Math Cheat Sheet: Usual types of questions and Tips, Cheat Sheet of Mathematics

Download Math Cheat Sheet: Usual types of questions and Tips and more Cheat Sheet Mathematics in PDF only on Docsity! www.drfrostmaths.com C4 Cheat Sheet Chapter Usual types of questions Tips What can go ugly 1 – Partial Fractions  Be able to split a fraction whose denominator is a product of linear expressions, e.g. 2𝑥+3 𝑥(𝑥+1)  Be able to split a fraction where one (or more) of the factors in the denominator are squared, e.g. 2𝑥+3 𝑥2(𝑥+1)  Deal with top-heavy fractions where the highest power in the denominator is greater or equal to the highest power in the denominator, e.g 𝑥2+2 𝑥(𝑥+1)  The textbook provides two methods for dealing with top heavy fractions. The algebraic long division method is miles easier! e.g. 𝑥2+2 𝑥(𝑥+1) = 𝑥2+2 𝑥2+𝑥 . Using long division we get a quotient of 1 and a remainder of −𝑥 + 2, thus: 𝑥2 + 2 𝑥2 + 𝑥 = 1 + −𝑥 + 2 𝑥(𝑥 + 1) Then split the −𝑥+2 𝑥(𝑥+1) into partial fractions as normal.  Don’t forget that when you have a squared factor in the denominator, you need two fractions in your partial fraction sum: 2 𝑥2(𝑥 + 1) ≡ 𝐴 𝑥 + 𝐵 𝑥2 + 𝐶 𝑥 + 1  When you have three unknowns it’s generally easiest to use substitution to get two of them (e.g. the 𝐴 and the 𝐵) then compare the coefficients of 𝑥2 to get the 𝐶. For the above example: 2 ≡ 𝐴𝑥(𝑥 + 1) + 𝐵(𝑥 + 1) + 𝐶𝑥2 We can see immediately, without needing to write out the expansion, that 0 = 𝐴 + 𝐶, by comparing 𝑥2 terms.  Forgetting the extra term when the denominator’s factors are squared.  Being sloppy at algebraic long division!  Be careful with substitution of negative values.  You may have to spot that you need to factorise the denominator first before expressing as partial fractions.  Not realising the fraction is top heavy and therefore trying to incorrectly do: 2𝑥2 𝑥(𝑥 + 1) = 𝐴 𝑥 + 𝐵 𝑥 + 1 2 – Parametric Equations  Know that 𝑑𝑦 𝑑𝑥 = ( 𝑑𝑦 𝑑𝑡 ) ( 𝑑𝑥 𝑑𝑡 ) (This makes sense as we have just divided numerator and denominator by 𝑑𝑡)  Be able to integrate parametric equations.  Be able to convert parametric equations into a single Cartesian one.  Note: You will NOT be asked to sketch parametric equations.  To convert parametric equations involving trig functions to Cartesian ones, the strategy is usually to make sin 𝑥 and cos 𝑥 the subject before using the identity sin2 𝑥 + cos2 𝑥 ≡ 1. Often squaring one of the parametric equations helps so that we have sin2 𝑥 and/or cos2 𝑥: 𝑥 = √3 sin 2𝑡 𝑦 = 4 cos2 𝑡 𝑥 = 2√3 sin 𝑡 cos 𝑡 𝑥2 = 12 sin2 𝑡 cos2 𝑡 𝑥2 = 12(1 − cos2 𝑡) cos2 𝑡 𝑥2 = 12 (1 − ( 𝑦 4 ) 2 ) 𝑦 4  Hitting a dead end converting parametric equations to Cartesian. See tips on left.  