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→∞ (1 + 0.5/x)3x. Give a decimal value.
f(x) = (1 + 0.5/x)3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 3x)/sin(x), with x in radians.
f(x) = ln(1 + 3x)/sin(x)x → 0
L =
3.DerivativesMedium
Let f(x) = x2·e0.5x. Find f′(2).
f(x) = x2·e0.5xx = 2
f′(2) =
4.DerivativesHard
The tangent line to f(x) = 0.5x2 + 5x − 4 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 + 5x − 4x = 1
b =
5.IntegralsMedium
The region under y = √(2x) from x = 0 to x = 2 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(2x)0 ≤ x ≤ 2
V =
6.IntegralsHard
Evaluate I = ∫₀2.5 x²·eˣ dx.
∫₀2.5 x²eˣ dx
I =
7.Complex numbersMedium
Find the argument θ of z = 5 − j7 in degrees, as the principal value between −180° and 180°.
z = 5 − j7
θ = °
8.Complex numbersHard
One root of z³ − 3z + 52 = 0 is z₁ = 2 + j3. The coefficients are real. Find the real root r.
z₁ = 2 + j3
r =
9.Differential equationsMedium
Solve y″ + 3y′ + 2y = 0 with y(0) = 5 and y′(0) = 0. Find y(2).
y″ + 3y′ + 2y = 0y(0) = 5y′(0) = 0
y(2) =
10.Differential equationsHard
A system obeys y″ + 4y′ + 20y = 40 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1.5); angles in radians.
y″ + 4y′ + 20y = 40y(0) = 0y′(0) = 0
y(1.5) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−1 2 −5; 3 −5 −1; 4 −3 −2]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−1 2; −5 6]
λ =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (−4, −2, −4) and b = (−3, 0, 1).
a = (−4, −2, −4)b = (−3, 0, 1)
a·b =
14.Vectors: dot & cross productHard
For a = (−1, 5, −6) and b = (−1, 0, 1), find the x-component (a × b)x of the cross product a × b.
a = (−1, 5, −6)b = (−1, 0, 1)
(a × b)x =
15.ProbabilityMedium
A bag holds 6 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
6 / 4
P =
16.ProbabilityHard
A game costs €3 to play. You win €15 with probability 0.2 and €60 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€15: p = 0.2€60: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 29, 54, 19, 60, 15, 10.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 12. What proportion P of values is above 54.8?
μ = 50σ = 12x = 54.8
P =
19.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 482 and 540 h? Give P as a decimal.
μ = 500σ = 40x1 = 482x2 = 540
P =
20.Normal distributionHard
Hole diameters D are normal with mean 25.02 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.02 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
3Solutions
1.Limits
Answer: L = 4.482
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).
Take the logarithm
ln L = limx→∞ 3·x·ln(1 + 0.5/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 3·0.5 = 1.5
Undo the logarithm
L = e1.5 = 4.482
Answer
L = 4.482
2.Limits
Answer: L = 3
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).
L'Hôpital's rule
L = limx→0 [3/(1 + 3x)]/[cos x]
Substitute x = 0
L = 3/1 = 3
Answer
L = 3
3.Derivatives
Answer: f′(2) = 16.31
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′ = e0.5x·(2x + 0.5·x2)
Substitute
f′(2) = 2.718·(4 + 2) = 16.31
Answer
f′(2) = 16.31
4.Derivatives
Answer: b = −4.5
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(1) = 1.5
f′(x) = x + 5 → f′(1) = 6
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 1.5 − 6·1 = −4.5
Answer
b = −4.5
5.Integrals
Answer: V = 12.57
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∫₀2 2x dx
Integrate
V = π·2·2²/2 = 12.57
Answer
V = 12.57
6.Integrals
Answer: I = 37.59
Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.
First round
∫x²eˣ dx = x²eˣ − ∫2x eˣ dx
Second round
∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)
Limits
12.18·(3.25) − 1·2 = 37.59
Answer
I = 37.59
7.Complex numbers
Answer: θ = −54.46 °
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.
Quadrant
a = 5, b = −7 → IV
Reference angle
α = atan(7/5) = 54.46°
Argument
θ = −α = −54.46°
Answer
θ = −54.46°
In polar form z = 8.602∠−54.46°.
8.Complex numbers
Answer: r = −4
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.
Check with the sum of the roots: z₁ + z₂ + r = 4 + (−4) = 0 = −(0).
9.Differential equations
Answer: y(2) = 1.262
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.
Characteristic roots
(r + 1)(r + 2) = 0 → r = −1, −2
Constants from the initial values
A + B = 5, −1A − 2B = 0
A = 10, B = −5
Evaluate
y(2) = 10·e−2 + (−5)·e−4 = 1.262
Answer
y(2) = 1.262
10.Differential equations
Answer: y(1.5) = 1.918
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.
Steady value and roots
yss = 2, r = −2 ± 4j
Constants from the initial values
y(0) = 2 + A = 0 → A = −2
y′(0) = −2A + 4B = 0 → B = −1
Evaluate
y(1.5) = 2·(1 − e−3(cos 6 + 0.5·sin 6)) = 1.918
Answer
y(1.5) = 1.918
11.Matrices & determinants
Answer: det A = −58
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 = 7, M₁₂ = −2, M₁₃ = 11
Cofactor expansion
det A = (−1)·7 − 2·(−2) + (−5)·11 = −58
Answer
det A = −58
12.Matrices & determinants
Answer: λ = 4
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.
Characteristic equation
λ² − (5)λ + (4) = 0
Solve
λ = (5 ± √(9))/2 → λ = 1, 4
λ = 4
Answer
λ = 4
13.Vectors: dot & cross product
Answer: a·b = 8
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.
Positive: the angle between the vectors is less than 90°.
14.Vectors: dot & cross product
Answer: (a × b)x = 5
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 = 5·1 − (−6)·0 = 5
(a × b)y = azbx − axbz = (−6)·(−1) − (−1)·1 = 7
(a × b)z = axby − aybx = (−1)·0 − 5·(−1) = 5
Requested component
(a × b)x = 5
Answer
(a × b)x = 5
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: P = 0.3333
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 = 6/10 · 5/9 = 30/90 = 0.3333
Answer
P = 0.3333
16.Probability
Answer: E(G) = 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: 12 (p = 0.2), 57 (p = 0.05), −3 (p = 0.75)
Expected value
E(G) = 12·0.2 + 57·0.05 − 3·0.75 = 3 €
Answer
E(G) = 3 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 21.05
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 = 0.4
Standard normal table
Φ(0.4) = 0.6554
P = 1 − 0.6554 = 0.3446
Answer
P = 0.3446
19.Normal distribution
Answer: P = 0.515
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.
z-scores
z₁ = (482 − 500)/40 = −0.45
z₂ = (540 − 500)/40 = 1
Area between
P = Φ(1) − Φ(−0.45) = 0.8413 − 0.3264 = 0.515
Answer
P = 0.515
20.Normal distribution
Answer: P = 0.7881
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).