Thomas WHitham Sixth Form

FurtherFurther PurePure MathematicsMathematics

UnitUnit FP3FP3

 Hyperbolic Functions

 Further Integration  Reduction Formulae 2

 Numerical methods  Polars

 Hyperbolic Functions

1 x x 1 x  x Definitions cosh x  2 e  e  sinh x  2 e  e  e x  e x e 2x 1 1 e 2x tanh x    e x  e x e 2x 1 1 e 2x 1 1 1 sinh x sech x  , cosech x  , tanh x   cosh x sinh x coth x cosh x Graphs

y  cosh x y  sinh x

x y  sinh xe x e x e e x y y  tanh x Horizontal x asymptotes

Example Solve the equation cosh x  2

1 x  x 2x x 2 e  e   2  e  4e 1  0 3 4  12  e x   2  3  x  ln2  3 2 x  ln2  3 positive root and

 2  3  1   x  ln2  3  ln2  3    ln  ln2  3  2  3  2  3 Derivatives f x sinh x cosh x tanh x sech x cosech x coth x f x cosh x sinh x sec 2h x  tanh xsech x  cosech x coth x cosech 2 x

These can all be obtained via the definitions –the first three are given in the formula booklet. Learn how to produce them! Identities Proofs require the use of definitions. 1 1 Example 2cosh 2 x 1  2. e x  e x 2 1  e 2x  e 2x  21 4 2 1  e 2x  e 2x   cosh 2x 2 Obtain hyperbolic identities from trig equivalent using Osborne’s rule Viz “product of two sines invloves a change in sign” Example sinhA  B  sinh Acosh B  cosh Asinh B coshA  B  cosh Acosh B  sinh Asinh B (sign change!) 1 tanh 2 A  sech 2 A (sign change!) 4 The following are given in the formula booklet cosh 2 x  sinh 2 x  1, sinh 2x  2cosh xsinh x , cosh 2x  cosh 2 x  sinh 2 x . Why I don’t know, since all the rest you have to learn, or learn how to produce. sinh x Integrals tanh xdx  dx  lncosh x C   cosh x Integrals of sinh x and cosh x follow from reversing table of derivatives. All three of these are given in the formula booklet. Inverse Hyperbolic Functions Graphs 1 1 1 sinh x tanh x y cosh x y y

0 0 x x 0 1 x

Log forms cosh 1 x  lnx  x 2 1 x  1 sinh 1 x  lnx  x 2 1 all x 1 x  tanh 1 x  1 ln  1  x  1 2 1 x  Be able to produce any of these e 2 y 1 Example y  tan 1 x  x  tanh y  e 2 y 1 1 x e 2 y 1x  e 2 y 1  e 2 y 1 x  1 x  e 2 y  1 x 5 1 x  1 x  2y  ln  y  1 ln  1 x  2 1 x  These are not given explicitly in the booklet but see later in the note on integration. Derivatives f x sinh 1 x cosh 1 x tanh 1 x f x 1 1 1 2 1 x 2 x 2 1 1 x Be able to produce each dy Example y  tan 1 x tanh y  x sech 2 x  1 dx dy 1 1 1    dx sech 2 y 1 tanh 2 y 1 x 2

They are all given in the formula booklet. Integrals The table of integrals can be reversed, and log forms used. 6 The booklet gives the following more general results

f x  f xdx

1  1  x  2 2 sinh    lnx  x  a  a 2  x 2  a 

1  1  x  2 2 cosh    lnx  x  a  x 2  a 2  a 

1 1  1  x  1 a  x tanh    ln   a  x  a a 2  x 2 a  a  2a a  x  So, in this table you find inverse hyperbolic functions in logarithmic form! Methods: By analogy with trig techniques. Example  sinh 3 xdx put u  cosh x

 du  sinh xdx  sinh 3 xdx   sinh 2 xsinh xdx   cosh 2 x 1du   u 2 1du  u 3  u  c

1 3  3 cosh x  cosh x  c 7 sinh 2dx  1 cosh 2x 1 dx  1 sinh 2x  1 x c Example   2   4 2 1 sinh 1 xdx  sinh 1 x.1dx  sinh 1 x.x  x. dx Example    2 x 1

