C4 Cheat Sheet - DrFrostMaths
C4 Cheat SheetChapter1 β PartialFractionsUsual types of questions Be able to split a fraction whosedenominator is a product of2π₯ 3linear expressions, e.g. π₯(π₯ 1)Tips The textbook provides two methods for dealing with topheavy fractions. The algebraic long division method is mileseasier!π₯ 2 2and a remainder of π₯ 2, thus:π₯2 2 π₯ 2 1 2π₯ π₯π₯(π₯ 1) π₯ 2Then split theinto partial fractions as normal.π₯Deal with top-heavy fractionswhere the highest power in thedenominator is greater or equalto the highest power in thedenominator, e.g π₯ 2 2π₯(π₯ 1) 2 β ParametricEquations Know thatππ¦ππ₯ ππ¦)ππ‘ππ₯( )ππ‘((This makes sense as we have just dividednumerator and denominator by ππ‘) Be able to integrate parametricequations.Be able to convert parametricequations into a single Cartesianone.www.drfrostmaths.comπ₯ 2 2e.g. π₯(π₯ 1) π₯ 2 π₯. Using long division we get a quotient of 1Be able to split a fraction whereone (or more) of the factors inthe denominator are squared,2π₯ 3e.g. 2 (π₯ 1) π₯(π₯ 1)Donβt forget that when you have a squared factor in thedenominator, you need two fractions in your partial fractionsum:2π΄ π΅πΆ 2 2π₯ (π₯ 1) π₯ π₯π₯ 1When you have three unknowns itβs generally easiest to usesubstitution to get two of them (e.g. the π΄ and the π΅) thencompare the coefficients of π₯ 2 to get the πΆ. For the aboveexample:2 π΄π₯(π₯ 1) π΅(π₯ 1) πΆπ₯ 2We can see immediately, without needing to write out theexpansion, that 0 π΄ πΆ, by comparing π₯ 2 terms.Note: You will NOT be asked to sketch parametric equations.To convert parametric equations involving trig functions toCartesian ones, the strategy is usually to make sin π₯ andcos π₯ the subject before using the identity sin2 π₯ cos2 π₯ 1. Often squaring one of the parametric equations helps sothat we have sin2 π₯ and/or cos 2 π₯:π₯ 3 sin 2π‘π¦ 4 cos 2 π‘π₯ 2 3 sin π‘ cos π‘π₯ 2 12 sin2 π‘ cos 2 π‘π₯ 2 12(1 cos 2 π‘) cos 2 π‘π¦ 2 π¦π₯ 2 12 (1 ( ) )44What can go ugly Forgetting the extra term when thedenominatorβs factors are squared. Being sloppy at algebraic long division! Be careful with substitution of negativevalues. You may have to spot that you need tofactorise the denominator first beforeexpressing as partial fractions. Not realising the fraction is top heavy andtherefore trying to incorrectly do:2π₯ 2π΄π΅ π₯(π₯ 1) π₯ π₯ 1 Hitting a dead end converting parametricequations to Cartesian. See tips on left.ππ₯Forgetting to multiply by ππ‘ whenintegrating parametric equations.Remember that the ππ₯ in π¦ ππ₯ can beππ₯replaced with ππ‘ ππ‘, which is easy toremember, as the ππ‘βs cancel if we think ofππ₯ and ππ‘ just as quantities.
