Matemáticas Puras Avanzadas 3
Pearson Edexcel · International A-Level · Further Mathematics Tema 3 8:08 Narración en inglés · Subtítulos en inglés + 中文 quemados en pantalla
Capítulos
Transcripción
Paper WFM03: 1 hour 30 minutes, 75 marks, the last pure unit, with P1 to P4 and FP1 assumed.
WFM03 试卷：1 小时 30 分钟，75 分，是最后一个纯数单元，默认已学 P1 到 P4 和 FP1。
It adds hyperbolic functions and their inverses, the ellipse and hyperbola, reduction formulae, arc length and surfaces, vector products with lines and planes, and matrices up to eigenvalues.
它增加了双曲函数及其反函数、椭圆和双曲线、递推公式、弧长和旋转曲面、向量积与直线和平面，以及到特征值为止的矩阵。
The hyperbolic functions are built from exponentials.
双曲函数由指数函数构成。
cosh is the catenary: even, with its minimum at 0, 1.
cosh 是悬链线：偶函数，最小值在 (0, 1)。
sinh is odd, through the origin.
sinh 是奇函数，过原点。
And tanh, sinh over cosh, is trapped between the lines y equals plus and minus 1.
tanh 等于 sinh/cosh，被夹在直线 y = ±1 之间。
Show that cosh 6x plus cosh 4x equals 2 cosh 5x cosh x.
证明 cosh 6x + cosh 4x = 2cosh 5x cosh x。
Write both in exponentials, and pair e to the 6x with e to the 5x times e to the x.
都写成指数形式，把 e^{6x} 与 e^{5x}·e^{x} 配对。
Then cosh 6x plus cosh 4x equals 8 cosh x collapses to cosh 5x equals 4, because cosh is never zero: x is plus or minus a fifth of ln of 4 plus root 15.
于是方程 cosh 6x + cosh 4x = 8cosh x 化为 cosh 5x = 4，因为 cosh 从不为零：x = ±(1/5)ln(4 + √15)。
Solve 2 cosh x plus sinh x equals 2.
解 2cosh x + sinh x = 2。
The definitions give 3 e to the x plus e to the minus x equals 4, a quadratic in e to the x: 3 e to the 2x minus 4 e to the x plus 1 equals 0.
由定义得 3e^x + e^{−x} = 4，是关于 e^x 的二次方程：3e^{2x} − 4e^x + 1 = 0。
It factors, so x is 0 or minus ln 3.
因式分解，所以 x = 0 或 x = −ln 3。
sinh and tanh are one-to-one, and cosh is one-to-one for x at least 0, so each has an inverse, and each inverse has a logarithmic form you must be able to derive.
sinh 和 tanh 是一一对应的，cosh 在 x ≥ 0 时一一对应，所以它们都有反函数，每个反函数都有一个你必须会推导的对数形式。
Set w equal to sinh x.
令 w = sinh x。
Multiplying by 2 e to the x gives a quadratic: e to the 2x minus 2w e to the x minus 1 equals 0.
乘以 2e^x 得到二次方程：e^{2x} − 2we^x − 1 = 0。
The negative root is rejected because e to the x is positive, so x is ln of w plus root w squared plus 1.
因为 e^x 为正，舍去负根，所以 x = ln(w + √(w² + 1))。
The ellipse has parameters a cos t, b sin t; the hyperbola, a sec t, b tan t.
椭圆的参数式是 (a cos t, b sin t)；双曲线是 (a sec t, b tan t)。
The eccentricity e ties both to the focus-directrix property: the distance to the focus is e times the distance to the directrix.
离心率 e 把两者都联系到焦点–准线性质：到焦点的距离等于 e 乘到准线的距离。
Only the b squared formula differs.
只有 b² 的公式不同。
For the hyperbola x squared over 64 minus y squared over 49 equals 1: a squared is 64 and b squared is 49, so 49 equals 64 times e squared minus 1.
