Pure Mathematics 1
A-Level Mathematics Topic 1 28:42 English narration · English + 中文 subtitles burned in
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Look at a great suspension bridge.
看一座宏伟的悬索桥。
Its huge cable is not a random shape — it hangs in a perfect parabola, a curve you can write with a single equation.
它那巨大的缆索并不是随意的形状——它悬垂成一条完美的抛物线， 一条你能用一个方程写出的曲线。
The world is full of curves like this: the path of a ball, the height on a turning wheel, the growth of savings.
世界上到处都是这样的曲线：球的轨迹、转轮上的高度、 存款的增长。
Pure Mathematics, part one, is the toolkit that describes them all.
纯数学第一部分，就是描述这一切的工具箱。
Eight tools.
八件工具。
Let's tour them.
我们来逐一游览。
This is the algebra and calculus core of your A-Level.
这是你 A-Level 课程里代数与微积分的核心。
It splits into eight connected ideas — from quadratics and functions, through geometry and trigonometry, to the two great engines: differentiation and integration.
它分成八个彼此相连的主题—— 从二次式与函数，经过几何与三角，直到两大引擎：微分与积分。
Here is the whole map, in one lesson.
这一整张地图，就在这一节课里。
We start with the quadratic — any curve shaped like this parabola.
我们从二次式开始——任何形如这条抛物线的曲线。
The most useful trick is completing the square: rewriting it as a squared bracket, plus a number.
最有用的技巧是配方： 把它改写成一个平方括号，加上一个数。
Watch it fold together.
看它是怎样合拢的。
The moment you do, the graph gives up its lowest point, its vertex, for free.
你一旦做到， 图像就免费交出它的最低点——顶点。
That one form solves the equation and finds the turning point at once.
这一个形式，既解出方程，又同时找到转折点。
Completing the square rewrites a quadratic as a number times a bracket squared, plus a constant.
配方就是把二次式改写成:一个数乘以一个平方括号，再加上一个常数。
Why bother? A squared bracket can never be negative.
为什么值得这么做? 因为平方括号永远不会是负的。
So the smallest value of the whole expression happens when the bracket is zero.
所以整个式子的最小值，出现在括号为零的时候。
Read the vertex straight off: the turning point sits at minus p, comma, q.
顶点可以直接读出来:转折点就在负 p 与 q 处。
The same form also solves the equation — take the square root of both sides, and finish in two lines.
同一个形式也能解方程—— 两边开平方，两行就写完。
Here is the exam version.
来看考试版本。
Write nine x squared, minus thirty-six x, plus eight in completed-square form.
把九 x 平方减三十六 x 加八写成配方形式。
First take the nine outside the first two terms.
先把九从前两项里提出来。
Inside the bracket, halve the coefficient of x: minus four halves to minus two, so we write x minus two, all squared, minus four.
在括号里，把 x 的系数减四折半得到减二，于是写成 x 减二 的平方再减四。
Now multiply the nine back in — minus four times nine is minus thirty-six — and add the eight.
现在把九乘回去——减四乘九是减三十六——再加上八。
So the answer is nine, bracket x minus two, squared, minus twenty-eight.
所以答案是 九乘 x 减二 的平方再减二十八。
Which tells you the least value is minus twenty-eight, and it happens when x is two.
这也告诉你最小值是减二十八，并且发生在 x 等于二时。
How many times does a parabola cross the horizontal axis?
抛物线与横轴相交几次？
One number decides: the discriminant, b squared minus four a c.
由一个数决定：判别式，b 的平方减去四 a c。
If it is positive, the curve cuts the axis twice — two real roots.
如果它为正，曲线两次穿过横轴——两个实根。
If it is exactly zero, the curve just touches — one repeated root.
如果它恰好为零，曲线只是相切—— 一个重根。
And if it is negative, the curve floats clear of the axis — no real roots at all.
如果它为负，曲线整个悬在横轴之上——完全没有实根。
Now a classic.
来看一道经典题。
For which values of k does three k x squared, plus bracket k plus eight, x, plus three, have two distinct real roots? First name the coefficients: a is three k, b is k plus eight, and c is three.
k 取什么值时，三 k x 平方加 括号 k 加八 x 加三，有两个相异实根? 先写清系数:a 是三 k，b 是 k 加八，c 是三。
Two distinct roots means the discriminant is greater than zero.
两个相异实根，就意味着判别式大于零。
Expand carefully and collect the terms: k squared, minus twenty k, plus sixty-four, is greater than zero.
仔细展开并合并同类项:k 平方减二十 k 加六十四大于零。
That factorises into k minus four, times k minus sixteen.
它可以分解成 k 减四 乘 k 减十六。
So k is less than four, or k is greater than sixteen.
所以 k 小于四，或者 k 大于十六。
One last guard: a cannot be zero, so k is not zero.
最后还有一道防线: a 不能为零，所以 k 不等于零。
You have three ways to solve a quadratic equation.
解二次方程有三条路。
Factorising is fastest, when it works.
因式分解最快，前提是它行得通。
Completing the square always works, and hands you the vertex as a bonus.
