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

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.

We know from single maths that if the discriminant of a quadratic equation is positive, it has two solutions, if the discriminant = 0, there is one repeated root, and if it is negative there are no roots. Wrong. If the discriminant is negative, the quadratic has two imaginary roots, given in terms of i, the square root of -1: i = √-1 A complex number is a sum of a real and an imaginary number, given in the form a + b i, where a is the real number and b the coefficient of i. Often, complex numbers are represented by the letter z. Adding and Subtracting Complex Numbers To add or subtract complex numbers, treat the real and imaginary terms separately: (a+b i) + (c+d i) = (a+c) + (b+d)i A few examples: 3 + (2+4i) = (3+2) + (4i) = 5+4i (1+2i) - 4i = (1) + (2-4)i = 1-2i (2-2i) + (6+i) = (2+6) + (-2+1)i = 8-i Multiplying Complex Numbers To multiply an imaginary number by a real constant, just expand the brackets: k(a+b i) = ka+kb i For example: 3(2+4i) = 3(2) + 3(4i) = 6+12i To multiply two imaginary numbers, expand like any quadratic, using the fact that: i² = -1 For example: (2+3i)(3+4i) = 6 + 8i +9i +12i² = 6 + 17i - 12 = -6+17i Complex Conjugation Just like conjugate pairs when rationalising a denominator, the conjugate of a complex number has the opposite sign: For any complex number z = a +bi, the conjugate is given as z* = a -bi The complex conjugate is given an asterisk. For any complex number z, the product of z and z* is a real number For example, if z = 4+5i: z* = 4-5i zz* = (4+5i)(4-5i) = 16 + 20i - 20i -25i² = 16 + 25 = 41 Solving Complex Equations Complex roots always come in conjugate pairs. Therefore, if the discriminant of a quadratic is negative, if z is one of the roots, z* will be the other root. For example, the roots of the equation x² - 6x + 13 = 0 are x = (3+2i) and x = (3-2i) Often, equations with complex roots will use z instead of x. The same rule applies to cubic and quartic graphs: If a cubic graph has just one real root, it will have two complex roots, z and z* If a quartic graph has two real roots or one real repeated root, then it will also have two complex roots, z and z* If a quartic graph has no real roots, it will have two complex conjugate pairs as roots: z, z*, w and w* All cubic graphs must have a total of three roots, real or complex All quartic graphs must have a total of four roots, real or complex Argand Diagrams Complex numbers can be represented graphically on an Argand diagram. The x-axis is the real axis, Re, and the y-axis is the imaginary axis, Im. To plot a complex number, treat it as a coordinate: The complex number z = (a + bi) has coordinates (a, b) on an Argand diagram This can be represented as a position vector, at an angle, θ, to the positive real axis (generally in radians): Like any vector, the modulus (the magnitude) can be found using Pythagoras' theorem: |z| = √(a²+b²) and tan of the angle is given as: tan(θ) = b/a Modulus-Argument Form The magnitude of the position vector for a complex number, r, is known as the modulus, and the angle it makes with the positive real axis, arg z, is called the argument. Therefore, complex numbers can also be expressed in modulus-argument form: z = r(cos(θ) + i sin(θ)) r = |z| arg z = θ Multiplying & Dividing Mod-Arg Loci A locus of points is a collection of points that can be anywhere along a particular line. These can be represented with complex