MEI Core 4. Algebra. Section 1: The general binomial expansion. The general binomial expansion

MEI Core 4 Algebra Section : The general binomial epansion Notes and Eamples These notes contain subsections on The general binomial epansion Harder eamples Finding approimations The general binomial epansion In Section 5 of Polynomials, you came across the binomial epansion which n can be used to epand the epression, where n is a positive integer: n n( n ) n( n )( n )...!! n When n is a positive integer, this epansion always has a finite number of terms, since eventually there will be a factor of 0 in the numerator. However, this epansion can also be used for any value of n, including negative numbers and fractions. In cases where n is not a positive integer, there will never be a zero coefficient, so the epansion will have an infinite number of terms. This means that the epansion is only valid for cases where values of decrease as r increases, i.e. for - < <. Otherwise, the terms of the epansion would get bigger and bigger and the sum of the terms would increase without limit. r Eample Epand ( ) up to the term in ³, stating the values of for which the epansion is valid. 4 4 5 6 0... ( )... The epansion is valid for - < <. MEI, 08/06/09 /5

MEI C4 Algebra Section Notes and Eamples The epansion still works when the second term in the bracket is a multiple of. You just need to replace in the epansion with the multiple of, and remember that, for eample, ()³ is not ³ but 8³. Also, be careful with the range of for which the epansion is valid in such cases. Look carefully at Eample below. Eample Epand up to the term in ³, stating the values of for which the epansion is valid. ( ) ( )... 4 8... 8 6... The epansion is valid for - < < i.e.. When the second term involves a negative, be very careful with signs. Eample Find the first four terms in the epansion of the epansion is valid....... The epansion is valid for - < <., stating the values of for which Harder eamples When you used the binomial epansion in Core for values of n which were positive integers, you could epand epressions of the form a b n, starting with the term n a and ending with the term n b. You cannot do this when n is MEI, 08/06/09 /5

MEI C4 Algebra Section Notes and Eamples n not a positive integer: you can only epand epression of the form. However, it is still possible to epand epressions like a b n by taking out a factor n a first like this: n n b n a b a a. You can then epand the epression in the bracket using the binomial epansion. Be very careful it is easy to make mistakes in this kind of work. This method is shown in the net eample. Eample 4 Find the first four terms in the epansion of 4, stating the values of for which the epansion is valid. 4 4 4 4 4... 4 4 4 4... 8 8 6 6 64... 8 8 04... 4 8 8 04... 4 64 5 The epansion is valid for 4 i.e. 4 4 The net eample shows a more complicated epansion. Eample 5 Find the first three terms of the epansion of which the epansion is valid., stating the values of for MEI, 08/06/09 /5

MEI C4 Algebra Section Notes and Eamples ( )... 8 4...... ( )... 8... When multiplying out the brackets, ignore any terms in powers of greater than ².... 8... 8...... 8 The epansion for The epansion for 8... is valid for i.e. is valid for For both conditions to be true, then must satisfy the condition. You can see further eamples using the Flash resource The binomial epansion. For some practice in finding the range of validity for binomial epansions, try the binomial puzzle. Finding approimations The binomial epansion can be used for finding the approimate value of a function, by substituting an appropriate value for and evaluating the first few terms of the epansion. The more terms are used, the better the approimation. Eample 6 (i) Find the first four terms of the binomial epansion of (8 ). (ii) Use your result from (i) to find the value of 8. correct to si decimal places. 8 8 (i) (8 ) 8 8 MEI, 08/06/09 4/5

MEI C4 Algebra Section Notes and Eamples 5... 8 8! 8! 8... 4 576 447... 8 4 576 447... 88 076 (ii) Let = 0. 0. 0. 0. (8 0.... 88 076 0. Using the first two terms 8..008 0. 0. Using the first three terms 8..008986 88 0. 0. 0. Using the first four terms 8..008987 88 076 Adding the fourth term does not affect the sith decimal place so 8..00899 correct to si decimal places. MEI, 08/06/09 5/5

MEI Core 4 Algebra Section : Review of algebraic fractions Notes and Eamples These notes contain subsections on Simplifying algebraic fractions Multiplying and dividing algebraic fractions Adding and subtracting algebraic fractions Solving equations involving algebraic fractions Simplifying algebraic fractions You are familiar with the idea of cancelling to simplify numerical fractions: for eample, 9 can be simplified to by dividing both the numerator and 4 the denominator by. The same technique can be used in algebra. Remember that cancelling involves dividing, not subtracting. Eample 6y y Simplify 0y 6y y y( y ) 0 y 0 y y 5 y is a common factor of both top and bottom It is very important to remember that you can only cancel if you can divide each term in both the numerator and denominator by the same epression. In this case, don t be tempted to divide by ²y although this is a factor of both ²y and 0²y, it is not a factor of 6y³. In a case like this, it is best to factorise the top first, so that it is easier to see the factors. Multiplying and dividing algebraic fractions As with numerical fractions, you multiply algebraic fractions by multiplying the numerators and multiplying the denominators. Sometimes you can cancel before carrying out a multiplication, to make the numbers simpler: e.g. 9. You can do the same with algebraic 4 fractions. When multiplying, remember to use brackets where appropriate. MEI, 0/0/0 /

MEI C4 Algebra Section Notes and Eamples Eample a a Simplify a a Again, factorise where possible first. a a a ( a ) a a a a 6a a (a + ) is a common factor of both top and bottom Notice that you cannot cancel a here, as it is not a factor of a +. Remember that to divide fractions, you take the reciprocal of the second fraction and then multiply, in the same way as you do for numerical fractions: 4 7 7 7 4 6 Eample Simplify ( )( ) ( ) ( + ) and are both factors of both the top and the bottom Adding and subtracting algebraic fractions Algebraic fractions follow the same rules as numerical fractions. When adding or subtracting, you need to find the common denominator, which may be a number or an algebraic epression. Eample 4 Simplify 5 (i) 4 6 (ii) MEI, 0/0/0 /

MEI C4 Algebra Section Notes and Eamples (i) The common denominator is, as, 4 and 6 are all factors of. 5 8 0 4 6 8 0 (ii) The common denominator is ². Solving equations involving algebraic fractions If you need to solve an equation involving algebraic fractions, it is often best to start by multiplying through by all epressions in the denominators, to clear the fractions. Eample 5 Solve the equation ( ) The common denominator is ( + ). Multiply through by this epression. ( ) ( ) ( ) ( ) ( ) ( ) 4 4 Multiply each term by ( + ) MEI, 0/0/0 /

MEI Core 4 Algebra Section : Partial Fractions Notes and Eamples These notes contain subsections on Fractions with linear factors in the denominator Fractions with quadratic factors in the denominator Fractions with repeated linear factors in the denominator Using partial fractions with the binomial epansion In Section, you practised adding and subtracting algebraic fractions. e.g. 4. ( )( ) Sometimes it is useful to reverse this process: i.e. to epress a complicated fraction as the sum of two or more simpler ones. This process is called finding partial fractions. You will be looking only at proper fractions: i.e. algebraic fractions where the degree (or order) of the numerator is less than the degree of the denominator. Remember that the degree of a polynomial epression is the highest power of in the epression. Fractions with linear factors in the denominator p q Any fraction of the form can be written in the partial fractions ( a b )( c d ) A B form a b c d. There cannot be any terms in in the numerators of the partial fractions, as this would mean that the fractions would be improper. To find the unknown constants A and B in the partial fractions, you first multiply through by the common denominator to clear the fractions. There are then two basic methods to find the constants: substitute any two values for to give two equations in A and B (by choosing the values carefully you can make this very easy) multiply out and equate coefficients. MEI, /05/0 /7

