate
V5
x3
|Coes a
86
UF
The points P(ap*, 2ap) and O(aq*, 2aq) lie on the parabola
with equation y? = 4ax. The angle POO = 90°, where O is the
origin.
(a) Prove that pg = —4.
Given that the normal at P to the parabola has equation
Vo
= ao A 2ap
(b) write down an equation of the normal to the parabola
at Q.
(c) Show that these two normals meet at the point R, with
coordinates (ap* + aq* — 2a, 4a(p + q)).
(d) Show that, as p and g vary, the locus of R has equation
y =16ax = 96a"
87
[E]
Prove that
artanh A — artanh B = artanh
where A and B are real.
A= B
!— AB
[E]
Review exercise
88
Show that for all values of m, the straight lines with
equations
y = mx + \/(b* + a’m’) are tangents to the ellipse
ase
with equation =
[E]
a. a7= = 1,
89
Prove that, for all real x,
=|
90
[E]
Prove that the equation of the chord joining the points
|
(«.°)and. QO = (ca.| on the rectangular hyperbola
P
q
with equation xy = c? is
x + pay = c(p + 4)
If this chord passes through the point (cp + cq — c,c), show
that the tangents at P and Q meet on the line with equation
yes ke
[E]
129
De
<—-eoc
»
Oe on ep
or
nal we
=pqtit
/<
an
ie
we
1
|
|
7
iat
ed
oe
<8
@
Ee
==>
oo? oe
@
6d
*
3
=a8 2
rat i
-
=)
a
;
rere
|
een
li
ance
rT
die evi 8)
© pam |) tim
ee
a
n
r
7
¢,.—
&®
7
Ss oa
=
Oimeys.
Ope ww OS Ses
|
'
7
7
.
-
‘
Na
i
ume:
o Wie
lo
sy.
ee
eei
-
-
a
<<?»
Qa
=
—
oe@-—~==
22
-
- @)*')
Veco.
e®
exit
.
6446
a
_—
.
@
9
oper
~_
.
Ln ad fa
ea
|
hea
» sibel atnpienettor iar
= « =
a4 3:8
;
er
a?
av
:
I © Ge Se
ee
|=
Examination style paper
Answer all questions
L.
Time 90 minutes
Given that y = x”, prove that
a
ax
=
dy oe
el
7
k
(7 marks)
x
Given that sinh x = tant, 0 <t <4, prove that
(a) tanhx = sint.
(b) e* = sect + tant.
(4 marks)
(3 marks)
For the relation rote — 6x+ 13) =1,
(a) find |»dx in terms of x.
(4 marks)
3
(b) Hence evaluate iydx giving your answer in the form res
23
q
where p and q are integers to be found.
(5 marks)
Prove that the line with equation y = mx + c is a tangent to the
E
kee
ae
eee
hyperbola with equation—a:
— a = lif,and onlyif, am =b +c.
(9 marks)
The curve C has equations
x=f+sni,p= 1—cost,0
4 Ps 27,
where ¢ is a parameter. At the points O and A on C, t = 0 and
respectively and the radius of curvature of C at A 1s r.
t =
Show that
(a) (?+y)y= 2cos+t,
(4 marks)
(b) r =8.
(4 marks)
The arc OA of curve C is rotated completely about the x-axis.
(c) Find, in terms of x, the area of the surface generated.
(5 marks)
132
6.
Examination style paper FP2
The line with equation y=mx—2 meets the ellipse with
equation x° + 4)" = 16 at the points P and Q.
(a) Find the coordinates of the mid-point M of PQ, giving each
coordinate in terms of m.
(8 marks)
(6) As m varies find, in cartesian form, an equation of the locus
of M.
(6 marks)
Given that
Tt
Wee |x" sin5 x dx,
0
show that for n > 2,
I, = 4n(x)""! — 4n(n — 1)I,_>.
Evaluate J, in terms of z.
(8 marks)
(8 marks)
Answers
Edexcel accepts no responsibility whatsoever for the accuracy or method of working in the answers given for
examination questions.
Exercise
1A
coth x
>
(a)
T(e* — e~*)
(c)
ee 3
e 465?
(b)
a
=3)
(d) dev? + ev?)
