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• Composites

• A-Level Maths Cheat Sheet

This is absolutely not a cheat sheet... We do not endorse cheating at all! It's just a better name than "List of vital but forgettable equations for A Level Maths" The Formula Book for Edexcel A-Level Maths and Further Maths can be seen here and downloaded here: Trigonometry Cosine Rule a² = b² + c² - 2bc cosA Area of a Triangle Area = ½ ab sinC Radians 1° = π/180 1 rad = 180/π Small Angle Approximations sinθ ≈ θ tanθ ≈ θ cosθ ≈ 1 - θ²/2 Trig Identities sin² θ + cos² θ ≡ 1 tan θ ≡ sin θ / cos θ 1 + tan² x ≡ sec² x 1 + cot² x ≡ cosec² x Angle Addition Formulae sin(A ± B) ≡ sinA cosB ± cosA sinB cos(A ± B) ≡ cosA cos B ∓ sinA sinB tan(A ± B) ≡ (tanA ± tanB) / (1 ∓ tanA tanB) Double Angle Formulae sin2A ≡ 2 sinA cosA cos2A ≡ cos²A - sin²A ≡ 2cos²A - 1 ≡ 1 - 2sin²A tan2A ≡ (2 tanA) / (1 - tan²A) R-Addition Formula a sinx ± b sinx can be expressed as R sin(x ± α) a cosx ± b sinx can be expressed as R cos(x ∓ α) Where: R cos α = a R sin α = b R = √(a² + b²) Differentiation & Integration From First Principles: Product Rule: Quotient Rule: Integration by Parts Trigonometric Integration ∫ xⁿ dx = (xⁿ⁺¹) / (n+1) + c ∫ e ˣ dx = e ˣ + c ∫ 1/x dx = ln|x| + c ∫ cos(x) dx = sin(x) + c ∫ sin(x) dx = -cos(x) + c ∫ tan(x) dx = ln|sec(x)| + c ∫ sec²(x) dx = tan(x) + c ∫ cosec(x) cot(x) dx = -cosec(x) + c ∫ cosec²(x) dx = -cot(x) + c ∫ sec(x) tan(x) dx = sec(x) + c ∫ f'(ax+b) dx = f(ax+b) / a + c Series Expansions The Binomial Expansion: The Taylor Series: The Maclaurin Expansion: Common Expansions:

• The Second Law of Thermodynamics

• Entropy & The Second Law

• Elasticity & Collisions

According to Hooke's Law, tension, T, is directly proportional to extension, x, in elastic strings and springs: T = kx The constant of proportionality, k, depends on two things: the modulus of elasticity, λ, of the material, and its unstretched, natural length, l: Note: natural length is noted with a lowercase 'L' - do not confuse this with an uppercase 'i'. The font of these note sheets is not massively helpful here, so sorry! Tension, T, is a force so is measured in Newtons. Extension and length are both distances, measured in metres. This gives the modulus of elasticity must also be measured in Newtons. Strings can be compressed as well as stretched. In this instance, Hooke's law still applies but the force in the spring, thrust (helpfully also written as T), acts the other way. Modelling For all questions, strings and springs are modelled as light. This means you do not take into account its weight. See the notes sheet on modelling in mechanics for single maths - the same rules apply. Elastic Potential Energy When an elastic string or spring is stretched or compressed, it stores elastic potential energy within it. This is equal to the work done in stretching/compressing the string/spring. The work done is represented by the area under a force-extension graph: Using the equation for the area of a right-angled triangle, A = ½ab, we can find the equation for the elastic potential energy stored in an elastic system: Worked Example Elastic Collisions in One Dimension Newton's law of restitution is used to solve problems where two particles collide directly: Direct impact means the the particles are moving along the same straight line when they collide. This means the collision is "head-on", so to speak. Newton's law of restitution depends on the coefficient of restitution, e, between the two colliding objects: 0 ≤ e ≤ 1 In a perfectly elastic collision, e = 1, so the two particles rebound at the same speeds In a perfectly inelastic collision, e = 0, so the particles coalesce (join together) When there are two unknowns in a problem, you can use the principle of conservation of linear momentum to solve simultaneously: Collisions with a Wall When a particle collides with a wall, the same principles apply. The only difference is, the wall does not move. Therefore, Newton's law of restitution is just taken as: Kinetic Energy The definition of a perfectly elastic collision (when e = 1) is a collision in which both momentum and kinetic energy are conserved. When e ≠ 1, the collision is not perfectly elastic, so some kinetic energy is lost. This can be calculated easily once you know the masses of each particle and their initial and final speeds: When an object collides with a wall, only the object experiences a change in kinetic energy Oblique Collisions of Spheres with Surfaces An oblique impact is when the collision does not act normally to the surfaces: While the collision may not be perpendicular to the surface, the impulse of the particle acts perpendicular to the surface. This means that the component of the velocity of the particle that acts parallel to the surface remains unchanged. v cos(β) = u cos(α) The components perpendicular to the surface are parallel to the impulse. Therefore, we can apply Newton's law of restitution: v sin(β) = eu sin(α) e is still the coefficient of restitution between the particle and the wall Dividing the second equation by the first eliminates u and v, leaving only the angles and e: tan(β) = e tan(α) Using sin² + cos² = 1 we can eliminate the angles to leave only u and v: Angle of Deflection The angle of deflection is the total angle through which the path of the particle changes. It is given by α + β Vectors Sometimes, the velocities will be given as vectors. If the surface is parallel to one of the unit vectors, this makes the question easier. To determine the angle of deflection, θ, if the velocities before and after are given as vectors u and v respectively, use the scalar product, u . v = |u| |v| cos(θ) Oblique Collisions of Spheres with Spheres In addition to collisions between spheres and surfaces, you need to be able to solve problems where two spheres collide in two dimensions: The impulse on each sphere acts along the line of centres of the two spheres Like with oblique collisions between spheres and surfaces: the component of the velocities perpendicular to the impulse is unchanged through the impact Newton's Law of restitution applies to the parallel components of the velocities the principle of conservation of momentum applies parallel to the line of centres

