This handout covers Topic 1: Further Pure Mathematics 进阶纯数学 1. It adds new algebra, matrices, polar coordinates, more vectors, and proof by induction.
A-Level Further Mathematics
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1 Further Pure Mathematics 1
1.1
Roots of polynomial equations
Syllabus
- recall and use the relations between the roots and coefficients of polynomial equations
- use a substitution to obtain an equation whose roots are related in a simple way to those of the original equation
Source: Cambridge International syllabus
For a polynomial equation, the roots 根 are linked to the coefficients 系数. For $ax^2 + bx + c = 0$ with roots $\alpha, \beta$:
$$\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}.$$For a cubic $ax^3 + bx^2 + cx + d = 0$ with roots $\alpha, \beta, \gamma$:$$\sum\alpha = -\frac{b}{a}, \qquad \sum\alpha\beta = \frac{c}{a}, \qquad \alpha\beta\gamma = -\frac{d}{a}.$$To find an equation whose roots are changed in a simple way, use a substitution 代换 (for example, put $w = \alpha + 1$).Worked example. The equation $x^2 - 5x + 6 = 0$ has roots $\alpha, \beta$. Find the equation with roots $\alpha + 1, \beta + 1$.
Here $\alpha + \beta = 5$ and $\alpha\beta = 6$. The new sum is $(\alpha + 1) + (\beta + 1) = 7$, and the new product is $(\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = 6 + 5 + 1 = 12$. So the new equation is
$$x^2 - 7x + 12 = 0.$$Vocabulary TrainEnglish Chinese Pinyin Further Pure Mathematics 进阶纯数学 jìn jiē chún shù xué roots 根 gēn coefficients 系数 xì shù substitution 代换 dài huàn 1.2
Rational functions and graphs
Syllabus
- sketch graphs of simple rational functions, including the determination of oblique asymptotes, in cases where the degree of the numerator and the denominator are at most 2
- understand and use relationships between the graphs of $y = f(x)$, $y^2 = f(x)$, $y = \dfrac{1}{f(x)}$, $y = |f(x)|$ and $y = f(|x|)$
Source: Cambridge International syllabus
A rational function 有理函数 is a fraction of two polynomials. When the top has higher degree than the bottom, the graph has an oblique asymptote 斜渐近线 (a slanted line the curve approaches); find it by dividing out. You should also be able to relate the graph of $y = f(x)$ to those of $y^2 = f(x)$, $y = \dfrac{1}{f(x)}$, $y = |f(x)|$ and $y = f(|x|)$.
Far from the origin the curve hugs the slant line $y=x$; near $x=0$ it runs off along the vertical asymptote.Vocabulary TrainEnglish Chinese Pinyin rational function 有理函数 yǒu lǐ hán shù oblique asymptote 斜渐近线 xié jiàn jìn xiàn 1.3
Summation of series
Syllabus
- use the standard results for $\sum r$, $\sum r^2$ and $\sum r^3$ to find related sums
- use the method of differences to obtain the sum of a finite series
- recognise, by direct consideration of a sum to $n$ terms, when a series is convergent, and find the sum to infinity in such cases
Source: Cambridge International syllabus
Learn the standard results:
$$\sum_{r=1}^{n} r = \tfrac12 n(n+1), \qquad \sum_{r=1}^{n} r^2 = \tfrac16 n(n+1)(2n+1), \qquad \sum_{r=1}^{n} r^3 = \tfrac14 n^2(n+1)^2.$$The method of differences 差分法 sums a series by cancelling middle terms: if each term is $f(r) - f(r+1)$, almost everything cancels. From the sum to $n$ terms you can see whether a series is convergent 收敛 and, if so, find its sum to infinity 无穷和.Worked example. Find $\displaystyle\sum_{r=1}^{n} (2r + 1)$.
$$\sum_{r=1}^{n}(2r + 1) = 2\sum_{r=1}^{n} r + \sum_{r=1}^{n} 1 = 2\cdot\tfrac12 n(n+1) + n = n(n+1) + n = n(n+2).$$Vocabulary TrainEnglish Chinese Pinyin method of differences 差分法 chā fēn fǎ convergent 收敛 shōu liǎn sum to infinity 无穷和 wú qióng hé 1.4
Matrices
Syllabus
- carry out operations of matrix addition, subtraction and multiplication, and recognise the terms zero matrix and identity (or unit) matrix
- recall the meaning of 'singular' and 'non-singular' as applied to square matrices and, for $2\times 2$ and $3\times 3$ matrices, evaluate determinants and find inverses of non-singular matrices
- understand and use the result $(AB)^{-1} = B^{-1}A^{-1}$ for non-singular matrices
- understand the use of $2\times 2$ matrices to represent certain geometric transformations in the $x$-$y$ plane (including the relationship between $A$ and $A^{-1}$, products $AB$ as composed transformations, the area scale factor as the determinant, and finding the matrix of a given transformation or sequence)
- understand the meaning of 'invariant' as applied to points and lines under transformations represented by matrices, and solve simple problems involving invariant points and invariant lines
Source: Cambridge International syllabus
A matrix 矩阵 is a rectangular block of numbers. You can add, subtract and multiply matrices (multiplication is row $\times$ column). The zero matrix 零矩阵 has every entry $0$, and the identity matrix 单位矩阵 $I$ leaves any matrix unchanged under multiplication.