MEI C4 Algebra Section Notes and Eamples Substitution is often the most efficient method, but in some cases equating coefficients may be an easier way to find one or more of the unknown constants. Eample 9 Epress ( )( ) in partial fractions. A B The partial fractions must be of the form. 9 A B So ( )( ). Multiplying both sides by ( + )( ): 9 = A( ) + B( + ) This equation is true for all possible values of. This means that you can substitute any number for and get a true statement. Although any number will do, you can make it easier by choosing carefully. By choosing = -, the term in B will disappear, making it easy to find A. Substitute = - -0 = -5A A = To find B, you could choose = so that the term in A disappears. However, if you prefer to avoid using fractions, you can choose any value for and use the fact that A =. Perhaps the simplest choice is = 0. Substitute = 0-9 = (-) + B -9 = -6 + B B = - Hence 9 ( )( ) In some cases you may need to factorise the denominator before you start. Eample Epress in partial fractions. MEI, /05/0 /7

MEI C4 Algebra Section Notes and Eamples First factorise the denominator:. ( )( ) A B ( )( ) Multiplying through by ( )( + ): = A( + ) + B( ) Substitute = = A A Substitute = - = -B B Hence ( )( ) ( ) ( ) Fractions with quadratic factors in the denominator So far the eamples you have looked at have involved fractions with denominators which can be fully factorised into two or more linear factors (i.e. factors of the form a + b). Net you will look at fractions whose denominators contain one or more quadratic factor which cannot be factorised into linear factors. If the denominator contains a quadratic factor of the form a² + b + c which cannot be factorised into linear factors, this produces a partial fraction of the A B form. Since the denominator is of degree, then the numerator a b c must be either a constant (degree 0) or linear (degree ), for the fraction to be proper. The numerator is therefore written as A + B to cover either possibility (either A or B could turn out to be zero). In general, a fraction of the form p q r ( a b c)( d e) can be written in the A B C partial fractions form, and this can be etended to involve a b c d e any number of linear and quadratic factors. Eample Epress ( )( ) in partial fractions. A B C The partial fractions must be of the form. A B C ( )( ) MEI, /05/0 /7

MEI C4 Algebra Section Notes and Eamples Multiplying through by (² + )( ): + = (A + B)( ) + C(² + ) Substituting = : Substituting = 0: 5 = 5C C = = -B + C = -B + B = 0 Equating coefficients of ²: 0 = A + C 0 = A + A = - Hence ( )( ) Substituting = removes the term involving A and B, allowing C to be found easily. Substituting = 0 removes the term in A, allowing B to be found easily You can now use the method of equating coefficients. Alternatively, substitute any value for to find A. Fractions with repeated linear factors in the denominator This method needs to be adapted slightly when there is a repeated linear p q A B factor in the denominator. Trying to write in the form ( a b) a b a b p q A B does not work as this will just give ( a b) a b. p q However, any fraction of the form ( a b) A B. a b ( a b) can be written in the form Eample 4 Epress 44 in partial fractions. Factorising the denominator: 4 4 ( ) A B ( ) ( ) Multiplying through by ( )²: = A( ) + B Substituting = : 5 = B Equating coefficients of : = A 5 Hence ( ) ( ) MEI, /05/0 4/7

MEI C4 Algebra Section Notes and Eamples If there are other linear factors in the denominator, then treat them in the p q r same way as before. In general, any fraction of the form can ( a b) ( c d) be written in the form A B C a b, and this can be etended to ( a b) c d any number of factors in the denominator. Eample 5 Epress 7 () ( ) in partial fractions. 7 A B C ( ) ( ) ( ) Multiplying through by ( + )²( + ): ² + 7 = A( + )( + ) + B( + ) + C( + )² Substituting = -: 8 + 7 = C(-6 + )² 5 = 5C C = Substituting = : 4 7 B( ) 5 5 B B = Substituting = 0: 7 = A + B + C 7 = A + 9 + A = - A = - 7 Hence ( ) ( ) ( ) For etra practice in finding partial fractions, try the interactive questions Partial Fractions. The Mathcentre video Partial fractions looks at eamples of finding different types of partial fractions, including some which involve improper fractions, which are not included in the C4 specification. Using partial fractions with the binomial epansion The following eample shows how partial fractions can be used to simplify the working if you want to find a binomial epansion. MEI, /05/0 5/7

MEI C4 Algebra Section Notes and Eamples For eample, if you want to find the binomial epansion of up to ( )( ) the term in ³, you could write it as, apply the binomial epansion to each bracket separately and then multiply out the brackets, discarding any terms of order greater than ³. However, Eample 6 shows how using partial fractions can make the work much easier, as in this case you have to add two algebraic epressions rather than multiplying. Eample 6 Epand up to the term in ³, stating the range of values of for which ( )( ) the epansion is valid. A B ( )( ) Multiplying through by ( )( ): A( ) B( ) Substituting = : = B Substituting = : = A Both parts must be written in ( )( ) the form k before the epansion can be carried out. ( )( ) ( )( )( ) ( )( ) ( ) ( )...... This epansion is valid for - < <. Epand the two parts separately ( )( ) ( )( )( ) ( )...... 4 8 This epansion is valid for ( )( )... 4 8...... 4 8 6... 7 5 4 8 6 The epansion is valid for - < <. The epansion is valid for values of for which both parts are valid, i.e. both - < < and - < < must be true. MEI, /05/0 6/7

MEI C4 Algebra Section Notes and Eamples Note: another application of partial fractions is in integration. You will look at this in chapter 0. MEI, /05/0 7/7

MEI Core 4 Trigonometry Section : Trigonometric identities and equations Notes and Eamples In this unit you learn to solve more comple trigonometric equations. These notes contain subsections on The reciprocal trigonometric functions Trigonometric identities The compound angle formulae The double angle formulae The reciprocal trigonometric functions In addition to using the sine, cosine and tangent functions you will need to know three more functions. These are the reciprocal trigonometric functions and are defined as: cosec sin sin 0 sec cos cos 0 cot tan tan 0 Remember the reciprocal of a number or function is divided by that number or function Use the rd letter of each function as a memory aid: e.g. rd letter of cosec is s and cosec θ = sin You can use the Geogebra resource Si trigonometric functions to trace out the graphs of the sin, cos, tan, sec, cosec and cot functions, and look at the relationships between them. You may also find the Mathcentre video Trigonometric functions: cosec, sec, cot useful. The graphs of these functions are given in the tetbook (page84) make sure that you are familiar with them. MEI, /05/0 /7

MEI C4 Trigonometry Section Notes and Eamples Eample Write down the eact value of cot 0. cot0 tan0 y = tan θ has period of 80 so tan 0 = tan ( 60 ). y = tan θ has rotational symmetry about the origin so tan ( 60 ) = tan 60 So cot0 Trigonometric identities Remember: An identity is true for all values of θ. In Core you learned the following trigonometric identities: sin tan cos sin cos e.g. sin 0 tan 0 cos0 By dividing the identity sin cos through by cos² and by sin² two further identities are obtained: tan sec cot cosec These two new identities, together with sin cos, are often called the Pythagorean identities since they are derived using Pythagoras theorem. Eample In ΔABC, angle A = 90 and cosec B =. (i) Find angles B and C. (ii) Find tan C and sec C (iii) Show that tan C sec C (i) cosec B sin B B 0 B must be an acute angle as otherwise the sum of the angles would be greater than 80 MEI, /05/0 /7

MEI C4 Trigonometry Section Notes and Eamples C 80 90 0 60 (ii) tan 60 cos 60 sec60 So tan C and secc (iii) L.H.S. tan C R.H.S. 4 sec C 4 So tan C sec C as required. For practice in using the reciprocal trigonometric functions and the Pythagorean identities, try the Trigonometric identities puzzle. In Core you saw how trigonometric identities can be used in solving equations. These new identities can be used in the same way. Eample Solve cosec cot for 80 80. cosec cot cot cot cot cot 0 Let cot 0 Factorising: ( )( ) 0 or cot or cot The identity cot cosec links cosec andcot. Substitute this into the equation. This is a quadratic in cot. You can factorise this directly and get (cot θ - )(cot θ + ) = 0, or use the approach shown. cot tan 6.6 tan cot tan 45 tan MEI, /05/0 /7