 xsinh 1 x  x 2 1  c

 Further Integration will require knowledge of standard integrals such as in the afore going section and also recall the following, which are given in the booklet. f x  f xdx 1  x  sin 1   a 2  x 2  a  From FP2 1 1  x  tan 1   a 2  x 2 a  a  1 1  x  a  ln  -readily proven using partial x 2  a 2 2a  x  a  fractions Type  e ax cosbxdx and similar require two applications of parts 8 Example

  cos 2x    cos 2x  I  e x sin 2xdx  e x      e x dx   2    2 

  1 e x cos 2x  1 e x cos 2xdx 2 2 

1 x 1  x  sin 2x  sin 2x x    2 e cos 2x  2 e     e dx   2   2    1 e x cos 2x  1 e x sin 2x  1 e x sin 2xdx 2 4 4 

1  x 1  I   4 e cos 2x  sin 2x 4 I 5 1 x 4 I   4 e cos 2x  sin 2x 1 x I   5 e cos 2x  sin 2x

1  x  Type dx and similar use a t  tan  substitution  a  bcos x  2   x  1 t  tan  5  3cos x dx 2 2 Example    31 t   5  3cos x  5    x  2  1 2    1 t   dt  2 sec  dx 2 2   8  2t  2   x  2 1 1 t 2  1 tan 2  dx 1 t dx   dt  2   5  3cos x  8  2t 2 1 t 2    2 1  dx  dt  dt 1 t 2  4  t 2

1 1  t   2 tan    C  2 

1 1  1  x   2 tan  2 tan   C   2  9 1 Type dx  ax 2  bx  c When ax 2  bx  c factorises use partial fractions and then integrate – covered in C4, FP1, FP2 Otherwise express ax 2  bx  c in the form ax  d 2  e by completing the square. Recall of the following will be useful, otherwise a substitution will have to be used in the same cases.  If  f xdx  Fx  C2 - and not in  then f ax  bdx  1 Fax  bformula booklet  a 

1 1 1 1  x  2  Example dx  dx  tan    C  x 2  4x  8  x  22  4 2  2  1 Example dx  x 2  4x  2 1 1 1 1 2  2  2  x  4x  2 x  2  2 x  22   2 x  2  2x  2  2 1 1 1 1 Using the cover up rule =  .  . 2 2 x  2  2 2 2 x  2  2 1 1 1  dx   lnx  2  2 lnx  2  2  x 2  4x  2 2 2 2 2

1  x  2  2     ln   C 2 2  x  2  2  10 1 Type dx -complete the square  ax 2  bx  c

1 1  x  1  dx  dx  sinh 1  2   C Example 2 2  3   x x 1  1 3   x  2   4  2  Or, as a log, using the given result in the booklet

1 1 2 3 I  lnx  2  x  2   4  C

1 2  lnx  2  x  x 1 C 1 1 1 Example I  dx  dx  1 2x 3x 2 3  4 1 2   9  x  3 

1  x  1  1  3x 1  sin 1  3   C  sin 1    C  2  3  3  3  2  Type  ax 2  bx  cdx -complete the square. A trig or hyperbolic substitution will follow.

2 Example I   x 2  6x 10dx   x  3 1dx

Let x  3  sinh y  dx  cosh ydy

x  32 1  sinh 2 y 1  cosh y

 I  cosh 2 ydy  1 1 cosh 2ydy  1 y  1 sinh 2y  C   2 2 4 11 1 1 1  2 sinh x  3 2 sinh y cosh y  C

1 2 1 2  2 lnx  3 x  6x 10 2 x  3 1 x  3  C

1 2 1 2  2 lnx  3 x  6x 10 2 x  3 x  6x 10  C

2 Example I   x 2  6x  6dx   x  3  3dx

Let x  3  3 cosh u  dx  3 sinh udu

x  32  3  3cosh 2 u  3  3 sinh u I   3 sinh u 3 sinh udu  3 sinh 2 udu 1 1  34 sinh 2u  2 u C

3 3 1  x  3  2 sinh u coshu  2 cosh    C  3  2 3  x  3  x  3 3 2    1   lnx  3 x  6x  6 2  3   3  2