For a $2\times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant 行列式 is $\det A = ad - bc$. If $\det A \neq 0$ the matrix is non-singular (otherwise it is a singular matrix 奇异矩阵), and the inverse matrix 逆矩阵 is
$$A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.$$For a product, $(AB)^{-1} = B^{-1}A^{-1}$. A $2\times 2$ matrix can represent a geometric transformation 几何变换 of the plane: the determinant gives the area scale factor, and a product $AB$ means "do $B$, then $A$". Points or lines that do not move are called invariant points 不变点 and invariant lines 不变直线.
The matrix sends the unit square to a parallelogram whose area is the determinant.Worked example. Find the inverse of $A = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}$.
$\det A = 3\times 4 - 1\times 2 = 10$, so
$$A^{-1} = \frac{1}{10}\begin{pmatrix} 4 & -1 \\ -2 & 3 \end{pmatrix}.$$Vocabulary TrainEnglish Chinese Pinyin matrix 矩阵 jǔ zhèn zero matrix 零矩阵 líng jǔ zhèn identity matrix 单位矩阵 dān wèi jǔ zhèn determinant 行列式 háng liè shì singular matrix 奇异矩阵 qí yì jǔ zhèn inverse matrix 逆矩阵 nì jǔ zhèn geometric transformation 几何变换 jǐ hé biàn huàn invariant points 不变点 bù biàn diǎn invariant lines 不变直线 bù biàn zhí xiàn 1.5
Polar coordinates
Syllabus
- understand the relations between Cartesian and polar coordinates, and convert equations of curves from Cartesian to polar form and vice versa
- sketch simple polar curves, for $0 \leqslant \theta < 2\pi$ or $-\pi < \theta \leqslant \pi$ or a subset of either interval
- recall the formula $\tfrac{1}{2}\int r^2\,d\theta$ for the area of a sector, and use this formula in simple cases
Source: Cambridge International syllabus
A spiral staircase: spirals are described naturally using polar coordinates.Polar coordinates 极坐标 give a point by its distance $r$ from the origin and its angle $\theta$. They link to Cartesian coordinates 直角坐标 by
$$x = r\cos\theta, \qquad y = r\sin\theta, \qquad r^2 = x^2 + y^2.$$
Polar coordinates $(r,\theta)$ convert to Cartesian by $x=r\cos\theta$ and $y=r\sin\theta$.You should sketch simple polar curves 极坐标曲线, and find the area of a sector with
$$\text{area} = \tfrac12\int r^2\,d\theta.$$
The cardioid $r=1+\cos\theta$, plotted straight from $r$ as a function of $\theta$.Worked example. Convert the polar equation $r = 4\cos\theta$ to Cartesian form.
Multiply both sides by $r$: $r^2 = 4r\cos\theta$, so $x^2 + y^2 = 4x$. Completing the square gives $(x - 2)^2 + y^2 = 4$, a circle of radius $2$ centred at $(2, 0)$.
Vocabulary TrainEnglish Chinese Pinyin polar coordinates 极坐标 jí zuò biāo Cartesian coordinates 直角坐标 zhí jiǎo zuò biāo polar curves 极坐标曲线 jí zuò biāo qū xiàn 1.6
Vectors
Syllabus
- use the equation of a plane in any of the forms $ax + by + cz = d$, $\mathbf{r}\cdot\mathbf{n} = p$ or $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}$, and convert equations of planes from one form to another
- recall that the vector product $\mathbf{a}\times\mathbf{b}$ can be expressed as $|\mathbf{a}||\mathbf{b}|\sin\theta\,\hat{\mathbf{n}}$ or in component form
- use equations of lines and planes, with scalar and vector products, to solve problems on distances, angles and intersections (line-in/parallel-to/meeting a plane, foot of perpendicular to a plane, angle between line and plane or between two planes, line of intersection of two planes, shortest distance between skew lines, common perpendicular to skew lines)
Source: Cambridge International syllabus
Forces like wind and water are vectors — they have both size and direction.In three dimensions a plane 平面 can be written as $ax + by + cz = d$, or $\mathbf{r}\cdot\mathbf{n} = p$, or $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}$. Besides the scalar product, there is the vector product 向量积 of two vectors 向量:
$$\mathbf{a}\times\mathbf{b} = |\mathbf{a}|\,|\mathbf{b}|\sin\theta\;\hat{\mathbf{n}},$$which gives a vector at right angles to both. With scalar and vector products you can find distances, angles, and where lines and planes meet — including the shortest distance between skew lines 异面直线.
$\mathbf{a}\times\mathbf{b}$ is perpendicular to both vectors; its length equals the area of the parallelogram they span.Worked example. Find $\mathbf{a}\times\mathbf{b}$ for $\mathbf{a} = \mathbf{i} + \mathbf{j}$ and $\mathbf{b} = \mathbf{j} + \mathbf{k}$.
$$\mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = \mathbf{i}(1) - \mathbf{j}(1) + \mathbf{k}(1) = \mathbf{i} - \mathbf{j} + \mathbf{k}.$$Vocabulary TrainEnglish Chinese Pinyin plane 平面 píng miàn vector product 向量积 xiàng liàng jī vectors 向量 xiàng liàng skew lines 异面直线 yì miàn zhí xiàn 1.7
Proof by induction
Syllabus
- use the method of mathematical induction to establish a given result
- recognise situations where conjecture based on a limited trial followed by inductive proof is a useful strategy, and carry this out in simple cases
Source: Cambridge International syllabus
Mathematical induction 数学归纳法 proves a result for every positive integer $n$ in two steps:
- Base case: show the result is true for $n = 1$.