=
(e) 4(e"-e-*)
2
By
(e+e
(f) (==
ei]
3(x? cosh 3x + x? sinh 3x)
l
144
re h haem
Hoe
x
x cos
sinh hae
cosh x — x sinh x
16
)
(cosh x x)e
1.818
(b)
—1.818
(c)
+0.962
(d)
+1.444
(e)
0.973
(f)
—0.805
la}
2.31
(b)
0.7172
(c)
—0.9640
(d)
-—0.5211
(e)
11.59
(f)
0.9287
= 2x cosech’
a 2x +a 3 coth 2x
ib)
yS el
x= 0
—cosech (x?) (2x? coth x* + 1)
Bae,
cosh? x
a
esinhx
x sinh x — cosh x
(a)
eos
x?
Bh
3 cosh? x sinh x eos *
eke
pe
0.72, 0.48
secd, sind
7494
Ma
2X
2a
2.37
2AN
2 +In2
28
0, —In3
29
—In3, —In 10
30.
(a?
— b’)
32
—4In5
33 +41n(3 + 2v2)
35 0.481, 0.881
34 In}
36 41,3,3,3
(a)
x?cosec?@— y?sec*@=1
(b)
x2 sech? t + y? cosech? t = |
2sinh2x
3
3sech? 3x
4
—2sech 2x tanh 2x
sech x(sech x + tanh x)
tanh x — sech x
V2
y-3= —2(x —In2)
(In 3, 0), y = 3(x
— In3)
yssatx=—Ind
=(3e? — €?)
max. (—0.96, 11.18),
Exercise 2A
1
x2
ce
21.
38
sinh 2x + 2x cosh 2x
min. (0.96, —11.18)
2
$cosh}x
“2= 3sinh 3x sin x + cosh 3x cos x
DG
d*y
dn = 8cosh 3x sin x + 6sinh 3x cos x
5
—4+cosech Ixcoth}x
6
e*(coshx + sinh x)
7
6sinh 3x cosh 3x
9
- i cosech?(In x)
3
y =—0,90) = O.181x = 1,5)
8
3 tanh” x sech” x
y — 0.905 = —5.534(x — 1.5)
l
]
B55
134
35
37
39
Answers
=|
=|
6
xJ/(1 — x?)
2
SS
x/(1 + x?)
38
4a
4
a
Ox =x)
x
arcosh x + ——_——
ee
v(x* — 1)
a
;
atcesmx+—_
2 =
(b)
-e* arccos x= _——
cc)
pice Ae lasea
———————
(a)
4/(l ey
V/(1 — x?)
x
©)
oe
a0 athe
a
ee
(arsinh : x)
ee
(arctan x)
=
eee
artanh x
2
x
(a)
Oy ot
ae
ae
2
See sta EES
joel
CC alony2 ear
1
—1
b
42
Seer?
eee
2xJ/(1 — x2),/(arsech x)
43
2xe* arsinh x + —————
2
e*
ee
Ka/(x*= 1)
Oe
seal a ts (x ds)
JV(1 ac xe)
=|
oa)
+arcosh x— Fae
44
.
(arcosh x)*
J(k2 — x?)
(ie
sane
erx = sec x
fo
Ser
m ee
arsinh ey (UA) = ee
/(1 + x*)(arsinh x)
Exercise 3A
48
y-x=0,y-In(l+ V2)= yr — 1
1 }sinh 2x
50
y-4in7=-Z(x-3
y—yin
16 (x —4)
3
52.
(0.3)
57
—45sechxtanhx
4 fe 45x
(bt) y+x+1=0
ae li
eds |
(Constant omitted in indefinite integration
neon)
56
(In2, 4)
60 (a) (-1,0)
(ey
©)
3
(bo) Zaresinx
Vd = x?)
SSS
Xf (x2 — 1)
pe
(b)
—6x
aes
ee
Td
— 9x4)
,
3
(x + 2)/(x? + 4x + 3)
4
pe
eee
erent
Co
x{1 + (in x)’]
—2coshix
=
6
VI i= 9?)
2 cosh 3x
he
2B
—]
2 (a) eeere
ae)ees
©)
x/(x?2 — k?)
—k
ee
(a)
—]r
(bb)
= Tig
a7
;
Exercise
1+ x?
