the Further Mathematics network www.fmnetwork.org.uk V 07 REVISION SHEET FP (MEI) CALCULUS The main ideas are: Calculus using inverse trig functions & hperbolic trig functions and their inverses. Maclaurin series Differentiating the Inverse Trig Functions Before the eam ou should know: That ou can differentiate the trig functions, the hperbolic trig functions and their inverses. That ou can appl the standard rules for differentiation (product rule, quotient rule and chain rule) to functions which involve the above. That ou can integrate trig functions and hperbolic trig functions. That ou can integrate, arcsin(), arccos(), arctan(), arccot(), arsinh(), arcosh() etc using integration b parts. Your trig identities and hperbolic function identities and how to use them in integration problems. Particularl get familiar with useful substitutions to make for certain problems..5 arccos( ) d d arctan( ) d d + -.5 -.5.5 arcsin( ) d d -.5 0.5.5-0 - 0 It is important to be aware of what the range is for each of these, namel: arcsin, 0 arccos, arctan Standard Calculus of Inverse Trig and Hperbolic Trig Functions arcsin( ) d d arccos( ) d d arctan( ) d d arsinh( ) d d + + ar cosh( ) d d arctan + c + a a a a arcsin + c a a ar cosh + c a + a ar sinh + c a Disclaimer: Ever effort has gone into ensuring the accurac of this document. However, the FM Network can accept no responsibilit for its content matching each specification eactl.
the Further Mathematics network www.fmnetwork.org.uk V 07 Calculus using these functions The eamples below are ver tpical and show most of the common tricks. Note details of all substitutions have been omitted, make sure ou understand how to do them in this case and also in the case of a definite integral. + d arsinh d c + + + ( + ) + d d d arcsin arcsin c c + 5 9 5 + + 9 + d d arcosh c + + + 0 ( ) d arcosh( ) d Some useful integration tricks Splitting up an integration: e.g. (to see this use the chain rule, set z 5 5 5 + 5 5 d d + d + + + and then d d dz ). d dz d B inspection: e.g. Since ln( + ) gives + when differentiated, we have d ln( + ) + c or + since ( + ) gives ( + ) when differentiated, we have d + + c + Using clever substitutions: e.g. the substitution u sinh( ) will help ou with + d. Maclaurin Series The Maclaurin epansion for a function f() as far as the term in n looks as follows. n ( n) f ( ) f (0) f (0) f (0) f (0)... f (0) + + + + +!! n! The Maclaurin series is obtained b including infinitel man terms (i.e. not terminating the sum as above). It ma onl be valid for certain values of. Eamples include: 5 7 sin + +... which is valid for all, 8...! 5! 7! + + + + which is valid onl when <, note that this second eample is the same as the binomial epansion of ( ). Useful tips You can find the Maclaurin series of, e.g. f(), b taking the series for f() and replacing the s with. If g is the derivative of f then ou can find the Maclaurin series for g b differentiating the one for f term b term. Likewise, if g is the integral of f then ou can find the Maclaurin series for g b integrating the one for f term b term (caution don t forget the constant of integration, this will be g(0)). Disclaimer: Ever effort has gone into ensuring the accurac of this document. However, the FM Network can accept no responsibilit for its content matching each specification eactl.
the Further Mathematics network www.fmnetwork.org.uk V 07 REVISION SHEET FP (MEI) COMPLEX NUMBERS The main ideas are: De Moivre s Theorem and its applications Eponential notation Using both of the above to get formulae b summing C+jS series n th roots of comple numbers Before the eam ou should know: How to multipl and divide comple numbers in polar form. What de Moivre s theorem is and how to appl it. About the eponential notation j e θ j cosθ + jsinθ, z + j re θ How to appl de Moivre s theorem to finding multiple angle formulae and to summing series. About the n nth roots of unit, including how to represent them on an Argand diagram. De Moivre s Theorem De Moivre s Theorem states that ( ) this are shown below. n cosθ + jsinθ cosnθ + jsin nθ for an integer n. Some applications of ( ) Eample Evaluate + j. Solution The first thing to do is to write + jin polar form. This is just + j cos + jsin Therefore ( + j) ( ) cos + jsin ( cos + jsin ) ( cos + jsin) ( + 0) Note: in eample on the right it is tpical to be asked to go on to integrate sin θ. De Moivre s theorem can also be used to epress multiple angles in terms of powers of the trig functions in a ver straightforward wa. Eample Epress sin θ in terms of multiple angles. Solution If z cosθ + jsinθ then jsinθ z z. So ( j) sin θ ( z z ) 5 5 z z z + 5z z 0z z + 5z z zz + z z + z ( z + z ) + 5( z + z ) 0 cosθ cosθ + 0cosθ 0 Therefore, sin θ cosθ cosθ + 0cosθ 0 0 cosθ + cosθ 0cosθ sin θ. 0 cos θ + cos θ 5cos θ Eponential notation for comple numbers j Eponential notation begins with e θ cosθ + jsinθ. This means that an comple number, z, can be written j in polar form as z + j re θ where r is the modulus of z and θ is the argument of z. Disclaimer: Ever effort has gone into ensuring the accurac of this document. However, the FM Network can accept no responsibilit for its content matching each specification eactl.