- Inductive step: assume it is true for $n = k$, then prove it for $n = k + 1$.
If both steps work, the result is true for all $n$. Often you first make a conjecture 猜想 (a sensible guess) from a few cases, then confirm it by inductive proof 归纳证明.
Worked example. Prove that $\displaystyle\sum_{r=1}^{n} r = \tfrac12 n(n+1)$.
Base case $n = 1$: the left side is $1$ and the right side is $\tfrac12(1)(2) = 1$. True. Inductive step: assume $\displaystyle\sum_{r=1}^{k} r = \tfrac12 k(k+1)$. Then
$$\sum_{r=1}^{k+1} r = \tfrac12 k(k+1) + (k+1) = (k+1)\left(\tfrac{k}{2} + 1\right) = \tfrac12 (k+1)(k+2).$$This is the formula with $n = k + 1$, so by induction it holds for all $n$.Vocabulary TrainEnglish Chinese Pinyin mathematical induction 数学归纳法 shù xué guī nà fǎ conjecture 猜想 cāi xiǎng inductive proof 归纳证明 guī nà zhèng míng -
2 Further Pure Mathematics 2
This handout covers Topic 2: Further Pure Mathematics 进阶纯数学 2. It adds hyperbolic functions, eigenvalues, new differentiation and integration, de Moivre's theorem, and methods for differential equations.
2.1
Hyperbolic functions
Syllabus
- understand the definitions of the hyperbolic functions $\sinh x$, $\cosh x$, $\tanh x$, $\operatorname{sech} x$, $\operatorname{cosech} x$, $\coth x$ in terms of the exponential function
- sketch the graphs of hyperbolic functions
- prove and use identities involving hyperbolic functions
- understand and use the definitions of the inverse hyperbolic functions and derive and use the logarithmic forms
Source: Cambridge International syllabus
A hanging cable forms a catenary — the curve of the hyperbolic cosine.The hyperbolic functions 双曲函数 are built from the exponential function:
$$\cosh x = \frac{e^x + e^{-x}}{2}, \qquad \sinh x = \frac{e^x - e^{-x}}{2}, \qquad \tanh x = \frac{\sinh x}{\cosh x}.$$
$\cosh$ is the average of $e^x$ and $e^{-x}$; $\sinh$ is half their difference.They obey identities much like the trigonometric ones, the main one being $\cosh^2 x - \sinh^2 x = 1$. The inverse hyperbolic functions 反双曲函数 have a logarithmic form 对数形式, for example $\sinh^{-1} x = \ln\!\left(x + \sqrt{x^2 + 1}\right)$.
Worked example. Show that $\cosh^2 x - \sinh^2 x = 1$.
$$\cosh^2 x - \sinh^2 x = \frac{(e^x + e^{-x})^2 - (e^x - e^{-x})^2}{4} = \frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4} = \frac{4}{4} = 1.$$Vocabulary TrainEnglish Chinese Pinyin Further Pure Mathematics 进阶纯数学 jìn jiē chún shù xué hyperbolic functions 双曲函数 shuāng qū hán shù inverse hyperbolic functions 反双曲函数 fǎn shuāng qū hán shù logarithmic form 对数形式 duì shù xíng shì 2.2
Matrices
Syllabus
- formulate a problem involving the solution of 3 linear simultaneous equations in 3 unknowns as a matrix equation, or vice versa
- understand the cases concerning the consistency or inconsistency of 3 linear simultaneous equations, relate them to the singularity of the matrix, solve consistent systems, and interpret geometrically in terms of lines and planes
- understand the terms 'characteristic equation', 'eigenvalue' and 'eigenvector' as applied to square matrices
- find eigenvalues and eigenvectors of $2\times 2$ and $3\times 3$ matrices
- express a square matrix in the form $QDQ^{-1}$, where $D$ is a diagonal matrix of eigenvalues and $Q$ is a matrix whose columns are eigenvectors, and use this expression
- use the fact that a square matrix satisfies its own characteristic equation
Source: Cambridge International syllabus
You can write three linear equations in three unknowns as a single matrix equation 矩阵方程 $A\mathbf{x} = \mathbf{b}$. If $A$ is non-singular there is one solution; if $A$ is singular the equations are either inconsistent (no solution) or have infinitely many.
For a square matrix $A$, the characteristic equation 特征方程 is $\det(A - \lambda I) = 0$. Its solutions are the eigenvalues 特征值 $\lambda$, and for each one the vector $\mathbf{v}$ with $A\mathbf{v} = \lambda\mathbf{v}$ is an eigenvector 特征向量. You can then write $A = QDQ^{-1}$, where $D$ is a diagonal matrix 对角矩阵 of eigenvalues and the columns of $Q$ are the eigenvectors.
Multiplying an eigenvector by $A$ only stretches it; a general vector is turned to a new direction.Worked example. Find the eigenvalues of $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$.