6ee h =
abcgies
Son
ee
(b)
earctan x
Or mire
ees
x
5 sinh* h2 :x (or 1zcosh 2x)
7 ~41sinh2x
8 !sinh2x —~
5;
9 5tanh2x
10
x —45tanh2x
11
x—2coth}x
12
—4 cosech 2x
13.
—1 3 sech3x
14
arcsin}a
15
arsinh 4x
16
arcosh $x
17. 4arctan}
x
2
18
iln se te |
4
19
arcsin4x
20
4arcsinix
21} arcosh +x
22
4 arsinh x
1
:
a
a)
Answers
23
x-2
xqarctan 3x
24
4 arcosh 4x
23
4arctan (:;
:
25;il arsinh
4x
26
arctane 2x
a
Surciane
27
$Incosh 2x
28
+ 2sinh 3x
33.
—Larctan (=
= *)
29
2sinh¢x —%
30
x —1ttanh Sx
34
jarctan (3 tan x)
SL.
1,81
S27
34
1.57
a?
40
)
24
c+2
arcsin (:: )
35
arctan (3e*)
i, ae
36
0.0358
35 0.881
36 0.446
39 0.458
40 0.202
41 0.220
0.161
38
1.10
39
64.0
42
0.384
43
0.768
44
4.93
1.05
41
44.1
42
16.6
45
0.902
46
0.128
47
0.682
43 0.67
PA
Adies ; 452.02
0255
37
48 17.2
0.326
38
49 0.470
135
0.256
50 0.524
ex + ]
46
(a)
1.25
(6)
47 2.80
4.25
49 0.632
50 2n
Exercise 3C
(Constant omitted in indefinite integration
Exercise 3B
questions)
at
1
Drees
3
=
5
1
EO,
ju:
omitted in indefinite integration
,
Lin cae
questions)
1
a
fin :
3
arcsin(x — 1)
4
arctan (A)
Z
|
x
p
2
2x+1
arcosh (x —1)
;
2
]
1
:
4 ee arcsiti|x.)
v2
(«v/3)
x+3
1+2x
J; arcosh (x/2)2
6 ale
zl
anf
5
arsinh (fel )
q
6 fln
(=a)
AAS
7
In|1 + sinh x|
9
—4cos2x +4cos? 2x
ee
10
$(x +4) V(x — 2)
8
In(coshx — 1)
9
esinh
10
In|x?+6x+1|
;
11
.
$In|x?
— 4x —2
11
Stan 2x — x
13
In|jax + bx?|
:
12
ie*(cos 2x + 2sin 2x)
13
Ox =4./x44)e"*
15
5(arcsin x)
;
v3
F
12
(x*+1)
14
5(arsinh x
;
a
17
v3
‘
o
8 AGx4aG—3°
3
14 2ev*
l
16
2,/(2x?
+ 7x + 3)
18
arctan(x + 4)
ieee
19
2(x? — 9) — 3arcosh 3 x
20
l
1
4\/(x* a + 16) — arsinh
7x
21
}arctan(2x
+ 1)
22
iin =
V10 arctan (2 v5)
.
aresin
(2x+1
a
15
)
“—Jinkk
3) 36-3)
x—2 |
a
7
' rw
_Laresin
18
ggarctan xy/S
19
a arsinh x\/5
20
5 arcosh x
21
arctan
22
jaresin( _
23
2In(cosh 5x)
5}
l
.
(2x-1
)
1
AG +
5
D
136
Answers
=
24
26
x — ttanh 5x
25
,
3x+2sinx+4 sin2x
s
27
x + 8In|x — 4|
29
=> Gin|x| — 1)
it
———
1 + cosx
63
68
Z+4in2
64 =+!
(2-2
(1+ 23x7)?(3x7-2)
67
1
4in3
(a)
arsinhx
(b)
Y(x?4+1)
(c)
dxV(x? + 1) — 5 arsinh x
(a)
4In
e*
28
In (5)
esr
2+
tani:
x a
2 — tansx
(b) 4 arctan (2 tan;x)
_
e
esin x
tan x(In tan |x| — 1)
Exercise 3D
$(x + 1)2(x — 2)
In |x? — 9] + 21n
I
Re
l
:
Zs
3567, 0
x=3
Ain2yY — Gn2y + 122 —6
x+3
—1
eee
3x :
n—-2, 45 = 9, en
(a) 4
43X SiN 5X+ 35 C08 5x
(b) asn
—2(1-x *
xarsinh 4x — /(4 + x?)