numbers on an Argand diagram. The distance between two complex numbers, z₁ and z₂, is given as |z₂ - z₁| If, instead of defining both complex numbers, you just define one and give one general complex number, you get the locus of a circle: |z - z₁| = r is a circle with centre z₁ and radius r This can also be given as |z - (a+bi)| = r, where r is the radius and (a, b) the centre. You can find the locus of points for a perpendicular bisector of a line segment between two points: The perpendicular bisector of a line segment between two complex numbers, z₁ and z₂, is given as |z - z₁| = |z - z₂| Using the argument of a complex number, z₁, we can express the locus of points for a half line. This is a line from, but not including, the fixed point z₁ at an angle of arg(z-z₁): arg(z-z₁) = θ is a half line from but not including the point z₁ at an angle of θ to the horizontal. Regions Loci can be used to represent regions on an Argand diagram: Exponential Form Complex numbers can also be expressed in exponential form, z = r e^iθ, using Euler's relation: Euler's relation is derived from the Maclaurin series expansions of sin(θ), cos(θ) and eˣ. See Further Maths Notes Sheet on Series. This leads to the exponential form for a complex number, z: Where r = |z| and arg z = θ just like in modulus-argument form. Multiplying & Dividing The same rules as for modulus-argument can be applied to multiplying and dividing complex numbers in exponential form: De Moivre's Theorem Euler's relation can be used to find powers of complex numbers that are in the modulus-argument form. (r (cos(θ) + i sin(θ))ⁿ = rⁿ(cos(nθ) + i sin(nθ)) For any integer n This can be proven with Euler's relation: It can also be proven using proof by induction (See the Notes Sheet on this) Trigonometric Identities De Moivre's theorem, combined with the binomial expansion, can be used to derive trigonometric identities. To drive an equation for sin(nθ) or cos(nθ): To drive an equation for sinⁿ(θ) or cosⁿ(θ): There are four standard results you need to know: An example of how to apply these: nth Roots of a Complex Number Just as a real number, x, has two square roots, ±√x, any complex number has n distinct nth roots. If z and w are non-zero complex numbers and n is a positive integer, the equation zⁿ = w has n distinct roots. We can use de Moivre's theorem to find the solutions to zⁿ = w by taking into account the fact that its argument is repeated every 2π. For any complex number z = r(cos(θ) + i sin(θ)), z = r(cos(θ + 2kπ) + i sin(θ + 2kπ)) Where k is any integer When zⁿ = 1 In general, the solutions to zⁿ = 1 are z = cos(2kπ / n) + i sin(2kπ / n) for k = 1, 2, 3, ..., n These are known as the nth roots of unity. If n is a positive integer, then there is an nth root of unity ω such that: the nth roots of unity are 1, ω², ω³, ω⁴, ..., ωⁿ‾¹ 1, ω, ω², ω³, ω⁴, ..., ωⁿ‾¹ form the vertices of a regular n-gon 1 + ω² + ω³ + ω⁴ + ... + ωⁿ‾¹ = 0 Using nth Roots Geometrically The nth roots of any complex number, z, form the vertices of a regular n-gon with the origin at its centre The size and orientation of the polygon depend on the complex number z. If you know the coordinates of a single vertex, you can find the others by rotating that point about the origin by 2π/n. This is the same as multiplying by the nth roots of unity. If z₁ is one of the roots of the equation zⁿ = w, and 1, ω, ω², ω³, ω⁴, ..., ωⁿ‾¹ are the roots of unity, then the roots of zⁿ = w are z₁, z₁ω, z₁ω², z₁ω³, z₁ω⁴, ..., z₁ωⁿ‾¹