The characteristic equation is
$$\det\begin{pmatrix} 2 - \lambda & 1 \\ 1 & 2 - \lambda \end{pmatrix} = (2 - \lambda)^2 - 1 = 0 \;\Rightarrow\; 2 - \lambda = \pm 1.$$So $\lambda = 1$ or $\lambda = 3$. (The eigenvector for $\lambda = 3$ is $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and for $\lambda = 1$ is $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$.)Vocabulary TrainEnglish Chinese Pinyin matrix equation 矩阵方程 jǔ zhèn fāng chéng characteristic equation 特征方程 tè zhēng fāng chéng eigenvalues 特征值 tè zhēng zhí eigenvector 特征向量 tè zhēng xiàng liàng diagonal matrix 对角矩阵 duì jiǎo jǔ zhèn 2.3
Differentiation
Syllabus
- differentiate hyperbolic functions and differentiate $\sin^{-1}x$, $\cos^{-1}x$, $\sinh^{-1}x$, $\cosh^{-1}x$ and $\tanh^{-1}x$
- obtain an expression for $\dfrac{d^2y}{dx^2}$ in cases where the relation between $x$ and $y$ is defined implicitly or parametrically
- derive and use the first few terms of a Maclaurin's series for a function
Source: Cambridge International syllabus
You can now differentiate the hyperbolic functions and the inverse functions $\sin^{-1}x$, $\tan^{-1}x$, $\sinh^{-1}x$ and so on. You can also find $\dfrac{d^2y}{dx^2}$ for curves given implicitly or parametrically.
A Maclaurin's series 麦克劳林级数 writes a function as a power series:
$$f(x) = f(0) + f'(0)\,x + \frac{f''(0)}{2!}\,x^2 + \frac{f'''(0)}{3!}\,x^3 + \cdots$$For example $e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots$
Adding more terms makes the polynomial match $e^x$ over a wider stretch around $x=0$.Vocabulary TrainEnglish Chinese Pinyin Maclaurin's series 麦克劳林级数 mài kè láo lín jí shù 2.4
Integration
Syllabus
- integrate hyperbolic functions and recognise integrals of functions of the form $\dfrac{1}{\sqrt{a^2 - x^2}}$, $\dfrac{1}{a^2 + x^2}$ and $\dfrac{1}{\sqrt{x^2 - a^2}}$, integrating associated functions using trigonometric or hyperbolic substitutions
- derive and use reduction formulae for the evaluation of definite integrals
- understand how the area under a curve may be approximated by areas of rectangles, and use rectangles to estimate or set bounds for the area or to derive inequalities or limits concerning sums
- use integration to find arc lengths (Cartesian, parametric or polar) and surface areas of revolution about one of the axes
Source: Cambridge International syllabus
Learn these standard integrals:
$$\int \frac{1}{a^2 + x^2}\,dx = \frac{1}{a}\tan^{-1}\frac{x}{a} + C, \qquad \int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \sin^{-1}\frac{x}{a} + C.$$A trigonometric or hyperbolic substitution handles related forms. A reduction formula 递推公式 links an integral $I_n$ to $I_{n-1}$, so you can work down step by step. Integration also gives the arc length 弧长 of a curve and the surface area of revolution 旋转曲面面积 when a curve is turned about an axis.Worked example. Find $\displaystyle\int \frac{1}{\sqrt{4 - x^2}}\,dx$.
Here $a^2 = 4$, so $a = 2$ and
$$\int \frac{1}{\sqrt{4 - x^2}}\,dx = \sin^{-1}\frac{x}{2} + C.$$Vocabulary TrainEnglish Chinese Pinyin reduction formula 递推公式 dì tuī gōng shì arc length 弧长 hú zhǎng surface area of revolution 旋转曲面面积 xuán zhuǎn qū miàn miàn jī 2.5
Complex numbers
Syllabus
- understand de Moivre's theorem, for a positive or negative integer exponent, in terms of the geometrical effect of multiplication and division of complex numbers
- prove de Moivre's theorem for a positive integer exponent
- use de Moivre's theorem for a positive or negative rational exponent to express trigonometric ratios of multiple angles in terms of powers, to express powers of $\sin\theta$ and $\cos\theta$ in terms of multiple angles, in the summation of series, and in finding and using the $n$th roots of unity
Source: Cambridge International syllabus
Self-similar patterns like Romanesco broccoli arise from iterating functions in the complex plane.A complex number 复数 in polar form is $z = r(\cos\theta + i\sin\theta)$. De Moivre's theorem 棣莫弗定理 says that for any integer $n$,
$$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta.$$It is used to expand $\cos n\theta$ and $\sin n\theta$ in powers, to sum series, and to find the $n$ roots of unity 单位根 (the solutions of $z^n = 1$, equally spaced around the unit circle).
The five solutions of $z^5=1$ are equally spaced $72^\circ$ apart on the unit circle.Worked example. Use de Moivre's theorem to express $\cos 3\theta$ in terms of $\cos\theta$.