! cosh? x sinh x + 3 cosh x sinh x +3x + C
x arctan 3x — éIn(1 + 9x?)
720 — 1957e7!
Se*(cos x + sin x)
1/3 +5 In(2+ V3)
sinh x + sinh? x
35sin 4x + jsin 2x + 3x
|12 arctan
er?
4
Bxe=D
+ arcosh (sla
)
5)
Exercise 3E
+cos?4x — 2cos4+x
45
—1(2x? +5)°
47 21n(sinh3x)
3 In|tanh
xsinhx —coshx
x|
1
Sue
2
39
2
35i72
Tog 7
3 (52
55g (52 - 152)
4 28:
5 r= 4acos
6sin’ 0
7 +(x‘ — 48x? + 384)
6 (a)
5, Bn
‘tan? x — tanx+x
£/2+2In(1 + V2)
10
#
—icot? 2x — Scot 2x
%
52 je? +t-te?
Exercise
3F
aes
1 364
56
2
5 in (342
a
4
58
28
8 iV3+3n
9 3(5V75
— 22)
59
5In2
60
0.530
10
/2(e2
— 1)
11
15+21n4
61
g
62
26/73 —1)
12 2+4sinh4
13
n(2 + sinh2)
5x — 3
In(7 + 5V2)
m+2
2
3 41n3
4 3 wi
6 2u
1 22
Answers
14
162\/5
15
16 nfV2+In(l1+V2))
18 §n(5V/5 — 1)
20
Exercise 4B
17 Bn
19 $x
32
2
(a)
127
Lee
ah
ara
ae
ee
‘ 2
(d) be
x2
-
¢Vv3
(b) 3
2
; 2
2
; 2D
(c)
a
ee!
(a)
(44,0),x=+%
(+V3,0), x= =5
21 (b) 2[sinh8 +8 — 4sinh4]
3
2
(b) 2x
8
22
(a)
24
2nab
(b)
:
(c)
Exercise 4A
i ay Poe
:
3
3 @) F
(on
en a
xo=
Loy a,
(b)
+V3
(d)
4y=(x-1)/
(c)
V2
6/0), 4380
(yee
(b)
(0, Da
(c)
(0, —4),y-—4=0
(d)
Ue
Bead
(a)
y=s(xt+
iD) 2x+y=
(b)
y=2x4+4,x+2y = 18
AS)
3°
2 = 2
ee
(ce)
Or Sh
I
vay
5
(a)
es
Oy?
GW
tes
3
yO
x+2yS3
4x —9y = 35
a
(d)
2x +9y
1 +X#1
=
(0)
(3;POSTS yee ‘
(c)
(10cost, 2 sin?)
:
it 5“coer 2 nS
]
4
p-t= ~—(x-T)
4
a
pe
=
= ES
ae
ee alaezi
ae
16
BES
2
sint
25
%)
:
— 8)
(c)
ee
(x-1)+ (2) =1
; +9y?= 324
12 (x +25)?
7 x+y=3,(9, -6)
Set s(t 0), 05-10)
14.923
14 x+2y=2
9
(ZA); 16x + 64y = 33
15
ie
ee
ae
ie 49" ee
100
14
(2a, 4a); x(x — a) + y(v — 2a) = 0
16
Ai+4=0
18
(4, 3);8y—12x+51=0
20
x+y+a=0
2aes
cost
(b) aie!
y—# -—t=(2t+1)~%-t+1)
—
2y
;
ae ‘ y
faa
2t
0— 2
+ er)
2
(c) y— 322 =—161(x +42)
Sa
(a)
(b)
(c)
7 (a) (3cost, 2sin?)
ie.