3 β BinomialExpansion Expanding out an expression ofthe form (1 ππ₯)π , where π isnegative or fractional.Expanding out an expression ofthe form (π ππ₯)π , where πneeds to be factorised out first.Finding the product of twoBinomial expansions, e.g.11 1 π₯ (1 π₯)2 (1 π₯) 2 1 π₯ π(π 1)(1 ππ₯)π 1 π(ππ₯) (ππ₯)22!π(π 1)(π 2)(ππ₯)3 3!Your expression may be a binomial expansion in disguise, e.g.13 11 22 ( 2π₯)2 2(1( 2π₯) 2π₯) 1 ( ) 22! 1 2π₯1 When the first term is not 1, you have to factorise thisnumber out, raised to the power outside the brackets. e.g.12 115(4 5π₯)2 42 (1 π₯)41 5 2 [1 ( π₯) ]2 4Ensure the outer brackets are maintained till the very end,when you expand them out.When finding the product of two expansions, then if youneeded up to the π₯ 2 term, then you only need to find up tothe π₯ 2 term in each of the two expansions. Only considerthings in the expansion which are up to π₯ 2 . e.g.111 π₯ 1 π₯ (1 π₯)2 (1 π₯) 21 π₯ 1 π₯111(1 π₯)2 1 π₯ π₯ 228113 (1 π₯) 2 1 π₯ π₯ 228111113(1 π₯)2 (1 π₯) 2 (1 π₯ π₯ 2 ) (1 π₯ π₯ 2 )282813111 1 π₯ π₯2 π₯ π₯2 π₯2282481 1 π₯ π₯22www.drfrostmaths.comMany things! Lack of brackets when squaring/cubingthings, e.g. you need (2π₯)3 8π₯ 3 not 2π₯ 3 With say (3 4π₯) 1, forgetting to raise the3 you factor out to the power of -1. Forgetting to put the factorial in thedenominators of the Binomial coefficients (a common error is instead of ) 33!Being careless in using your calculator whensimplifying coefficients.Be ridiculously careful with signs!Accidentally forgetting the minus in the1power when expanding say (π₯ 1)2
4Differentiation Appreciate that π¦ π π₯represents βexponential growthβwhen π 1, and βexponentialdecayβ when 0 π 1 (andfrom C3, know the graphs foreach).πKnow that ππ₯ (π π₯ ) π π₯ ln π(proof unlikely to be asked for)Be able to differentiateπππ¦implicitly, e.g. ππ₯ (π¦ 2 ) 2π¦ ππ₯and subsequently be able toππ¦make ππ₯ the subject.Be able to set up differentialequations, e.g. understand thatβthe temperature falls at a rateproportional to its currenttemperatureβ could beππrepresented as ππ‘ ππConnect different derivativesππ΄ππ΄ππ‘involving rates, e.g. ππ₯ ππ‘ ππ₯ Example of implicit differentiation (which involves collectingππ¦the terms on one side and factorising it out):ππ₯ππ¦βGiven that π₯π¦ 2 2π¦ π₯ 2 , find .βππ₯Differentiating both sides with respect to π₯:π₯ (2π¦ππ¦ππ¦) π¦2 2 2π₯ππ₯ππ₯ππ¦(2π₯π¦ 2) 2π₯ π¦ 2ππ₯ππ¦ 2π₯ π¦ 2 ππ₯ 2π₯π¦ 2 They particularly love use of the product rule!A βdifferential equationβ is an equation involving both somevariables and derivatives involving those variables, e.g. a mixππ¦of π₯, π¦ and ππ₯. βSolvingβ this equation means to obtain anequation only involving the variables, and not the derivatives.Whenever you see the word βrateβ, think /ππ‘.βA circleβs radius increases at a rate of 2cm/s. Find the rate ofincrease of its area when the radius is 10cm.βFirst note the variables involved: π΄, π and because weβreππ΄talking about rates, π‘. We need to find . Since derivativesππ‘behave pretty much like normal fractions, first write thefollowing product with the ππ΄ and ππ‘ copied into thediagonals:ππ΄ ππ΄ ππ‘ ππ‘Then fill the remaining diagonals with the remaining variable,ππ΄ππ΄ππππ: ππ‘ππππ‘ππOne value, in this case ππ‘ 2, is always given. The other weneed to form some formula, in this case π΄ ππ 2 (and oftenusing simple geometry to find an area of volume), anddifferentiate:ππ΄ππ΄ 2ππππThus when π 10, ππ 2 π 10 20πππ΄Thus: ππ‘ 20π 2 40πwww.drfrostmaths.com A classic is to accidentally treat π₯ or π¦ asconstants rather than variables, whendifferentiating implicitly. Note thatππ(ππ₯) π if π is a constant, but (π₯π¦) ππ₯ππ¦π₯ ππ₯ππ₯ π¦ by the product rule, and not just π¦.When differentiating implicitly, you mightππ¦πforget to put the ππ₯, e.g. ππ₯ (π¦ 2 ) 2π¦rather than the correct 2π¦ ππ¦ππ₯Exponential functions do not behave likepolynomials when differentiated. e.g.ππ(π₯ 3 ) 3π₯ 2 , but (3π₯ ) 3π₯ ln 3, andππ₯ππ₯absolutely not π₯ 3π₯ 1 !Many students often get their equationwrong when connecting rates of change,often say dividing instead of multiplying, orvice versa. If you use the βfill in thediagonalsβ tip on the left this will unlikely bea problem.