对双曲线 x²/64 − y²/49 = 1：a² = 64，b² = 49，所以 49 = 64(e² − 1)。
e is root 113 over 8, and the foci sit at plus and minus root 113, inside the branches, between the asymptotes.
e = √113/8，焦点在 ±√113，位于两支内侧、渐近线之间。
At P, 8 sec theta, 7 tan theta, differentiate parametrically.
在 P(8sec θ, 7tan θ) 处用参数求导。
The tangent meets the y-axis at Q, 0, minus 7 cot theta; the normal meets it at R, 0, 113 over 7 tan theta.
切线与 y 轴交于 Q(0, −7cot θ)；法线交于 R(0, (113/7)tan θ)。
The circle on diameter QR meets the x-axis where x squared is minus q r, which is 113 for every theta: both foci lie on it.
以 QR 为直径的圆与 x 轴交于 x² = −qr 处，对每个 θ 都等于 113：两个焦点都在圆上。
Substitute y equals m x plus c into the conic and force a zero discriminant.
把 y = mx + c 代入圆锥曲线，令判别式为零。
For the ellipse, c squared equals a squared m squared plus b squared.
椭圆：c² = a²m² + b²。
For the hyperbola it is minus b squared, so the gradient must be steeper than the asymptotes.
双曲线是减 b²，所以斜率必须比渐近线更陡。
At a parametric point, just differentiate.
在参数点上，直接求导即可。
sinh differentiates to cosh, and cosh to sinh, with no minus sign.
sinh 的导数是 cosh，cosh 的导数是 sinh，没有负号。
tanh gives sech squared.
tanh 的导数是 sech²。
The inverses give 1 over root x squared plus 1, 1 over root x squared minus 1, and 1 over 1 minus x squared.
反函数的导数分别是 1/√(x² + 1)、1/√(x² − 1) 和 1/(1 − x²)。
Differentiate arsinh x plus arsinh of 1 over x, for x positive.
对 x > 0 求 arsinh x + arsinh(1/x) 的导数。
The chain rule gives the second term, and x positive turns root x squared into x.
链式法则给出第二项，x 为正使 √(x²) 变成 x。
The derivative is x minus 1 over x root 1 plus x squared, zero at x equals 1, where y is 2 ln of 1 plus root 2.
导数是 (x − 1)/(x√(1 + x²))，在 x = 1 处为零，此时 y = 2ln(1 + √2)。
For 1 over root 4 x squared plus 8x plus 9, complete the square: 4 times x plus 1 squared plus five quarters, which gives a half arsinh of 2x plus 2 over root 5.
对 1/√(4x² + 8x + 9)，先配方：4 ，得到 (1/2)arsinh((2x + 2)/√5)。
For arcosh 3x, integrate by parts with u the inverse function and 1 as the rest.
对 arcosh 3x，用分部积分，u 取反函数，其余取 1。
Four integrals to know cold.
四个必须熟记的积分。
1 over root x squared plus a squared gives arsinh; with minus a squared, arcosh.
1/√(x² + a²) 得 arsinh；减 a² 时得 arcosh。
Without the root, x squared plus a squared gives arctan, and x squared minus a squared gives a logarithm.
没有根号时，x² + a² 得 arctan，x² − a² 得对数。
Take u as x to the n and v as two thirds root 3x minus 2.
取 u = xⁿ，v = (2/3)√(3x − 2)。
Writing the leftover root as 3x minus 2 over the root exposes I n and I n minus 1, so 3 plus 6n times I n equals 4n I n minus 1 plus 8 times 6 to the n minus 2.
把剩下的根式写成 (3x − 2)/√(3x − 2)，就出现了 Iₙ 和 Iₙ₋₁，所以 (3 + 6n)Iₙ = 4nIₙ₋₁ + 8·6ⁿ − 2。
From I 1 equals 6, 15 I 2 is 334, and I 3 is 3322 over 35.
