Decide which form suits the operation: rectangular a + jb for adding and subtracting, polar r∠θ for multiplying, dividing, powers and roots.
Multiply out with j² = −1; divide by multiplying the top and bottom by the conjugate of the denominator.
Convert with r = √(a² + b²) and θ from a sketch of the point (check the quadrant), and back with a = r cos θ, b = r sin θ.
For powers use De Moivre, (r∠θ)ⁿ = rⁿ∠nθ; for zⁿ = R∠φ list all n roots R1/n∠((φ + 360°k)/n).
Differential equations
Exponential growth/decay
dy/dt = ky → y = y0ekt
Half-life
k = ln 2 / t½
Newton's law of cooling
T(t) = Ta + (T0 − Ta)e−kt
Integrating factor
y′ + p(x)y = q(x): μ = e∫p dx, (μy)′ = μq
Mixing tank
dm/dt = r·c − (r/V)·m
Matrices & determinants
2×2 determinant
det = ad − bc
2×2 inverse
A⁻¹ = (1/det)·[d −b; −c a]
Product
(AB)ij = Σk AikBkj
Eigenvalues
det(A − λI) = 0
Cramer’s rule
xi = det(Ai)/det(A)
Vectors: dot & cross product
Magnitude
|a| = √(ax² + ay² + az²)
Unit vector
û = a/|a|
Components from an angle
Fx = F·cos θ, Fy = F·sin θ
Dot product
a·b = axbx + ayby + azbz = |a||b|cos θ
Scalar projection of a onto b
a·b/|b|
Work
W = F·d
Cross product
a × b = (aybz − azby, azbx − axbz, axby − aybx)
Areas
parallelogram |a × b|, triangle ½|a × b|
Moment about O
MO = r × F, r = A − O
Volume of a parallelepiped
V = |a·(b × c)|
Method
Write every vector in components; for a force given by a magnitude and an angle, use F·cos θ and F·sin θ.
For lengths and directions use the magnitude √(x² + y² + z²) and the unit vector a/|a|.
Use the dot product for angles, projections, perpendicularity and work: a·b = |a||b|cos θ.
Use the cross product for perpendicular vectors, areas and moments: write the three components carefully, keeping the order a × b.
Probability
Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k
Descriptive statistics
Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ
Normal distribution
z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ
Method
Convert the value to a z-score: z = (x − μ)/σ.
Look up Φ(z), the area to the left of z.
For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
For a percentile, work backwards: find z for the area, then x = μ + z·σ.
2Practice problems
1.LimitsMedium
Find the limit L = limx→∞ (5x2 − 3x + 3)/(5x2 + 2x + 1).
f(x) = (5x2 − 3x + 3)/(5x2 + 2x + 1)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 2x)/(4x²), with x in radians.
f(x) = (1 − cos 2x)/(4x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x2·ex. Find f′(1).
f(x) = x2·exx = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 − 6x2 + 3. Find the x-coordinate x of the local minimum.
f(x) = x3 − 6x2 + 3
x =
5.IntegralsMedium
The region under y = √(2x) from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(2x)0 ≤ x ≤ 1
V =
6.IntegralsHard
Find the area A of the region between the parabola y = 1 − x² and the x-axis.
y = 1 − x²
A =
7.Complex numbersMedium
Given z₁ = 2∠70° and z₂ = 2∠−95°, find the real part of w = z₁·z₂.
z₁ = 2∠70°z₂ = 2∠−95°
Re(w) =
8.Complex numbersHard
Use De Moivre's theorem to find the imaginary part of z⁷ for z = 1 + j.
z = 1 + jn = 7
Im(z⁷) =
9.Differential equationsMedium
Solve y″ + 2y′ + y = 0 with y(0) = 1 and y′(0) = 0. Find y(1).
y″ + 2y′ + y = 0y(0) = 1y′(0) = 0
y(1) =
10.Differential equationsHard
A tank holds 400 L of water in which 5 kg of salt is dissolved. Brine with 0.1 kg of salt per litre flows in at 4 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 23 kg of salt?