5 - Vectors(In rough descending order of howfrequently they appear in exams) Find the point of intersection oftwo lines or prove that two linesdo not intersect. Find the angle between twolines. Finding a missing π₯/π¦/π§ value ofa point on a line. Find the length of a vector or thedistance between two points. Find the nearest point on a lineto a point not on the line (oftenthe origin) β note: not in yourtextbook! Show lines are perpendicular. Show a point lies on a line. Show 3 points are collinear (i.e.lie on the same straight line) Find the area of a rectangle,parallelogram or triangle formedby vectors. Find the equation of a line. Find the reflection of a point in aline.www.drfrostmaths.com When you see the π, π, π unit vectors used in an examquestion, never actually use this notation yourself: alwaysjust write all vectors in conventional column vector form.Almost always draw a suitable diagram. This will beparticularly helpful when you need to find the area of someshape (typically the last part of a question).When finding the area of a shape, you can almost always useyour answers from previous parts of the questions, includinglengths of vectors and angles between two vectors.1Remember that area of non-right angled triangle 2 ππ sin πΆwhere the angle πΆ appears between the two sides π and π. Aparallelogram can be cut in half to form two congruent nonright angled triangles (i.e. multiply by 2).To show 3 points π΄, π΅, πΆ are collinear, just show that βββββπ΄π΅ is aβββββmultiple of π΅πΆ (i.e. vectors are parallel).βShow 3π 3π 2π lies on the line with vector equation π π 3π 4π π‘(π π)β31 π‘i.e. Show (3) lies on ( 3 ). Equating 3 1 π‘ to π‘ 2.24 π‘Then 4 π‘ 4 2 2, so π¦ and π§ components are same.βLet π1 : π (9π 13π 3π) π(π 4π 2π)Given point π΄ has positive vector 4π 16π 3π and π lies onπ1 such that π΄π is perpendicular to π1 , find π.β9 ππ1 : ( 13 4π ) 3 2πNote that the direction vector of the line, and the vector βββββππ΄are perpendicular. π is just a point on the line so can be9 πrepresented as ( 13 4π ) for some specific π we need to 3 2πfind.1Direction vector of π1 is ( 4 ) 2 When finding the angle between two lines,accidentally using the full vectorrepresentation of the line (in your dotproduct), and not just the direction1 π‘component, e.g. using ( 2 ) instead of3 2π‘ 1just the correct ( 0 ).π‘Making sign errors when subtractingvectors, particularly when subtracting anexpression involving a negative. Correctly:π‘11 π‘(2) ( 2 ) ( 4 )7 2π‘32π‘ 4Once finding out π and π‘ (or π and π) whensolving simultaneous equation to find theintersection of two lines, forgetting to showthat these satisfy the remaining equation.Forgetting the square root when finding themagnitude of a vector.