Take the real part of $(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta$:
$$\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta = \cos^3\theta - 3\cos\theta(1 - \cos^2\theta) = 4\cos^3\theta - 3\cos\theta.$$Vocabulary TrainEnglish Chinese Pinyin complex number 复数 fù shù de Moivre's theorem 棣莫弗定理 dì mò fú dìng lǐ roots of unity 单位根 dān wèi gēn 2.6
Differential equations
Syllabus
- find an integrating factor for a first order linear differential equation, and use it to find the general solution
- recall the meaning of 'complementary function' and 'particular integral', and that the general solution is the sum of the two
- find the complementary function for a first or second order linear differential equation with constant coefficients
- find a particular integral for a first or second order linear differential equation in cases where a polynomial, $ae^{bx}$ or $a\cos px + b\sin px$ is a suitable form
- use a given substitution to reduce a differential equation to a linear equation with constant coefficients or to a first order separable equation
- use initial conditions to find a particular solution, and interpret a solution in terms of a problem modelled by the equation
Source: Cambridge International syllabus
For a first order linear equation $\dfrac{dy}{dx} + P(x)\,y = Q(x)$, multiply by the integrating factor 积分因子 $\mu = e^{\int P\,dx}$. The left side then becomes $\dfrac{d}{dx}(\mu y)$, so you can integrate directly.
For a linear equation with constant coefficients, the general solution 通解 is the sum of two parts: the complementary function 余函数 (the solution of the equation with the right side set to $0$, found from the auxiliary equation) and a particular integral 特积分 (any one solution of the full equation).
A solution curve follows the slope field, staying tangent to the little segments everywhere.Worked example. Solve $\dfrac{dy}{dx} + 2y = e^x$.
The integrating factor is $\mu = e^{\int 2\,dx} = e^{2x}$. Multiplying through gives $\dfrac{d}{dx}\!\left(y\,e^{2x}\right) = e^{3x}$, so
$$y\,e^{2x} = \tfrac13 e^{3x} + C \;\Rightarrow\; y = \tfrac13 e^x + C e^{-2x}.$$Vocabulary TrainEnglish Chinese Pinyin integrating factor 积分因子 jī fēn yīn zi general solution 通解 tōng jiě complementary function 余函数 yú hán shù particular integral 特积分 tè jī fēn -
3 Further Mechanics
This handout covers Topic 3: Further Mechanics 进阶力学. It extends mechanics to projectiles, rigid bodies, circular motion, elastic strings, variable forces and collisions. Take $g = 10\ \text{m s}^{-2}$.
3.1
Motion of a projectile
Syllabus
- model the motion of a projectile as a particle moving with constant acceleration and understand any limitations of the model
- use horizontal and vertical equations of motion to solve problems on projectile motion, including the magnitude and direction of the velocity at a given time or position, the range on a horizontal plane and the greatest height reached
- derive and use the Cartesian equation of the trajectory of a projectile, including problems in which the initial speed and/or angle of projection may be unknown
Source: Cambridge International syllabus
Water jets follow parabolic paths — the classic projectile motion under gravity.A projectile 抛射体 moves freely under gravity, so it has constant acceleration 匀加速 $g$ downwards and no horizontal acceleration. Treat the horizontal and vertical motions separately. If it is launched at speed $u$ and angle $\alpha$:
$$\text{horizontal: } x = u\cos\alpha\,\cdot t, \qquad \text{vertical: } y = u\sin\alpha\,\cdot t - \tfrac12 g t^2.$$Eliminating $t$ gives the Cartesian equation 直角坐标方程 of the trajectory 轨迹 (the path), which is a parabola. The range on level ground is $\dfrac{u^2\sin 2\alpha}{g}$ and the greatest height is $\dfrac{u^2\sin^2\alpha}{2g}$.
The path is a parabola; the launch speed $u$ splits into a steady horizontal part and a vertical part slowed by gravity.Worked example. A ball is thrown at $u = 20\ \text{m s}^{-1}$ at $30^\circ$ to the horizontal. Find the range and greatest height.
$$\text{range} = \frac{20^2\sin 60^\circ}{10} = 40\times 0.866 = 34.6\ \text{m}, \qquad \text{height} = \frac{20^2\sin^2 30^\circ}{2\times 10} = \frac{400\times 0.25}{20} = 5\ \text{m}.$$Vocabulary TrainEnglish Chinese Pinyin Further Mechanics 进阶力学 jìn jiē lì xué projectile 抛射体 pāo shè tǐ constant acceleration 匀加速 yún jiā sù Cartesian equation 直角坐标方程 zhí jiǎo zuò biāo fāng chéng trajectory 轨迹 guǐ jì 3.2
Equilibrium of a rigid body
Syllabus
- calculate the moment of a force about a point
- use the result that the effect of gravity on a rigid body is equivalent to a single force at the centre of mass, and identify the centre of mass of a uniform body using symmetry
- use given information about the position of the centre of mass of a triangular lamina and other simple shapes
- determine the position of the centre of mass of a composite body by considering an equivalent system of particles
- use the principle that a rigid body in equilibrium under coplanar forces has zero vector sum of forces and zero sum of moments about any point, and the converse
- solve problems involving the equilibrium of a single rigid body under coplanar forces, including those involving toppling or sliding
Source: Cambridge International syllabus
The moment 力矩 of a force about a point is force $\times$ perpendicular distance; it measures turning effect. The weight of a body acts at its centre of mass 质心, which you can find using symmetry 对称 for a uniform shape, or by treating a composite body as a set of particles.