配方总能用，而且顺手给你顶点。
The quadratic formula never fails: minus b, plus or minus the square root of b squared minus four a c, all over two a.
求根公式万无一失:负 b，加或减 b 平方减四 a c 的平方根，整个除以二 a。
Notice the discriminant sitting inside that square root — it is the same number, doing the same job.
注意判别式就藏在那个根号里面——同一个数，做着同一件事。
Pick the route the question points at, and show every line.
按题目指的方向选路，并且每一步都写出来。
Inequalities trip people up, so always sketch.
不等式最容易出错，所以一定要画草图。
Take bracket k minus four, times bracket k minus sixteen, greater than zero.
看 括号 k 减四 乘 括号 k 减十六 大于零。
The roots are four and sixteen, and the parabola opens upwards.
两个根是四和十六，而抛物线开口向上。
Outside the roots the curve is up in the air, so it is positive there.
在两根之外，曲线高高在上，所以那里是正的。
Between the roots it dips below the axis, so that stretch is negative.
在两根之间，它下沉到横轴以下，所以那一段是负的。
We wanted positive, so the answer is two separate pieces: k less than four, or k greater than sixteen.
我们要的是正， 所以答案是两段:k 小于四，或者 k 大于十六。
Never write four is less than k is less than sixteen here — that is the negative part.
千万不要在这里写四小于 k 小于十六—— 那正是负的那一段。
Next, a pair of simultaneous equations where one is straight and one is curved.
接下来是一组联立方程，其中一个是直线，一个是曲线。
Geometrically, you are looking for where the line meets the curve.
从几何上看， 你要找的就是直线与曲线相交的地方。
The method is substitution.
方法是代入。
Rearrange the linear equation for one letter — here, y equals x plus one.
把线性方程解出一个字母—— 这里是 y 等于 x 加一。
Put that expression into the quadratic wherever y appears.
凡是出现 y 的地方，都换成这个表达式。
You now have a single quadratic in x, which factorises to give x equals two or x equals minus one.
于是你得到一个只含 x 的二次方程，分解后给出 x 等于二或 x 等于负一。
Feed each one back into the line to find its partner.
再把每个 x 代回直线，求出它对应的 y。
Always give the answers as coordinate pairs.
答案一定要写成坐标对。
Some equations are quadratics in disguise.
有些方程是伪装的二次式。
Look at x to the fourth, minus five x squared, plus four, equals zero.
看 x 的四次方减五 x 平方加四等于零。
The powers are four, two and zero — each one double the quadratic pattern.
指数是四、二、零——每一个都是二次式指数的两倍。
So let u stand for x squared.
所以让 u 代表 x 平方。
The equation becomes u squared minus five u plus four equals zero, which factorises to u minus one, times u minus four.
方程就变成 u 平方减五 u 加四等于零，分解为 u 减一 乘 u 减四。
So u is one, or u is four.
所以 u 是一，或者 u 是四。
Now go back to x: x squared equals one gives plus or minus one, and x squared equals four gives plus or minus two.
现在回到 x:x 平方等于一给出正负一， x 平方等于四给出正负二。
Four solutions.
一共四个解。
The very same trick returns in trigonometry.
同样的技巧会在三角函数里再出现一次。
Next, functions — rules that turn each input into exactly one output.
接下来是函数——把每个输入变成唯一一个输出的规则。
The exam loves transformations: how a graph moves when you change its equation.
考试很爱考变换： 当你改动方程时，图像如何移动。
Add to the output, and it slides up.
给输出加一个数，它向上平移。
Add to the input, and it slides sideways.
给输入加一个数， 它向侧面平移。
Multiply, and it stretches.
乘一个数，它就被拉伸。
Learn these four moves, and you can reshape any curve on sight.
学会这四种动作，你就能一眼重塑任何曲线。
Let's be precise, because these words earn marks.
我们把话说准，因为这些词是给分点。
A function is a rule that sends each input to exactly one output.
函数是一条规则，把每个输入送到唯一一个输出。
The domain is the set of inputs you are allowed to use.
定义域是你被允许使用的那些输入。
The range is the set of outputs the rule actually produces.
值域是这条规则实际产生的那些输出。
A function is one-one when different inputs always give different outputs — no output is used twice.
当不同的输入总是给出不同的输出时，函数是一一对应的——没有输出被用过两次。
That matters, because only a one-one function can be reversed.
这很重要，因为只有一一对应的函数才可以反过来。
Restricting the domain is how a question makes a curve one-one.
限制定义域， 就是题目让一条曲线变成一一对应的办法。
Composition means doing one function after another.
复合就是把两个函数一个接一个地做。
Read f g of x from the inside out: do g first, then feed its output into f.
f g of x 要从里往外读:先做 g， 再把它的输出喂给 f。
The order matters — g f is usually a completely different function, so never swap them.
顺序很重要—— g f 通常是完全不同的函数，所以绝不能对调。
And there is a condition the exam checks: the composite f g only exists when the range of g fits inside the domain of f.