Thus:49 π 5 πβββββππ΄ ( 16 ) ( 13 4π ) ( 3 4π ) 3 3 2π2π1 5 π( 3 4π ) ( 4 ) 02π 2 5 π 12 16π 4π 01π 3119 933Thus: π 13 4(13) 14 1316 - Integration Integrate a large variety ofexpressions. See the βintegrationcheat sheetβ overleaf. But bycategory:o Integrating trigfunctions, includingreciprocal functions andsquared functionssin2 π₯, cos2 π₯ , sec 2 2π₯,etc.o Integrating by βreversechain ruleβ (also knownas βintegration byinspectionβ).o Integrating by a givensubstitution.o Integration by parts.o Integrating by use ofpartial fractions.o Integrating top heavyfractions by algebraicdivision. 3 2( )3 )2 3( 3)One often forgotten integration is exponential functions suchas 2π₯ . Differentiating has effect of multiplying by ππ of thebase, and thus integrating divides by it. i.e.π π₯(2 ) ln 2 2π₯ππ₯ 2π₯ ππ₯ 1 π₯2 πΆln 2Know the two double angle formulae for πππ like the back ofyour hand, for use when integrating sin2 π₯ or cos 2 π₯In general, know your integrals of all the βtrig squaresβ, i.e.sin2 π₯ , cos2 π₯ , tan2 π₯ , πππ ππ 2 π₯, sec 2 π₯ , cot 2 π₯For integration by βreverse chain ruleβ, always βconsiderβsome sensible expression to differentiate, then adjust for thefactor difference. e.g. (4 3π₯)5 ππ₯Then your working might be:βConsider π¦ (4 3π₯)6 . Then www.drfrostmaths.com(ππ¦ 6(4 3π₯)5 ( 3) 18(4 3π₯)5ππ₯1 (4 3π₯)5 ππ₯ (4 3π₯)6 πΆ18For integration by substitution, the official specification saysβExcept in the simplest of cases, the substitution will begiven.βRemember that starting with the substitution, say π’ π₯ 2 1, it helps to make π₯ the subject, except in some cases wherethereβs a trigonometric substitution, e.g. if π’ sin π₯ 1, butsin π₯ appears in the expression to integrate, then we mightWhere to start! One big problem is just not knowing whatmethod to use to integrate a particularexpression. The cheat sheet overleaf shouldhelp, as should lots of practice of a varietyof expressions! Similarly getting stuck on integration bysubstitution, because you canβt get thewhole original expression only in terms ofthe new variable (π‘ or otherwise). Perhaps the all-time biggest mistake isforgetting to consider the effects of chainrule. e.g. Accidentally doing cos 2π₯ ππ₯ sin 2π₯ Sign errors when integrating/differentiatingtrig functions. Other than sin and cos, becareful about cot/cosec:π(cot π₯) πππ ππ 2 π₯ππ₯thus πππ ππ 2 π₯ ππ₯ cot π₯ πΆA common one: Forgetting about the chainrule when integrating expressions of theform (π ππ₯)π , see (4 3π₯)5 ππ₯example.
