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→∞ (√(9x2 − 9x) − 3x).
f(x) = √(9x2 − 9x) − 3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e1.5x − 1 − 1.5x)/x².
f(x) = (e1.5x − 1 − 1.5x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x3·e2x. Find f′(2).
f(x) = x3·e2xx = 2
f′(2) =
4.DerivativesHard
Let f(x) = x3 + 9x2 + 24x − 2. Find the x-coordinate x of the local maximum.
f(x) = x3 + 9x2 + 24x − 2
x =
5.IntegralsMedium
The region under y = 2x from x = 0 to x = 3 is rotated about the x-axis. Find the volume V of the solid (a cone).
y = 2x0 ≤ x ≤ 3
V =
6.IntegralsHard
Find the area A of the region enclosed by y = 2x2 and y = 2x.
y = 2x2y = 2x
A =
7.Complex numbersMedium
Given z₁ = 3∠150° and z₂ = 3∠20°, find the real part of w = z₁/z₂.
z₁ = 3∠150°z₂ = 3∠20°
Re(w) =
8.Complex numbersHard
Find the real part of (3 + j3)⁴.
z = 3 + j3n = 4
Re(z⁴) =
9.Differential equationsMedium
Solve y″ + 7y′ + 10y = 0 with y(0) = 3 and y′(0) = -2. Find y(0.5).
y″ + 7y′ + 10y = 0y(0) = 3y′(0) = -2
y(0.5) =
10.Differential equationsHard
A tank holds 1000 L of water in which 10 kg of salt is dissolved. Brine with 0.05 kg of salt per litre flows in at 2 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 40 min?
V = 1000 Lm0 = 10 kgc = 0.05 kg/Lr = 2 L/min
m = kg
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [−3 −4; 5 4]B = [−4 4; −6 1]
c11 =
12.Matrices & determinantsHard
Solve the system 4x − 5y = −9 and −5x + y = −6. Find x.
4x − 5y = −9−5x + y = −6
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (−7, −3, 7) onto b = (5, 0, 3), that is, the component of a in the direction of b.
a = (−7, −3, 7)b = (5, 0, 3)
compb a =
14.Vectors: dot & cross productHard
A force F = (20, −30, −60) N acts at point A(−2, −3, 4). Coordinates are in metres. Find the z-component Mz of the moment of the force about point O(1, −1, −3).
F = (20, −30, −60) NA(−2, −3, 4) mO(1, −1, −3) m
Mz = N·m
15.ProbabilityMedium
A bag holds 5 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
5 / 9
P =
16.ProbabilityHard
A game costs €3 to play. You win €12 with probability 0.25 and €30 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€12: p = 0.25€30: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 41, 33, 16, 24, 16, 25.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 12. What proportion P of values is above 36?
μ = 60σ = 12x = 36
P =
19.Normal distributionMedium
X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 111.4 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 111.4 s
P =
20.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 85% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g85%
x = g
3Solutions
1.Limits
Answer: L = −1.5
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) = −9x/(√(9x2 − 9x) + 3x)
Divide by x
f(x) = −9/(√(9 + (−9)/x) + 3)
Let x → ∞
L = −9/(3 + 3) = −1.5
Answer
L = −1.5
2.Limits
Answer: L = 1.125
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
(1.5·e1.5x − 1.5)/(2x) → 0/0
Second application
2.25·e1.5x/2
Substitute x = 0
L = 2.25·1/2 = 1.125
Answer
L = 1.125
3.Derivatives
Answer: f′(2) = 1529
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′ = e2x·(3x2 + 2·x3)
Substitute
f′(2) = 54.6·(12 + 16) = 1529
Answer
f′(2) = 1529
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 + 18x + 24 = 3·(x2 + 6x + 8)
Solve f′(x) = 0
x = −4 or x = −2
Which one is which
f″(x) = 6x + 18: negative at a maximum, positive at a minimum.
f″(−4) = −6, f″(−2) = 6
x = −4
Answer
x = −4
5.Integrals
Answer: V = 113.1
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∫₀3 4x² dx
Integrate
V = π·4·3³/3 = 113.1
Answer
V = 113.1
Check: a cone of radius 6 and height 3 has (1/3)πr²h = 113.1.
6.Integrals
Answer: A = 0.3333
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.
Intersection points
2x2 = 2x → x = 0, x = 1
Integrate upper minus lower
A = [x2 − 0.6667x3]01 = 0.3333
Answer
A = 0.3333
7.Complex numbers
Answer: Re(w) = −0.6428
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 / 3 = 1
θ = 150° − 20° = 130°
Rectangular form
Re(w) = 1·cos(130°) = −0.6428
Im(w) = 1·sin(130°) = 0.766
Re(w) = −0.6428
Answer
Re(w) = −0.6428
8.Complex numbers
Answer: Re(z⁴) = −324
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.
Powers step by step
z² = (3 + j3)(3 + j3) = j18
z³ = (j18)(3 + j3) = −54 + j54
z⁴ = (−54 + j54)(3 + j3) = −324
Check with De Moivre
z = 4.243∠45° → z⁴ = 324∠180°
Re(z⁴) = −324
Answer
Re(z⁴) = −324
9.Differential equations
Answer: y(0.5) = 1.485
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 + 2)(r + 5) = 0 → r = −2, −5
Constants from the initial values
A + B = 3, −2A − 5B = -2
A = 4.333, B = −1.333
Evaluate
y(0.5) = 4.333·e−1 + (−1.333)·e−2.5 = 1.485
Answer
y(0.5) = 1.485
10.Differential equations
Answer: m = 13.08 kg
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.
Set up the equation
dm/dt = 2·0.05 − (2/1000)·m = 0.1 − 0.002·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 50 + (−40)·e−t/500
Substitute t
m(40) = 50 + (−40)·0.9231 = 13.08 kg
Answer
m = 13.08 kg
11.Matrices & determinants
Answer: c11 = 36
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.
Row times column
c11 = (−3)·(−4) + (−4)·(−6) = 36
Answer
c11 = 36
12.Matrices & determinants
Answer: x = 1.857
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 = 4·1 − (−5)·(−5) = −21
Replace the x column
det Aₓ = (−9)·1 − (−5)·(−6) = −39
x = −39/(−21) = 1.857
Answer
x = 1.857
y = 3.286
13.Vectors: dot & cross product
Answer: compb a = −2.401
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 = (−7)·5 + (−3)·0 + 7·3 = −14
Length of b
|b| = √(5² + 0² + 3²) = 5.831
Scalar projection
compb a = a·b/|b| = −14/5.831 = −2.401
Answer
compb a = −2.401
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: Mz = 130 N·m
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.
Position vector from O to A
r = A − O = (−2, −3, 4) − (1, −1, −3) = (−3, −2, 7) m
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 = 5/14 · 4/13 = 20/182 = 0.1099
Answer
P = 0.1099
16.Probability
Answer: E(G) = 1.5 €
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: 9 (p = 0.25), 27 (p = 0.05), −3 (p = 0.7)
Expected value
E(G) = 9·0.25 + 27·0.05 − 3·0.7 = 1.5 €
Answer
E(G) = 1.5 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 9.786
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 = −2
Standard normal table
Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772
Answer
P = 0.9772
19.Normal distribution
Answer: P = 0.1587
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 sum
μ = 105 s, σ = √(5² + 4²) = 6.403 s
Standardise
z = (111.4 − 105)/6.403 = 1
Upper tail
P = 1 − Φ(1) = 1 − 0.8413 = 0.1587
Answer
P = 0.1587
20.Normal distribution
Answer: x = 244.8 g
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.