1 2 3 2  2 x  6x  6x  3 2 lnx  3 x  6x  6 C

2 Example I   5  4x  x 2 dx   9  x  2 dx Let x  2  3sin u  dx  3cosudu

9  x  22  9  9sin 2 u  3cosu 12 I  3cosu3cosudu  9 cos 2 udu  9 1 cos 2u 1du  2 9 sin 2u    u  C 2  2  9  sin u cosu  u C 2  2  9 x  2  x  2  1  x  2    1    sin     C 2 3 3 3       x  2   1 x  2 5  4x  x 2  9 sin 1    C 2 2  3 

 Arc length of a curve

2 y x2  dy  Cartesian form l  1   dx x 1 dx   l

2 2 t2  dy   dx  Parametric form l       dx t 1 dt dt     x x x 1 2 NB Formulae are given in booklet Example Find the length of the curve y  cosh x from x  0 to x  a dy y  cosh x   sinh x dx 13 2  dy   1    1 sinh 2 x  cosh x  dx 

a a  l  cosh xdx  sinh x0  sinh a  sinh 0  sinh a  0 Example The astroid has parametric equations x  a cos3 

, y  asin 3  . It consists of 4 equal arcs as shown. a Find the length of one of the arcs.

0 -a a  At a,0   0 , and at 0,a   -a 2

 2 2  dy   dx   length of arc in first quadrant = 2      d  0  d   d 

2 2  dy   dx  2 2 and        3a cos2  sin    3asin 2  cos   d   d   9a 2 cos 4  sin 2   9a 2 sin 4  cos 2   9a 2 cos 2  sin 2  cos 2   sin 2    9a 2 cos 2  sin 2 

  3a  length of arc = 2 3acossind  2 sin2d  0 2  0

 3a 1 2 3a 1 1  3a       cos 2        2  2  0 2 2  2  2

 Area of surface of revolution about 0 x 14 2 x2  dy  Cartesian form A  2y 1   dx x1  dx 

2 2 t2  dy   dx  Parametric form A  2y      dx t1  dt   dt 

2 y2  dx  For revolution about 0y the formulae become A  2x 1   dx y1  dy 

2 2 t2  dy   dx  and A  2x      dx t1  dt   dt  Example For the curve y  cosh x from x  0 to x  a

2 a  dy  a Surface area = 2y 1    2 cosh x cosh xdx  0  dx  0

a a  sinh 2x   1 cosh 2xdx   x   0    2  0

 sinh 2a    a    2 

Example For the astroid in the first quadrant, rotating about 0x, then

 2 2  dy   dx  Surface area generated = 2 2y      dx  0  dt   dt  15 

 2 2asin 3 .3a cos sind  0 

 6a 2 2 sin 4 .cosd Sub in sin if you  0 need to  2 2 1 5   6a  sin   5  0  6a 2

 Reduction formulae Generally we use integration by parts

n Example If I  ln x dx prove that I  xln xn  nI n  n n1

n n n1 1 I n  ln x 1dx  ln x x  x.nln x . dx   x  xln xn  ln xn1 dx  xln xn  nI  n1

 1 Example If 4 n show that I   I and I n  tan d n n2  0 n 1

find the value of I 6

  

4 n 4 2 n2 4 2 n2 I n  tan d  tan  tan d  sec  1tan d  0  0  0



 4 sec 2 tan n2   tan n2  d  0  4  1 n1  1   tan    I n2   I n2 n 1  0 n 1 16 1 I   I 6 5 4 1 I   I 4 3 2 1 I   I 2 1 0 

4  I 0  1d     0 4  1     2  I  1 I 4   1    2 4 3  4  4 3 1  2  13  I 6        5  4 3  15 4

 Maclaurin and Taylor series The Maclaurin series expansion is based upon an assumption that such an expansion is possible and might be subject to convergence conditions. x 2 x3 f x  f 0 f 0x  f 0  f 0  ...... [in booklet] 2! 3! This is readily proved by writing