Be able to differentiateparametric equations:ππ₯ π¦ ππ₯ π¦ππ‘ππ‘Calculate volumes of revolutionboth for normal and parametricequations:π π π¦ 2 ππ₯ππ₯π π π¦2ππ‘ππ‘Solve differential equations.ππ¦e.g. ππ₯ π₯π¦ π₯Trapezium Rule as per C2, butnow with C3/C4 expressions tointegrate. You will frequently beasked to compare the actualerror and the estimated areausing the rule, and thepercentage error. make sin π₯ the subject instead.Differentiate and make ππ₯ the subject also, then ensureoriginal expression is only in terms of new variable.Donβt feel as if you need to memorise a separate formula forππ₯parametric volumes of revolution, since ππ₯ ππ‘ clearly byππ‘the fact that the ππ‘βs cancel.You have to change the limits whenever you do either of: (a)parametric integration or (b) integration by substitution,because youβre integrating in terms of a new variable.πThis is more use for STEP, but remember that π π(π₯)ππ₯ Remember that constants differentiate toπnothing, i.e. (π 2 ) 0 not 2π! π π(π₯), useful when the limits are the wrong way round.You can tidy things up sometimes using π(π₯)ππ₯ π(π₯)ππ₯, since the -1 can be factored out the integral.For integration by parts, if you ever have to IBP twice, writethe second integral as a separate result first beforesubstituting it in after. This is to avoid sign errors and keepthings tidy. e.g. Workings might be: Similarly ln π is a constant. π₯ ln 2 woulddifferentiate to ln 2.If π’2 π₯ 1 is the substitution, youβredoing unnecessary work if you then squareroot. Differentiating implicitly:ππ’2π’ 1ππ₯ππ₯ 2π’ ππ’This is much much tidier!Forgetting to change your limits for eitherparametric integration or integration bysubstitution!But note that in integration by substitution,once youβve changed back to the originalvariable (probably π₯), you should use theoriginal limits.Donβt try and use integration by parts if youcan use βintegration by inspectionβ.2e.g. For π₯ π π₯ , then integration by partswould lead to a dead end.For differential equations, ensure theπ variable at the top of the π matches whatyouβve moved to the LHS. e.g. Ifππ‘ π2π‘ππthen itβs the π‘ you want on the LHS.π π₯ 2 cos π₯ ππ₯π’ π₯2ππ’ 2π₯ππ₯2ππ£ cos π₯ππ₯π£ sin π₯2 π₯ sin π₯ ππ₯ π₯ sin π₯ 2π₯ sin π₯ ππ₯βFor 2π₯ sin π₯ ππ₯:βπ’ 2π₯ππ£ sin π₯ππ₯ππ’ 2 π£ cos π₯ππ₯ 2π₯ sin π₯ ππ₯ 2π₯ cos π₯ 2 cos π₯ ππ₯ 2π₯ cos π₯ 2 cos π₯ ππ₯ 2π₯ cos π₯ 2 sin π₯2 π₯ cos π₯ ππ₯ π₯ 2 sin π₯ ( 2π₯ cos π₯ 2 sin π₯) π₯ 2 sin π₯ 2π₯ cos π₯ 2 sin π₯ πΆwww.drfrostmaths.com ππ₯
Note the nice double negative tidying up trick towards theend.ππ¦If youβre solving ππ₯ π₯π¦ π¦, then you need the π¦ (orππ¦whatever variable appears at the top of ) on the LHS. Thisππ₯is always achieved by a division or multiplication, which mayrequire factorisation first:ππ¦1 ππ¦ π¦(π₯ 1) π₯ 1ππ₯π¦ ππ₯11 ππ¦ π₯ 1 ππ₯ln π¦ π₯ 2 π₯ πΆπ¦21 2 www.drfrostmaths.com1 2π¦ π 2π₯ π₯ πΆ π΄π 2π₯ π₯Note in the above example, we let some new constant π΄ π πΆ to help tidy things up. If we had ln π₯ πΆ on the righthand-side, weβd make πΆ ln π΄ so that ln π₯ ln π΄ ln(π΄π₯).Similarly if we had ln π¦ π₯ πΆ, and hence π¦ π π₯ πΆ π π₯ π πΆ , we could make π΄ π πΆ .In differential equations, ensure you separate the RHS intothe form π(π₯)π(π¦) first so that you are able to divide byππ¦ππ¦1π(π¦), e.g. ππ₯ π₯ π₯π¦ ππ₯ π₯(1 π¦) 1 π¦ ππ¦ π₯ ππ₯In differential equations, if youβre given initial conditions(note, often π‘ 0 is often implied for the initial condition),then itβs generally easier to plug them in to work out yourconstant of integration sooner rather than later.