而且考试会检查一个条件:只有当 g 的值域落在 f 的定义域之内时，复合函数 f g 才存在。
If g can produce a number that f is not allowed to take, the composite breaks.
如果 g 可能产生一个 f 不允许接收的数，复合就断了。
To find an inverse, write y equals f of x, make x the subject, then swap the letters.
求反函数的做法是:写 y 等于 f of x，把 x 解出来，然后交换字母。
Watch it on a real exam function: f of x is bracket x plus three, squared, minus twelve, defined for x greater than or equal to zero.
来看一个真实的考试函数:f of x 等于 括号 x 加三 的平方减十二， 定义在 x 大于或等于零上。
Add twelve to both sides.
两边加十二。
Now take the square root — and here is the marking point: choose the positive root, because x is at least zero, so x plus three is at least three.
现在开平方——给分点就在这里: 要取正的那个根，因为 x 至少是零，所以 x 加三至少是三。
Subtract three, and swap the letters.
再减三，然后交换字母。
The inverse is the square root of x plus twelve, minus three.
反函数就是 x 加十二 的平方根，再减三。
Remember that the domain and the range swap over too.
别忘了定义域和值域也会互换。
There is a lovely picture behind that algebra.
这段代数背后有一幅很漂亮的图。
The graph of the inverse is the reflection of the graph of f in the line y equals x — its mirror image.
反函数的图像，就是 f 的图像关于直线 y 等于 x 的镜像。
Swapping the letters is exactly what reflecting in that line does to a point: a comma b becomes b comma a.
交换字母，正是关于那条线做反射对一个点所做的事:a 逗号 b 变成 b 逗号 a。
Two useful consequences follow.
由此得到两个有用的结论。
The domain of f becomes the range of the inverse, and the range of f becomes its domain.
f 的定义域变成反函数的值域，f 的值域变成它的定义域。
And where the curve crosses the mirror line, f of x equals x, which is a quick way to find those special points.
而在曲线与那条镜像线相交处，f of x 等于 x。
Now pin the four transformations down.
现在把四种变换钉死。
Add a to the whole function, and the graph is a translation up by a.
给整个函数加 a，图像向上平移 a。
Add a inside the bracket, and it translates left by a — inside moves go backwards, which is the classic trap.
在括号内加 a， 它向左平移 a——括号里的移动是反着来的，这是经典的坑。
Multiply the whole function by a, and it stretches in the y direction by factor a.
把整个函数乘以 a， 它在 y 方向按倍数 a 伸缩。
Multiply x by a inside, and it stretches in the x direction by factor one over a.
在括号内把 x 乘以 a，它在 x 方向按 一比 a 伸缩。
When two of them are combined, the order can change the result, so in your answer state the type, the direction and the amount, every time. It is one of the easiest marks on the whole paper.
当两种变换结合时，顺序可能改变结果，所以在答案里，每次都要写清类型、方向和大小。
Now put curves on a grid.
现在把曲线放到坐标网格上。
A straight line is captured by its gradient — its steepness — the change in height over the change across.
一条直线由它的斜率抓住——也就是它的陡峭程度—— 即纵向变化除以横向变化。
Two lines are parallel when their gradients match, and perpendicular when the gradients multiply to give minus one.
两条直线斜率相等时平行，斜率之积为负一时垂直。
A circle is just as tidy: a centre and a radius.
圆同样简洁：一个圆心加一条半径。
And here is the fact that cracks most circle questions — a tangent always meets the radius at a right angle.
还有一个能破解大多数圆题的事实—— 切线总是与半径成直角相交。
A straight line has three standard forms, and questions swap between them.
直线有三种标准写法，题目会在它们之间来回切换。
Y equals m x plus c shows the gradient and the y-intercept at a glance.
y 等于 m x 加 c 一眼就看出斜率和 y 截距。
The point-gradient form, y minus y one equals m times x minus x one, is fastest when you know one point and the gradient.
点斜式，y 减 y 一 等于 m 乘 括号 x 减 x 一， 在已知一点和斜率时最快。
And a x plus b y plus c equals zero is the tidy form some questions demand.
而 a x 加 b y 加 c 等于零，是某些题目要求的整齐写法。
All three describe the same line.
三种写的都是同一条直线。
Pick the form that matches what you were given, then rearrange at the end if the question asks for it.
按你手上已有的条件选写法，如果题目要求，最后再整理成它想要的形式。
A circle often arrives expanded, like x squared plus y squared, minus six x, plus ten y, minus twenty-seven, equals zero.
圆经常以展开的样子出现，比如 x 平方加 y 平方减六 x 加十 y 减二十七等于零。
Complete the square twice: once in x, once in y.
配方两次:一次对 x，一次对 y。
In x, x squared minus six x becomes bracket x minus three, squared, minus nine.
对 x，x 平方减六 x 变成 括号 x 减三 的平方再减九。
In y, y squared plus ten y becomes bracket y plus five, squared, minus twenty-five.
对 y，y 平方加十 y 变成 括号 y 加五 的平方再减二十五。
Move every constant to the right hand side.
把所有常数都搬到等号右边。
Twenty-seven plus nine plus twenty-five is sixty-one.