= 20x
]
y=
y=+V2
3
(by y= 10t = (x — 5¢”’)
(dd)
*) ;
(2) at(0, #1),y= £5
(c) yt+2x=—16, 2y—x = 48
(d) 16x+y+32=0,x—
l6y +259 =0
5
(0,44),
(
(b) y =16x +16
2 @)
oD,
18
(44, 0x
(2x — a)
a
Py ae ce
aa
= 22,
4y* bag
be
Sx
Sy
1 no
137
138
Answers
Exercise 4C
a ae)OY
SORE of LE
ee
(c)
a
ce
ae
eur
Og
|
(dye
17
ceae:Oy
ey a
(a)
l6a
is 1 1V6
7
64aV/3
(b)
57
19
—2/2
3
4
(a~ ++ p):
Soe 2
QabV/2
3a)
4G)
2
(ii) (ze 2, 0)
(iil) 225
(Be
@
p= 494 (v)
3
y= tH
(a)
3x-4y=1
(by
ay
—
yp
ey
a4
(d)
x
+a
=!
()
xy
yo
=!
(0)
Vf
x
9-55
ee
Cee:
ex
8(x? +4)
y=
= 2e-
13)
3y2
19
a
b2
(—-acosht, asinhé),
2
16 = 20"
= 26) 2 _,
:
t=In(2+
V3)
]
10
(a)
arctan (e*/3) ay
12
(b)
Kez
13.
C—/(1 =x); arcsinx +./(1 — x2) +C
15
=
>
2ln3
x=1n(3+ V6), y= In — V6)
CNC
(by 4yer
I
plea
eee6
22
In3
23
(a)
24
+1n3
arsinhl
25°
bee
a
28 (a)
y)
oo
e*
Se
2
+4 e*
e-*
29
(a)
3a
30°
“fa)i (72
8.96
68
l
(b)
$na’
(b) Zarsinh 1—}\/2
31 (a) arsinh (>*)=e
\
arctan
SS
Se sin
36 (b) 6— 16e!
5
y= 2(sinw —wcecosyw)
13 (a) —3
14 (a) 6y3
26
——“<— — (b) —_;
SS
odd function
33
8avy3
5
ee
s=In(secy
+ tan)
(a)
arsinhl —/2+1
am
2
12
}
(b)
32 (a) arsinh}
8 1(13)
3
18
Exercise 4D
1 s=jtany
152
10 3sinT
Wi
17 (b) 0
(b)
7
18
x
a
(Db)
n/3
Be
6
tanh
1
oy =
Cy
8(x +2)
i
(—3, —18)
(a)
(b) =
2—x
ex =: 1
(positive x-branch only)
oy
ay
+ ile3
2
ped
12
x
11
=
(e)
10x
Y=
x+y=3
6x — 5y = 16
eo,
—
Zz.
l
5x+6y+7=0
1 lea
ty=x+at?;ty=x-a
y= tx¥P
4x+3y=18
(c)
1
4 (a) an
(uu) (el, 0)
(ii) x = +2 (iv)
6
Review exercise
a Bey
(ii) (0, +3¥V2)
(ii)
()
AGW)
39
(a)
x— 1
Fan
3-3
a
—2
a
+ C
(b) :
35
4a
38 (b) 1In3
a
]
1
9
dary
40
2(In2)° — 3(In 2) + 31n2 -2
(b)
4
41
10a,/5
42
2y=x-+4a; (4a, — 4a); area = Sa?
(b) 333
ise
2)
2
V(1 — 4x?)
(b)
1+ x?
Answers
45
(a) (iii) #
72
Lil
(ey 0) acini
74
ay
(c)
a
(a)
-—4sech3x+C
1
47
(a)
——;
1 — x?
(c)
0.131
49
p=l+/2
50
(b) on
51
0.479
=
54
|
L(v2+!
tanhx
656
(b) 2xcosh(x’)
¥
7tis
(c) 4n2
(a) -$4-x°F+C
78
al2./5 +In(2 + /5)|
(b)
2arcsintx +1x(4—22 +
wD
6 — 2e
ze
3
80 #42
Sve.
The
83
(b) 4,5
10
19
84 (b) 1—Je* cos?
86 (b) y+ qx = 2aq + aq?
