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 + 2/x)2x. Give a decimal value.
f(x) = (1 + 2/x)2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 5x)/(x²), with x in radians.
f(x) = (1 − cos 5x)/(x²)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(3x2 + 1). Find f′(−2).
f(x) = ln(3x2 + 1)x = −2
f′(−2) =
4.DerivativesHard
The tangent line to f(x) = 2x2 + 4x − 4 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = 2x2 + 4x − 4x = 3
b =
5.IntegralsMedium
Evaluate I = ∫01 2x·(x² + 1)1 dx.
f(x) = 2x·(x² + 1)1[0, 1]
I =
6.IntegralsHard
Evaluate I = ∫₀1.5 x²·eˣ dx.
∫₀1.5 x²eˣ dx
I =
7.Complex numbersMedium
Write z = −3 − j5 in exponential form r·ejθ. Give θ in radians, between −π and π.
z = −3 − j5
θ = rad
8.Complex numbersHard
Solve z² − 4z + 20 = 0. The roots are a ± jb; find the real part a.
p = −4q = 20
a =
9.Differential equationsMedium
A cup of coffee at 80 °C is left in a room at 19 °C. After 29 min its temperature is 30 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 80 °CTa = 19 °CT(29) = 30 °C
k = min⁻¹
10.Differential equationsHard
A system obeys y″ + y′ + 16.25y = 16.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + y′ + 16.25y = 16.25y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [5 −5 −5; −5 −2 −1; −2 0 −4]
det A =
12.Matrices & determinantsHard
Solve the system −4x − 2y = 0 and −4x + 3y = 5. Find x.
−4x − 2y = 0−4x + 3y = 5
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−3, −3, 6) and b = (−4, 0, −1), in degrees.
a = (−3, −3, 6)b = (−4, 0, −1)
θ = °
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(−2, 3, 3), B(−3, 2, −1) and C(0, 2, 2). Coordinates are in metres.
A(−2, 3, 3)B(−3, 2, −1)C(0, 2, 2)
A = m²
15.ProbabilityMedium
A bag holds 5 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
5 / 6
P =
16.ProbabilityHard
A game costs €10 to play. You win €20 with probability 0.2 and €200 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €10€20: p = 0.2€200: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 57, 38, 31, 22, 36, 46.
n = 6
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 15. Find the z-score z of a score of 88.
μ = 100σ = 15x = 88
z =
19.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 57 and 85.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 57x2 = 85.5
P =
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 = 54.6
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→∞ 2·x·ln(1 + 2/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 2·2 = 4
Undo the logarithm
L = e4 = 54.6
Answer
L = 54.6
2.Limits
Answer: L = 12.5
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
5·sin 5x/(2x) → 0/0
Second application
25·cos 5x/2
Substitute x = 0
L = 25/2 = 12.5
Answer
L = 12.5
3.Derivatives
Answer: f′(−2) = −0.9231
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 for the logarithm
f′(x) = 6x/(3x2 + 1)
Substitute
f′(−2) = −12/13 = −0.9231
Answer
f′(−2) = −0.9231
4.Derivatives
Answer: b = −22
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(3) = 26
f′(x) = 4x + 4 → f′(3) = 16
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 26 − 16·3 = −22
Answer
b = −22
5.Integrals
Answer: I = 1.5
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.
Substitution
u = x² + 1, du = 2x dx
I = ∫12 u1 du
Integrate and evaluate
I = [u2/2]12 = (4 − 1)/2 = 1.5
Answer
I = 1.5
6.Integrals
Answer: I = 3.602
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
4.482·(1.25) − 1·2 = 3.602
Answer
I = 3.602
7.Complex numbers
Answer: θ = −2.111 rad
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
r = √(9 + 25) = 5.831
Argument in radians
θ = −121° = −121·π/180 = −2.111 rad
Exponential form
z = 5.831·e−j2.111
Answer
θ = −2.111 rad
8.Complex numbers
Answer: a = 2
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.
Discriminant
p² − 4q = 16 − 80 = −64 < 0
Quadratic formula
z = (4 ± j√64)/2 = 2 + j4, 2 − j4
Answer
a = 2
Answer
a = 2
9.Differential equations
Answer: k = 0.05907 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−29k = 11/61
Solve for k
k = ln(61/11)/29 = 0.05907 min⁻¹
Answer
k = 0.05907 min⁻¹
10.Differential equations
Answer: ymax = 1.675
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 = 1, r = −0.5 ± 4j
Constants from the initial values
y(0) = 1 + A = 0 → A = −1
y′(0) = −0.5A + 4B = 0 → B = −0.125
Peak
t = π/4 = 0.7854
ymax = 1·(1 + e−0.5·π/4) = 1.675
Answer
ymax = 1.675
11.Matrices & determinants
Answer: det A = 150
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 = 8, M₁₂ = 18, M₁₃ = −4
Cofactor expansion
det A = 5·8 − (−5)·18 + (−5)·(−4) = 150
Answer
det A = 150
12.Matrices & determinants
Answer: x = −0.5
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 = (−4)·3 − (−2)·(−4) = −20
Replace the x column
det Aₓ = 0·3 − (−2)·5 = 10
x = 10/(−20) = −0.5
Answer
x = −0.5
y = 1
13.Vectors: dot & cross product
Answer: θ = 78.58 °
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 = (−3)·(−4) + (−3)·0 + 6·(−1) = 6
Magnitudes
|a| = √((−3)² + (−3)² + 6²) = 7.348
|b| = √((−4)² + 0² + (−1)²) = 4.123
Angle
cos θ = 6/(7.348·4.123) = 0.198
θ = arccos(0.198) = 78.58°
Answer
θ = 78.58°
14.Vectors: dot & cross product
Answer: A = 4.975 m²
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.
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 = 5/11 · 4/10 = 20/110 = 0.1818
Answer
P = 0.1818
16.Probability
Answer: E(G) = −2 €
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: 10 (p = 0.2), 190 (p = 0.02), −10 (p = 0.78)
Expected value
E(G) = 10·0.2 + 190·0.02 − 10·0.78 = −2 €
Answer
E(G) = −2 €
A negative value means you lose money on average.
17.Descriptive statistics
Answer: s = 12.11
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 = (88 − 100)/15 = −0.8
Answer
z = −0.8
19.Normal distribution
Answer: P = 0.768
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 = (57 − 65)/10 = −0.8
z₂ = (85.5 − 65)/10 = 2.05
Area between
P = Φ(2.05) − Φ(−0.8) = 0.9798 − 0.2119 = 0.768
Answer
P = 0.768
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.