由 I₁ = 6 得 15I₂ = 334，I₃ = 3322/35。
Rotating y equals cos 2x, from 0 to pi over 4, through 2 pi about the x-axis sweeps a surface.
把 y = cos 2x 从 0 到 π/4 绕 x 轴旋转 2π，得到一个曲面。
Its area is 2 pi times the integral of y root 1 plus y dash squared.
它的面积是 2π 乘 y√(1 + y′²) 的积分。
With 2 sin 2x equal to sinh theta, the cos 2x factors cancel exactly, leaving pi over 2 times the integral of cosh squared theta.
令 2sin 2x = sinh θ，cos 2x 的因子恰好抵消，剩下 (π/2)∫cosh²θ dθ。
Between 0 and arsinh 2, that is pi over 4 times 2 root 5 plus ln of 2 plus root 5.
在 0 到 arsinh 2 之间，结果是 (π/4)(2√5 + ln(2 + √5))。
The vector product is perpendicular to both factors, and its length is the parallelogram area.
向量积垂直于两个因子，它的长度是平行四边形的面积。
The scalar triple product is a volume, zero when the vectors are coplanar.
混合积是体积，向量共面时为零。
And a point's distance to a plane is the modulus of x dot n minus p, over the length of n.
点到平面的距离是 |x·n − p| 除以 n 的长度。
The plane's normal is the cross product of its directions, 2 times 8, 2, 11, so the plane is r dot 8, 2, 11 equals 25.
平面的法向量是其两个方向向量的叉积，等于 2(8, 2, 11)，所以平面是 r·(8, 2, 11) = 25。
Its line of meeting with x minus y plus z equals 7 has direction 13, 3, minus 10.
它与 x − y + z = 7 的交线方向是 (13, 3, −10)。
Projecting the offset onto the common normal gives the shortest distance, 11 over root 434.
把两点之差投影到公共法向量上，得最短距离 11/√434。
A v equals lambda v holds row by row: each component of A v is lambda times the matching component of v, not lambda itself.
Av = λv 逐行成立：Av 的每个分量等于 λ 乘 v 的对应分量，而不是 λ 本身。
One row gives lambda, the other rows give the unknowns, and the trace and determinant give the remaining eigenvalues.
一行给出 λ，其余行给出未知数，迹和行列式给出其余特征值。
With eigenvector 1, 2, 1: row 1 gives 3, so lambda is 3.
特征向量是 (1, 2, 1)：第一行给出 3，所以 λ = 3。
Rows 2 and 3 must give 6 and 3: a is 2 and b is minus 1.
第二、三行必须给出 6 和 3：a = 2，b = −1。
The trace is 2 and the determinant minus 6, so the other two eigenvalues sum to minus 1 and multiply to minus 2: they are 1 and minus 2.
迹是 2，行列式是 −6，所以另外两个特征值之和为 −1、之积为 −2：它们是 1 和 −2。
Expand along row 1: the determinant is minus 36 minus 32k, so A is singular when k is minus 9 over 8.
沿第一行展开：行列式是 −36 − 32k，所以 k = −9/8 时 A 奇异。
Otherwise the inverse is the adjugate over the determinant, and the first row times the first column reproduces the determinant: a free check.
否则逆矩阵是伴随矩阵除以行列式，第一行乘第一列又得到行列式：一个免费的检验。
Check yourself.
自我检测。
cosh differentiates to sinh.
cosh 的导数是 sinh。
arsinh x is ln of x plus root x squared plus 1.
arsinh x = ln(x + √(x² + 1))。
For the hyperbola, b squared is a squared times e squared minus 1.
双曲线中 b² = a²(e² − 1)。
A zero triple product means coplanar.
混合积为零表示共面。
And the eigenvalues sum to the trace and multiply to the determinant.
特征值之和等于迹，之积等于行列式。