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 + 3/x)2x. Give a decimal value.
f(x) = (1 + 3/x)2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 3x)/(2x²), with x in radians.
f(x) = (1 − cos 3x)/(2x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(2x), with x in radians. Find f′(2).
f(x) = x·sin(2x)x = 2 rad
f′(2) =
4.DerivativesHard
The tangent line to f(x) = 0.5x2 + 3x + 6 at x = −2 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 + 3x + 6x = −2
b =
5.IntegralsMedium
Evaluate I = ∫24 2/x dx.
f(x) = 2/x[2, 4]
I =
6.IntegralsHard
The region between the line y = 2x and the parabola y = 0.5x² is rotated about the x-axis. Find the volume V of the solid.
y = 2xy = 0.5x²
V =
7.Complex numbersMedium
Given z₁ = 5∠30° and z₂ = 0.5∠−50°, find the argument θ of w = z₁·z₂ in degrees, between −180° and 180°.
z₁ = 5∠30°z₂ = 0.5∠−50°
θ = °
8.Complex numbersHard
Exactly one of the points A–D is a root of z⁶ = 729∠60°. Click it.
w = 729∠60°n = 6
1A (3∠60°)
2B (3∠−140°)
3C (5.4∠−170°)
4D (3∠−170°)
9.Differential equationsMedium
Solve y″ + 6y′ + 5y = 0 with y(0) = 2 and y′(0) = 2. Find y(1.5).
y″ + 6y′ + 5y = 0y(0) = 2y′(0) = 2
y(1.5) =
10.Differential equationsHard
A system obeys y″ + 4y′ + 29y = 145 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 4y′ + 29y = 145y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [2 4 4; 3 4 −2; 3 4 −5]
det A =
12.Matrices & determinantsHard
Solve the system −2x − y = −6 and −4x − 6y = −2. Find x.
−2x − y = −6−4x − 6y = −2
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (6, −4, −6) and b = (k, −1, 1) are perpendicular.
a = (6, −4, −6)b = (k, −1, 1)
k =
14.Vectors: dot & cross productHard
A force F = (−30, 50, 50) N acts at point A(−1, −2, −4). Coordinates are in metres. Find the z-component Mz of the moment of the force about point O(2, 0, −3).
F = (−30, 50, 50) NA(−1, −2, −4) mO(2, 0, −3) m
Mz = N·m
15.ProbabilityMedium
In how many ways N can a committee of 3 people be chosen from a group of 15? The order does not matter.
n = 15k = 3
N =
16.ProbabilityHard
A game costs €1 to play. You win €5 with probability 0.3 and €10 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €1€5: p = 0.3€10: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 11, 25, 16, 58, 20, 44, 48.
n = 7
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 10. Find the z-score z of a score of 155.
μ = 170σ = 10x = 155
z =
19.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 462 and 542 h? Give P as a decimal.
μ = 500σ = 40x1 = 462x2 = 542
P =
20.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 80% of them lie (the 80th percentile).
μ = 100σ = 15P = 80%
x = points
3Solutions
1.Limits
Answer: L = 403.4
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 + 3/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 2·3 = 6
Undo the logarithm
L = e6 = 403.4
Answer
L = 403.4
2.Limits
Answer: L = 2.25
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
3·sin 3x/(4x) → 0/0
Second application
9·cos 3x/4
Substitute x = 0
L = 9/4 = 2.25
Answer
L = 2.25
3.Derivatives
Answer: f′(2) = −3.371
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 and chain rule
f′(x) = sin(2x) + 2x·cos(2x)
Substitute
f′(2) = −0.7568 + 4·(−0.6536) = −3.371
Answer
f′(2) = −3.371
4.Derivatives
Answer: b = 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.
Point and slope
f(−2) = 2
f′(x) = x + 3 → f′(−2) = 1
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 2 − 1·(−2) = 4
Answer
b = 4
5.Integrals
Answer: I = 1.386
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.
Antiderivative
F(x) = 2·ln x
Evaluate at the limits
I = 2·(ln 4 − ln 2) = 2·ln 2 = 1.386
Answer
I = 1.386
6.Integrals
Answer: V = 107.2
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.
Intersection points
2x = 0.5x² → x = 0, x = 4
Washer method
V = π∫₀4 ((2x)² − (0.5x²)²) dx
Integrate
V = π(4·4³/3 − 0.25·4⁵/5) = π·34.13 = 107.2
Answer
V = 107.2
7.Complex numbers
Answer: θ = −20 °
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| = 5 · 0.5 = 2.5
θ = 30° + (−50)° = −20°
Argument
θ = −20°
Answer
θ = −20°
8.Complex numbers
Answer: D (3∠−170°)
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.
Modulus and arguments of the roots
|z| = 3
θ = 10°, 70°, 130°, −170°, −110°, −50°
The point
D = 3∠−170°
Answer
D (3∠−170°)
9.Differential equations
Answer: y(1.5) = 0.6688
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 + 5) = 0 → r = −1, −5
Constants from the initial values
A + B = 2, −1A − 5B = 2
A = 3, B = −1
Evaluate
y(1.5) = 3·e−1.5 + (−1)·e−7.5 = 0.6688
Answer
y(1.5) = 0.6688
10.Differential equations
Answer: ymax = 6.423
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 = 5, r = −2 ± 5j
Constants from the initial values
y(0) = 5 + A = 0 → A = −5
y′(0) = −2A + 5B = 0 → B = −2
Peak
t = π/5 = 0.6283
ymax = 5·(1 + e−2·π/5) = 6.423
Answer
ymax = 6.423
11.Matrices & determinants
Answer: det A = 12
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 = −12, M₁₂ = −9, M₁₃ = 0
Cofactor expansion
det A = 2·(−12) − 4·(−9) + 4·0 = 12
Answer
det A = 12
12.Matrices & determinants
Answer: x = 4.25
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 = (−2)·(−6) − (−1)·(−4) = 8
Replace the x column
det Aₓ = (−6)·(−6) − (−1)·(−2) = 34
x = 34/8 = 4.25
Answer
x = 4.25
y = −2.5
13.Vectors: dot & cross product
Answer: k = 0.3333
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 equals zero
a·b = 6·k + (−4)·(−1) + (−6)·1 = 0
−2 + 6·k = 0
Solve for k
k = −(−2)/6 = 0.3333
Answer
k = 0.3333
14.Vectors: dot & cross product
Answer: Mz = −210 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 = (−1, −2, −4) − (2, 0, −3) = (−3, −2, −1) 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.
Combinations
C(15, 3) = (15·14·13)/(3·2·1) = 455
Answer
N = 455
16.Probability
Answer: E(G) = 0.7 €
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 = 0.3), 9 (p = 0.02), −1 (p = 0.68)
Expected value
E(G) = 4·0.3 + 9·0.02 − 1·0.68 = 0.7 €
Answer
E(G) = 0.7 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 18.1
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 = (155 − 170)/10 = −1.5
Answer
z = −1.5
19.Normal distribution
Answer: P = 0.6821
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 = (462 − 500)/40 = −0.95
z₂ = (542 − 500)/40 = 1.05
Area between
P = Φ(1.05) − Φ(−0.95) = 0.8531 − 0.1711 = 0.6821
Answer
P = 0.6821
20.Normal distribution
Answer: x = 112.6 points
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.