MEI C4 Trigonometry Section Notes and Eamples Since y tan has a period of 80 any other solutions can be found by adding/subtracting 80 to the principal value. So the other solutions are: 6.6 80 5.4 and 45 80 5 So the values of θ for which cosec cot are -5.4, -45, 6.6 and 5. the value that your calculator gives you. The Mathcentre video Trigonometric identities shows the derivation of these identities and their use in solving equations. The compound angle formulae Some further useful identities are the compound angle formulae: sin( A B) sin Acos B cos Asin B sin( A B) sin Acos B cos Asin B cos( A B) cos Acos B sin Asin B cos( A B) cos Acos B sin Asin B tan A tan B tan( A B) ( A B) 90,70,... tan Atan B tan A tan B tan( A B) ( A B) 90,70,... tan Atan B Remember tan θ is undefined for these values. These are proved on page 88 of the tetbook. You can use the Geogebra resource The compound angle formulae to see a geometrical proof of the formulae for sin (A + B) and cos (A + B). The Mathcentre video The compound angle formulae also shows the geometrical derivation of the formulae, and gives some applications. Eample 4 Write cos75 as a surd. cos 75 cos(45 0 ) Using the compound-angle formula cos( A B) cos Acos B sin Asin B we have: MEI, /05/0 4/7

MEI C4 Trigonometry Section Notes and Eamples cos(45 0 ) cos 45cos0 sin 45sin 0 cos45 sin 45 cos0 sin 0 The compound angle formulae can also be used to solve equations. Eample 5 Solve sin(60 θ) cos θ for 0 θ 60 Use the compound-angle formula sin( A B) sin Acos B cos Asin B sin(60 θ) cos θ sin60 cos θ cos60 sin θ cos θ cos θ sin θ cos θ Rearrange to collect like terms: Factorise the LHS: Now you can divide by cosθ : Make tanθ the subject: So θ 65 or 45 cos θ cos θ sin θ cos θ sin θ sin θ cos θ tan θ θ 5 Remember you can only solve an equation which has either tan θ, cos θ or sin θ as the subject. Whenever you have an equation which is a miture of sines and cosines see if you can use the identity sin θ tan θ cos θ To find all the solutions in range you need to add on multiples of 80. The double angle formulae In the case where A = B, the compound angle formulae become the double angle formulae. MEI, /05/0 5/7

MEI C4 Trigonometry Section Notes and Eamples sin A sin Acos A cos A cos A sin A sin cos A tan A tan A A 45,5,... tan A A These three epressions for cos A are all equivalent Remember tan A is undefined for these values. The proofs of these are given on page 9 of the tetbook. You can use the Geogebra resource The double angle formulae to see a geometrical proof of the formulae for sin A and cos A. The Mathcentre video The double angle formulae looks at the derivation and use of the formulae. Eample 6 Use the double angle identities to show that sin θ tan θ cos θ. tan θ You are aiming for: tan θ tan θ Working with the LHS: sin θ sin θ cosθ cos cos sin θ θ θ sin θcosθ cos cos cos θ sin θ cosθ sin θ cos θ tan θ tan θ tan θ θ θ sin θ cos θ Only work with one side of the identity. tan θ In order to get in the denominator, divide both the numerator and the denominator by cos θ : Remember so long as you multiply both the top and the bottom of a fraction by the same thing you don t change it. So sin θ tan θ as required. cos θ MEI, /05/0 6/7

MEI C4 Trigonometry Section Notes and Eamples In this eample you need to use a double-angle formula to solve an equation giving your answer in radians. Eample 7 Solve cos θ 5cos θfor 0θ π Look to see which of the three formula for cosθ is best. This one means that you end up with an equation just in terms of cosθ. Use the double-angle identity cos A cos A Substituting: cos θ 5cos θ cos θ 5cos θ Rearranging: cos θ5cos θ 0 Factorising: (cos θ)(cos θ) 0 cosθ or cos θ cosθ π θ or θ 4π This is a quadratic in cosθ. cos has no solutions The second solution is found by subtracting from π. For practice in using the compound angle formulae and the double angle formulae, try the Trigonometric identities puzzle. MEI, /05/0 7/7

MEI Core 4 Trigonometry Section : More trigonometric equations Notes and Eamples In this unit you learn to convert epressions in the form asin θ bcos θ to the form r cos( θ α) or r sin( θ α). This is a useful skill as it enables you to solve equations and sketch curves that you previously wouldn t have been able to. These notes contain subsections on The forms r cos ( ) and r sin ( ) Solving equations The forms r cos( ) and r sin( ) Try using a computer package or graphic calculator to sketch graphs of the form y asin bcos, with various values of a and b. You should find that all the graphs of this form are the same shape as a sine or cosine graph, but translated by various amounts in the direction, and stretched by various amounts in the y direction. This suggests that any epression of the form asin bcos may be written in the form r sin( ) or r cos( ) for particular values of r and. You can eplore this idea using the Geogebra resource a sin + b cos. The tetbook shows you (pages 0 0) how the compound angle formulae can be used to do this. The results can be summarised as follows: asin θ bcosθ r sin( θ α) asin θ bcosθ r sin( θ α) a b where r a b, cos α and sin α a cosθ bsin θ r cos( θ α) r r a cosθ bsin θ r cos( θ α) However, it is best not to try to learn and apply the formula above as it is easy to get muddled. A better approach is to use the compound angle formulae and then compare coefficients, as shown in Eample. MEI, /05/0 /

MEI C4 Trigonometry Section Notes and Eamples The Mathcentre video The form r cos ( + a) looks at finding this form and using it to solve equations and find maimum and minimum values. Eample (i) Find the positive value of r and the acute angle α for which sin 4cos rsin( ) (ii) Sketch the curve with the equation y sin 4cos. (i) sin 4 cos r sin( ) r(sin cos cos sin ) r sin cos r cos sin sin 4cos rsin cos r cos sin r cos and r sin 4 4 cos and sin r r Compare coefficients of cos and sin This information can be illustrated on a right-angled triangle, which allows you to find the values of r and. 4 r α So: sin 4cos 5sin( 5. ) r 4 5 and 4 tan 5. (ii) y sin 4cos y 5sin( 5. ) The minimum value that 5sin( 5. ) has is 5. The maimum value that 5sin( 5. ) has is 5. Remember y = sin θ lies between and The graph of y 5sin( 5. ) is a translation of the graph y sin by the vector 5. and a one-way stretch parallel to the y ais scale factor 5. 0 MEI, /05/0 /

MEI C4 Trigonometry Section Notes and Eamples 5 y -5. 90 90 80 70 60 5 For practice in using these forms, try the Alternative form dominoes. You need to match up equivalent epressions, and the dominoes will eventually form a closed loop. Solving equations Eample Solve the equation sin 4cos for 0 60 From Eample, you can write sin 4cos as y 5sin( 5. ) sin 4cos 5sin( 5.) sin( 5.) 0.6 5. 6.9 or 4. 6. or 90 You need to solve sin θ = 0.6 and then subtract 5. from each solution. 6. is outside the range, so add 60 to give a solution of 4.7 The solutions are = 90 or 4.7 to d.p. MEI, /05/0 /