;
128’ 256
:
55 (12, 3), (—3, —12)
85
(b) 4[/2+In(1 + 2)
(c)
1—/24+1n(1+
v2)
58
(a) =
61
(b) wVv5
62
63
ie ee 2A
(b) Dawa
ep
59 +11n3
(d) Ins35
Examination
82 in}
style paper FP2
3
(a)
1
S
i)
t
6
) Kee
(c) 44
(a) (0, 0), (3a, 0); (a, + 3a)
(b)
66
In
71
(a).
2a
64 (c) 1
9 S('+4—-<)
+ 3y = 16
(b)
In2
arsinh (=) +C
4m?+ 1
(b)
x*+4y?+8y=0
162° — 384 + 768
4m?+ |
(b)
In?
139
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o¢,*
List of symbols and notation
The following notation will be used in all Edexcel examinations.
is an element of
is not an element of
the set with elements x,,X>,...
the set of all x such that...
the number of elements in set A
the empty set
the universal set
the complement of the set A
the set of natural numbers, {1, 2, 3, ...}
thé setofuntegers, (0;a2 luck 2, +35...)
the set of positive integers, {1, 2, 3,...}
the set of integers modulo n, {0, 1, 2,...,n—1}
the set of rational numbers (2peZ
q
Qe z*\
the set of positive rational numbers, {x € Q: x > 0}
the set of positive rational numbers and zero, {x € Q: x > 0}
the set of real numbers
the set of positive real numbers, {x € R: x > 0}
the set of positive real numbers and zero, {x € R: x > 0}
the set of complex numbers
the ordered pair x, v
the cartesian product of sets
A and B,
Ax
B= {(a, b):a€ A, b € BS
is a subset of
is a proper subset of
union
intersection
the closed interval, {x € R:a<x <b}
the interval {x ER:a<x<b}
the interval {x €R:a<x<b}
the open interval {x ER: a<x<b}
y is related to x by the relation R
y is equivalent to x, in the context of some equivalence relation
is equal to
is not equal to
is identical to or is congruent to
142
List of symbols and notation
NE
YQWVVAAR
IR
is approximately equal to
is isomorphic to
is proportional to
is less than
is less than or equal to, is not greater than
is greater than
is greater than or equal to, is not less than
infinity
p and gq
p or q (or both)
not p
p implies q (if p then q)
p is implied by g (if ¢g then p)
p implies and is implied by q (p is equivalent to q)
there exists
for all
a plus b
a minus b
a multiplied by b
a divided by b
Oh =P Gb) 2 000
3 Oe
(Oh) 2S hy 7S oa
7S OF
the positive square root of a
the modulus of a
n factorial
the binomial coefficient
|
ey,
forne Zt
n(n—1)...(n—r+])
r!
forne Q
the value of the function f at x
f is a function under which each element of set A has an image in set
B
the function f maps the element x to the element y
the inverse function of the function f
the composite function of f and g which is defined by
(go f)(x) or gfix) = g(f(x))
the limit of f(x) as x tends to a
an increment of x
the derivative of y with respect to x
List of symbols and notation
143
the nth derivative of y with respect to x
the first, second, ... mth derivatives of f(x) with respect to x
the indefinite integral of y with respect to x
b
|ydx
the definite integral of » with respect to x between the limits
x =a
and.
x =)
OV
Ox
the partial derivative of V with respect to x
the first, second, .. . derivatives of x with respect to f
e
e*, exp x
log, x
Inx, log. x
Ig x, logig x
sin, cos,
tan
cosec, sec, Cot
arcsin, arccos, arctan
arccosec,
arcsec, arccot
sinh, cosh, tanh
cosech, sech, coth
arsinh, arcosh, artanh,
arcosech, arsech, arcoth
base of natural logarithms
exponential function of x
logarithm to the base a of x
natural logarithm of x
logarithm to the base 10 of x
the circular functions
the inverse circular functions
the hyperbolic functions
the inverse hyperbolic functions
square root of —1
a complex number, z= x +1y
the teal part of 2. Ke
z=
the imaginary part of z, Im z=y
the modulus of z, |z| = Mie +y*)
the argument of z, arg z= arctan
Bs
the complex conjugate of z, x —1y
a matrix M
the inverse of the matrix M
the transpose of the matrix M
the determinant of the square matrix M
the vector a
the vector represented in magnitude and direction by the directed line
segment AB
a unit vector in the direction of a
unit vectors in the directions of the cartesian coordinate axes
the magnitude of a
.