2 3 f x  a0  a1 x  a2 x  a3 x  ... etc and using successive differentiation with x  0 at each step. Example Let f x  sin x f 0  0 f x  cos x f 0  1 f x  sin x f 0  0 f x  cos x f 0  1 f x  sin x f 0  0 and a pattern is clearly established 17 x 3 x 5 x 7  sin x  x     .... 3! 5! 7! This series is valid for all x   and given in radians. Expansions for e x , sinh x,cosh x,cos x,ln1 x,ln1 x, tan 1 x and more are readily found using similar application. However note that a power series expression for ln x is not possible since ln 0 is not defined, and neither are derivatives of ln x at x  0 Taylor series again with assumptions that expansions are indeed possible x 2 x 3 (i) f a  x  f a f ax  f a  f a  .... 2! 3! Example To find the expansion of cosa  x then f x  cos x f a  cos a f x  sin x f a  sin a f x  cos x f a  cos a f x  sin x f a  sin a the pattern now established x 2 x 3 cosa  x  cos a  sin ax  cos a  sin a  .... 2! 3! Note that putting a  0 in this expansion x 2 x 4 cos x  1   ..... 2! 4!

Also note that putting a  0 in the expansion f a  x leads to the

Maclaurin of f x . 18 Example To find the expansion for a  xn then f x  x n f a  a n f x  nx n1 f a  na n1 f x  nn 1x n2 f a  nn 1a n2 etc.. x 2 a  xn  a n  na n1  nn 1a n2  ... 2! readily recognisable as the binomial expansion. The series can be used for approximation. For example in the expansion

  for cosa  x, aforegoing, putting a  (60) and x  (1) gives 3 180 an approximate value for cos 61° . The more terms taken, the better the approximation.

2 3           So °          180     180  cos 61  cos  sin    cos   sin   .... 3  3 180   3  2!  3  3! Etc.. (check after 3 terms and compare with your calculator value). In such approximations, the smaller the value of x the more rapid the convergence. (ii) A second form is given by x  a2 x  a3 f x  f a x  a f a f a f a .... 2! 3! Note that putting a  0 in this expansion also leads to the Maclaurin expansion for f x 19 Example Recall that we had no series expansion for ln x using Maclaurin. However an expansion for ln x in terms of powers of x  a can be found using this second form of Taylor. With f x  ln x , f a  ln a 1 1 f x  , f a  x a 1 1 f x   , f a   x 2 a 2 2 2 f x  , f a  x 3 a 3 etc.. 1 1 x  a2 2 x  a3  ln x  ln a  x  a   ... a a 2 2! a 3 3! Approximations When x is close to a such that x  a is small, and consequently successive powers of x  a [eg x  a2 , x  a3 ] become smaller and smaller approximate values can be found Example In the a foregoing an approximate value of ln1.01 can

be found by letting x  1.01, a  1 and x  a  0.01

1 1 0.012 2 0.013 ln1.01  ln1 0.01   ... 1 12 2! 13 3!   0  0.01 0.00005  0.000000333...  0.0099503 Compare with calculator value 20 Other approximations (i) A linear approximation for f x when x is close to a can be

found by ignoring x  a2 and higher powers, for then f x  f a x  a f a  Example Find the linear approximation for tan x when x is close to . 4    f x  tan x f a  tan a  f    tan  1  4  4    f x  sec 2 x f a  sec2 a  f    sec 2  2.  4  4

     f x  1  x  2. f x  1  2x  4  2  {As an exercise, find the equation of the tangent to y  tan x at x  4

 and find that it is y  1  2x } 2

(ii) A quadratic approximation for f x when x is close to a can be

found by ignoring x  a3 and higher powers, for then

x  a2 f x  f a x  a f a f a ... 2! Example Find the quadratic approximation for cos x when x is close to  21 f x  cos x f a  cos a f    cos  1 f x  sin x f a  sin a f    sin  0 f x  cos x f a  cos a f    cos  1

x  2  f x  1 2 Further work on infinite series Term by term differentiation and/or integration can be useful. d 1 Example Since ln1 x  dx 1 x 1 And  1 x  x 2  x 3  x 4  ..... 1 x x 2 x 3 x 4 Then ln1 x  x     ....  A 2 3 4 When x  0 ln1 x  0  A  0

x 2 x 3 x 4  ln1 x  x     .... 2 3 4

1 d 1 1 2  1 2 3 4 5 6 Example sin x   1 x  2  1 x  x  x  .. dx 1 x 2 2 8 16

x3 3 5  sin 1 x  x   x 5  x 6  ....  A 6 40 112 when x  0 sin 1 x  0  A  0 x3 3 5  sin 1 x  x   x 5  x 6  .... 6 40 112 22  Approximate solutions to equations using iterative methods The intermediate value theorem states that if f x is continuous in the

