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)x. Give a decimal value.
f(x) = (1 − 2/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 4x)/sin(3x), with x in radians.
f(x) = ln(1 + 4x)/sin(3x)x → 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 + 5 at x = 2 crosses the y-axis at y = b. Find b.
f(x) = −x2 − 5x + 5x = 2
b =
5.IntegralsMedium
Evaluate I = ∫₁2.5 x·ln x dx.
∫₁2.5 x ln x dx
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 4 − x² and the x-axis.
y = 4 − x²
A =
7.Complex numbersMedium
Find the argument θ of z = −2 + j7 in degrees, as the principal value between −180° and 180°.
z = −2 + j7
θ = °
8.Complex numbersHard
Use De Moivre's theorem to find the real part of z⁵ for z = 1 + j√3.
z = 1 + j√3n = 5
Re(z⁵) =
9.Differential equationsMedium
Solve y″ + 7y′ + 10y = 0 with y(0) = 5 and y′(0) = 6. Find y(1.5).
y″ + 7y′ + 10y = 0y(0) = 5y′(0) = 6
y(1.5) =
10.Differential equationsHard
A system obeys y″ + y′ + 16.25y = 32.5 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 = 32.5y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c22 (row 2, column 2).
A = [−2 1; −2 3]B = [−1 0; −5 −3]
c22 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−3 2; −5 4]
λ =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (−6, 0, −2) onto b = (5, −1, −2), that is, the component of a in the direction of b.
a = (−6, 0, −2)b = (5, −1, −2)
compb a =
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (−5, 0, 1), b = (3, −1, 3) and c = (−2, 0, 2).
a = (−5, 0, 1)b = (3, −1, 3)c = (−2, 0, 2)
V =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4p(x): 0.2, 0.25, 0.25, 0.3
σ =
16.ProbabilityHard
A game costs €2 to play. You win €6 with probability 0.3 and €40 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €2€6: p = 0.3€40: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 14, 28, 24, 59, 17.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 25. What proportion P of values is above 450?
μ = 500σ = 25x = 450
P =
19.Normal distributionMedium
The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 12 of them exceeds 484.2 g. Round z to two decimals.
μ = 40 gσ = 1.2 gn = 12> 484.2 g
P =
20.Normal distributionHard
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
3Solutions
1.Limits
Answer: L = 0.1353
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→∞ x·ln(1 − 2/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·(−2) = −2
Undo the logarithm
L = e−2 = 0.1353
Answer
L = 0.1353
2.Limits
Answer: L = 1.333
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 [4/(1 + 4x)]/[3·cos 3x]
Substitute x = 0
L = 4/3 = 1.333
Answer
L = 1.333
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 = 9
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) = −9
f′(x) = −2x − 5 → f′(2) = −9
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −9 − (−9)·2 = 9
Answer
b = 9
5.Integrals
Answer: I = 1.551
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.
Choose u and dv
u = ln x, du = dx/x; dv = x dx, v = x²/2
Antiderivative
∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4
Limits
(3.125·ln 2.5 − 1.5625) − (0 − 0.25) = 1.551
Answer
I = 1.551
6.Integrals
Answer: A = 10.67
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
4 − x² = 0 → x = ±2
Integrate
A = [4x − x³/3]−22 = 5.333 − (−5.333) = 10.67
Answer
A = 10.67
7.Complex numbers
Answer: θ = 105.9 °
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 = −2, b = 7 → II
Reference angle
α = atan(7/2) = 74.05°
Argument
θ = 180° − α = 105.9°
Answer
θ = 105.9°
In polar form z = 7.28∠105.9°.
8.Complex numbers
Answer: Re(z⁵) = 16
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.
Polar form
z = 1 + j√3 = 2∠60°
De Moivre
z⁵ = (2)⁵∠300° = 32∠−60°
Rectangular form
z⁵ = 32·(cos −60° + j sin −60°) = 16 − j27.71
Re(z⁵) = 16
Answer
Re(z⁵) = 16
9.Differential equations
Answer: y(1.5) = 0.5115
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 = 6
A = 10.33, B = −5.333
Evaluate
y(1.5) = 10.33·e−3 + (−5.333)·e−7.5 = 0.5115
Answer
y(1.5) = 0.5115
10.Differential equations
Answer: ymax = 3.35
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 = −0.5 ± 4j
Constants from the initial values
y(0) = 2 + A = 0 → A = −2
y′(0) = −0.5A + 4B = 0 → B = −0.25
Peak
t = π/4 = 0.7854
ymax = 2·(1 + e−0.5·π/4) = 3.35
Answer
ymax = 3.35
11.Matrices & determinants
Answer: c22 = −9
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 = (−2)·0 + 3·(−3) = −9
Answer
c22 = −9
12.Matrices & determinants
Answer: λ = 2
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 − (1)λ + (−2) = 0
Solve
λ = (1 ± √(9))/2 → λ = −1, 2
λ = 2
Answer
λ = 2
13.Vectors: dot & cross product
Answer: compb a = −4.747
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 = (−6)·5 + 0·(−1) + (−2)·(−2) = −26
Length of b
|b| = √(5² + (−1)² + (−2)²) = 5.477
Scalar projection
compb a = a·b/|b| = −26/5.477 = −4.747
Answer
compb a = −4.747
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: V = 8
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 b × c
(b × c)x = bycz − bzcy = (−1)·2 − 3·0 = −2
(b × c)y = bzcx − bxcz = 3·(−2) − 3·2 = −12
(b × c)z = bxcy − bycx = 3·0 − (−1)·(−2) = −2
Dot product with a
a·(b × c) = (−5)·(−2) + 0·(−12) + 1·(−2) = 8
V = |8| = 8
Answer
V = 8
15.Probability
Answer: σ = 1.108
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.
Mean
μ = 1·0.2 + 2·0.25 + 3·0.25 + 4·0.3 = 2.65
Mean of the squares
E(X²) = 1·0.2 + 4·0.25 + 9·0.25 + 16·0.3 = 8.25
Variance
Var(X) = 8.25 − 2.65² = 1.228
σ = √1.228 = 1.108
Answer
σ = 1.108
16.Probability
Answer: E(G) = 0.6 €
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), 38 (p = 0.02), −2 (p = 0.68)
Expected value
E(G) = 4·0.3 + 38·0.02 − 2·0.68 = 0.6 €
Answer
E(G) = 0.6 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 17.98
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 = 0.1562
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 total
μT = 480 g, σT = 1.2·√12 = 4.157 g
Standardise
z = (484.2 − 480)/4.157 = 1.01
Probability
P = 1 − Φ(1.01) = 0.1562
Answer
P = 0.1562
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).