eee
the magnitude of AB
144
List of symbols and notation
a.b
axb
the scalar product of a and b
the vector product of a and b
Bad 2 akGee ie
AUB
ANB
P(A)
A
P(A|B)
events
Nor
XS Vt
D5
ete:
ele.
ee. We wn
Fis t2,--p(x)
union of the events 4 and B
intersection of the events A and B
probability of the event A
complement of the event A
probability of the event A conditional on the event B
random variables
values of the random variables X, Y, R, etc
observations
frequencies with which the observations x), x2,... occur
probability function P(X = x) of the discrete random variable XY
probabilities of the values x,, x»,... of the discrete random variable
X
the value of the probability density function of a continuous random
variable X
the value of the (cumulative) distribution function P(¥ < x) of a
continuous random variable X¥
expectation of the random variable XY
expectation of g(X)
variance of the random variable XY
probability generating function for a random variable which takes the
values(0 Sly 2. aac
binomial distribution with parameters n and p
normal distribution with mean p and variance o7
population mean
population variance
population standard deviation
sample mean
unbiased estimate of population variance from a sample,
l
pe
Ss aac
re
y (x; — Xx)
probability density function of the standardised normal variable with
distribution N(0, 1)
eG
so
SS
COV (Xr)
corresponding cumulative distribution function
product-moment correlation coefficient for a population
product-moment correlation coefficient for a sample
covariance of X¥ and Y
index
answers to exercises
133-139
areas by integration
regions
55—S7
surfaces
61-62-65
astroid
56—S7
asymptotes
hyperbola
90-91, 95
rectangular hyperbola
95
auxiliary circle of ellipse 85
cartesian equations
ellipse 79
hyperbola
89, 93
parabola
71
tangents to parabola
74-76
catenary
104
centre of hyperbola
89
completing the square
34
conjugate axis of hyperbola
89
coordinate systems
71-111
key points
109-111
curve sketching
71-111
curves
astroid
56-57
catenary
104
cycloid
105-106
ellipse 79-81—82-83-84-86
gradient angle
101
hyperbola
89-92-94
intrinsic coordinates
101-103-107
intrinsic equation
102, 104-106
length
59-60-61
parabola
71-76
properties
71-111
radius of curvature
102-103-104, 106-107
rectangular hyperbola
95-96-97—-98
cycloid
105-106
derivatives
hyperbolic functions
13-15-16
inverse hyperbolic functions
17-18
inverse trigonometric functions
20—21-22-23
standard forms
27-30
differentiation
hyperbolic functions
13-25
key points 24-25
directrix
ellipse 81-82
hyperbola
92-93
parabola
72
eccentric angle of ellipse 85
eccentricity
ellipse 81-82
hyperbola
92-93
rectangular hyperbola
95
ellipse 79-81—82-83-84-86
auxiliary circle 85
cartesian equation
79
cartesian equation
82
directrix
81-82
eccentric angle 85
eccentricity
81-82
focal distance property
82-83
focus 81-82
focus-directrix property
81-82
major axis 80
minor axis 80
normals, equations of 80-81
tangents, equations of 80-81
examination style paper
131-132, answers 139
examination technique
34, 40
exponential functions defining hyperbolic functions
focal distance property of ellipse
focus
ellipse 81-82
hyperbola
92-93
parabola
72
focus-directrix property
ellipse 81-82
hyperbola
92-93
parabola
72
gradient angle of curves
graphs
astroid
56
82-83
101
catenary
104
ellipse 80
hyperbola
90
hyperbolic functions
2-4
inverse hyperbolic functions
parabola
72
rectangular hyperbola
95
hyperbola
16, 89-92-94
asymptotes 90-91, 95
cartesian equation
89, 93
centre
89
conjugate axis 89
directrix
92-93
7-8
146
|ndex
eccentricity
92-93
focus 92-93
focus-directrix property
92-93
normals, equations of 91-92
rectangular
95-96-9798
tangents, equations of 91-92
transverse axis 89
hyperbolic functions
1—12
defined with exponential functions
derivatives 13-15-16
differentiation
13-25
graphs 2-4
identities 4-6—7
integration
27-69
integration using identities 30-32
key points
12
name
16
Osborn’s rule 6-7