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

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:

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:

This is a bit of an odd notes sheet - its just an amalgamation of all that extra integration stuff: improper integrals the mean value of a function inverse trig partial fractions volumes of revolution Improper Integrals In single maths, you learn how to integrate to find the area enclosed between a curve and the x-axis between two fixed limits. An improper integral is when one or both limits are infinite, or when the function is undefined at a certain point in the given interval. An improper integral does not always exist: if it does, it is convergent; if it does not, it is divergent. One Infinite/Undefined Limit To see if an improper integral with one infinite and one finite limit is convergent you need to substitute the infinite limit for t, and consider what happens as the function tends towards infinity: If the integral part with t in it did not tend to 0, but instead tended to ∞, the function would be divergent. Two Infinite/Undefined Limits When both limits are infinite or undefined, you need to separate the integral into two improper integrals each with one infinite and one finite limit. The finite limit of both parts must be the same. It is generally good to use a simple value such as x=0, as this makes workings a lot simpler. Even if only one of the two separated integrals does not converge, the whole integral will not converge to a defined value. Mean Value of a Function It is possible to find an average for the area enclosed under a graph over a fixed interval. The height of this average area is known as the mean value of the function: The mean can be transformed along with the graph it represents: Inverse Trigonometric Functions It is possible to differentiate inverse trigonometric functions implicitly. This leads to standard results that are incredibly useful for both differentiation and integration: Differentiating Inverse Trigonometric Functions It is possible to derive these differentiation functions by differentiating implicitly. Integrating Inverse Trigonometric Functions It is possible to derive the function for the second integral by substitution. Integrating with Partial Fractions If the denominator of a partial fraction has a square x-term in it, you cannot use the standard method to find the numerator coefficients. Instead, you must write the coefficient of the partial fraction with a quadratic denominator as a linear function in x: These partial fractions can then often be integrated using the inverse trig results above. Volumes of Revolution The mathematical equivalent of pottery, integration can be used to find volumes of circular shapes, where the sides are defined by equations of lines integrated for a full rotation about a coordinate axis. For example, the line y=r, where r is a positive integer, becomes a cylinder of radius r when revolved about the x-axis. About the x-axis To calculate the volume of revolution for any function y = f(x) that is rotated 2π radians about the x-axis between x=a and x=b, use the formula: About the y-axis To calculate the volume of revolution for any function y = f(x) that is rotated 2π radians about the y-axis between y=a and y=b, use the formula: Adding and Subtracting Volumes Often, for hollow object or complex shapes, you will need to subtract one volume from another to find a sort of shell. Similarly, you may need to add volumes. This is exactly the same as finding the area between different curves, just in 3D. Common examples include cones and cylinders: A cylinder of height h and radius r has volume πr²h A cone of height h and base radius r has volume ⅓πr²h Parametric Volumes of Revolution There is no need to convert parametric equations into Cartesian form to find their volumes of revolution. Instead, a variant of the chain rule can be used. This means you will need to differentiate one of the equations, depending on whether the revolution is around the x- or y-axis. To calculate the volume of revolution for parametric equations x = f(t), y = g(t) that are rotated 2π radians about the x-axis between x=a and x=b, use the formula: Where p and q are the values of t when x=a and x=b respectively To calculate the volume of revolution for parametric equations x = f(t), y = g(t) that are rotated 2π radians about the y-axis between y=a and y=b, use the formula: Where p and q are the values of t when y=a and y=b respectively

The equation of a line can be represented by vectors in 3D. The line that passes through A and R can be written as: r = a + λb where a is the position vector of a known point on the line b is the direction vector of the line (a vector parallel to the line) r is the position vector of any arbitrary point on the line λ is a scalar parameter If you need to find the equation of a line from just two points, C and D, and you do not know the direction vector, use the fact that the vector between the two points (d - c) is the direction vector of the line: The equation of this line is given as r = c + λ(d-c) Equations of 3D line in Cartesian Form If you need to find the equation of a straight three dimensional line in Cartesian form (in terms of x, y and x), convert using the vectors a and b from the vector equation r = a + λb. Equations of a Plane in 3D The vector equation of a plane is given as: r = a + λb + μc where a is the position vector of a known point, A, in the plane b and c are non-parallel, non-zero vectors in the plane r is the position vector of any arbitrary point, R, in the plane λ and μ are scalar parameters Normal Vector The normal vector of a plane is used to describe the direction of the plane. It is the vector that is exactly perpendicular to the plane. Equation of a 3D plane in Cartesian Form You can use the normal vector of a plane to write a Cartesian equation describing the plane. Where the normal vector, n = ai + bj + ck, the Cartesian equation of the plane is given as: ax + by + cz = d a, b, c and d are all constants Scalar Product (a.k.a "Dot Product") The scalar product, a . b, is given by the magnitude of the two vectors a and b, and by the angle between them, θ: a . b = |a| |b| cos(θ) If θ = 90° (the two vectors are perpendicular), then the scalar product will equal zero. If a and b are parallel, a . b = |a| |b| If the two vectors are identical (both are a), the scalar product will equal |a|² In Cartesian Form Scalar Product Equation of a Plane The equation of a plane can also be written as the scalar product of the normal vector to the plane and the position vector of any arbitrary point on the plane: r . n = k r is the position vector of any arbitrary point on the plane n is the normal vector of the plane k is a scalar constant for the plane, where k = a . n for a specific point in the plane with position vector a Angles between Lines & Planes Angle Between Two Lines The acute angle between two intersecting lines, with direction vectors a and b, is given as above. |a| |b| represents the magnitude of each vector - use 3D Pythagoras for this. To find the obtuse angle, subtract the acute angle from 180°. Angle Between a Line and a Plane The acute angle between the line r = a + λb and the plane r . n = k is given as above. again, |b| |n| represents the magnitude of each vector - use 3D Pythagoras for this. Angle Between Two Planes The acute angle between the plane r . n₁ = k₁ and the plane r . n₂ = k₂ is given as above. Points of Intersection When you have the vector equations of two lines, you can see if they intersect or not. If the two lines are parallel, they never intersect. Write out the two equations as column vectors Write each row of the column vectors as simultaneous equations involving λ and μ Solve two of the three equations for λ and μ See if these values are consistent with the third equation There are a number of possible outcomes: If all three equations are consistent with the values of λ and μ, the lines intersect at those values of λ and μ If two of the three equations have solutions, but the third does not match, the lines do not intersect. If none of the three equations can be solved simultaneously, the lines do not intersect. If the two lines are neither parallel nor intersect, they are skew Finding Perpendiculars Shortest Perpendicular Distance Between a Point and a Line Shortest Perpendicular Distance Between Two Non-Intersecting Lines Shortest Perpendicular Distance Between a Point and a Plane Unlike the above two examples, there is a fixed equation that can be used to find the perpendicular distance between a point and a plane: All you have to do is substitute in the values.