MEI Core 4 Parametric Equations Section : Using Parametric Equations Notes and Eamples These notes contain subsections on The definition of a parametric equation Sketching a parametric curve Finding the cartesian equation of a curve The parametric equation of a circle The definition of a parametric equation An equation like y = 5 + or y sin 4cos or y is called a cartesian equation. A cartesian equation gives a direct relationship between and y. In parametric equations and y are both defined in terms of a third variable (parameter) usually t or. t For eample are a pair of parametric equations y t cos and are also a pair of parametric equations. y sin Parametric equations can be used for a complicated curve which doesn t have a simple Cartesian equation. Sketching a parametric curve To sketch a curve given its parametric equations follow these steps. Step Step Step Step 4 Make a table like this one: t or y Choose values of t or (these will be usually be given to you) Work out the corresponding values of and y using the parametric equations. Plot the (, y) coordinates. Join them up in a smooth curve. These eamples shows you how to do this. MEI, 0/0/09 /6

MEI C4 Parametric Section Notes and Eamples Eample A curve has the parametric equations Sketch the curve for 4 t 4, t y t. t -4 - - - 0 4 - -9-6 - 0 6 9 y 6 9 4 0 4 9 6 6 y 4 0 8 6 4 0 5 5 0 Eample A curve has the parametric equations cos θ, y sin θ. Sketch the curve for 0 θ 60 θ 0 0 60 90 0 50 80 0 40 70 00 0 60.7 0-0.7 - -0.7 0.7 y 4 4.7 5 4.7 4.7.7 6 y 4 4 MEI, 0/0/09 /6

MEI C4 Parametric Section Notes and Eamples Finding the cartesian equation of a curve To find the cartesian equation of a curve from its parametric equations you need to eliminate the parameter t or θ. It is not always possible to find the Cartesian equation of a curve defined parametrically. There are essentially methods which you will need to use depending on the parametric equations. Method Use this method when it is straightforward to make t the subject of one equation. Step Step Step Make t the subject of one of the parametric equations. Substitute your equation for t into the other parametric equation. Simplify. Eample Find the cartesian equation of the curve defined by the parametric equations t, y t Step Make t the subject of t : Step Step Substitute into Simplify: y t : t y y ( ) y 9 Sometimes it is easier to rearrange both equations to give an epression for t, as shown in the net eample. Eample 4 Find the cartesian equation of the curve defined by the parametric equations t t, y. t t MEI, 0/0/09 /6

MEI C4 Parametric Section Notes and Eamples t t ( t ) t t t t t t( ) t t y t y( t ) t ty y t ty t y t( y ) y y t y Equating the two epressions for t: y y ( y ) y( ) y y y y y y( ) y Method Use this method when it is not straightforward to make t the subject of one equation. Step Check whether adding/subtracting the two equations will result in a rd equation in which t can easily be made the subject. If not, look to rearrange the equations (e.g. squaring, cross multiplying) so this can be done. Step Then proceed as method. Method Use this method when you have trigonometric functions in the parametric equations. Step Step Step Find an identity which connects the two trigonometric functions. Rearrange the parametric equations so that you can substitute them into the trig identity. Simplify. Eample 5 Find the cartesian equation of the curve defined by the parametric equations 4 cos θ, y cos θ MEI, 0/0/09 4/6

MEI C4 Parametric Section Notes and Eamples The identity which connects cosθ and cosθ is cos θ cos θ 4 cos θ cos θ 4 y cos θ cos θ y Substitute and into y ( 4) Simplify y 4( 4) The parametric equation of a circle A circle with radius r and centre (0, 0) has parametric equations: rcos y rsin A circle with radius r and centre (a, b) has parametric equations: a r cos y b r sin Eample 6 (i) Find the cartesian equation of the curve given by the parametric equations cosθ (ii) Sketch the curve. y sin θ (i) The trig identity sin θcos θ links sinθ and cosθ. Make cosθ the subject of cos θ : cosθ Make sinθ the subject of y sin θ : sin θ y Substitute and into sin θcos θ : So So y ( ) ( y) 9 9 ( ) ( y ) 9 (ii) Need to sketch a circle of radius and centre (, -) MEI, 0/0/09 5/6

MEI C4 Parametric Section Notes and Eamples 4 y ( ) ( y ) 9 4 4 5 6 4 5 6 Eample 7 (i) Write down the parametric equations of the curve (ii) Sketch the curve. (i) This is a circle radius 4 and centre (0, 0) So the parametric equations are 4cosθ (ii) 6 y y 4sin θ y 6 4 4cosθ y 4sin θ 6 4 4 6 4 6 To test your understanding of this section, try recreating the Parametric Equations Pictures, either on your own or with a group of friends. MEI, 0/0/09 6/6

MEI Core 4 Parametric Equations Section : Parametric Differentiation Notes and Eamples These notes contain subsections on Finding the gradient of a curve given by parametric equations Finding the equation of the tangent and normal to a curve Finding the turning points of a curve Finding the gradient of a curve given by parametric equations You can use the chain rule d y d d y t d dt d to find the gradient function, d y, of a curve defined by parametric d equations. The chain rule can be rewritten as Multiplying by d t is the dy d dy dt d d, d dt 0 same as dividing by d dt dt Eample 5cosθ A circle has the parametric equations y 4 5sin θ Find the gradient of the circle at the point with parameter θ. d 5cos θ 5sin θ dθ dy y 4 5sin θ 5cos θ dθ dy dy Using the chain rule dθ d d dθ dy 5 d cos θ 5 sin θ dy cos θ So cot θ d sin θ Dividing by 0 is undefined. MEI, 0/0/0 /5

MEI C4 Parametric Section Notes and Eamples Finding the equation of the tangent and normal to a curve Finding d y gives you the gradient of the tangent to the curve at the point with d parameter t (or θ). You can then use y y m( ) to find the equation of the tangent. The gradient of the normal is perpendicular to each other. dy d since the tangent and normal are m normal m tangent Eample 4cosθ An ellipse is defined by the parametric equations y sin θ (a) Find the equation of the tangent to the ellipse at the point with parameter θ. (b) Find the equation of the normal to the ellipse at the point (, ) (a) First you need to find the gradient function of the curve d 4cos θ 4sin θ dθ dy y sinθ cos θ dθ dy dy Using the chain rule dθ d d dθ dy cos θ d 4sin θ dy cosθ So d sin θ Now use y y m( ) where 4cos θ, y sinθ and to find the equation of the tangent. m cosθ sin θ So: Multiply both sides by sinθ : Epanding the brackets: Rearranging: cos θ y sin θ ( 4cos θ) sin θ ysin θ 4sin θ cos θ( 4cos θ) sin 4sin cos 4cos y θ θ θ θ sin cos 4cos 4sin y θ θ θ θ Now cos θsin θ so 4cos θ4sin θ 4 So the equation of the tangent is ysin θ cos θ 4 MEI, 0/0/0 /5

MEI C4 Parametric Section Notes and Eamples (b) We need to find the value of the parameter at the point (, ) 4cosθ Now the curve is y sin θ π 5π So solving 4cos θ cos θ θ, π π and sin θ sin θ θ, π so the value of the parameter at (, ) is θ The gradient of the tangent is d y cos θ d sin θ So the gradient of the normal is sin θ cosθ When π θ the gradient of the normal is from part (a) sin π cos π Now use y y m( ) where, y and m to find the equation of the tangent. So y ( ) Epanding the brackets y 4 Simplifying y m normal m tangent Finding the turning points of a curve At a turning point d y 0 d. So to find the turning points Find an epression for the gradient function d y d Put your epression equal to 0 Solve the equation to find the value of the parameter at the turning point You can identify the nature of any turning points by eamining the sign of d y d just before and after the turning point. MEI, 0/0/0 /5