the Further Mathematics network www.fmnetwork.org.uk V 07 Eample (of using the eponential notation and De Moivre s theorem to sum C+jS.) jθ jθ + e + e 5+ cosθ. i. Show that ( )( ) sinθ sin θ sin θ sin θ ii. Let S + +... cosθ cosθ cosθ cosθ B considering C js where C + +... show that Solution ( )( ) i) + e + e + e + e + e e jθ jθ jθ jθ jθ jθ jθ jθ ( e e ) 5+ + ( θ θ) ( θ θ ) ( θ θ θ θ) 5+ cos + jsin + cos( ) + jsin( ) 5+ cos + jsin + cos jsin 5+ (cos θ ) 5+ cos θ. This is just a geometric series with first term and common ratio jθ jθ jθ sinθ S. 5 + cosθ cosθ sinθ cosθ sinθ cosθ sinθ ii) C js j + + j j +... cosθ + jsinθ cosθ + jsin θ cosθ + jsin θ +... + cosθ + jsinθ ( cosθ + jsinθ) ( cosθ + jsinθ) + +... cosθ + jsinθ cosθ + jsinθ cosθ + jsinθ + +... jθ jθ jθ e e e +... + j e θ. Its sum is given b e e e + +.... jθ jθ e + e Now the real and imaginar parts of need to be calculated. Part i) is useful because it gives j + e θ jθ jθ + e ( + e ) ( + cosθ jsinθ ) C js jθ jθ jθ e e e. + + + 5+ cosθ 5+ cosθ sinθ Equating imaginar parts gives S. 5 + cosθ n th roots of comple numbers imaginar ais + j The non-zero comple number r(cosθ + jsin θ ) has n different nth roots, which are: z 5 θ + k θ + k r n cos + jsin n n, where k 0,,,, n. - z z real ais nth roots of comple numbers are best thought about geometricall, the diagram shows the 5 th roots of +j. z z 5 You should be able to epress these roots in polar form using the eponential notation. - Disclaimer: Ever effort has gone into ensuring the accurac of this document. However, the FM Network can accept no responsibilit for its content matching each specification eactl.
the Further Mathematics network www.fmnetwork.org.uk V 07 REVISION SHEET FP (MEI) HYPERBOLIC TRIG FUNCTIONS The main ideas are: Definitions of the hperbolic trig functions and their inverses. Working with the hperbolic trig functions Identities involving hperbolic trig functions The Hperbolic Trig Functions These are defined as: e e e + e sinh( ), cosh( ), sinh( ) e e tanh( ). cosh( ) e + e ln0 ln0 0 e e For eample, sinh(ln0) 0 99. 0 Before the eam ou should know: e e e + e The definitions sinh( ), cosh( ), sinh( ) e e tanh( ) cosh( ) e + e That ou can prove that arccosh( ) ln( + ),arsinh( ) ln( + + ) + artanh( ) ln Your trig identities and hperbolic function identities, eperience will tell ou when it is best to work in the eponential form when dealing with equations. How to prove hperbolic identities from the definitions e e e + e sinh( ),cosh( ), it s worth practicing indices for this. The Inverse Hperbolic Trig Functions Just as the hperbolic trig functions are defined in terms of e, their inverses can be epressed in term of logs. In + fact arcosh( ) ln( + ), arsinh( ) ln( + + ), artanh( ) ln. You should be able to prove (and use) all of these. Here is the proof that arcosh( ) ln( ) +. e + e Let ar cosh( ), then cosh( ). Rearranging this gives 0 e + e. Multipling this b e gives 0 e e +. This is a quadratic in e and using the formula for the roots of a quadratic gives ± e ±. Taking logs gives ar cosh( ) ln( ± ). The value corresponding to the minus sign is rejected here, ou should know wh. These epressions can be used to give eact values of the inverse hperbolic trig functions in term of logs. For 5 5 5 5 eample, arcosh ln + ln + ln(). 9 Disclaimer: Ever effort has gone into ensuring the accurac of this document. However, the FM Network can accept no responsibilit for its content matching each specification eactl.