二十七加九加二十五是六十一。
So the centre is three, minus five, and the radius is the square root of sixty-one.
所以圆心是三、负五，半径是六十一的平方根。
Here is the diameter question.
来看直径题。
P is one comma one, Q is seven comma eleven, and they are the two ends of a diameter.
P 是一逗号一，Q 是七逗号十一，它们是一条直径的两个端点。
The centre must be the midpoint, so average the coordinates: four comma six.
圆心一定是中点，所以把坐标取平均:四逗号六。
The radius is half the length of P Q.
半径是 P Q 长度的一半。
The steps across and up are six and ten, so P Q squared is thirty-six plus one hundred, which is one hundred and thirty-six.
横向和纵向的位移是六和十，所以 P Q 的平方是三十六加一百，等于一百三十六。
Half of the root of one hundred and thirty-six is the root of thirty-four.
一百三十六的根的一半，就是三十四的根。
So the circle is x minus four squared, plus y minus six squared, equals thirty-four.
所以这个圆是 x 减四 的平方加 y 减六 的平方等于三十四。
For circles, degrees are clumsy.
对于圆，角度制很笨拙。
Mathematicians prefer the radian.
数学家更喜欢弧度。
One radian is the angle whose arc is exactly as long as the radius.
一弧度，就是弧长恰好等于半径时， 圆心处的那个角。
Half a turn is pi radians.
半圈是派弧度。
The payoff is two beautifully simple formulas: an arc length is the radius times the angle, and a sector's area is one half r squared times the angle — but only when the angle is measured in radians.
回报是两个极其简洁的公式：弧长等于半径乘以角度， 扇形面积等于二分之一 r 的平方乘以角度——但前提是角度以弧度来量。
Radians and degrees have to convert instantly.
弧度和角度必须能立刻互换。
The one fact to remember is that pi radians equals one hundred and eighty degrees.
要记住的那一个事实是:派弧度等于一百八十度。
So to go from degrees to radians, multiply by pi over one hundred and eighty.
所以从度换到弧度，乘以 派 除以一百八十。
To come back, multiply by one hundred and eighty over pi.
换回来，乘以 一百八十 除以派。
Learn the common angles by heart: thirty degrees is pi over six, forty-five is pi over four, sixty is pi over three, ninety is pi over two, and a full turn is two pi.
常用角度要背下来:三十度是六分之派，四十五度是四分之派，六十度是三分之派， 九十度是二分之派，一整圈是二派。
In circular measure, and in the calculus of trig functions, radians are the default — degrees are the exception, so convert early and stay in radians.
在弧度制的计算里，以及三角函数的微积分里， 弧度是默认的——用度才是例外。
Two formulas do most of the work here.
这里主要靠两个公式。
The arc length is the radius times the angle.
弧长等于半径乘以角度。
The sector area is one half r squared times the angle.
扇形面积等于二分之一 r 平方乘以角度。
Both of them need the angle in radians.
两个公式都要求角度用弧度。
Now draw the chord that joins the two ends of the arc.
现在画出连接弧两端的弦。
It cuts the sector into a triangle and a segment.
它把扇形切成一个三角形和一个弓形。
The triangle has area one half r squared sine theta.
三角形的面积是二分之一 r 平方乘 sin theta。
So the segment is the sector minus the triangle: one half r squared, bracket theta minus sine theta.
所以弓形就是扇形减去三角形: 二分之一 r 平方乘 括号 theta 减 sin theta。
That single line answers a whole family of exam questions.
这一行公式，能回答一大类考题。
Let's use it.
我们来用一下。
A sector has an angle of two thirds pi radians.
一个扇形的圆心角是三分之二派弧度。
Show that the segment cut off by the chord has area about zero point six one four r squared.
证明被弦截下的弓形面积约为零点六一四 r 平方。
Substitute into the formula.
代入公式。
Two thirds pi is two point zero nine four four.
三分之二派是二点零九四四。
And the sine of two thirds pi is the sine of sixty degrees, which is zero point eight six six zero.
而三分之二派的正弦，就是六十度的正弦，等于零点八六六零。
Subtracting leaves one point two two eight four, and half of that is zero point six one four two.
相减得到一点二二八四，它的一半是零点六一四二。
So the segment area is about zero point six one four r squared, as required.
所以弓形面积约为零点六一四 r 平方，正如所求。
Notice we never needed a value for r.
注意我们从来不需要 r 的具体数值。
Spin a point around a circle, and watch its height.
让一个点绕圆旋转，看它的高度。
It rises and falls, tracing the sine curve — the same gentle wave a Ferris wheel gives you.
它上下起伏，画出正弦曲线—— 就是摩天轮带给你的那种和缓波动。
Cosine is the very same wave, just shifted along.
余弦是完全相同的波，只是沿着平移了一段。
These functions repeat forever, and they obey one golden identity: sine squared plus cosine squared always equals one.
这些函数永远重复，并且遵守一条黄金恒等式：正弦的平方加余弦的平方，永远等于一。
Most trig problems come down to using that.
