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→∞ (√(x2 − 10x) − x).
f(x) = √(x2 − 10x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e−0.5x − 1 + 0.5x)/x².
f(x) = (e−0.5x − 1 + 0.5x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x3·ex. Find f′(0.5).
f(x) = x3·exx = 0.5
f′(0.5) =
4.DerivativesHard
The tangent line to f(x) = x2 − 5x − 3 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = x2 − 5x − 3x = 1
b =
5.IntegralsMedium
The region under y = 0.5x from x = 0 to x = 2 is rotated about the x-axis. Find the volume V of the solid (a cone).
y = 0.5x0 ≤ x ≤ 2
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
Write z = 3∠50° in rectangular form a + jb and give the imaginary part b.
r = 3θ = 50°
b =
8.Complex numbersHard
Find the imaginary part of (−2 − j)⁵.
z = −2 − jn = 5
Im(z⁵) =
9.Differential equationsMedium
Solve y″ + 7y′ + 10y = 0 with y(0) = 5 and y′(0) = -2. Find y(2).
y″ + 7y′ + 10y = 0y(0) = 5y′(0) = -2
y(2) =
10.Differential equationsHard
A system obeys y″ + 4y′ + 8y = 16 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 4y′ + 8y = 16y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c22 (row 2, column 2).
A = [1 −1; −5 −4]B = [−1 −4; 4 −2]
c22 =
12.Matrices & determinantsHard
Solve the system x + 5y = 2 and 6x + y = −4. Find x.
x + 5y = 26x + y = −4
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (−3, 1, −2) onto b = (−3, 2, 1), that is, the component of a in the direction of b.
a = (−3, 1, −2)b = (−3, 2, 1)
compb a =
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (5, 2, −1) and b = (1, 2, −2) and has a positive z-component. Find its y-component n̂y.
a = (5, 2, −1)b = (1, 2, −2)n̂z > 0
n̂y =
15.ProbabilityMedium
A bag holds 4 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
4 / 9
P =
16.ProbabilityHard
A game costs €3 to play. You win €15 with probability 0.15 and €150 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€15: p = 0.15€150: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 13, 42, 19, 29, 58, 27.
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 26?
μ = 50σ = 12x = 26
P =
19.Normal distributionMedium
X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 33.3 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 33.3 min
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 99% of them lie (the 99th percentile).
μ = 500σ = 40P = 99%
x = h
3Solutions
1.Limits
Answer: L = −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/(√(x2 − 10x) + x)
Divide by x
f(x) = −10/(√(1 + (−10)/x) + 1)
Let x → ∞
L = −10/(1 + 1) = −5
Answer
L = −5
2.Limits
Answer: L = 0.125
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
(−0.5·e−0.5x + 0.5)/(2x) → 0/0
Second application
0.25·e−0.5x/2
Substitute x = 0
L = 0.25·1/2 = 0.125
Answer
L = 0.125
3.Derivatives
Answer: f′(0.5) = 1.443
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′ = ex·(3x2 + 1·x3)
Substitute
f′(0.5) = 1.649·(0.75 + 0.125) = 1.443
Answer
f′(0.5) = 1.443
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(1) = −7
f′(x) = 2x − 5 → f′(1) = −3
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −7 − (−3)·1 = −4
Answer
b = −4
5.Integrals
Answer: V = 2.094
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 0.25x² dx
Integrate
V = π·0.25·2³/3 = 2.094
Answer
V = 2.094
Check: a cone of radius 1 and height 2 has (1/3)πr²h = 2.094.
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: b = 2.298
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.
Components
a = 3·cos(50°) = 1.928
b = 3·sin(50°) = 2.298
Rectangular form
z = 1.928 + j2.298
b = 2.298
Answer
b = 2.298
8.Complex numbers
Answer: Im(z⁵) = −41
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 = (−2 − j)(−2 − j) = 3 + j4
z³ = (3 + j4)(−2 − j) = −2 − j11
z⁴ = (−2 − j11)(−2 − j) = −7 + j24
z⁵ = (−7 + j24)(−2 − j) = 38 − j41
Check with De Moivre
z = 2.236∠−153.4° → z⁵ = 55.9∠−47.17°
Im(z⁵) = −41
Answer
Im(z⁵) = −41
9.Differential equations
Answer: y(2) = 0.1403
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 = 5, −2A − 5B = -2
A = 7.667, B = −2.667
Evaluate
y(2) = 7.667·e−4 + (−2.667)·e−10 = 0.1403
Answer
y(2) = 0.1403
10.Differential equations
Answer: ymax = 2.086
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 ± 2j
Constants from the initial values
y(0) = 2 + A = 0 → A = −2
y′(0) = −2A + 2B = 0 → B = −2
Peak
t = π/2 = 1.571
ymax = 2·(1 + e−2·π/2) = 2.086
Answer
ymax = 2.086
11.Matrices & determinants
Answer: c22 = 28
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
c22 = (−5)·(−4) + (−4)·(−2) = 28
Answer
c22 = 28
12.Matrices & determinants
Answer: x = −0.7586
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 = 1·1 − 5·6 = −29
Replace the x column
det Aₓ = 2·1 − 5·(−4) = 22
x = 22/(−29) = −0.7586
Answer
x = −0.7586
y = 0.5517
13.Vectors: dot & cross product
Answer: compb a = 2.405
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)·(−3) + 1·2 + (−2)·1 = 9
Length of b
|b| = √((−3)² + 2² + 1²) = 3.742
Scalar projection
compb a = a·b/|b| = 9/3.742 = 2.405
Answer
compb a = 2.405
14.Vectors: dot & cross product
Answer: n̂y = 0.7373
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
(a × b)x = aybz − azby = 2·(−2) − (−1)·2 = −2
(a × b)y = azbx − axbz = (−1)·1 − 5·(−2) = 9
(a × b)z = axby − aybx = 5·2 − 2·1 = 8
Normalise
The z-component of a × b is already positive.
|a × b| = √((−2)² + 9² + 8²) = 12.21
n̂ = (−2, 9, 8)/12.21 = (−0.1638, 0.7373, 0.6554)
n̂y = 0.7373
Answer
n̂y = 0.7373
15.Probability
Answer: P = 0.07692
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 = 4/13 · 3/12 = 12/156 = 0.07692
Answer
P = 0.07692
16.Probability
Answer: E(G) = 0.75 €
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 = 0.15), 147 (p = 0.01), −3 (p = 0.84)
Expected value
E(G) = 12·0.15 + 147·0.01 − 3·0.84 = 0.75 €
Answer
E(G) = 0.75 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 16.35
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 = 0.3015
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 = 32 min, σ = √(1.5² + 2²) = 2.5 min
Standardise
z = (33.3 − 32)/2.5 = 0.52
Upper tail
P = 1 − Φ(0.52) = 1 − 0.6985 = 0.3015
Answer
P = 0.3015
20.Normal distribution
Answer: x = 593.1 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.