MEI C4 Parametric Section Notes and Eamples Sign of d y d +ve 0 -ve maimum Sign of d y d -ve 0 +ve minimum Sign of d y d +ve 0 +ve -ve 0 -ve Point of inflection Eample Find the turning points of the curve defined by the parametric equations 4 t, y t t and identify their nature. d t dt 4 dy y t t 4t 6t dt Using the chain rule dy dy d d dt 4t 6t At a turning point d y 0 d So t (t ) 0 t = 0 or t So there are turning points at the points with parameters t = 0 or t 4 Substitute t = 0 and t into the parametric equations t, y t t to find the coordinates of the turning points: At t = 0: = - and y = 0 (-, 0) is a turning point 7 7 At t : and, is a turning point y 6 6 Now eamine the sign of dy 4 6 d t t just before and after each turning point. At t = 0 = - At t Value of t t 0. t 0 t 0. Value of -. -0.9 Sign of d y -ve 0 -ve d Value of t t.4 t.5 t.6 Value of 0.4 0.5 0.6 Sign of d y -ve 0 +ve d So there is a point of inflection at (-, 0) and a minimum at, 7 6 Check that you have the points just before and after = - MEI, 0/0/0 4/5

MEI C4 Parametric Section Notes and Eamples You may also like to look at the Parametric differentiation video. For some interesting etension work, try finding and identifying the stationary t points of the graph. Then use a computer to sketch the graph. You 4 y t t may be surprised by the result! MEI, 0/0/0 5/5

MEI Core 4 Further techniques for integration Section : Volumes of revolution Notes and Eamples These notes contain subsections on Solids of revolution formed by rotation about the -ais Solids of revolution formed by rotation about the y-ais Solids of revolution formed by rotation about the -ais y y = f() a b The diagram above shows the solid of revolution formed when the section of the curve y = f() between = a and = b is rotated through 60 about the ais. The volume of the solid is given by b V y d a The derivation of this formula is given on pages 54 55 in the tetbook. Eample A solid is formed by rotating the part of the graph of 60 about the -ais. Find the volume of the solid. y between = and = through MEI, 0/0/09 /

MEI C4 Integration Section Notes and Eamples Volume ( ) d 4 4 d 5 4 5 8 4 5 5 4 5 y d Substitute y = ² into the formula It is a good idea to leave your answer as a multiple of Solids of revolution formed by rotation about the y-ais y d y = f() c The diagram above shows the solid of revolution formed when the section of the curve y = f() between y = c and y = d is rotated through 60 about the y ais. The volume of the solid is given by d V dy c This formula is obtained in a similar way to the one for rotation about the -ais (see page 57 in the tetbook). Notice that in this case the integration is carried out with respect to y rather than and the limits of integration are the y-coordinates rather than the -coordinates. Eample A solid is formed by rotating the part of the graph of 60 about the y-ais. Find the volume of the solid. y between = and = through MEI, 0/0/09 /

MEI C4 Integration Section Notes and Eamples y y When =, y = When =, y = 8 8 Volume dy 8 y dy 8 4 y 6 5 You need in terms of y to be substituted into the formula The limits of integration need to be the y-coordinates. Substitute y into the formula You may also like to look at the Volumes of solids of revolution video. MEI, 0/0/09 /

MEI Core 4 Further techniques for integration Section : General Integration Notes and Eamples These notes contain subsections on Using partial fractions in integration Putting all the integration techniques together Numerical integration Using partial fractions in integration The technique of splitting a fraction into partial fractions can be used to integrate some functions involving fractions, as the following two eamples illustrate. Eample Find d The integrand splits into two partial fractions, like this: A B ( )( ) Multiplying through by ( + )( ): = A( ) + B( + ) Substituting = : 4B B Substituting = : 4A A So 4 4 d d 4 4 d 4 ln ln c 4 4 ln c 4 4 4 MEI, 4/0/ /8

Eample Evaluate MEI C4 Integration Section Notes and Eamples d, giving your answer eactly. A B C ( ) Multiplying through by ²( + ): + = A( + ) + B( + ) + C Substituting = : 4C C Substituting = 0: A A Equating coefficients of : 0 B C B So 4 4 d d 4 ln ln( ) 4 4 4 ln ln 4 ln ln ln ln 4 ln ln 4 4 4 You may also like to look at the Integrating algebraic fractions video. Putting all the integration techniques together The weight of integration facts and techniques can be a little daunting! Choosing the right technique to solve each problem is not always obvious. However, there are two basic ways to become confident at integration: Know your standard derivative and integral results Wherever possible, do integrals by inspection. Here is a summary of the integration techniques so far, and when to use them: MEI, 4/0/ /8

MEI C4 Integration Section Notes and Eamples Standard integrals Learn these! Function Integral Function Integral n n sin cos c c for n - n cos sin c ln c e e c sec tan c Simple variations of these should also be learned: Function e k sin k cosk tan k Integral e k c k cos k c k sin k c k sec k c k Eample Find the following integrals: (i) sin d (ii) e d (i) sin d cos c (ii) e d e c Integration by inspection Other variations of the standard functions can be integrated by guessing what the answer is and differentiating this. If the result is within a constant multiple of the integral, adjust accordingly. Functions which can be integrated by inspection can be recognised. An integral of df the form g[f ( )] d (i.e. the product of a function of f() and the derivative of f()) d can be integrated by inspection. The integral with be of the form of g( ) d MEI, 4/0/ /8

MEI C4 Integration Section Notes and Eamples A particular type of integral which you should recognise is part (iii) of Eample 4). f ( ) d ln f( ) c (see f( ) Eample 4 Find using integration by inspection: (i) d (ii) (i) d The integral is of the form sin cos d (iii) d The integrand is a product of and (a multiple of the derivative of + ²). (a function of + ²), (ii) (iii) d ( ) ( ) d So ( ) which is times the integral. d ( ) c sin cos d The integral is of the form cos. d (cos ) cos sin d sin cos which is times the integral. So sin cos d cos c d The integral is of the form ln. d ln( ) 4 d 4 which is 4 times the integral. So d ln( ) 4 c The integrand is a product of cos (a function of cos ) and sin, (a multiple of the derivative of cos ). The integrand is of the form f ( ) f ( ), where f ( ) With practice you may find that you can do the differentiation in your head and make the necessary adjustment. MEI, 4/0/ 4/8

MEI C4 Integration Section Notes and Eamples Integration by substitution Many of the simpler eamples can be done by inspection, as can be seen by the following eample, in which eamples (i), (ii) and (iii) are as in Eample 4. Eample 5 Find using integration by substitution: (i) d (ii) sin cos d (iii) d (iv) d 0 (i) d du u d d du d u du u du u c c (ii) sin cos d du u cos sin d d du sin sin cos d sin u du sin u du u cos c c (iii) d u du d 4 d d u 4 MEI, 4/0/ 5/8

MEI C4 Integration Section Notes and Eamples d du u d u 4. u ln uc 4 4 c 4 ln( ) (iv) 0 d du u d ddu When = 0, u = When =, u = 4 = u 4 d ( u ) u du 0 4 5 5u u 8 5 6 4 5 6 5 ( u u )du 4 For definite integration, the limits of integration must be changed from values of to values of u. Integration by parts This technique comes directly from the product rule for differentiation, and so is often appropriate for dealing with integrals which are products, often with another function of, e.g. cos, e, ln. The formula for integration by parts is dv du u d uv v d d d Usually the simpler function (often ) is taken to be u. However, integrals which are products including ln are dealt with slightly differently by taking u = ln. Eample 6 Find using integration by parts: (i) cos d (i) cos d (ii) ln d MEI, 4/0/ 6/8

MEI C4 Integration Section Notes and Eamples du Let u d dv cos v sin d cos d sin sin d sin sin d sin cos c 9 (ii) ln d du Let u ln d dv v d ln d ln d ln d ln c 9 Numerical integration Not all functions can be integrated. For functions which cannot be integrated, you can find an approimate value for a definite integral by using a numerical method such as the trapezium rule. The accuracy of the approimation in using the trapezium rule depends on the number of strips the larger the number of strips, the closer the approimation approaches the true value of the integral. The trapezium rule is given by: h A y y y y... y 0 n n where h is the width of each strip.. Eample 7 Use the trapezium rule with (i) 4 strips, (ii) 8 strips to evaluate d, giving your 0 answers to 4 decimal places. Given that the eact area to 4 decimal places is.40, find the percentage error for your answers in (i) and (ii). Comment on your answers. (i) 4 strips: 0 0.5.5 f().00000 0.948 0.707 0.47809 0. MEI, 4/0/ 7/8