大多数三角题，归根结底都是在用它。
You must be able to sketch all three graphs from memory.
三条图像你都必须能凭记忆画出来。
Sine starts at zero, rises to one, and comes back down — a wave that stays between minus one and one and repeats every three hundred and sixty degrees.
正弦从零开始，升到一，再降回来—— 这条波始终在负一和一之间，并且每三百六十度重复一次。
Cosine is the same wave, only it starts at one.
余弦是同一条波，只是从一开始。
Tangent is different: it climbs from zero, shoots off to infinity just before ninety degrees, and repeats every one hundred and eighty.
正切不一样:它从零往上爬，在快到九十度时冲向无穷，并且每一百八十度重复一次。
And those vertical lines it never touches are asymptotes.
而那些它永远碰不到的竖线，叫渐近线。
Mark the intercepts and the peaks on your sketch — examiners give marks for the right shape.
在草图上标出交点和峰值—— 形状画对是有分的。
Nine exact values are worth memorising, because a calculator is sometimes banned and a surd answer is sometimes demanded.
有九个精确值值得背下来，因为有时不许用计算器，有时又要求写成根式。
Sine of thirty is one half; sine of forty-five is one over root two; sine of sixty is root three over two.
三十度的正弦是二分之一;四十五度的正弦是一比根二;六十度的正弦是二分之根三。
Cosine runs the same list backwards: root three over two, one over root two, one half.
余弦把这张表倒过来念:二分之根三、一比根二、二分之一。
Tangent is sine divided by cosine, so it gives one over root three, then one, then root three.
正切是正弦除以余弦，所以依次得到一比根三、一、根三。
If you forget them, draw the two special triangles and read the sides off. They take ten seconds.
如果忘了，就画出那两个特殊三角形，直接把边长读出来。
Two identities carry the whole topic.
两个恒等式撑起整个主题。
First, tangent theta is identically sine theta over cosine theta.
第一，tan theta 恒等于 sin theta 除以 cos theta。
Second, sine squared theta plus cosine squared theta is identically one — which is just Pythagoras on a triangle with hypotenuse one.
第二，sin theta 的平方加 cos theta 的平方恒等于一—— 这只是斜边为一的三角形上的勾股定理。
The three-bar sign means true for every angle, not merely for a special one.
三条横线的符号表示对每个角都成立， 而不只是对某个特殊角成立。
Their job is to rewrite an equation until it contains only one trig function.
它们的用途是:把方程改写到只剩一个三角函数。
Note the notation too: sine to the minus one of x means the angle, and your calculator returns just one of them, the principal value. You have to find the other angles yourself.
也注意记号:x 的反正弦表示那个角，而计算器只返回其中一个，即主值。
Now a full trigonometric equation.
现在来一道完整的三角方程。
Solve six sine theta equals one plus two over sine theta, between minus one hundred and eighty and one hundred and eighty degrees.
解 六 sin theta 等于 一 加 二除以 sin theta， 范围在负一百八十度到一百八十度之间。
Multiply every term by sine theta to clear the fraction.
每一项都乘以 sin theta，把分母清掉。
That leaves a quadratic in sine theta: six sine squared theta, minus sine theta, minus two, equals zero.
剩下的是一个关于 sin theta 的二次方程:六 sin theta 平方减 sin theta 减二等于零。
Factorise it like any quadratic.
像解普通二次方程一样分解。
So sine theta is two thirds, or sine theta is minus one half.
于是 sin theta 等于三分之二，或者 sin theta 等于负二分之一。
Each value gives two angles inside the interval: two thirds gives forty-one point eight degrees and one hundred and thirty-eight point two; minus one half gives minus thirty and minus one hundred and fifty.
每个值在区间内都给出两个角:三分之二给出四十一点八度和一百三十八点二度; 负二分之一给出负三十度和负一百五十度。
Four solutions — and the marks are for finding all of them.
一共四个解——分数就在于把它们全部找出来。
A series is the sum of a pattern.
级数是一种规律的求和。
In a geometric progression, each term multiplies by the same ratio.
在等比数列里，每一项都乘以同一个公比。
If that ratio is smaller than one, the terms shrink toward nothing, and the endless sum settles on a finite total — watch the pieces fill the square.
如果这个公比小于一，各项就趋向于零，而这无穷的和会稳定在一个有限的总量上—— 看这些小块填满正方形。
There is also the binomial expansion, which powers up a bracket using the numbers of Pascal's triangle.
此外还有二项展开式，它用帕斯卡三角形里的数，把一个括号展开成幂。
The binomial expansion powers up a bracket without multiplying it out by hand.
二项展开式能把一个括号升幂展开，而不用手动去乘。
Bracket a plus b, all to the n, equals a to the n, plus n choose one, a to the n minus one, times b, plus n choose two, a to the n minus two, b squared, and so on.
括号 a 加 b 的 n 次方， 等于 a 的 n 次方，加 n 选一 乘 a 的 n 减一次方 乘 b，加 n 选二 乘 a 的 n 减二次方 乘 b 平方，依此类推。
The powers of a fall while the powers of b rise, and in every single term they add up to n.
a 的指数在下降，b 的指数在上升，而在每一项里它们加起来都等于 n。
The coefficient n choose r is n factorial over r factorial times n minus r factorial — and it is exactly the number you would read off row n of Pascal's triangle, which is why those two methods always agree.