Forgetting to multiply by 𝑑𝑥 𝑑𝑡 when integrating parametric equations. Remember that the 𝑑𝑥 in ∫𝑦 𝑑𝑥 can be replaced with 𝑑𝑥 𝑑𝑡 𝑑𝑡, which is easy to remember, as the 𝑑𝑡’s cancel if we think of 𝑑𝑥 and 𝑑𝑡 just as quantities. www.drfrostmaths.com 3 – Binomial Expansion  Expanding out an expression of the form (1 + 𝑘𝑥)𝑛, where 𝑛 is negative or fractional.  Expanding out an expression of the form (𝑎 + 𝑘𝑥)𝑛, where 𝑎 needs to be factorised out first.  Finding the product of two Binomial expansions, e.g. √1 + 𝑥 √1 − 𝑥 → (1 + 𝑥) 1 2(1 − 𝑥)− 1 2  (1 + 𝑘𝑥)𝑛 = 1 + 𝑛(𝑘𝑥) + 𝑛(𝑛−1) 2! (𝑘𝑥)2 + 𝑛(𝑛 − 1)(𝑛 − 2) 3! (𝑘𝑥)3 +⋯  Your expression may be a binomial expansion in disguise, e.g. 1 √1 − 2𝑥 = (1 − 2𝑥)− 1 2 = 1 + (− 1 2 ) (−2𝑥) + − 1 2 × − 3 2 2! (−2𝑥)2 +⋯  When the first term is not 1, you have to factorise this number out, raised to the power outside the brackets. e.g. (4 + 5𝑥) 1 2 = 4 1 2 (1 + 5 4 𝑥) 1 2 = 2 [1 + 1 2 ( 5 4 𝑥) +⋯] Ensure the outer brackets are maintained till the very end, when you expand them out.  When finding the product of two expansions, then if you needed up to the 𝑥2 term, then you only need to find up to the 𝑥2 term in each of the two expansions. Only consider things in the expansion which are up to 𝑥2. e.g. √ 1 + 𝑥 1 − 𝑥 = √1 + 𝑥 √1 − 𝑥 = (1 + 𝑥) 1 2(1 − 𝑥)− 1 2 (1 + 𝑥) 1 2 ≈ 1 + 1 2 𝑥 − 1 8 𝑥2 (1 − 𝑥)− 1 2 ≈ 1 + 1 2 𝑥 + 3 8 𝑥2 (1 + 𝑥) 1 2(1 − 𝑥)− 1 2 ≈ (1 + 1 2 𝑥 − 1 8 𝑥2) (1 + 1 2 𝑥 + 3 8 𝑥2) = 1 + 1 2 𝑥 + 3 8 𝑥2 + 1 2 𝑥 + 1 4 𝑥2 − 1 8 𝑥2 = 1 + 𝑥 + 1 2 𝑥2 Many things!  Lack of brackets when squaring/cubing things, e.g. you need (2𝑥)3 = 8𝑥3 not 2𝑥3  With say (3 + 4𝑥)−1, forgetting to raise the 3 you factor out to the power of -1.  Forgetting to put the factorial in the denominators of the Binomial coefficients (a common error is □ 3 instead of □ 3! )  Being careless in using your calculator when simplifying coefficients.  Be ridiculously careful with signs!  Accidentally forgetting the minus in the power when expanding say 1 (𝑥+1)2 www.drfrostmaths.com 𝑃𝐴⃗⃗⃗⃗ ⃗ = ( 4 16 −3 ) − ( 9 + 𝜆 13 + 4𝜆 −3 − 2𝜆 ) = ( −5 − 𝜆 3 − 4𝜆 2𝜆 ) Thus: ( −5 − 𝜆 3 − 4𝜆 2𝜆 ) ⋅ ( 1 4 −2 ) = 0 −5 − 𝜆 + 12 − 16𝜆 − 4𝜆 = 0 𝜆 = 1 3 Thus: 𝑃 = ( 9 + 1 3 13 + 4( 1 3 ) −3 − 2( 1 3 ) ) = ( 9 1 3 14 1 3 −3 2 3) 6 - Integration  Integrate a large variety of expressions. See the ‘integration cheat sheet’ overleaf. But by category: o Integrating trig functions, including reciprocal functions and squared functions sin2 𝑥, cos2 𝑥 , sec2 2𝑥, etc. o Integrating by ‘reverse chain rule’ (also known as ‘integration by inspection’). o Integrating by a given substitution. o Integration by parts. o Integrating by use of partial fractions. o Integrating top heavy fractions by algebraic division.  One often forgotten integration is exponential functions such as 2𝑥. Differentiating has effect of multiplying by 𝑙𝑛 of the base, and thus integrating divides by it. i.e. 