hyperbolic functions, inverse
derivatives
17-18
graphs
7-8
integration by parts 36-38
logarithmic form
8-9
differentiation
24-25
hyperbolic functions
integration
67-69
12
intrinsic coordinates
Osborn’s rule
12
radius of curvature
111
standard forms
1
identities
hyperbolic functions 4—6~—7
integration
30-32
trigonometric
30
integrals
standard forms
27-30, 40, 67-68
integration
applications
55-57
area of a surface 61-62-65
area of regions
55-57
by parts 36-38, 40, 48-53
hyperbolic functions
27-69
hyperbolic functions using identities
30-32
identities
30-32, 40
key points 67-69
length of a curve
59-60-61
partial fractions
34-35, 40
reduction formulae
48-53
replacement by identical function
40
standard forms
27-30, 40, 67-68
strategy
40-45
substitution
36, 40
volumes of solids 55-57
integration by parts 36-38, 40, 48-53
intrinsic coordinates
101-103-107
key points
111
intrinsic equation of a curve
102, 104-106
inverse hyperbolic functions
derivatives
17-18
graphs
7-8
integration by parts 36-38
logarithmic form
8-9
inverse trigonometric functions
derivatives
20-—21-—22-23
integration by parts 36-38
Key points
coordinate systems
109-111
111
67-68
length of a curve
59-60-61
locus 71
logarithmic form of inverse hyperbolic functions
8-9
normals, equations of
ellipse 80-81
hyperbola
91—92
parabola
73-74
rectangular hyperbola
96-97
notation and symbols
141-144
Osborn’s
rule
6-7
parabola
71—76
cartesian equation
71
directrix
72
equations
71
focus 72
focus-directrix property
72
normals, equations of 73-74
other properties
73-74
tangent equation in cartesian form
74-76
tangents, equations of 73-74
parametric equations of rectangular hyperbola
partial fractions
integration
34-35, 40
properties of curves
71-111
quadratic expressions
34
radius of curvature
102-103-104, 106-107
key points
111
rectangular hyperbola
106
rectangular hyperbola
95—96—97-98
eccentricity
95
graphs
95
normals, equations of 96-97
parametric equations
96
radius of curvature
106
tangents, equations of 96-97
reduction formulae
integration 48-53
integration applications
55-57
trigonometric functions
48-53
review exercise
113-129, answers 138-139
standard forms
derivatives
27-30, 67-68
integrals 27-30, 40, 67-68
key points 67-68
strategy for integration
40-45
substitution in integration
36, 40
symbols and notation
141-144
96
Index
tangents, equations of
ellipse 80-81
hyperbola
91-92
hyperbola, cartesian form 91
parabolas
73-74
parabolas, cartesian form 74-76
rectangular hyperbola
96-97
transverse axis of hyperbola
89
trigonometric functions
derivatives of inverse
20—21—22-23
identities
30
reduction formulae
48—53
volumes of solids by integration
55—57
1
0086986
147
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—
==
HEINEMANN MODULAR MATHEMATICS
for
EDEXCEL AS AND A-LEVEL
Written by a practising Senior Examining Team, Heinemann Modular
Mathematics provides thorough, up-to-date preparation for the Edexcel
exams at AS and A-Level.
Further Pure Mathematics
2
Fae alte]alavarclerx-tct-)1
0) (s-laremagre)gele
le]amela-)ey-]e-14(e]amce) mel al]at a
/ Packed with student-friendly features from worked examples to
practice questions.
/ Features an exam-style practice paper to ensure students are
fully prepared.
Y Straightforward explanations of the key ideas make the maths
easier to tackle.
Y/Y Comprehensive exercises develop and reinforce concepts
Flare es\eqalalce[el-top
Also available from Heinemann:
Complete course textbooks:
Revision guides:
@ Core maths units
W@ Core maths units
@ Further maths units
ie Mechanics units
W Mechanics
units
W@ Statistics units
Statistics units
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™@ Decision maths units
Endorsed by
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To find out more about
Heinemann products, plus free
supporting resources, visit
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le laabevareateheve
hep gro
Hn
ISBN
0-435-51101-7
|app