MEI C4 Integration Section Notes and Eamples A 0.5 0. 0.948 0.707 0.47809.97 (ii) 8 strips: 0 0.5 0.5 0.75.5.5.75 f().0000 0.998 0.948 0.886 0.707 0.589 0.47809 0.9655 0. A 0.5 0. 0.998 0.948 0.886 0.707 0.589 0.47809 0.9655.400 % error with 4 strips =.97.40 00 0.5%.40 % error with 8 strips =.400.40 00 0.086%.40 The result with 8 strips is more accurate. MEI, 4/0/ 8/8

MEI Core 4 Vectors Section : Introduction to vectors Notes and Eamples These notes contain subsections on Vector in magnitude-direction form or component form Multiplying a vector by a scalar Adding and subtracting vectors Equal vectors and position vectors Unit vectors The vector equation of a line Converting between the vector and cartesian equations of a line Finding the intersection of two lines Vectors in magnitude-direction form or component form A vector quantity has both magnitude (size) and direction. A scalar quantity has magnitude only. Vectors may be written in bold, a, or underlined, a, or with an arrow above, a. Two vectors are equal if they have the same magnitude and direction. You need to be able to write down a vector in two different ways: Magnitude-direction form ( r, θ ) 5 This vector is (5, 5. ) 5. The angle, θ is measured in an anticlockwise direction from the positive ais. Component form The vector is epressed using the unit vectors i and j. i is a unit vector in the direction. j j is a unit vector in the y direction. i A unit vector has a magnitude of. MEI, /05/0 /

MEI C4 Vectors Section Notes and Eamples This vector is i + 4j or 4 The magnitude of a vector given in component form is found using Pythagoras s theorem. So the vector c ai bjhas magnitude: The magnitude of a vector is sometimes c a b called the modulus. A vector given in magnitude-direction form can be written in component form using the rule: r cosθ a ( r, θ) a r cos θi r sin θj r sin θ The following two eamples show you how to convert between the two forms. Eample Write the vectors: (i) (0, 70 ) (ii) (5, 0º) in component form. (i) Using the formula r cos θ a ( r, θ) a r cos θi r sin θj r sin θ (0, 70 ) = 0cos70i 0sin70 j =.4i 9.40j (ii) (5, 0º) = 5cos0 i 5sin 0j =.i.8 j Eample Write the vector: (i) 5i j (ii) i4j in magnitude-direction form. (i) The magnitude of the vector 5i j is 5 5 9 4 Use a sketch to help you find the direction: MEI, /05/0 /

MEI C4 Vectors Section Notes and Eamples θ The angle θ gives the direction of the vector. tan θ θ.0 5 So 5i j ( 4,.0 ) (ii) The magnitude of the vector i4j is ( ) ( 4) 4 6 0 Use a sketch to help you find the direction: θ 80 Remember the direction is measured in an anticlockwise direction from the positive ais. The angle θ + 80º gives the direction of the vector. 4 tan θ θ 6.4 so the direction is 6.4 + 80º= 4.4 So i 4 j ( 0, 4.4 ) For further practice in eamples like the one above, use the interactive questions Magnitude of a vector. Multiplying a vector by a scalar To multiply a vector by a scalar (number) simply multiply each of the components by the scalar. Note: when the scalar is positive the direction of the vector remains the same but the length (or magnitude) of the vector increases by the same factor. when the scalar is negative the direction of the vector is reversed and again the length (or magnitude) of the vector increase. MEI, /05/0 /

MEI C4 Vectors Section Notes and Eamples Eample a i j (i) Find 4a (ii) Find the value of a (iii) Write down the value of 4a (i) 4a 4(i j) 8i j (ii) a ( ) 4 9 (iii) 4a 4 a 4 Eample 4 a 5i 7j Find a. This is the same as multiplying by -. Just reverse the signs! a 5i 7j So a 5i 7j Adding and subtracting vectors To add/subtract vectors simply multiply add/subtract the i components and then the j components. Adding two or more vectors is called finding the resultant. Eample 5 (i) Find the resultant of (5i 7 j) and( i j ) 9 5 (ii) Work out 8 (i) To find the resultant you need to add the vectors. (5i 7 j) ( i j) i 5j You can see this more clearly in this diagram: -5j 5i -7j The resultant is shown by a double headed arrow. i j -i (ii) 9 5 4 8 5 MEI, /05/0 4/

MEI C4 Vectors Section Notes and Eamples The Geogebra resource Adding and subtracting vectors demonstrates the geometrical interpretation of vector addition and subtraction. You may also find the Mathcentre video Introduction to vectors useful. For further practice in vector arithmetic, use the interactive questions Operations on vectors. Equal vectors and position vectors Two vectors are equal if they have the same magnitude and direction. They do not have to be in the same place! Eample 6 The diagram shows a parallelogram ABCD. A B E D DA a AE b AB c (a) Find in terms of a, b and c the vectors: (i) CB (ii) BC (iii) BD. (b) Find two equivalent epressions for AC. C (a) (i) CB DA a (ii) BC CB a (iii) BD BA AD BD c a (b) AC AB BC AC ca Also AC AE b A position vector is a vector which starts at the origin. So if two position vectors are equal they will be in the same place! For eample the point A (5, -) has the position vector OA 5ij. MEI, /05/0 5/

MEI C4 Vectors Section Notes and Eamples You need to know that AO OA AB OA OB So AB OB OA The mid-point, M, has position vector: OM OA AB You can see the reason for these results more clearly in this diagram: y A AB M AO OA OM B O OB Eample 7 The points A and B have coordinates (, 4) and (5, -) respectively. (i) Write down the position vectors OA and OB. (ii) Find the vector AB. (iii) Find the position vector of the mid-point, M of AB. (i) OA = 4 OB = 5 (ii) 5 AB OB OA 4 5 (iii) OM OA AB 4 5 4 MEI, /05/0 6/

MEI C4 Vectors Section Notes and Eamples A unit vector has a magnitude of. i and j are eamples of unit vectors. Unit vectors You need to be able to find a unit vector which has the same direction as a given vector, a. You do this by: Finding the magnitude of the vector, a Dividing a by its magnitude, a Say a hat. The unit vector of a is written a Eample 7 Find the unit vector in the direction of a a ( ) 4 9 a i j The vector equation of a line To write down the cartesian equation of a line you need to know: the coordinates of one point on the line the gradient of the line (or the coordinates of nd point) To write the vector equation of a line you need to know: the position vector of one point on the line the direction of the line (or the position vector of a nd point) The general vector equation of a line AB is: This is the position vector of a point on the line. This is like the +c part of a cartesian equation. r OA λab This gives the direction vector of the line. This is like the m part of a cartesian equation. MEI, /05/0 7/

MEI C4 Vectors Section Notes and Eamples Each different point on the line corresponds to a different value of the parameter. You can see a demonstration of this using the Geogebra resource The vector equation of a line. Note: the vector equation of a line (like its cartesian equivalent) can be written in many different forms. If A and B are the points with position a and b, you can write the vector equation of a line as: which is the same as: r a λ( b a ) r ( λ) a λb Eample 8 Find the vector equation of the line joining the points A (5, -) and B (, 4). 5 a and b 4 r a λ( b a ) 5 5 So r λ 4 5 4 r λ 6 You can also write this as: 5 r λ or even r λ Moving in the direction 4 left and 6 up is the same as moving in the direction left and up If you start at the point (5, -) and move left and up you get to (, ) Eample 9 Find the vector equation of the line through the points A (-4, -) parallel to 4i + j. 4 a and the line is in the direction 4 MEI, /05/0 8/