𝑑 𝑑𝑥 (2𝑥) = ln 2 ⋅ 2𝑥 ∫ 2𝑥 𝑑𝑥 = 1 ln 2 2𝑥 + 𝐶  Know the two double angle formulae for 𝑐𝑜𝑠 like the back of your hand, for use when integrating sin2 𝑥 or cos2 𝑥  In general, know your integrals of all the ‘trig squares’, i.e. sin2 𝑥 , cos2 𝑥 , tan2 𝑥 , 𝑐𝑜𝑠𝑒𝑐2𝑥, sec2 𝑥 , cot2 𝑥  For integration by ‘reverse chain rule’, always ‘consider’ some sensible expression to differentiate, then adjust for the factor difference. e.g. ∫(4 − 3𝑥)5𝑑𝑥 Then your working might be: “Consider 𝑦 = (4 − 3𝑥)6. Then 𝑑𝑦 𝑑𝑥 = 6(4 − 3𝑥)5 × (−3) = −18(4 − 3𝑥)5 ∴ ∫(4 − 3𝑥)5𝑑𝑥 = − 1 18 (4 − 3𝑥)6 + 𝐶  For integration by substitution, the official specification says “Except in the simplest of cases, the substitution will be given.”  Remember that starting with the substitution, say 𝑢 = 𝑥2 + 1, it helps to make 𝑥 the subject, except in some cases where there’s a trigonometric substitution, e.g. if 𝑢 = sin𝑥 + 1, but sin 𝑥 appears in the expression to integrate, then we might Where to start!  One big problem is just not knowing what method to use to integrate a particular expression. The cheat sheet overleaf should help, as should lots of practice of a variety of expressions!  Similarly getting stuck on integration by substitution, because you can’t get the whole original expression only in terms of the new variable (𝑡 or otherwise).  Perhaps the all-time biggest mistake is forgetting to consider the effects of chain rule. e.g. Accidentally doing ∫cos 2𝑥 𝑑𝑥 = sin 2𝑥  Sign errors when integrating/differentiating trig functions. Other than sin and cos, be careful about cot/cosec: 𝑑 𝑑𝑥 (cot 𝑥) = −𝑐𝑜𝑠𝑒𝑐2𝑥 thus ∫ 𝑐𝑜𝑠𝑒𝑐2𝑥 𝑑𝑥 = −cot 𝑥 + 𝐶  A common one: Forgetting about the chain rule when integrating expressions of the form (𝑎 + 𝑏𝑥)𝑐, see ∫(4 − 3𝑥)5𝑑𝑥 example. www.drfrostmaths.com  Be able to differentiate parametric equations: ∫𝑦 𝑑𝑥 = ∫𝑦 𝑑𝑥 𝑑𝑡 𝑑𝑡  Calculate volumes of revolution both for normal and parametric equations: 𝑉 = 𝜋∫ 𝑦2 𝑑𝑥 𝑉 = 𝜋∫𝑦2 𝑑𝑥 𝑑𝑡 𝑑𝑡  Solve differential equations. e.g. 𝑑𝑦 𝑑𝑥 = 𝑥𝑦 + 𝑥  Trapezium Rule as per C2, but now with C3/C4 expressions to integrate. You will frequently be asked to compare the actual error and the estimated area using the rule, and the percentage error. make sin 𝑥 the subject instead. Differentiate and make 𝑑𝑥 the subject also, then ensure original expression is only in terms of new variable.  Don’t feel as if you need to memorise a separate formula for parametric volumes of revolution, since 𝑑𝑥 = 𝑑𝑥 𝑑𝑡 𝑑𝑡 clearly by the fact that the 𝑑𝑡’s cancel.  You have to change the limits whenever you do either of: (a) parametric integration or (b) integration by substitution, because you’re integrating in terms of a new variable.  