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→∞ (√(4x2 − 10x) − 2x).
f(x) = √(4x2 − 10x) − 2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 4x)/(3x²), with x in radians.
f(x) = (1 − cos 4x)/(3x²)x → 0
L =
3.DerivativesMedium
Let f(x) = (3x − 1)4. Find f′(0).
f(x) = (3x − 1)4x = 0
f′(0) =
4.DerivativesHard
The tangent line to f(x) = −x2 − 4x + 4 at x = 2 crosses the y-axis at y = b. Find b.
f(x) = −x2 − 4x + 4x = 2
b =
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 = 3∠−60° and z₂ = 0.5∠65°, find the argument θ of w = z₁/z₂ in degrees, between −180° and 180°.
z₁ = 3∠−60°z₂ = 0.5∠65°
θ = °
8.Complex numbersHard
Use De Moivre's theorem to find the real part of z⁹ for z = 2 + j2.
z = 2 + j2n = 9
Re(z⁹) =
9.Differential equationsMedium
A can of soda at 5 °C is taken out of the fridge into a room at 27 °C. After 19 min its temperature is 18 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 5 °CTa = 27 °CT(19) = 18 °C
k = min⁻¹
10.Differential equationsHard
Solve y″ + y′ + 25.25y = 0 with y(0) = 3 and y′(0) = 0 (released from rest). Find y(0.5); angles in radians.
y″ + y′ + 25.25y = 0y(0) = 3y′(0) = 0
y(0.5) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [5 −2 5; −5 5 −5; 4 5 −3]
det A =
12.Matrices & determinantsHard
Solve the system −3x − 4y = 11 and −5x + 3y = 7. Find x.
−3x − 4y = 11−5x + 3y = 7
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (5, 0, −3) and b = (1, 3, 4), in degrees.
a = (5, 0, −3)b = (1, 3, 4)
θ = °
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−2, −3, −5) and b = (0, −3, −2) and has a positive z-component. Find its x-component n̂x.
a = (−2, −3, −5)b = (0, −3, −2)n̂z > 0
n̂x =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2, 3p(x): 0.25, 0.25, 0.2, 0.3
σ =
16.ProbabilityHard
A game costs €1 to play. You win €5 with probability 0.1 and €20 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €1€5: p = 0.1€20: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 55, 42, 31, 56, 42, 19, 40.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 75. What proportion P of values is above 387.5?
μ = 500σ = 75x = 387.5
P =
19.Normal distributionMedium
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 10 of them exceeds 251.6 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 10> 251.6 kg
P =
20.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 80% of them lie (the 80th percentile).
μ = 178σ = 7P = 80%
x = cm
3Solutions
1.Limits
Answer: L = −2.5
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).
Multiply by the conjugate
f(x) = −10x/(√(4x2 − 10x) + 2x)
Divide by x
f(x) = −10/(√(4 + (−10)/x) + 2)
Let x → ∞
L = −10/(2 + 2) = −2.5
Answer
L = −2.5
2.Limits
Answer: L = 2.667
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
4·sin 4x/(6x) → 0/0
Second application
16·cos 4x/6
Substitute x = 0
L = 16/6 = 2.667
Answer
L = 2.667
3.Derivatives
Answer: f′(0) = −12
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.
Chain rule
f′(x) = 4·(3x − 1)3·3
Substitute
f′(0) = 4·(−1)3·3 = −12
Answer
f′(0) = −12
4.Derivatives
Answer: b = 8
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.
Point and slope
f(2) = −8
f′(x) = −2x − 4 → f′(2) = −8
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −8 − (−8)·2 = 8
Answer
b = 8
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: θ = −125 °
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| = 3 / 0.5 = 6
θ = −60° − 65° = −125°
Argument
θ = −125°
Answer
θ = −125°
8.Complex numbers
Answer: Re(z⁹) = 8192
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 = 2 + j2 = 2√2∠45°
De Moivre
z⁹ = (2√2)⁹∠405° = 11585∠45°
Rectangular form
z⁹ = 11585·(cos 45° + j sin 45°) = 8192 + j8192
Re(z⁹) = 8192
Answer
Re(z⁹) = 8192
9.Differential equations
Answer: k = 0.04704 min⁻¹
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.
Ratio of temperature differences
e−19k = -9/-22
Solve for k
k = ln(-22/-9)/19 = 0.04704 min⁻¹
Answer
k = 0.04704 min⁻¹
10.Differential equations
Answer: y(0.5) = −1.732
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.
Complex roots
r = −0.5 ± 5j
Constants
A = 3, B = 0.5·3/5 = 0.3
Evaluate
y(0.5) = e−0.25(3·cos 2.5 + 0.3·sin 2.5) = −1.732
Answer
y(0.5) = −1.732
11.Matrices & determinants
Answer: det A = −105
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 = 10, M₁₂ = 35, M₁₃ = −45
Cofactor expansion
det A = 5·10 − (−2)·35 + 5·(−45) = −105
Answer
det A = −105
12.Matrices & determinants
Answer: x = −2.103
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 = (−3)·3 − (−4)·(−5) = −29
Replace the x column
det Aₓ = 11·3 − (−4)·7 = 61
x = 61/(−29) = −2.103
Answer
x = −2.103
y = −1.172
13.Vectors: dot & cross product
Answer: θ = 103.6 °
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 = 5·1 + 0·3 + (−3)·4 = −7
Magnitudes
|a| = √(5² + 0² + (−3)²) = 5.831
|b| = √(1² + 3² + 4²) = 5.099
Angle
cos θ = −7/(5.831·5.099) = −0.2354
θ = arccos(−0.2354) = 103.6°
Answer
θ = 103.6°
14.Vectors: dot & cross product
Answer: n̂x = −0.7804
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.
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.
Mean
μ = 0·0.25 + 1·0.25 + 2·0.2 + 3·0.3 = 1.55
Mean of the squares
E(X²) = 0·0.25 + 1·0.25 + 4·0.2 + 9·0.3 = 3.75
Variance
Var(X) = 3.75 − 1.55² = 1.348
σ = √1.348 = 1.161
Answer
σ = 1.161
16.Probability
Answer: E(G) = −0.3 €
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: 4 (p = 0.1), 19 (p = 0.01), −1 (p = 0.89)
Expected value
E(G) = 4·0.1 + 19·0.01 − 1·0.89 = −0.3 €
Answer
E(G) = −0.3 €
A negative value means you lose money on average.
17.Descriptive statistics
Answer: s = 12.96
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 = −1.5
Standard normal table
Φ(−1.5) = 0.06681
P = 1 − 0.06681 = 0.9332
Answer
P = 0.9332
19.Normal distribution
Answer: P = 0.1562
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the total
μT = 250 kg, σT = 0.5·√10 = 1.581 kg
Standardise
z = (251.6 − 250)/1.581 = 1.01
Probability
P = 1 − Φ(1.01) = 0.1562
Answer
P = 0.1562
20.Normal distribution
Answer: x = 183.9 cm
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.