V = 400 Lm0 = 5 kgc = 0.1 kg/Lr = 4 L/min
t = min
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [1 −1 −1; 1 −4 3; 4 5 1]
det A =
12.Matrices & determinantsHard
Solve the system −6x − 4y = −3 and 2x − y = 3. Find x.
−6x − 4y = −32x − y = 3
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (4, −4, −2) onto b = (5, −6, 2), that is, the component of a in the direction of b.
a = (4, −4, −2)b = (5, −6, 2)
compb a =
14.Vectors: dot & cross productHard
For a = (6, −3, 2) and b = (4, 4, −3), find the y-component (a × b)y of the cross product a × b.
a = (6, −3, 2)b = (4, 4, −3)
(a × b)y =
15.ProbabilityMedium
A bag holds 3 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
3 / 8
P =
16.ProbabilityHard
A game costs €2 to play. You win €8 with probability 0.3 and €100 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €2€8: p = 0.3€100: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 34, 38, 28, 21, 32, 51.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 15. What proportion P of values is below 79?
μ = 70σ = 15x = 79
P =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.7887. Find z to two decimals.
A = 0.7887
z =
20.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 5% of them lie (the 5th percentile).
μ = 500σ = 40P = 5%
x = h
3Solutions
1.Limits
Answer: L = 1
Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).
Divide by the highest power
(5 + (−3)/x + 3/x2)/(5 + 2/x + 1/x2)
Let x → ∞
L = 5/5 = 1
Answer
L = 1
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 0.5
Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).
First application
2·sin 2x/(8x) → 0/0
Second application
4·cos 2x/8
Substitute x = 0
L = 4/8 = 0.5
Answer
L = 0.5
3.Derivatives
Answer: f′(1) = 8.155
Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.
Product rule
f′(x) = u′v + uv′ = ex·(2x + 1·x2)
Substitute
f′(1) = 2.718·(2 + 1) = 8.155
Answer
f′(1) = 8.155
4.Derivatives
Answer: x = 4
Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.
Derivative
f′(x) = 3x2 − 12x = 3·(x2 − 4x)
Solve f′(x) = 0
x = 0 or x = 4
Which one is which
f″(x) = 6x − 12: negative at a maximum, positive at a minimum.
f″(0) = −12, f″(4) = 12
x = 4
Answer
x = 4
5.Integrals
Answer: V = 3.142
Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.
Disc method
V = π∫₀1 2x dx
Integrate
V = π·2·1²/2 = 3.142
Answer
V = 3.142
6.Integrals
Answer: A = 1.333
Definite integrals: ∫ₐᵇ f(x) dx is the signed area between the graph and the x-axis from a to b. Find an antiderivative F (with F′ = f) and evaluate F(b) − F(a). Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1). The area between two curves is ∫(upper − lower) dx.
Zeros
1 − x² = 0 → x = ±1
Integrate
A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333
Answer
A = 1.333
7.Complex numbers
Answer: Re(w) = 3.625
Polar form: z = r∠θ = r(cos θ + j sin θ) = r·ejθ, with r = |z| and θ = arg z. Back to rectangular form: a = r cos θ, b = r sin θ. The argument needs the quadrant: θ = atan(b/a) only when a > 0; for a < 0 add or subtract 180°. In polar form multiplication is easy: multiply the moduli and add the arguments; for division divide the moduli and subtract the arguments.
Modulus and argument of w
|w| = 2 · 2 = 4
θ = 70° + (−95)° = −25°
Rectangular form
Re(w) = 4·cos(−25°) = 3.625
Im(w) = 4·sin(−25°) = −1.69
Re(w) = 3.625
Answer
Re(w) = 3.625
8.Complex numbers
Answer: Im(z⁷) = −8
De Moivre and roots: (r∠θ)ⁿ = rⁿ∠nθ. The equation zⁿ = R∠φ has exactly n roots, all with modulus R1/n, at the arguments (φ + 360°·k)/n for k = 0, 1, …, n − 1: they lie evenly spaced on a circle, 360°/n apart. A quadratic z² + pz + q = 0 with p² < 4q has the complex roots z = −p/2 ± j√(q − p²/4); a polynomial with real coefficients always has its complex roots in conjugate pairs a ± jb.