interval a,b and f a and f b are opposite in sign the f x  0 has at least one root in that interval. This is formal statement of what you already knew, having met up with it in P2. Iterative method for an equation in the form x  f x Again already seen in P2, but the condition for convergence is required in this module

and that is f x  1 in the interval under consideration. y y “staircase” “cobweb”

x x x x x x x x 3 2 1 1 4 3 2 Example Show that x3  3x 15  0 has one real root and that this root is near to x  2 This equation can be written in any one of the forms 1 (i) x  15  x3  3 15 (ii) x  x 2  3 23 1 (iii) x  15  3x3

1 15  2 (iv) x    3  x  Which of these forms would be suitable for the basis of an iterative formula x  f x for solution of the equation.

Let gx  x 3  3x 15

gx  3x 2 1  1  No SPs  only one root ( say)

g1.9  2.441 change of sign therefore  g2.1  0.561 1.9    2.1 i.e. root near to x  2

1 (i) If f x  15  x 3 , f x  x 2 , f 2  4  Form not 3 convergent.  30x 15   60 (ii) If f x  , f x  2 , f 2   Form not x 2  3 x 2  3 49 convergent.

1  1 (iii) If f x  15  3x3 , f x  115  3x 2 , f 2  0.23..  This is convergent, hence would be suitable.

1 1  15  2 15  2 (iv) If f x    3 , f x    3 , f 2  0.883..  This is  x   x  convergent, but slower than (iii) since f 2 is closer to 1. 24 The Newton-Raphson method Suppose that x  a is an approximate root of the equation f x  0 and that a  h is the actual root where h is small.

x 2 Now f a  x  f a f ax  f a  ... (Taylor) 2! Put x  h h 2 f a  h  f a f ah  f a  ... 2! but f a  h  0 and if h is small so that h 2 and higher powers are ignored 0  f a f ah f a  h   f a f a Hence a closer approximation to the root is a  f a This formula can be derived from a diagram. Assume that f x is an increasing function near x  a

y f a Tangent at  f a d f a  d  f a Actual  Closera, f a approx = root f a f a a  d f a 0 x First approximation at Closer approximation 25

The next two diagrams show convergence for different gradients.

y y

This next diagram shows one example where the Newton-Raphson fails a1 and a2 are first and second approximations. y The first approximation is 0 1st 2nd 3rd x too far from the root, being 0 3rdon 2 thend wrong1st approx side ofx the maximum point.

0 x a1 a2 26 Vertical asymptote y Here is another example:

Again a1 too far from the root, and f x increasing rapidly so that a2 ends up at the wrong side of 0 a a x the asymptote. 1 2

Example It can be shown graphically that the equation xe x  3 has a root  which is approximately equal to 1. Let f x  xe x  3 Then f x  x 1e x

f a1  f 1 e  3 With a1  1, a2  a1   1  1  1.051819... f a1  f 1 2e

f a2  a3  a2   1.049911... etc. f a2  P.S The Newton-Raphson method is sometimes, more briefly known as Newton’s method. 27  Polar Coordinates (i) Px, y x, y are the Cartesian coordinates of P r r,  are the polar coordinates of P

 0x is called the radius vector and is 0 directed along OP   vectorial angle and is directed with anticlockwise positive. By convention polar coordinates r,  are given such that r  0 and       5  So, for example, the point P2.   4  5 4 Would be written having coordinates x 3  3  2  2,  4  4  P And, for example the point P  3  P 2,  would be written as  4  3 4  x     having coordinates 2,  4  4  2 P’

(ii) Relations between cartesian and polar coordinates x  r cos y  r sin y x 2  y 2  r 2 tan  x 28 (iii) Polar equations, some simple examples r  a is the circle centre 0 and P radius a. r  The Cartesian equivalent is given 0 x by r 2  a 2 i.e. x 2  y 2  a 2

r  2a cos is the circle centre x, y  a,0 and radius a. r  The Cartesian equivalent is given by a x r 2  2ar cos x 2  y 2  2ax x 2  2ax  y 2  0 x  a2  y 2  0

Which you would know anyway!

   is a ‘half line’ through 0, gradient tan .