Hyperbolic functions are similar to trigonometric functions, but are defined in terms of exponentials. There are three fundamental hyperbolic functions: sinh, cosh and tanh: Similarly, the reciprocal of each function exists: Hyperbolic Graphs For any value of x, sinh(-x) = -sinh(x) y = sinh(x) has no asymptotes For any value of x, cosh(-x) = -cosh(x) y = cosh(x) never goes below y=1 y = tanh(x) has asymptotes at y = ±1, and always stays between these Inverse Hyperbolic Functions Just like sin, cos and tan, the hyperbolic functions have inverses, arcsinh, arcosh and artanh: The graphs of these are their respective reflections in the line y=x: Hyperbolic Identities & Equations The same identities exist for hyperbolic functions as they do for trigonometric functions: sinh(A ± B) ≡ sinh(A) cosh(B) ± cosh(A) sinh(B) cosh(A ± B) ≡ cosh(A) cosh(B) ∓ sinh(A) sinh(B) Equations with a sinh² in them, however, are different: cosh²(x) - sinh²(x) ≡ 1 Note that here, the sinh² is negative (the trigonometric identity is sin² + cos² ≡ 1). This is known as Osborn's rule: According to Osborn's rule, when using trigonometric identities as hyperbolic identities, any sinh² must be multiplied by -1. Differentiating Hyperbolic Functions This is very similar to trigonometric functions: Note that the derivative of cosh(x) is positive sinh(x), not negative. The inverse functions differentiate as such: Integrating Hyperbolic Functions Simply the reverse of differentiation, but remember the " +c " and that the signs are different: The inverse functions can also be integrated: These standard results for when the equation you need to integrate does not have either (x²+1) or (x²-1) in the root in the denominator:

The Cartesian system in two-dimensions models points in terms of x and y. The polar system, however, models a point as a distance form the pole, r (generally the origin) at a certain angle from the initial line, θ (typically the positive horizontal axis). Yes, this is like the modulus-argument form of complex numbers and Argand diagrams. From the diagram, we can derive equations to convert between polar and Cartesian systems: r cos(θ) = x r sin(θ) = y Where θ is given by: θ = arctan(y/x) And r is defined using Pythagoras' theorem: r² = x² + y² Sketching Polar Curves To sketch a polar curve, use a graphical calculator or draw u a table of values for regular intervals of θ. This can be done quickly using the table function on the CASIO ClassWiz fx-991EX, and we recommend using π/6 as an interval. The curve in this example is known as a cardioid, due to its dimple. This is common for equations in the form r = a(p+qcos(θ)), but only if q ≤ p < 2q. When p ≥ 2q, there is no dimple, making it more egg-shaped: Areas Enclosed by Polar Curves The area of a sector of a polar curve can be calculated using integration. However, simply integrating r will not work. Instead: Of course, you an also calculate areas between polar curves. To do this, you need to find the angle at which they intersect. Tangents to Polar Curves To find tangents to a polar curve, you need to convert it into Cartesian form (one equation for x and one for y, both in terms of θ), using the formula at the top of this notes sheet. Then, you can differentiate parametrically. Standard results are:

A matrix is a system of elements within a pair of brackets. The size of the matrix is given as the number of rows and columns in it. A square matrix is one where there is an equal number of rows and columns A zero matrix is one in which all the elements are zero An identity matrix is a square matrix where the all the values on the diagonal from top left to bottom right are 1, and all other values are zero. This is noted as capital i, I with a subscript number afterwards to show its size. Working with Matrices Adding and Subtracting To add or subtract matrices, you add or subtract the corresponding elements in each matrix. You can only add or subtract matrices of the same size Multiplying by a Scalar To multiple a vector by a scalar, simply multiple each element in the vector by the scalar. This can be factorised out, too. Multiplying One Vector by Another Matrix multiplication is only possible if the number of columns in the first matrix equals the number of rows in the second matrix. The result of the multiplication is known as the product matrix, and will have the same number of rows as the first matrix and the same number of columns as the second matrix. Order of matrix multiplication matters For two matrices, A and B: In general, AB ≠ BA If AB exists, it does not mean BA exists To find the product of two matrices, multiply the elements in each row of the first matrix by the elements in each column of the second matrix Determinants The determinant of a square matrix is a scalar value that represents the matrix. It is only possible for square matrices. The determinant of a matrix, M, is written as det M, or as |M| A singular matrix has a determinant of zero. A non-singular matrix has a determinant that is not zero. To find the determinant of a 2x2 matrix: To find the determinant of a 3x3 matrix: Inverse Matrices The inverse of a matrix M is the matrix M¯¹ M M¯¹ = M¯¹ M = 1 Finding the Inverse of a 2x2 Matrix Find the determinant Write out the matrix where a and d are swapped, and both b and c are multiplied by -1 Multiply by one over the determinant If A and B are non-singular matrices, (AB)¯¹ = B¯¹A¯¹ Finding the Inverse of a 3x3 Matrix Find the determinant Replace each element with it matrix of minors & find the determinants Form the matrix of co-factors (switch every other sign) Find the transpose of the matrix of co-factors by switching rows and columns Multiply this by one over the determinant Since this is such a complicated thing to do, unless specifically needed, use the calculator function. On the CASIO ClassWiz fx-991EX: Click MODE Click 4: Matrix Click 1: Define Matrix A Click 3 twice to set size of matrix Fill in Matrix Click AC Click OPTN CLICK 3: Matrix A Click x¯¹ Click = Matrix Systems & Geometry It is possible to use matrices and their inverses to easily solve simultaneous equations. This example is consistent, meaning there is at least one set of values that satisfies all three equations simultaneously. The determinant is not zero, making the matrix non-singular. This is especially useful for 3D geometry, and the equations of planes: Linear Transformations A linear transformations are transformations with specific properties: they are made up of linear x, y and z terms only they have no non-variable terms they map the origin onto itself they can be represented by matrices Points and lines that do not change in a transformation are called invariant points/lines The origin is always an invariant point. Every point on an invariant line is an invariant point. Reflections The line of reflection is always an invariant line. In two-dimensions, there are standard reflections for the coordinate axes and y=±x. In three-dimensions, reflections ca happen in planes: For a linear transformation with matrix M, if the determinant is negative, the shape has been reflected Rotations For rotations about the origin, the only invariant point is the origin. In 3D, rotations can be about an entire axis: Enlargements & Stretches If a = b, the transformation is an enlargement of scale factor a. For a stretch that is only in the x-direction, the y-axis is an invariant line. For a stretch that is only in the y-direction, the x-axis is an invariant line. For stretches in both directions, there are no invariant lines and the only invariant point is the origin. For a linear transformation with matrix M, the determinant is the area scale factor. Successive & Inverse Transformations It is possible to apply multiple transformations, one after the other. In this instance, order of matrix multiplication is particularly important. For two linear transformations, represented by the matrices P and Q respectivley: PQ represents the transformation Q followed by P A linear transformation with matrix A can be undone with the transformation represented by the inverse matrix, A¯¹.