the Further Mathematics network www.fmnetwork.org.uk V 07 Graphs of the Hperbolic Trig Functions cosh( ) d sinh( ) d 8 tanh( ) d d sech ( ) - - sinh( ) d cosh( ) d - Graphs of the Inverse Hperbolic Trig Functions - - You must also know the graphs of the inverse hperbolic trig functions, arsinh, arcosh and artanh. As for an function these are obtained b reflecting the respective graphs of sinh, cosh and tanh in. The eamples of arsinh and arcosh are shown here. Notice that arcosh() is onl defined for greater than or equal to. Identities Involving Hperbolic Trig Functions Identities involving hperbolic trig functions include: cosh u sinh u, cosh( u) cosh u+ sinh u, sin( u+ v) sinh( u)cosh( v) + cosh( u)sinh( v ) The onl difference between a hperbolic trig identit and the corresponding standard trig identit is that the sign is reversed when a product of two sines is replaced b a product of two sinhs. This is called Osborn s Rule. You can prove an hperbolic trig identit using their definitions and should be able to do this for the eam. Equations Involving Hperbolic Trig Functions Eample Solve the equation cosh + 5sinh 0 giving our answer in terms of natural logarithms. Solution e + e e e cosh + 5sinh 0 + 5 0 8e + 8e 0 0 9e 0e + 0 9e e 0 or e e 9 ln or ln 9 ( )( ) Disclaimer: Ever effort has gone into ensuring the accurac of this document. However, the FM Network can accept no responsibilit for its content matching each specification eactl.
the Further Mathematics network www.fmnetwork.org.uk V 07 REVISION SHEET FP (MEI) INVESTIGATION OF CURVES The main ideas in this topic are: The cartesian, parametric and polar forms of equations Smmetr and periodicit of curves, asmptotes, nodes and loops Calculus techniques for curves in Cartesian, parametric and polar form Conic sections Defining a curve Tpes of equation: - Cartesian - Parametric - Polar To convert from parametric to Cartesian, ou need to eliminate the parameter. It ma be possible to obtain a simple relationship between the parameter, and. This can then be substituted into the equation for or. If the parametric form involves trig functions, ou ma be able to use identities like sin θ + cos θ and tan θ + sec θ To convert between polar and Cartesian form, use the relationships rcosθ, rsinθ and + r Features of curves The important features of curves to recognise are Smmetr and periodicit. Vertical, horizontal and oblique asmptotes. Cusps, loops and dimples. Nodes (points where a curve crosses over itself). Nodes (points where a curve crosses over itself). Before the eam ou should know: How to use our graphic calculator efficientl. How to convert between different forms of the equations. The links between the equation of a curve and its shape. Using calculus for curves given in Cartesian and parametric and polar form and understanding what the will show. The standard equations of conics in Cartesian and parametric form. Eample: convert the parametric equations sect, 5tant into (a) Cartesian form (b) polar form (a) sec t 5 tan t sec t tan t 5 + tan t + - 5 5 (b) using + r gives sec t + tan t r ( + tan t) + tan t r tan t + r The Hperbola: a sect, b tant has oblique asmptotes. Epiccloid: kacost a cos kt, kasint - asinkt This eample has K. It has 5 dimples. Smmetr and asmptotes Vertical asmptotes are the values of which make the denominator zero when the equation is in Cartesian form. Horizontal and oblique asmptotes depend on the behaviour of the curve as ±. The clue is in the orders of the numerator and denominator of the graph: If the order of the denominator is greater than the order of the numerator, then 0 as ±, and so the ais is a horizontal asmptote. If the order of the denominator is equal to the order of the numerator, then k for some constant k as, ± and the line k is a horizontal asmptote. If the order of the denominator is less than the order of the numerator, then numericall increases without limit as ± and there is an oblique asmptote. Disclaimer: Ever effort has gone into ensuring the accurac of this document. However, the FM Network can accept no responsibilit for its content matching each specification eactl.