This is more use for STEP, but remember that ∫ 𝑓(𝑥)𝑑𝑥 𝑎 𝑏 = −∫ 𝑓(𝑥) 𝑏 𝑎 , useful when the limits are the wrong way round.  You can tidy things up sometimes using −∫−𝑓(𝑥)𝑑𝑥 = +∫𝑓(𝑥)𝑑𝑥, since the -1 can be factored out the integral.  For integration by parts, if you ever have to IBP twice, write the second integral as a separate result first before substituting it in after. This is to avoid sign errors and keep things tidy. e.g. Workings might be: ∫𝑥2 cos 𝑥 𝑑𝑥 𝑢 = 𝑥2 𝑑𝑣 𝑑𝑥 = cos 𝑥 𝑑𝑢 𝑑𝑥 = 2𝑥 𝑣 = sin 𝑥 ∫𝑥2 sin 𝑥 𝑑𝑥 = 𝑥2 sin 𝑥 − ∫2𝑥 sin 𝑥 𝑑𝑥 “For ∫ 2𝑥 sin 𝑥 𝑑𝑥:” 𝑢 = 2𝑥 𝑑𝑣 𝑑𝑥 = sin 𝑥 𝑑𝑢 𝑑𝑥 = 2 𝑣 = − cos 𝑥 ∫2𝑥 sin 𝑥 𝑑𝑥 = −2𝑥 cos 𝑥 − ∫−2 cos 𝑥 𝑑𝑥 = −2𝑥 cos 𝑥 + ∫2 cos 𝑥 𝑑𝑥 = −2𝑥 cos 𝑥 + 2 sin 𝑥 ∴ ∫𝑥2 cos 𝑥 𝑑𝑥 = 𝑥2 sin 𝑥 − (−2𝑥 cos 𝑥 + 2 sin 𝑥) = 𝑥2 sin 𝑥 + 2𝑥 cos 𝑥 − 2 sin 𝑥 + 𝐶  Remember that constants differentiate to nothing, i.e. 𝑑 𝑑𝑥 (𝜋2) = 0 not 2𝜋!  Similarly ln 𝑐 is a constant. 𝑥 ln 2 would differentiate to ln 2.  If 𝑢2 = 𝑥 + 1 is the substitution, you’re doing unnecessary work if you then square root. Differentiating implicitly: 2𝑢 𝑑𝑢 𝑑𝑥 = 1 𝑑𝑥 = 2𝑢 𝑑𝑢 This is much much tidier!  Forgetting to change your limits for either parametric integration or integration by substitution! But note that in integration by substitution, once you’ve changed back to the original variable (probably 𝑥), you should use the original limits.  Don’t try and use integration by parts if you can use ‘integration by inspection’. e.g. For ∫𝑥 𝑒𝑥 2 , then integration by parts would lead to a dead end.  For differential equations, ensure the variable at the top of the 𝑑□ 𝑑□ matches what you’ve moved to the LHS. e.g. If 𝑑𝑡 𝑑𝑟 = 𝑟2𝑡 then it’s the 𝑡 you want on the LHS. www.drfrostmaths.com Note the nice double negative tidying up trick towards the end.  If you’re solving 𝑑𝑦 𝑑𝑥 = 𝑥𝑦 + 𝑦, then you need the 𝑦 (or whatever variable appears at the top of 𝑑𝑦 𝑑𝑥 ) on the LHS. This is always achieved by a division or multiplication, which may require factorisation first: 𝑑𝑦 𝑑𝑥 = 𝑦(𝑥 + 1) 1 𝑦 𝑑𝑦 𝑑𝑥 = 𝑥 + 1 ∫ 1 𝑦 𝑑𝑦 = ∫𝑥 + 1 𝑑𝑥 ln|𝑦| = 1 2 𝑥2 + 𝑥 + 𝐶 𝑦 = 𝑒 1 2 𝑥2+𝑥+𝐶 = 𝐴𝑒 1 2 𝑥2+𝑥  Note in the above example, we let some new constant 𝐴 = 𝑒𝐶 to help tidy things up. If we had ln 𝑥 + 𝐶 on the right- hand-side, we’d make 𝐶 = ln𝐴 so that ln 𝑥 + ln𝐴 = ln(𝐴𝑥). Similarly if we had ln 𝑦 = 𝑥 + 𝐶, and hence 𝑦 = 𝑒𝑥+𝐶 = 𝑒𝑥𝑒𝐶, we could make 𝐴 = 𝑒𝐶.  In differential equations, ensure you separate the RHS into the form 𝑓(𝑥)𝑔(𝑦) first so that you are able to divide by 𝑔(𝑦), e.g. 𝑑𝑦 𝑑𝑥 = 𝑥 + 𝑥𝑦 → 𝑑𝑦 𝑑𝑥 = 𝑥(1 + 𝑦) → 1 1+𝑦 𝑑𝑦 = 𝑥 𝑑𝑥  In differential equations, if you’re given initial conditions (note, often 𝑡 = 0 is often implied for the initial condition), then it’s generally easier to plug them in to work out your constant of integration sooner rather than later.