Polar form
z = 1 + j = √2∠45°
De Moivre
z⁷ = (√2)⁷∠315° = 11.31∠−45°
Rectangular form
z⁷ = 11.31·(cos −45° + j sin −45°) = 8 − j8
Im(z⁷) = −8
Answer
Im(z⁷) = −8
9.Differential equations
Answer: y(1) = 0.7358
Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.
Double root
(r + 1)² = 0 → r = −1
Constants
A = y(0) = 1
y′(0) = B − 1A → B = 0 + 1 = 1
Evaluate
y(1) = (1 + 1·1)·e−1 = 0.7358
Answer
y(1) = 0.7358
10.Differential equations
Answer: t = 72.21 min
First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e∫p dx: (μy)′ = μq, so μy = ∫μq dx + C.
Result
m(t) = 40 + (−35)·e−t/100
Solve for t
e−t/100 = (40 − 23)/(40 − 5) = 0.4857
t = 100·ln(2.059) = 72.21 min
Answer
t = 72.21 min
11.Matrices & determinants
Answer: det A = −51
Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.
2×2 minors of the first row
M₁₁ = −19, M₁₂ = −11, M₁₃ = 21
Cofactor expansion
det A = 1·(−19) − (−1)·(−11) + (−1)·21 = −51
Answer
det A = −51
12.Matrices & determinants
Answer: x = 1.071
Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.
Determinant of the coefficient matrix
det A = (−6)·(−1) − (−4)·2 = 14
Replace the x column
det Aₓ = (−3)·(−1) − (−4)·3 = 15
x = 15/14 = 1.071
Answer
x = 1.071
y = −0.8571
13.Vectors: dot & cross product
Answer: compb a = 4.961
Dot product:a·b = axbx + ayby + azbz = |a|·|b|·cos θ, a scalar. It gives the angle between two vectors, cos θ = a·b/(|a|·|b|), and it is zero when they are perpendicular. The scalar projection of a onto b is a·b/|b|, and a constant force F moving through a displacement d does work W = F·d.
Dot product
a·b = 4·5 + (−4)·(−6) + (−2)·2 = 40
Length of b
|b| = √(5² + (−6)² + 2²) = 8.062
Scalar projection
compb a = a·b/|b| = 40/8.062 = 4.961
Answer
compb a = 4.961
14.Vectors: dot & cross product
Answer: (a × b)y = 26
Cross product:a × b = (aybz − azby, azbx − axbz, axby − aybx) is perpendicular to both vectors, and |a × b| = |a|·|b|·sin θ is the area of the parallelogram they span (the triangle has half of it). Order matters: b × a = −(a × b). The moment of a force F applied at point A about point O is MO = r × F, with r = A − O.
Cross product, component by component
(a × b)x = aybz − azby = (−3)·(−3) − 2·4 = 1
(a × b)y = azbx − axbz = 2·4 − 6·(−3) = 26
(a × b)z = axby − aybx = 6·4 − (−3)·4 = 36
Requested component
(a × b)y = 26
Answer
(a × b)y = 26
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: P = 0.05455
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
First and second draw
P = 3/11 · 2/10 = 6/110 = 0.05455
Answer
P = 0.05455
16.Probability
Answer: E(G) = 5.4 €
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Net outcomes
G: 6 (p = 0.3), 98 (p = 0.05), −2 (p = 0.65)
Expected value
E(G) = 6·0.3 + 98·0.05 − 2·0.65 = 5.4 €
Answer
E(G) = 5.4 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 10.14
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Standardise
z = 0.6
Standard normal table
Φ(0.6) = 0.7257
P = 0.7257
Answer
P = 0.7257
19.Normal distribution
Answer: z = 1.25
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.
Area to the left of z
Φ(z) = (1 + 0.7887)/2 = 0.8944
Read the z-table backwards
Find 0.8944 inside the table and read off z.
z = 1.25
Answer
z = 1.25
20.Normal distribution
Answer: x = 434.2 h
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.