MEI C4 Vectors Section Notes and Eamples r a λ( b a ) 4 4 So r λ 4 r λ r λ 0 Converting between the vector and cartesian equations of a line You need to be able to convert between the vector and cartesian equations of a line. To convert from a Cartesian equation to a vector equation: Step : Step : Find the position vector of any point on the line, Find a vector in the same direction as the gradient. The following eample shows how to do this. Eample 0 Find the vector equation of the line y 0 Step : When = 0 then y = so the point with position vector is on the line. Step : The gradient is which is in the same direction as the vector So the vector equation is: 0 r λ right and up gives a gradient of = To convert from a vector equation to a Cartesian equation: Step : Write r, y Step : Use the above to write down equations for and y, Step : Make the subject of each equation, Step 4: Equate the two equations, Step 5: Tidy up. MEI, /05/0 9/

MEI C4 Vectors Section Notes and Eamples The following eample shows how to do this. Eample Find the cartesian equation of the line r λ Step : r so λ y y Step : Reading across: λ Step : Step 4: y λ λ λ y y λ λ y This is the position vector of the general point on the line. You have eliminated λ and have an equation in cartesian form. Step 5: Cross multiply: y ( y) ( ) Epand brackets: ( y ) ( ) y 9 So: y 0 For further practice in eamples like the one above, use the interactive questions Cartesian equation of a line. Finding the intersection of two lines You need to be able to convert between the vector and cartesian equations of a line. Step : Equate the two equations, Step : Write down two equations in and, Step : Solve these equations simultaneously, Step 4: Substitute back into the original equations. The following eample will show you how to do this. Eample Find the position vector of the point where the lines: r 4 λ and r μ intersect. MEI, /05/0 0/

MEI C4 Vectors Section Notes and Eamples Step : 4 λ μ Step : Reading across: : 4λ μ y: λ μ Step : 4 8 4λ 4 μ The two lines meet at the point where they are equal to each other. You now have two equations and two unknowns. Solve them simultaneously to find either λ or μ. - 4λ μ - 8 4λ 4 μ 5 0 6 μ So: 5 6 μ μ μ Step 4: μ r So the lines intersect at the point with position vector. MEI, /05/0 /

MEI Core 4 Vectors Section : The scalar product Notes and Eamples These notes contain subsections on The scalar product Finding the angle between two vectors The scalar product The scalar product (or dot product) of two vectors a ai aj and b bi bj is written as aband is found by: Say a dot b ab a b a b Eample 5 Work out a b Using the formula a b a b a b 5 5 ( ) ( ) ( ) 5 5 6 For further practice in eamples like the one above, use the interactive questions Scalar product. You may also find the Mathcentre video The scalar product useful. Finding the angle between two vectors The angle between two vectors is found using the formula: This is also written as: cos a b ab a b a b cos MEI, /05/0 /

MEI C4 Vectors Section Notes and Eamples The net eample shows you how to use this formula to find the angle between two vectors. Eample Work out the angle between the vectors a i j and b i 4j Use the formula cos a b. ab ab ( ) 4 0 4 a b ( ) 4 9 4 6 7 Substituting into cos a b : ab 0 cos 0.676... 7 So.to d.p. For further practice in eamples like the one above, use the interactive questions Angle between two vectors. The angle between two lines is the angle between the two direction vectors. Eample Work out the angle between the lines r 0 4 and r We need to work out the angle between 4 and cos a b ab ab ( 4) 6 4 4 a ( 4) 9 6 5 5 using the formula b 4 5 MEI, /05/0 /

MEI C4 Vectors Section Notes and Eamples Substituting into cos a b : ab cos 0.788... 5 5 So 79.7to s.f. When two lines are perpendicular to each other then: ab 90 cos 0 0 ab ab If 0 then ab 0 ab This is a very important result and means that to show that two vectors are perpendicular you only need to show that the scalar product is 0. Eample 4 Show that the vectors 4 and 6 are perpendicular. If two vectors a ai aj and b bi bj are perpendicular then ab 0 6 6 ( 4) 0as required. 4 MEI, /05/0 /

MEI Core 4 Vectors Section : Vectors in three dimensions Notes and Eamples These notes contain subsections on -D coordinates Magnitude of a -D vector Vector equation of a line in -D Angle between two -D vectors Cartesian equation of a line in -D Special cases of a cartesian equation of a line in -D -D coordinates Much the same methods can be applied to solve -D problems using vector geometry as were covered in units and. In this unit you will learn how to solve problems in -D. -D coordinates can be plotted on a grid like this one: z P(, 5, 6) y The coordinates are always given in the order (, y, z). Magnitude of a -D vector The length of the vector a ai aj ak is: a a a a Eample The points A and B have coordinates (4, -, ) and (-,, ) respectively. Find AB. MEI, 0/0/09 /5

MEI C4 Vectors Section Notes and Eamples 4 OA and OB AB OB OA 4 5 AB 4 AB ( 5) 4 ( ) 5 6 4 For further practice in eamples like the one above, try the interactive questions Magnitude of a vector. To practice other operations on D vectors, use the interactive test Operations on D vectors (which includes adding, subtracting, multiplying by a scalar, and the scalar product). Vector equation of a line in -D The vector equation of a line in -D is the same as in -D: r OA λab Eample The points A and B have coordinates (-,, -) and (0, -, ) respectively. Find the equation of the line AB. 0 OA and OB AB OB OA 0 AB 4 4 The vector equation of a line is r OA λab So: r λ 4 4 MEI, 0/0/09 /5

MEI C4 Vectors Section Notes and Eamples Angle between two -D vectors The angle between two -D vectors is the same as in -D: cos a b ab To find the angle between two lines simply find the angle between their two direction vectors. Note: Two lines in -D may not touch but you can still work out the angle between them in the usual way. To find ABC you need to work out the angle between the vectors BA and BC. A Eample B Find the angle between the lines θ r λ and 0 C r 5 μ 0 Find the angle between the two direction vectors and cos a b ab. 0 0 ( ) ( ) 0 6 9 The length of is: ( ) 9 4 4 The length of 0 is: 0 ( ) 0 9 0 So cos 9 0.7606... 4 0 40.5 to.s.f. 0 using the formula: MEI, 0/0/09 /5

MEI C4 Vectors Section Notes and Eamples For further practice in eamples like the one above, try the interactive questions Angle between two vectors. Cartesian equation of a line in -D The vector equation of a line: a u r a λ u a u can be epressed in cartesian form as: a y a z a u u u Eample 4 Write down the cartesian equation of the line r λ 5 Substituting into the general form: a y a z a u u u we have: ( ) y z 5 this can be written as: y z 5 Note there are two = signs! Eample 5 Write down the vector equation of the line z y 4 5 Write the equation in the same form as: a y a z a u u u we have: z y 4 z 0 y 4 5 5 So the vector equation is: r 4 λ 0 5 MEI, 0/0/09 4/5

MEI C4 Vectors Section Notes and Eamples Special cases of a cartesian equation of a line in -D When the direction vector of a line in -D contains one zero the line is running parallel to either the -y plane, -z plane or the y-z plane. In this case, the Cartesian equation is written slightly differently as is shown in the net eample. Eample 6 4 0 Write down the cartesian equation of the line r λ Substituting into the general form: a y a z a u u u we have: 4 y ( ) z 0 this can be written as: ( y) z 4 and In the line the coordinate is always 4 as the line isn t moving in the direction this means that you would be dividing by 0 which is undefined so we write the equation of the line like this. The net eample shows you how to treat two zeros in the direction vector. Eample 7 0 Write down the cartesian equation of the line r λ 4 0 Substituting into the general form: a y a z a u u u we have: ( ) y z 0 4 0 this can be written as: and z In the line the coordinate is always - and the z coordinate is always as the line isn t moving in the or z directions this means that the y co-ordinate could be any value so we write the equation of the line like this. MEI, 0/0/09 5/5