系数 n 选 r 等于 n 的阶乘除以 r 的阶乘乘以 n 减 r 的阶乘—— 它正是你从帕斯卡三角形第 n 行读到的那个数。
Find the first three terms of bracket two minus p x, all to the fifth, in ascending powers of x.
求 括号 二减 p x 的五次方 按 x 的升幂排列的前三项。
Here a is two and b is minus p x — carry that minus sign, because it is where marks are lost.
这里 a 是二，b 是负 p x—— 一定要把这个负号带着，因为分就丢在这里。
Term one: two to the fifth is thirty-two.
第一项:二的五次方是三十二。
Term two: five, times two to the fourth, times minus p x, which gives minus eighty p x.
第二项:五乘二的四次方乘 负 p x，得到负八十 p x。
Term three: ten, times two cubed, times minus p x all squared.
第三项:十乘二的三次方乘 负 p x 的平方。
Squaring kills the minus sign, so this one is plus eighty p squared x squared.
平方把负号消掉了， 所以这一项是正八十 p 平方 x 平方。
So the answer is thirty-two, minus eighty p x, plus eighty p squared x squared.
所以答案是三十二减八十 p x 加八十 p 平方 x 平方。
There are two families of sequence.
数列有两大家族。
An arithmetic progression adds the same common difference at every step, so the nth term is a plus n minus one lots of d, and the sum of n terms is n over two, times two a plus n minus one d.
等差数列每一步都加上同一个公差，所以第 n 项是 a 加 n 减一 个 d， 前 n 项和是 n 除以二，乘以 二 a 加 n 减一 个 d。
A geometric progression multiplies by the same common ratio at every step, so the nth term is a times r to the n minus one, and the sum of n terms is a, times one minus r to the n, all over one minus r.
等比数列每一步都乘以同一个公比， 所以第 n 项是 a 乘 r 的 n 减一次方，前 n 项和是 a 乘 括号 一减 r 的 n 次方， 整个除以 一减 r。
Notice that both nth terms use n minus one, not n — the first term has not stepped yet.
注意两个通项公式用的都是 n 减一，而不是 n—— 第一项还没有走过任何一步。
When does an endless sum have a finite answer? Only when the terms shrink — that is, when the size of r is less than one.
无穷多项相加，什么时候才有有限的答案?只有当各项在缩小时——也就是 r 的大小小于一。
Then the progression is convergent — it converges — and the sum to infinity is a over one minus r.
这时数列收敛，无穷和等于 a 除以 一减 r。
Look at why: r to the n dies away to zero, so the bracket in the sum formula closes on one.
看看为什么:r 的 n 次方衰减到零， 于是求和公式里的括号趋近于一。
If the size of r is one or more, the terms do not shrink, and the sum runs away.
如果 r 的大小大于或等于一，各项就不缩小，和会跑掉。
Always check the condition before you use the formula, because stating it carries a mark.
用公式前一定要检查这个条件，因为写出它本身就有分。
A two-star exam question.
一道两星考题。
The third term of a geometric progression is eighteen.
等比数列的第三项是十八。
The sum of the first three terms is twenty-six.
前三项之和是二十六。
The common ratio is negative.
公比是负的。
Find the sum to infinity.
求无穷和。
From the third term, a r squared is eighteen, so a is eighteen over r squared.
由第三项，a r 平方等于十八，所以 a 等于十八除以 r 平方。
The sum of three terms is a, times one plus r plus r squared.
前三项之和是 a 乘 括号 一加 r 加 r 平方。
Substitute, multiply through by r squared, and tidy: eight r squared, minus eighteen r, minus eighteen, equals zero.
代入，两边乘以 r 平方，再整理: 八 r 平方减十八 r 减十八等于零。
Factorise: four r plus three, times r minus three.
分解:四 r 加三 乘 r 减三。
The ratio is negative, so r is minus three quarters, and a is thirty-two.
公比是负的，所以 r 等于负四分之三，而 a 等于三十二。
The sum to infinity is thirty-two over seven quarters, which is one hundred and twenty-eight over seven.
无穷和是三十二除以四分之七，等于七分之一百二十八。
Now the first engine of calculus: differentiation.
现在是微积分的第一台引擎：微分。
It measures the steepness of a curve at a single point.
它量的是曲线在某一点的陡峭程度。
The idea is to draw a chord between two points, then slide them closer and closer, until the chord becomes the tangent.
思路是：在两点之间画一条割线，然后让它们越靠越近，直到割线变成切线。
That limit is the gradient, the derivative.
那个极限就是斜率，也就是导数。
From it, the whole curve reveals how fast it is changing, everywhere.
有了它，整条曲线都会显露出它在每一处变化得有多快。
Here are the rules you will use every day.
这些是你每天都会用到的法则。
Bring the power down and drop it by one: the derivative of x to the n is n x to the n minus one.