• Momentum, Impulse, Work, Energy & Power

As you know, momentum, p, is defined as the product of an object's mass and velocity: p = mv The Impulse-Momentum Principle When an object collides with something else, it experiences a change in momentum. This is equal to the impulse experienced by that object in the collision: I = Δmv The impulse can also be calculated as the product of the force of the collision, F, and its duration, t: I = Δmv = Ft Therefore, the units of impulse are Newton-seconds, Ns. Often, the force will have to be calculated using Newton's second law (F=ma), or SUVAT equations need to be used to calculate the duration or change in momentum. The Principle of Conservation of Momentum According to Newton's third law, for every reaction there is an equal but opposite reaction. This means that when two particles collide, both experience the same magnitude of impulse, but in opposite directions. Since impulse equals the change in momentum, the two particles experience equal and opposite changes in momentum, Δp and -Δp. Overall, these two changes in momentum cancel out: the total momentum before impact equals the total momentum after the impact Mathematically, this can be expressed as: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ Work & Energy Work done is defined as the the amount of energy transferred from one form to another when a force causes movement. This is known as the work-energy principle. W = Fx Work done = force x distance moved. The force and motion must be in the same direction Because of this, you will often need to use trigonometry to find the correct components of forces: The units of energy are joules, J - so when the force is given in Newtons, N, and the distance in metres, m, the work done has units joules, J. Work Done Against Gravity & Gravitational Potential Energy Work done against gravity is given by the equation: W = mgh This can easily be derived from the work-done equation above: W = max: substitute F = ma into W = Fx W = mah: for work done against gravity, we tend to refer to the distance, x, as height, h W = mgh: acceleration due to gravity is g, 9.8 m/s² According to the Work-Energy Principle, work done is equal to energy transferred. Therefore, gravitational potential (G.P.E.) is also defined with this equation: G.P.E = mgh Kinetic Energy Kinetic Energy, KE, is given by the equation: KE = ½mv² This can also be derived from the work-done equation: a = (v²-u²) / 2s: rearrange v² = u ²+ 2as to make a the subject F = m(v²-u²) / 2s: substitute this into F = ma Fs = ½m(v²-u²): multiply both sides by displacement, s W = ½m(v²-u²): Fs = Fx = Work done ΔKE = ½m(v²-u²): Work done = the change in energy KE = ½mv² Power Power is defined as the rate at which work is done: P = Fv The units of power are Watts, W. If the force is given in Newtons and the velocity in m/s, the units of power are watts, 1 joule per second.