the Further Mathematics network www.fmnetwork.org.uk V 07 Using calculus For curves given in cartesian and parametric form calculus techniques are used to find the equations of tangents and normals; the maimum and minimum values of and the maimum and minimum distances of a curve from the origin. For curves given in polar form calculus techniques are used to find the maimum and minimum distance of the curve from the pole; the points on the curve where the tangent is parallel or perpendicular to the initial line. + 7+ Eample: determine the equations of the vertical and oblique asmptotes of the curve + Solution: Vertical asmptote + 0 ⅓ Oblique asmptote (A + B)( + ) +C + 7 + A + (B-A) + (C-B) + + Equating coefficients gives A, B, C. Hence + + so the oblique asmptote is the line + + Conic sections The conics can be thought of as the curves given b slicing through a double cone. The can be defined in terms of the locus of a point P in relation to a fied point S (the focus) and a fied line d (the directri). A conic is the locus of P such that the ratio of distance of P from S and the distance of P from d is a constant e (the eccentricit). Parabola e Cartesian form a c + Parametric form at at a Ellipse 0<e< b acost bsint a Hperbola e> b asect btant Rectangular hperbola e ct c t t t Eample: A curve has parametric equations,. + t + t (i) Use a graphical calculator to draw the curve for 0 t 0 (ii) The curve is part of a conic. Name the conic. (iii) Write down the co-ordinates of the centre point of the conic and of the points where it crosses the -ais. d t (iv) Show that for the curve S,. d t (v) Find the values of t at the points where the curve is parallel to the -ais, and the Cartesian coordinates of these points. Solution: i) ii) An ellipse iii) Centre point (0, ½); the curve cuts the ais at (0, 0) and (, 0) iv) d ( + t ) t.t dt ( + t ) t t ( + t ) ; d ( t + t ) t.t dt ( + t ) t ( + t ) d d v) d d when t 0 t ±. When t,, ½. When t -, -, ½. Co-ordinates are (, ½) and (-, ½) t ( + t ) t ( + t ) t t Disclaimer: Ever effort has gone into ensuring the accurac of this document. However, the FM Network can accept no responsibilit for its content matching each specification eactl.
the Further Mathematics network www.fmnetwork.org.uk V 07 REVISION SHEET FP (MEI) MATRICES The main ideas are: The inverse of a b matri Solving simultaneous equations in unknowns, and interpreting the solutions geometricall Finding eigenvalues and eigenvectors Diagonalisation and applications The Determinant of a matri If M a a a b b b c c c then Before the eam ou should know: How to find the determinant of a matri and how to invert a matri whose determinant is not zero. The language of matrices singular, non-singular, cofactor, adjugate. How to solve simultaneous equations in unknowns. What to epect when the underling coefficient matri is zero. How to interpret each case geometricall. What is meant b an eigenvalue and an eigenvector of a matri and how to find them. How to diagonalise a matri M where possible. In other words how to find a matri S such that S - MS is a diagonal matri. Applications of diagonalising matrices, such as finding large powers of a matri. The Cale Hamilton Theorem and its applications. ( ) ( ) ( ) det M a b c bc a bc bc + a bc b c. This is epanding b the first row, it is possible to epand b an row or column and ou should know how to do this. The Inverse of a matri a a a A B C If M b b b then M A B C c c c det M, A B C where A for eample is the determinant of the matri which is left after removing the column and row containing a from M, BB is the determinant of the matri which is left after removing the column and row containing b from M etc. Matrices and Simultaneous Equations The three simultaneous equations in three variables a+ b+ cz d a+ b+ cz d a+ b+ cz d d a a a are equivalent to the matri equation M d where M b b b. z d c c c Disclaimer: Ever effort has gone into ensuring the accurac of this document. However, the FM Network can accept no responsibilit for its content matching each specification eactl.