MEI Core 4 Vectors Section 4: Planes Notes and Eamples These notes contain subsections on Finding the equation of a plane Finding the intersection of a line and a plane Finding the distance of a point from a plane Finding the equation of a plane You can write down the equation of a plane in the form: n n y nz d 0 where d an a and a a is the position vector of a point on the plane a n and n n is a vector perpendicular to the plane. n Eample (i) Write down the equation of the plane through the point (4, 5, -) given that the vector is perpendicular to the plane. (ii) Verify that the point (, 4, -) also lies on the plane. (i) The equation of the plane is n n y n z d 0 where d an 4 n and a 5 4 an. 5 4 5 ( ) ( ) 8 5 6 Now d an d MEI, 0/0/09 /4

MEI C4 Vectors Section 4 Notes and Eamples Substituting into n n y n z d 0 gives: y z 0 All planes with these coefficients of, y and z will be parallel... the value of d locates the plane. and the vector will be perpendicular to each plane (ii) You need to verify that y z 0 when, y 4and z So: 4 ( ) 4 4 0 as required. For further practice in eamples like the one above, use the interactive resource Vector plane equation using the normal. Finding the intersection of a line and a plane The following eample shows you how to find the intersection of a line and a plane. Eample Find the point of intersection of the line y z 6 r λ and the plane 4 The general point on the line is: r y λ z 4 So reading across: λ y λ z 4λ Substitute these into the equation of the plane y z 6 : ( λ) ( λ) (4 λ) 6 Simplifying: MEI, 0/0/09 /4

MEI C4 Vectors Section 4 Notes and Eamples 4 λ 9 6λ 4 λ 6 7 λ 6 λ λ Now substitute λ into the equation of the line to find the position vector of the point of intersection. r y z 4 So the coordinates of the point of intersection are: (, -, ) Check this point lies on the plane y z 6 : ( ) 6 as required.. Finding the distance of a point from a plane The shortest distance of a point A to a plane is the distance AP where AP is a line perpendicular to the plane and P is a point on the plane. To do this: Step : Find the equation of the line through A perpendicular to the plane Step : Find the point of intersection, P, of the line and the plane Step : Find the distance AP. The following eample shows you how to do this. Eample Find the distance of the point A(-,, 4) from the plane y z. Step : The direction vector perpendicular to the plane is: So the equation of the line through A(-,, 4) is: r λ 4 Step : The general point on the line is: r y λ z 4 The coefficients of, y and z in the equation of the plane. MEI, 0/0/09 /4

MEI C4 Vectors Section 4 Notes and Eamples So reading across: λ y λ z 4 λ Substitute these into the equation of the plane y z : ( λ) ( λ) (4 λ) Simplifying: 9λ 4 4λ 4 λ 4λ 4λ 4 λ Now substitute λ into the equation of the line to find the position vector of the point of intersection. r y 0 z 4 5 So the coordinates of the point of intersection are: P(, 0, 5) Step : AP 0 5 4 AP ( ) 9 4 4 So the distance of the point A to the plane is 4 units. MEI, 0/0/09 4/4

MEI Core 4 Differential equations Section : Introduction to differential equations Notes and Eamples These notes contain subsections on Rates of change Differential equations Formulating differential equations Solving differential equations by direct integration Solving differential equations by separating the variables Rates of change The derivative d y represents the rate of change of y with respect to. When d time t is used instead of, then d y represents the rate of change of y. For dt eample: If is a displacement, then d is the rate of change of displacement, or dt the velocity; If v is a velocity, then d v is the rate of change of velocity, or the dt acceleration; If P represents the number of ants in a colony, then d P is the rate of dt change of population; if this is positive then the population is increasing, if negative then it is declining. If M represents the mass of a compound during a chemical process, then d M is the rate of change of mass. dt Differential equations Theory often leads us to formulate mathematical models for the rate of change of quantities. For eample, when a cup of hot water cools, the rate of cooling obviously depends on the temperature; when a car accelerates, the acceleration at faster speeds is slower than at slower speeds so the rate of change of velocity depends on the velocity; the rate of population growth depends on size of the population. MEI, 0/0/09 /5

MEI C4 Diff. equations Section Notes & Eamples In all these cases, we can model the situation by means of a differential equation. This is simply an equation which includes a derivative. If the derivative is of the first order, it is called a first order differential equation. Formulating differential equations If a is proportional to b, we can write this as a b, and deduce that a = kb, where k is a constant, called the constant of proportionality. Eample Formulate a differential equation for the following situations. (i) The acceleration of a particle is proportional to the square of its velocity. (ii) The rate of population growth is proportional to the size of the population at any time. (iii) The rate of cooling of a glass of water is proportional to the difference between its temperature and room temperature. (i) Let the velocity be v. The acceleration is the rate of change of velocity, d v dt. dv kv dt. (ii) (iii) Let P be the population size. The rate of population growth is d P dt. dp kp dt. Let T be the temperature of the water, and T 0 be room temperature. The rate of change of temperature is d T dt, so the rate of cooling is d T dt. dt This is proportional to T T 0, so k( T T0 ), where k > 0. dt Notice the negative sign! This will be crucial to the solution. Solving differential equations by direct integration You already know how to solve some differential equations, like the ones in the net eample. MEI, 0/0/09 /5

MEI C4 Diff. equations Section Notes & Eamples Eample Solve the differential equations: (i) d v t dt (ii) d y, given that when = 0, y =. d dv (i) t dt v t dt t c The solution is v t c (ii) dy y d d. ln c When = 0, y = ln c c The solution is y ln A solution like the one in part (i) of Eample, which involves an arbitrary constant (c), is called the general solution of the differential equation. In effect, the solution is a family of functions, one for each value of c. In part (ii) of Eample, the solution function must pass through the point = 0, y =, and this enables you to calculate the value of c. This gives a particular solution to the differential equation. Solving differential equations by separating the variables Now look at this differential equation: dv dt v, and when t = 0, v = 0 On the right of this equation is a function of v, not t. To get v, you need to integrate with respect to t, not v. You solve this problem like this: dv v dt dv v dt dv dt dt v dt. Integrate both sides with respect to t Note that you only need to add a constant of integration to one side of the equation. MEI, 0/0/09 /5

MEI C4 Diff. equations Section Notes & Eamples dv Now dt is the same as dv v dt or v v dv dt v t c. t c v When t = 0, v = 0 0 c 0 t t v 0 0 0 v 0t. v dv. In this technique, you arrange for the v s on the same side of the equations as the dv, and the t s on the same side as dt. This is called separating the variables. The first few steps can also be written more simply like this: dv v dt dv dt v, and so on Here are some more eamples. Eample The rate of growth of a population after t months P of ants is proportional to the number of ants. At time t = 0, there are 500 ants, and after month the population has risen to 750. Find the population after months. dp kp dt, where k is a constant. dp kdt P ln P kt c ktc P e e kt Ae e kt c Remember that the inverse of the natural logarithm function is the eponential function When dealing with eponentials, it is usual to simplify the solution by replacing the constant e c by a new constant A. When t = 0, P = 500: 500 = Ae k0 = A P 500e kt When t =, P = 750: 750 500e k e k.5 MEI, 0/0/09 4/5

MEI C4 Diff. equations Section Notes & Eamples k t P 500e When t =, t 500.5 P 500.5 687.5 Eample 4 Find the general solution of the differential equation d y y. d ( ) dy y d ( ) dy d y ( ) By partial fractions, d y d y ln y ln ln c ln ln ln A A ln A y The details of finding the partial fractions are omitted here. Replace c with a new constant ln A to give a simpler epression You can omit the modulus signs at this stage the value of A will deal with this. You can practice solving first order differential equations with separable variables using the interactive test Differential equations: first order (all the questions are of the same type). MEI, 0/0/09 5/5