A rigid body is in equilibrium 平衡 when two conditions both hold: the vector sum of the forces is zero, and the sum of the moments about any point is zero. A body may also be on the edge of toppling 翻倒 (turning over) or sliding 滑动 (slipping).
Taking moments about the pivot $A$ balances the turning effects: $F\times4 = 100\times2$.Worked example. A uniform beam $AB$ of length $4\ \text{m}$ and weight $100\ \text{N}$ rests on a pivot at $A$. A vertical force $F$ at $B$ keeps it horizontal. Find $F$.
Take moments about $A$ (the weight acts at the centre, $2\ \text{m}$ from $A$):
$$F\times 4 = 100\times 2 \;\Rightarrow\; F = 50\ \text{N}.$$Vocabulary TrainEnglish Chinese Pinyin moment 力矩 lì jǔ centre of mass 质心 zhì xīn symmetry 对称 duì chèn equilibrium 平衡 píng héng toppling 翻倒 fān dǎo sliding 滑动 huá dòng 3.3
Circular motion
Syllabus
- understand the concept of angular speed for a particle moving in a circle, and use the relation $v = r\omega$
- understand that the acceleration of a particle moving in a circle with constant speed is directed towards the centre, and use the formulae $r\omega^2$ and $\dfrac{v^2}{r}$
- solve problems which can be modelled by the motion of a particle in a horizontal circle with constant speed
- solve problems which can be modelled by the motion of a particle in a vertical circle without loss of energy
Source: Cambridge International syllabus
For a particle moving in a circle of radius $r$, the angular speed 角速度 $\omega$ links to the speed by $v = r\omega$. The acceleration points towards the centre — the centripetal acceleration 向心加速度 — with size
$$a = r\omega^2 = \frac{v^2}{r}.$$
The velocity points along the tangent; the acceleration points inward to the centre.In a horizontal circle 水平圆 the speed is constant. In a vertical circle 竖直圆 use energy conservation, because the speed changes with height.
Worked example. A particle moves in a horizontal circle of radius $2\ \text{m}$ with angular speed $3\ \text{rad s}^{-1}$. Find its speed and acceleration.
$$v = r\omega = 2\times 3 = 6\ \text{m s}^{-1}, \qquad a = r\omega^2 = 2\times 3^2 = 18\ \text{m s}^{-2}.$$Vocabulary TrainEnglish Chinese Pinyin angular speed 角速度 jiǎo sù dù centripetal acceleration 向心加速度 xiàng xīn jiā sù dù horizontal circle 水平圆 shuǐ píng yuán vertical circle 竖直圆 shù zhí yuán 3.4
Hooke's law
Syllabus
- use Hooke's law as a model relating the force in an elastic string or spring to the extension or compression, and understand the term modulus of elasticity
- use the formula for the elastic potential energy stored in a string or spring
- solve problems involving forces due to elastic strings or springs, including those where considerations of work and energy are needed
Source: Cambridge International syllabus
Hooke's law 胡克定律 says the tension in an elastic string or spring is proportional to its extension $x$:
$$T = \frac{\lambda x}{L},$$where $L$ is the natural length and $\lambda$ is the modulus of elasticity 弹性模量. Stretching stores elastic potential energy 弹性势能:$$E = \frac{\lambda x^2}{2L}.$$
Tension rises in proportion to extension; the shaded triangle is the stored elastic energy.Worked example. An elastic string of natural length $2\ \text{m}$ and modulus $50\ \text{N}$ is stretched by $0.5\ \text{m}$. Find the tension and the stored energy.
$$T = \frac{50\times 0.5}{2} = 12.5\ \text{N}, \qquad E = \frac{50\times 0.5^2}{2\times 2} = 3.125\ \text{J}.$$Vocabulary TrainEnglish Chinese Pinyin Hooke's law 胡克定律 hú kè dìng lǜ modulus of elasticity 弹性模量 tán xìng mó liàng elastic potential energy 弹性势能 tán xìng shì néng 3.5
Linear motion under a variable force
Syllabus
- solve problems which can be modelled as the linear motion of a particle under the action of a variable force, by setting up and solving an appropriate differential equation
Source: Cambridge International syllabus
When the force depends on position $x$, use acceleration in the form $a = v\dfrac{dv}{dx}$, which turns Newton's law into a differential equation 微分方程 relating $v$ and $x$.
Worked example. A particle of mass $8\ \text{kg}$ moves along a line under a variable force 变力 of size $(x^3 + 4x)\ \text{N}$ acting in the direction of motion. When $x = 0$, $v = 1$. Find $v$ in terms of $x$.