In single maths, first-order differential equations are the only ones looked at, and are solved by separating the variables: When dy/dx = f(x) g(y), you can say ∫ 1/g(y) dy = ∫ f(x) dx Move all the y terms to the left where the dy is Move all the x terms to the right, including the dx This allows you to integrate each side with respect to the variable on that side to solve the equation. You only need to add the ' +c ' to one side. Just like when integrating an indefinite function, the initial result is a general solution and could be anywhere along the y-axis (see section on indefinite integral functions above). To fond the particular solution, you need to know a coordinate point on the curve - sometimes this is called a boundary condition. Integrating Factor This is not covered in single maths, but is an important method of solving first-order differential equations where x and terms are multiplied by one another in one of the terms: Rearrange to be in the form dy/dx + P(x)y = Q(x) Find the integrating factor using the formula above Multiply the dy/dx by the integrating factor and by y Multiply the right hand side by the integrating factor & simplify (if you can) The middle term, P(x)y disappears Move the dx to the right hand side and integrate this side Rearrange to make y the subject - this is the general solution Find the particular solution by substituting in boundary conditions (if there are any) At further stage, we also look at second-order differential equations - these come in two types: homogenous and non-homogenous. Second-Order Homogeneous Differential Equations Second-order homogeneous differential equations can be solved when in the form: Homogeneous means it equals zero The solution in terms of y will have four constants that need to be found: λ and μ, which are found by solving the auxiliary equation A and B, which can only be found if you have boundary conditions The auxiliary equation is am² + bm + c = 0, and its solutions are λ and μ a, b, and c are the coefficients in the second-order homogenous differential equation above. The format of the solution in terms of y depends on the auxiliary equation: Second-Order Non-Homogeneous Differential Equations Second-order homogeneous differential equations can be solved when in the form: The left hand side of this equation is known as the corresponding homogeneous equation and its solution is called the complementary function. The right hand side of this equation is a function of x and its solution is called the particular integral. Non-homogeneous means the corresponding homogeneous equation equals a function of x To solve a second-order non-homogeneous differential equation, first solve the corresponding homogeneous equation as normal (see above), and then find the particular integral: Take the correct standard function Find dy/dx and d²y/dx² Substitute these into the initial equation Solve to find the constants in the standard function The form of the standard function depends on the form of f(x): When the particular integral is already in the complementary function, you need to multiply the particular integral by x For example, if f(x) is in the exponential form where k is the same as one of the roots in the auxiliary equation, multiply the particular integral by x. The complete solution to the non-homogenous second-order differential equation is given as the sum of the complementary function and the particular integral: Solution = complementary function + particular integral Note that in this example, the exponent in the particular integral is -x. If it were -2x or -3x, it would be the same as the solutions to the auxiliary equations, so we would have to multiply the particular integral by x. Simple Harmonic Motion Simple harmonic motion (SHM) is when a particle always accelerates to a fixed central point (or origin) on its line of motion. The acceleration is proportional to the displacement of the particle from the origin. It may be useful to check out the notes sheet on SHM from A-Level Physics before reading on. SHM can be modelled with second order differential equations, because: x is displacement v is velocity, displacement with respect to time a is acceleration, velocity with respect to time Therefore: Often, we represent this with dot notation, where the dot above a variable means that that variable is differentiated with respect to time. Since the acceleration is always proportional to displacement, and always directed to the origin, O, there must be a negative constant of proportionality. This is called the angular velocity, and is represented by ω². Terminology Amplitude is the maximum displacement of the particle from the origin Period is the time taken to make one complete oscillation: this is from the origin out once, through it again to the other side, and then back to the origin. If an object is oscillating, this generally means it is moving with SHM R-Addition Formula When x is given in the form A cos(ωt) + B sin(ωt), it can be written as r sin(ωt - α). Here, r is the amplitude of the SHM, and is found as: r² = a² + b² tan(α) = A/B Damped and Forced Harmonic Motion Harmonic motion is only simple if there are no external forces slowing it down or speeding it up. Damping If there is an external force slowing down the motion of the particle, it is known as a dampening force. This is modelled by second-order homogeneous differential equations with a dx/dt term in the middle: There are three separate cases, depending on the discriminant of the auxiliary equation: Forced Motion If energy is being put into the system, and the oscillating object is driven by an external force, it can be modelled with second-order non-homogeneous differential equations: Questions relating to these contexts are solved exactly as any second-order differential equations should be solved. Coupled First-Order Simultaneous Differential Equations Often in nature, the rate of change of one thing is dependent on the rate of change of another thing. A common example is how the population size of predator affects that of the prey, and vice versa: If the population of rabbit is large, the population of foxes increases If the population of foxes is large, the population of rabbits decreases In such an example, there are two variables (one for the size of each population), and so there are two differential equations to represent how each changes with respect to time. Naturally, these can be solved simultaneously. If r is the number of rabbits at time t, and f is the number of foxes at time t, we can write: If both f(t) and g(t) equal zero, the system is homogeneous. To solve coupled first-order differential equations, you need to eliminate one of the variables.