• Complex Numbers, Argand Diagrams & Loci

• Proof by Induction

Proof by induction is used to prove that a general statement is true for all positive integer values. All proofs by mathematical induction follow four basic steps: Prove that the general statement is true when n = 1 Assume the general statement is true for n = k Show that, if it is true for n = 1, the general statement is also true for n = k+1 Conclude that the general statement is true whenever n ∈ ℕ All four steps must be shown clearly in your workings: prove, assume, show, conclude. Proving Sums Often, questions will use the standard results for the sums of r, r² and r³. Regardless, the method for all sums is the same and follows the four steps above. Quick Tip For the 3rd step, it is generally best to write the last line out first using the general function - just substitute (k+1) into it. You know that this is the answer you want to reach, so use it as a target to help you. Proving Divisibility Results Again, follow the four standard steps for proof by induction. For divisibility results, make step 2 equal any general multiple of the divisor: Proof Using Matrices Exactly the same four steps apply:

• Series

Sigma notation is used to write series quickly. This describes each term of a series as an equation, where substituting in values for r gives that term in the series. There are standard sums of series formulae that are used all the time: The Sum of Natural Numbers To find the sum of a series of constant terms, use the formula: To find the sum of the first n natural numbers (positive integers): If the series does not start at r = 1, but at r = k, subtract one series from another, where the first goes from r = 1 to r = k, and the second from r = 1 to r = (k-1): Expressions in sigma notation can be rearranged to simplify complicated series: The Sum of Squares The first sum formula above, for constant terms, is linear. The second, for the sum of natural numbers, is quadratic. Therefore, the formula for the sum of square terms is cubic. To find the sum of the sum of the squares of the first n natural numbers: The Sum of Cubics Following through with this pattern, the formula to find the sum of the cubes of the first n natural numbers is quartic: The Method of Differences The method of differences is used when the expansion of a series leads to pairs of terms that cancel out, leaving only a few single terms behind. When the general term, u₁, of a series can be expressed in the form f(r) - f(r+1), the method of differences applies. This means that: To solve problems with the method of differences Write out the first few terms to see what cancels Write out the last few terms to see what cancels Add the left overs from the beginning and the end Often, you will need to use partial fractions Example The Maclaurin Series We often work with first and second order derivatives, but why stop there? Functions that can be written as an infinite sum of terms in the form axⁿ can be differentiated infinite times. We already know of a few such examples: the binomial expansion of 1 / (1-x) forms an infinite polynomial, 1 + x + x² + x³ + x⁴ + ... the binomial expansion of 1 / √(1+x) forms an infinite polynomial, 1 + ½ x - ⅟₈ x² + ⅟₁₆ x³ - ⅟₃₂ x⁴ + ... eˣ = 1 + x + ½ x² + ⅟₆ x³ + ⅟₂₄ x⁴ + ... This happens when x = 0 is substituted into increasing orders of derivatives, which, in turn, increases the power of the polynomial. This is only true if the function can be differentiated an infinite number of times, and if the series converges. When the above criteria are satisfied and each derivative has a finite value, the Maclaurin series expansion is valid: There are many standard versions of the Maclaurin series expansion that are useful to know, as they crop up a great deal:

• Roots of Polynomials

Won't lie, this topic isn't particularly fun. It's just a whole lot of algebra and equations that you need to know and get right. At least the topic is small. To be perfectly honest, the only challenge is writing everything out clearly, and not getting confused between the Greek and Latin alphabets: don't confuse a with α don't confuse b with β don't confuse g with γ don't confuse d with δ a, b, c, d & e are real constants α, β, γ, δ are roots of polynomials Quadratic Polynomials A quadratic equation in the form ax² + bx + c = 0 has two possible roots, α and β. These can be distinct and real, repeated (the same), or complex conjugates. The roots can be added or multiplied to give the following results: Squaring or cubing the roots gives different results: Cubic Polynomials A cubic equation in the form ax³ + bx² + cx + d = 0 has three possible roots, α, β and γ. The roots can be added or multiplied to give the following results: Squaring or cubing the roots gives different results: Quartic Polynomials A quartic equation in the form ax⁴ + bx³ + cx² + dx + e = 0 has four possible roots, α, β, γ and δ. The roots can be added or multiplied to give the following results: Squaring the roots gives different results: Linear Transformations of Roots If you know the sums and products of roots of a polynomial, you can find the equation of another polynomial whose roots are a linear transformation of the first. For example, if the equation x² - 2x + 3 = 0 has roots α and β, find the equation that has roots (α+2) and (β+2). This can be done in two ways: Use the roots of polynomial equations above Rearrange the roots and substitute into the first equation Method 1 (α+2) + (β+2) = α + β + 4 = -b/a + 4 = 6 (α+2)(β+2) = αβ + 2α + 2β + 4 = αβ + 2(α+β) + 4 = c/a - 2b/a + 4 = 11 So the equation is w² - 6w + 11 = 0 The coefficient of each term must be the same as the initial equation Method 2 Let w = x + 2 Rearrange for x: x = w - 2 Substitute into first equation: (w-2)² - 2(w-2) + 3 = 0 w² - 4w + 4 -2w + 4 +3 = 0 w² - 6w + 11 = 0