the Further Mathematics network www.fmnetwork.org.uk V 07 If det M 0, M is non-singular, M - eists and the equations have a unique solution. If det M 0, M is singular, M - does not eist and either: (i) (ii) the equations are inconsistent and have no solutions the equations are consistent and have infinitel man solutions. The three equations ma be regarded as the equations of three planes in three-dimensional place and ou should be able to interpret the solutions in terms of the configurations of these planes. Eigenvalues and Eigenvectors For a given matri M, an eigenvector is a vector which is mapped to a multiple of itself b M. For eample since 0 5 5 is an eigenvector of with eigenvalue 5. To find eigenvectors of a matri M (i) Form the characteristic equation: det(m - λi) 0 (ii) Solve the characteristic equation to find the eigenvalues, λ (iii) For each eigenvalue λ find an eigenvector v 0 b solving Mv λv Diagonalisation A diagonal matri is a square matri in which all the entries off the top left to bottom right diagonal are zero. If M is an matri and S is a matri whose columns are eigenvectors of M then, if S is invertible, S - MS is a diagonal matri. Suppose S - MS D where D is a diagonal matri. Then D n (S - MS) n (S - MS) (S - MS) (S - MS) S - M n S. This gives M n SD n S - which is an eas wa to calculate high powers of M. The Cale Hamilton Theorem The Cale Hamilton Theorem sas that a matri satisfies its own characteristic equation. The following eample shows what this means. λ Suppose M. The characteristic equation of M is 0 which is λ 5λ+ 0. λ Now we substitute λ M into the characteristic equation. This gives (notice how becomes I below). M 5M+ I 0 i.e. 0 0 5 0 0 0 0 5 + +. 0 5 8 5 0 0 0 0 The Cale Hamilton Theorem can be used to calculate high powers of a matri quickl. In the above eample we have M 5M I. This gives M ( 5M I)( 5M I) 5M 0M + I 5( 5M I) 0M + I 05M I Disclaimer: Ever effort has gone into ensuring the accurac of this document. However, the FM Network can accept no responsibilit for its content matching each specification eactl.
The main ideas are: What Polar Coordinates are Conversion between Cartesian and Polar Coordinates Curves defined using Polar Coordinates Calculating areas for curves defined using Polar Coordinates θ 5 θ 5 θ θ 5 θ 7 θ θ θ the Further Mathematics network www.fmnetwork.org.uk V 07 5, REVISION SHEET FP (MEI) POLAR COORDINATES θ 5 θ 8, 5,- θ θ θ θ 0 θ θ θ 5 θ θ θ,- θ θ 5 θ 7 θ Before the eam ou should know: How to change between polar coordinates (r, θ) and Cartesian coordinates (, ) use r cosθ, r sinθ, r + and tanθ. You ll need to be ver familiar with the graphs of sin, cos and tan and be able to give eact values of the trig functions for multiples of and. How to sketch a curve given b a polar equation. β The area of a sector is given b rd θ. How Polar Coordinates Work You will be familiar with using Cartesian Coordinates (, ) to specif the position of a point in the plane. Polar coordinates use the idea of describing the position of a point P b giving its distance r from the origin and the angle θ between OP and the positive -ais. The angle θ is positive in the anticlockwise sense from the initial line. If it is necessar to specif the polar coordinates of a point uniquel then ou use those for which r > 0 and < θ. It is sometimes convenient to let r take negative values with the natural interpretation that ( r, θ) is the same as (r, θ + ). α It is eas to change between polar coordinates (r, θ) and Cartesian coordinates (, ) since r cosθ, r sinθ, r + and tanθ. You need to be careful to choose the right quadrant when finding θ, since the equation tanθ alwas gives two values of θ, differing b. Alwas draw a sketch to check which one of these is correct. Disclaimer: Ever effort has gone into ensuring the accurac of this document. However, the FM Network can accept no responsibilit for its content matching each specification eactl.
the Further Mathematics network www.fmnetwork.org.uk V 07 The Polar Equation of a Curve The points (r, θ) for which the values of r and θ are linked b a function f form a curve whose polar equation is r f(θ). A good wa to draw a sketch of a curve is to calculate r for a variet of values of θ. Eample Sketch the curve which has polar equation r a(+ cos θ ) for θ, where a is a positive constant. Solution Begin b calculating the value or r for various values of θ. This is shown in the table. The curve can now be sketched. θ 0 r 0 a a a ( + ) a a 0 It s a good eercise to tr to spot the points given in the table above in polar coordinates on the curve shown here. (,0) For eample the point a ( + ) is here. The Area of a Sector The area of the sector shown in the diagram is Eample β rd α A curve has polar equation r a(+ cos θ ) for θ, θ where a is a positive constant. Find the area of the region enclosed b the curve. Solution The area is clearl twice the area of the sector given b 0 θ. Therefore the area is Note Even though r can be negative for certain values of θ, r is alwas positive, so there is no problem of negative areas as there is with curves below the -ais in cartesian coordinates. Be careful however when considering loops contained inside loops. ( ) ( ) rdθ a + cosθ dθ a + cosθ + cos θ dθ 0 0 0 0 ( cos cos ) a + θ + θ dθ a θ sinθ ( + ) a + + sin θ Disclaimer: Ever effort has gone into ensuring the accurac of this document. However, the FM Network can accept no responsibilit for its content matching each specification eactl. 0