把指数拿下来，再减一:x 的 n 次方的导数是 n 乘 x 的 n 减一次方。
That holds for any rational power, including fractions and negatives.
对任意有理次幂都成立，包括分数和负数。
Differentiate a sum one term at a time.
求和的导数就一项一项地求。
For a function inside a function, use the chain rule: differentiate the outside, keep the inside, then multiply by the derivative of the inside.
对于函数里面套函数，用链式法则:先对外层求导，里面保持不动，再乘以内层的导数。
And if you differentiate twice, you get the second derivative, written d two y by d x squared.
如果再求一次导，你得到二阶导数，写成 d 二 y 比 d x 平方。
The derivative has two immediate uses.
导数有两个立刻能用的用途。
First, the gradient of the curve at a point is the gradient of the tangent there, so put the x value into the derivative to get m, then use the point-gradient form to write the tangent's equation.
第一，曲线在某点的斜率，就是那里切线的斜率， 所以把 x 值代进导数求出 m，再用点斜式写出切线方程。
The normal is the line at right angles to the tangent at that same point, so its gradient is minus one over m.
法线是在同一点与切线成直角的直线， 所以它的斜率是 负一 除以 m。
Second, the sign of the derivative tells you which way the curve is going: positive means the function is increasing, negative means it is decreasing.
第二，导数的符号告诉你曲线往哪走: 正号表示函数在递增，负号表示它在递减。
A derivative is a rate of change, so calculus escapes geometry the moment time appears.
导数就是变化率，所以一旦时间出现，微积分就走出了几何。
If a balloon is being inflated, its radius grows at one rate per second and its volume grows at another.
如果一个气球正在被吹大， 它的半径每秒以某个速率增长，而体积以另一个速率增长。
The chain rule links them: d V by d t equals d V by d r, times d r by d t.
链式法则把它们连起来: d V 比 d t 等于 d V 比 d r 乘以 d r 比 d t。
Read it as a chain — write down the rate you are given, the rate you want, and the derivative that connects them.
把它当成一条链子来读—— 写下题目给你的那个速率、你想求的那个速率，以及把它们连起来的那个导数。
Then multiply.
然后相乘。
Units are worth a mark here, so keep them: centimetres cubed per second, not just a bare number.
这里单位是有分的，所以别丢掉。
The derivative is your key to the shape of a graph.
导数是你读懂图像形状的钥匙。
Where the gradient is zero, the curve is momentarily flat — a stationary point.
斜率为零的地方，曲线一瞬间是平的——这就是驻点。
That is where you find the peaks and the valleys, the maximum and the minimum.
峰与谷、极大值与极小值，都在那里。
To tell which is which, look at the second derivative: curving up gives a minimum, curving down gives a maximum.
要分辨哪个是哪个，就看二阶导数： 向上弯是极小值，向下弯是极大值。
Let's find a turning point properly.
我们来正规地求一个转折点。
The curve is y equals four root x, minus x, and it has a maximum.
曲线是 y 等于四根号 x 减 x，它有一个极大值。
Differentiate: four times a half is two, and the power drops to minus a half, so the derivative is two over root x, minus one.
求导:四乘二分之一是二，指数降到负二分之一，所以导数是 二比根号 x 再减一。
Set that equal to zero.
让它等于零。
Two over root x equals one, so root x equals two, and x equals four.
二比根号 x 等于一，所以根号 x 等于二，x 等于四。
That is the value asked for.
这就是所求的值。
To confirm it is a maximum, differentiate again — the second derivative is negative here, so the curve bends downwards.
要确认它是极大值，再求一次导——这里二阶导数是负的，所以曲线向下弯。
And if the second derivative had come out zero, the test fails: check the sign of the gradient just before and just after, because the point may be an inflexion.
如果二阶导数算出来是零，这个判别法就失效了:检查驻点前后一点点的斜率符号， 因为这个点可能是拐点。
The second engine is integration — the exact reverse of differentiation.
第二台引擎是积分——微分的精确逆运算。
Its great gift is area.
它最大的馈赠是面积。
To measure the space under a curve, slice it into thin rectangles, add them up, then make the slices thinner and thinner.
要量出一条曲线下方的空间，就把它切成一根根细长的矩形，加起来， 再让切片越来越细。
The sum becomes the integral.
这个和就变成了积分。
It turns a rough estimate into an exact answer, and it is how we find areas, and volumes of revolution.
它把粗略的估计变成精确的答案， 我们也正是用它来求面积，以及旋转体的体积。
Integration reverses differentiation, so the power rule runs backwards: raise the power by one, then divide by the new power.
积分是微分的逆运算，所以幂法则反过来跑:指数加一，再除以新的指数。
For a linear bracket there is a bonus shortcut: the integral of bracket a x plus b, to the n, is that bracket to the n plus one, divided by a times n plus one.
对于线性括号，还有一个额外的捷径:括号 a x 加 b 的 n 次方的积分， 等于这个括号的 n 加一 次方，除以 a 乘 n 加一。
You divide by the a as well, because the chain rule put it there.