Newton's law gives $8v\dfrac{dv}{dx} = x^3 + 4x$. Separating and integrating:
$$\int 8v\,dv = \int (x^3 + 4x)\,dx \;\Rightarrow\; 4v^2 = \tfrac14 x^4 + 2x^2 + C.$$At $x = 0$, $v = 1$ gives $C = 4$, so $4v^2 = \tfrac14 x^4 + 2x^2 + 4 = 4\left(\tfrac{x^2}{4} + 1\right)^2$. Hence $v = \tfrac14 x^2 + 1$.Vocabulary TrainEnglish Chinese Pinyin differential equation 微分方程 wēi fēn fāng chéng variable force 变力 biàn lì 3.6
Momentum
Syllabus
- recall Newton's experimental law and the definition of the coefficient of restitution, the property $0 \leqslant e \leqslant 1$, and the meaning of 'perfectly elastic' ($e = 1$) and 'inelastic' ($e = 0$)
- use conservation of linear momentum and/or Newton's experimental law to solve problems modelled as the direct or oblique impact of two smooth spheres, or the direct or oblique impact of a smooth sphere with a fixed surface
Source: Cambridge International syllabus
A Newton's cradle demonstrates conservation of momentum in collisions.In a collision, total momentum is conserved — the conservation of momentum 动量守恒. The bounciness is measured by Newton's experimental law 牛顿实验定律, which defines the coefficient of restitution 恢复系数 $e$:
$$e = \frac{\text{speed of separation}}{\text{speed of approach}}, \qquad 0 \leqslant e \leqslant 1.$$Here $e = 1$ is perfectly elastic (no energy lost) and $e = 0$ is inelastic (the bodies stay together).
The coefficient of restitution $e$ compares how fast the bodies separate with how fast they approached.Worked example. A sphere $A$ of mass $2\ \text{kg}$ moving at $5\ \text{m s}^{-1}$ hits a stationary sphere $B$ of mass $3\ \text{kg}$, with $e = 0.5$. Find the speeds afterwards.
Momentum: $2(5) = 2v_A + 3v_B$, so $2v_A + 3v_B = 10$. Restitution: $v_B - v_A = 0.5(5) = 2.5$. Solving together gives $v_A = 0.5\ \text{m s}^{-1}$ and $v_B = 3.0\ \text{m s}^{-1}$.
Vocabulary TrainEnglish Chinese Pinyin conservation of momentum 动量守恒 dòng liàng shǒu héng Newton's experimental law 牛顿实验定律 niú dùn shí yàn dìng lǜ coefficient of restitution 恢复系数 huī fù xì shù -
4 Further Probability & Statistics
This handout covers Topic 4: Further Probability & Statistics 进阶概率统计. It adds continuous distributions, small-sample inference, the chi-squared and non-parametric tests, and probability generating functions.
4.1
Continuous random variables
Syllabus
- use a probability density function which may be defined piecewise
- use the general result $E(g(X)) = \int g(x)f(x)\,dx$, where $f(x)$ is the probability density function of $X$ and $g(X)$ is a function of $X$
- understand and use the relationship between the probability density function (PDF) and the cumulative distribution function (CDF), and use either to evaluate probabilities or percentiles
- use cumulative distribution functions of related variables in simple cases
Source: Cambridge International syllabus
A continuous variable $X$ is described by a probability density function 概率密度函数 $f(x)$, which may be defined piecewise. The probability over a range is the area under $f$, and the mean of any function of $X$ is
$$E(g(X)) = \int g(x)\,f(x)\,dx.$$
The median sits where the area under $f(x)$ to its left is exactly $0.5$.The cumulative distribution function 累积分布函数 $F(x) = P(X \leqslant x)$ is the running total: $F(x) = \displaystyle\int_{-\infty}^{x} f(t)\,dt$, and $f(x) = F'(x)$. Use $F$ to find probabilities and percentiles 百分位数 (for example, the median 中位数 solves $F(x) = 0.5$).
The cumulative graph $F(x)$ rises from $0$ to $1$; the height $0.5$ is reached at the median.Worked example. A variable has $f(x) = \tfrac12 x$ for $0 \leqslant x \leqslant 2$. Find the median.
The cumulative distribution function is $F(x) = \displaystyle\int_0^x \tfrac12 t\,dt = \tfrac14 x^2$. Set $F(m) = 0.5$:
$$\tfrac14 m^2 = 0.5 \;\Rightarrow\; m^2 = 2 \;\Rightarrow\; m = \sqrt{2} = 1.41.$$Vocabulary TrainEnglish Chinese Pinyin Further Probability & Statistics 进阶概率统计 jìn jiē gài lǜ tǒng jì probability density function 概率密度函数 gài lǜ mì dù hán shù cumulative distribution function 累积分布函数 lěi jī fēn bù hán shù percentiles 百分位数 bǎi fēn wèi shù median 中位数 zhōng wèi shù 4.2
Inference using normal and t-distributions
Syllabus
- formulate hypotheses and apply a hypothesis test concerning the population mean using a small sample drawn from a normal population of unknown variance, using a $t$-test
- calculate a pooled estimate of a population variance from two samples
- formulate hypotheses concerning the difference of population means, and apply, as appropriate, a 2-sample $t$-test, a paired-sample $t$-test, or a test using a normal distribution
- determine a confidence interval for a population mean, based on a small sample from a normal population with unknown variance, using a $t$-distribution
- determine a confidence interval for a difference of population means, using a $t$-distribution or a normal distribution as appropriate
Source: Cambridge International syllabus
A Galton board: balls falling through pins pile up into the bell-shaped normal distribution.When a sample is small and the population variance is unknown, base your hypothesis test 假设检验 on the $t$-distribution instead of the normal. The same idea gives a confidence interval 置信区间 for the mean:
$$\bar{x} \pm t\,\frac{s}{\sqrt{n}},$$where $t$ comes from the $t$-tables with $n - 1$ degrees of freedom. To compare two populations, use a 2-sample or paired-sample $t$-test, after finding a pooled estimate 合并估计 of the shared variance when appropriate.