 The cartesian equivalent is given by x 0 tan  tan y  tan x y  tan x

(iv) Some curves expressed in polar form can be readily converted into cartesian form (as shown in (iii)) and sketches drawn . 29 Example r sin  2 is the line y  2 y

r 2  0 x

   Example r cos    2 y  4  2     rcos cos  sin sin   2 4 4   r 1 1  r cos.  r sin.  2  2 2 0 2 x x  y  2  4

2 Example  1 cos r y  2  r  r cos 2  x 2  y 2  x

 2  x  x 2  y 2

 2  x2  x 2  y 2 1 4  4x  x 2  x 2  y 2 x  y 2  4x 1

(v) Conversely the polar equation of curves expressed in cartesian form can be found 30 Example x 2  y 2  4 r 2 cos 2   r 2 sin 2   4 (a hyperbola) r 2 cos2   sin 2    4 r 2 cos 2  4 (vi) Curve tracing, as distinct from curve sketching, requires the use of ruler, compasses and protractor. Example For the cardioid r  a1 cos  obtain the points where tangents are perpendicular and parallel to the initial line. Show that the graph will be symmetrical about the initial line. Sketch the graph. dx For tangents perpendicular to initial line  0 d x  r cos  x  a1 cos cos  acos  cos 2   dx  a sin  2cos sin   asin 1 2cos  d dx  0   asin 1 2cos   0 d 2  sin  0 or cos   1    0,  2 3 dy For tangents parallel to initial line  0 d 1 y  r sin  y  a1 cos sin  asin  2 sin 2  dy  acos  cos 2   acos  2cos 2  1  a2cos 1cos 1 d dy  0  a2cos 1cos 1  0 d 31   cos  1 or cos  1     ,  2 3 For all , cos   cos hence since r is a function of cos only, the graph will be symmetrical about the initial line. Tabulation follows with inexact values given correct to 2dp.  0     2 3 5  6 4 3 2 3 4 6 cos 2 1.87 1.71 1.5 1 0.5 0.29 0.13 0

The feature at the origin is a cusp. The tangent at the origin is the initial line itself. Also the tangent at 2a,0 is perpendicular to 0x.

Example To sketch the graph of r  asin 2

   4 4 32 The graph of y  sin 2 tells us that there  Is symmetry about    4 With rotational symmetry about 0.

 0    7 5 3 12 8 2 12 8 4 2 0    7 5 3 6 4 3 6 4 4 sin 2 0 0.5 0.71 0 -0.5 -0.71 -1

NB Don’t expect all curves expressed in polar form to respond to techniques displayed in the a foregoing two examples. Whatever is required of you will be under guidance. Example r  a is a spiral 33

(vi) Sector Area  A  1 r 2d   2

 

Example For the cardioid r  a1 cos , the total area enclosed by the curve will be given by

 1 2 2 A  2 2 a 1 cos  d  0   a 2 1 2cos  cos2  d  0 34  2 3 1  a 2  2cos  2 cos 2 d  0  2 3 1   a   2sin  sin 2   2 4  0 3a 2  2

Example For one of the leaves of the four leaved rose r  asin 2



A  2 a 2 sin 2 2d  0



 a 2 2 1 1 cos 4 d  0 2   sin 4  2  1 a 2   2    4  0 1 2  4 a

(vii) Intersections of curves expressed in polar form. Example Find the points of intersection of r  a1 sin  and

1 r  2 a . Find also the area enclosed between the two graphs, outside the cardioid.

1 1 For intersections a  a 1 sin   sin   2 2  5     ,  6 6  Points of intersection are in the form r,  35  1    1 5   a,  and  a,   2 6   2 6  The sketch below helps.

  6 1 2 2 Shaded area = area of sector of circle  2  2 a 1 sin  d   2

  1 2 2 2 6 2  a .  a  1 2 sin  sin  d 2 3   2

 2  a 2 6 1   a  1 2sin  1 cos 2 d 3  2 2  2  a 3 1  6   a2  2cos  sin 2    3  2 4   2 a2   2 3 1  3  3  2        a         3  4 2 4  2   4  2 a 2  7 3    a    3  2 8  36  7 3     2    a  8 6 