你还要除以那个 a， 因为它是链式法则放进来的。
This fails when n is minus one, since dividing by zero is not allowed.
当 n 等于负一时这条法则失效，因为不能除以零。
And every indefinite integral needs plus C.
而且每个不定积分都要加常数 C。
If the question gives you one point on the curve, substitute it to pin C down.
如果题目给了曲线上的一个点，代进去就能把 C 定下来。
Add limits and the integral becomes a number.
加上上下限，积分就变成一个数。
Integrate, put the result in square brackets, then substitute the top limit and subtract the bottom limit.
先积分，把结果放进方括号， 再代入上限并减去下限的值。
No plus C is needed here — it cancels.
这里不需要加常数 C——它会互相抵消。
The area between a curve and the x axis, from p to q, is the integral of y with respect to x.
曲线与 x 轴之间、从 p 到 q 的面积，就是 y 对 x 的积分。
For the region between two curves, integrate top curve minus bottom curve, between the x values where they cross.
两条曲线之间的区域，就积分 上面的曲线减下面的曲线，范围取在它们相交的两个 x 值之间。
And a warning: where the curve dips below the axis, the integral comes out negative, so split the region at the crossing point.
还有一个警告:曲线落到横轴以下的地方，积分算出来是负的，所以要在交点处把区域分开。
Same curve as before, now for its area.
还是刚才那条曲线，现在求它的面积。
Integrate four root x minus x, from zero to four.
把 四根号 x 减 x 从零积到四。
Four root x is four x to the half, so raise the power to three halves and divide by three halves, which gives eight thirds x to the three halves.
四根号 x 就是 四 x 的二分之一次方，所以指数升到二分之三，再除以二分之三， 得到三分之八 x 的二分之三次方。
Minus x integrates to minus x squared over two.
减 x 积分成 减 x 平方除以二。
Now the limits.
现在代上下限。
At x equals four, root x is two, so x to the three halves is eight; eight thirds of eight is sixty-four over three, and sixteen over two is eight.
在 x 等于四时，根号 x 是二，所以 x 的二分之三次方是八;八的三分之八是三分之六十四， 而十六除以二是八。
At x equals zero everything vanishes.
在 x 等于零时一切都为零。
So sixty-four over three minus eight is forty over three.
所以三分之六十四减八，等于三分之四十。
The curve stays above the axis all the way from zero to four, so that number really is the area.
从零到四曲线始终在横轴上方，所以这个数就是面积。
Some integrals look impossible and still give a finite answer.
有些积分看起来不可能，却仍然给出有限的答案。
An improper integral either runs to infinity, or has an endpoint where the function blows up.
广义积分要么积到无穷， 要么在端点处函数发散。
Handle it as a limit.
把它当成一个极限来处理。
Take the integral of one over x squared, from one to infinity.
取 一除以 x 平方 从一积到无穷。
It integrates to minus one over x; evaluate up to b and you get one minus one over b.
它积分成 负一除以 x;代到 b，得到 一减一比 b。
Now let b grow without bound: the answer is exactly one.
现在让 b 无限增大:答案恰好是一。
An infinite region with a finite area.
一个无穷长的区域，却有有限的面积。
Similarly, one over root x from zero to one integrates to two, even though the curve shoots up at zero.
同样地，一比根号 x 从零到一积分得到二， 尽管曲线在零处冲向无穷。
Finally, spin a region and it sweeps out a solid.
最后，把一个区域旋转一圈，它会扫出一个立体。
Slice the solid across: each slice is a thin disc of radius y, so its face has area pi y squared.
把这个立体横着切开: 每一片都是半径为 y 的薄圆盘，所以它的截面面积是 派 y 平方。
Add the discs, and the volume about the x axis is pi times the integral of y squared, d x.
把这些圆盘加起来， 绕 x 轴的体积就是 派 乘以 y 平方 对 x 的积分。
Rotating about the y axis instead, it is pi times the integral of x squared, d y.
如果改成绕 y 轴， 就是 派 乘以 x 平方 对 y 的积分。
Try it on the region under root x from zero to four.
拿根号 x 从零到四下方的区域试一下。
Y squared is just x, so the volume is pi times the integral of x, which is eight pi. Square y first, then integrate — never the other way round.
y 平方就是 x，所以体积是 派 乘以 x 的积分，等于八派。
Before you go, four ways to protect your marks.
结束之前，四个保住分数的办法。
First, show every line of algebra — jumping steps loses method marks — and use the discriminant to count the roots.
第一，每一步代数都要写出来—— 跳步会丢方法分——并用判别式来数根的个数。
Second, always work in radians for arcs, and for the calculus of trig functions.
第二，涉及弧长以及三角函数的微积分， 一律用弧度。
Third, for turning points, set the derivative to zero, then classify with the second derivative.
第三，求转折点时，令导数为零，再用二阶导数分类。
Fourth, in integration, never forget the plus c, and use limits for area — remembering that area below the axis comes out negative.
第四，做积分时，千万别忘了加常数 C，求面积要用上下限——记住横轴下方的面积算出来是负的。