For a small sample the $t$-distribution is flatter with heavier tails, so its critical values are larger.Worked example. A sample of $n = 10$ has mean $\bar{x} = 50$ and standard deviation $s = 4$. Find a $95\%$ confidence interval for the mean (use $t = 2.262$ for $9$ degrees of freedom).
$$50 \pm 2.262\times\frac{4}{\sqrt{10}} = 50 \pm 2.86 \;\Rightarrow\; (47.1,\ 52.9).$$Vocabulary TrainEnglish Chinese Pinyin hypothesis test 假设检验 jiǎ shè jiǎn yàn confidence interval 置信区间 zhì xìn qū jiān pooled estimate 合并估计 hé bìng gū jì 4.3
Chi-squared tests
Syllabus
- fit a theoretical distribution, as prescribed by a given hypothesis, to given data
- use a $\chi^2$-test, with the appropriate number of degrees of freedom, to carry out the corresponding goodness of fit analysis
- use a $\chi^2$-test, with the appropriate number of degrees of freedom, for independence in a contingency table
Source: Cambridge International syllabus
A $\chi^2$-test (chi-squared test 卡方检验) compares observed counts $O$ with expected counts $E$ from a theoretical distribution 理论分布:
$$\chi^2 = \sum \frac{(O - E)^2}{E}.$$Compare this with a table value for the right number of degrees of freedom 自由度. Two uses: a goodness of fit 拟合优度 test (does the data follow the proposed model?), and a test for independence 独立性 of two variables in a contingency table 列联表.
The test rejects the model when $\chi^2$ exceeds the critical value, landing in the shaded $5\%$ tail.Worked example. Four equally likely categories give observed counts $20, 30, 25, 25$ (so each expected count is $25$). Test the fit at the $5\%$ level.
$$\chi^2 = \frac{(20-25)^2 + (30-25)^2 + 0 + 0}{25} = \frac{25 + 25}{25} = 2.$$With $4 - 1 = 3$ degrees of freedom the table value is $7.815$. Since $2 < 7.815$, do not reject the model.Vocabulary TrainEnglish Chinese Pinyin chi-squared test 卡方检验 kǎ fāng jiǎn yàn theoretical distribution 理论分布 lǐ lùn fēn bù degrees of freedom 自由度 zì yóu dù goodness of fit 拟合优度 nǐ hé yōu dù independence 独立性 dú lì xìng contingency table 列联表 liè lián biǎo 4.4
Non-parametric tests
Syllabus
- understand the idea of a non-parametric test and appreciate situations in which such a test might be useful
- understand the basis of the sign test, the Wilcoxon signed-rank test and the Wilcoxon rank-sum test
- use a single-sample sign test and a single-sample Wilcoxon signed-rank test to test a hypothesis concerning a population median
- use a paired-sample sign test, a Wilcoxon matched-pairs signed-rank test and a Wilcoxon rank-sum test, as appropriate, to test for identity of populations
Source: Cambridge International syllabus
A non-parametric test 非参数检验 makes no assumption that the data is normal, so it is useful when that assumption fails. The basic ones are:
- the sign test 符号检验: count how many values fall above and below a proposed median, and test those counts with a binomial model;
- the Wilcoxon signed-rank test 威尔科克森符号秩检验, which also uses the sizes of the differences, not just their signs;
- the Wilcoxon rank-sum test 威尔科克森秩和检验, for comparing two separate samples.
Vocabulary TrainEnglish Chinese Pinyin non-parametric test 非参数检验 fēi cān shù jiǎn yàn sign test 符号检验 fú hào jiǎn yàn Wilcoxon signed-rank test 威尔科克森符号秩检验 wēi ěr kē kè sēn fú hào zhì jiǎn yàn Wilcoxon rank-sum test 威尔科克森秩和检验 wēi ěr kē kè sēn zhì hé jiǎn yàn 4.5
Probability generating functions
Syllabus
- understand the concept of a probability generating function (PGF) and construct and use the PGF for given distributions
- use formulae for the mean and variance of a discrete random variable in terms of its PGF, and use these formulae to calculate the mean and variance of a given probability distribution
- use the result that the PGF of the sum of independent random variables is the product of the PGFs of those random variables
Source: Cambridge International syllabus
Dice: a starting point for probability and discrete random variables.The probability generating function 概率母函数 of a discrete variable $X$ is
$$G(t) = E(t^X) = \sum_x P(X = x)\,t^x.$$It packs the whole distribution into one function. The mean and variance come from its derivatives at $t = 1$: $E(X) = G'(1)$ and $\mathrm{Var}(X) = G''(1) + G'(1) - \big(G'(1)\big)^2$. Also, the PGF of a sum of independent variables is the product of their PGFs.Worked example. $X$ has $P(X=0) = 0.5$, $P(X=1) = 0.3$, $P(X=2) = 0.2$. Find $E(X)$ using the PGF.
Here $G(t) = 0.5 + 0.3t + 0.2t^2$, so $G'(t) = 0.3 + 0.4t$ and
$$E(X) = G'(1) = 0.3 + 0.4 = 0.7.$$Vocabulary TrainEnglish Chinese Pinyin probability generating function 概率母函数 gài lǜ mǔ hán shù