C4 Integration Cheat Sheetπ(π)πππ ππππ ππππ πHow to deal with itStandard resultStandard resultIn formula booklet, but usesin π₯ππ₯ which is of the form ππππ πππππ πππππ ππππππ ππππ ππππ ππ(π₯)sin π₯π β²(π₯)π(π₯)FormBk?NoNoYes11π₯ sin 2π₯2411π₯ sin 2π₯24tan π₯ π₯ ln πππ ππ π₯ cot π₯ Use partial fractions.Noππ(π π)ππππ πPower around denominator soNoππ(π π)πYesln π ππ π₯ YesNo!Yes (butπππ₯π§ π1 cot 2 π₯ πππ ππ 2 π₯Standard resultπ¦ π π₯ ln π¦ π₯ ln πThen differentiate implicitly.Standard result cot π₯ π₯ππ₯1ππ₯ln(π)ln π₯NoNoNoNoUse IBP, where π’ ln π₯ , ππ₯ 1www.drfrostmaths.comπ₯ ln π₯ π₯Noπ sin π₯π₯ ln π₯ 1 π₯1 1 π₯ 1π₯ 1Reverse chain rule. Of form ln π₯ ln π₯ 1 2 ln π₯ 2 1 ππ β²(π₯)π(π₯)NOT of form ππβ² (π₯)π(π₯). Rewrite asππππππ ππππ πππ πππ ππ ππ¬π’π§π π πππ πReverse chain rule (i.e. βConsiderπ¦ (π₯ 2 1) 1 " and differentiate.Fraction top heavy so do algebraicdivision first. Then split intoalgebraic fractions as4π₯ 2 1 (2π₯ 1)(2π₯ 1)For any function where βinnerfunctionβ is linear expression,divide by coefficient of π₯Use sensible substitution. π’ 2π₯ 1 or even better, π’2 2π₯ 1.Reverse chain rule.πππ ππ πππ ππ Use identities in C3 formulabooklet,1sin 3π₯ cos 2π₯ (sin 5π₯ cos 3π₯)ππ£FB?No1 (π₯ 2 1) 12π₯(π₯ 2 1) 2memorise)ππππ πππππOf form πβ² (π₯)π β² (π(π₯))Use algebraic division. π(π)π π ( constant)1 cos 4π₯8product.Yesln sec π₯ tan π₯ cot π₯tan π₯1sin 2π₯ cos 2π₯ sin 4π₯2Noππ₯By observation.By observation.π(π)How to deal with itπππ ππ πππ ππ For any product of sin and coswith same coefficient of π₯, usedouble angle.ππ¨π¬ π ππππ πππ πππ₯For both sin2 π₯ and cos 2 π₯ useidentities for cos 2π₯cos 2π₯ 1 2 sin2 π₯1 1sin2 π₯ cos 2π₯2 2cos 2π₯ 2 cos 2 π₯ 11 1cos 2 π₯ cos 2π₯2 21 tan2 π₯ sec 2 π₯tan2 π₯ sec 2 π₯ 1Would use substitution π’ πππ ππ π₯ cot π₯, but too hardfor exam.Would use substitution π’ sec π₯ tan π₯, but too hard forexam.cos π₯ππ₯ which is of the form ππππππ πππππ πcos π₯ππ β²(π₯) π(π)π π ( constant) cos π₯sin π₯ln sec π₯ 2Note: has never come up in an exam.12π₯ ln 1 2π₯ 21 ln 2π₯ 1 21 2π₯ 1π21 ln 1 3π₯ 331(2π₯ 1)2 (3π₯ 1)151 6sin π₯61cos 5π₯101 sin 3π₯6 Sort of
functions, including reciprocal functions and squared functions sin2 , cos2 ,sec22 , etc. o Integrating by βreverse chain ruleβ forgetting to consider the effects of chain (also known as βintegration by inspectionβ). o Integrating by a given substitution. o Integration by parts. o Integrating by use of partial fractions.
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cheat sheet for each student and have them glue it into their interactive notebooks. If you give one to each student, you could have them color the cheat sheet (If time is limited, I would skip or have students color at home). Please let me know if you have any questions about the cheat sheet! You can email me at mathindemand@hotmail.com.
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