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 + 0.5/x)x. Give a decimal value.
f(x) = (1 + 0.5/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 4x)/(4x²), with x in radians.
f(x) = (1 − cos 4x)/(4x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(x), with x in radians. Find f′(0.5).
f(x) = x·sin(x)x = 0.5 rad
f′(0.5) =
4.DerivativesHard
Let f(x) = x3 − 1.5x2. Find the x-coordinate x of the local minimum.
f(x) = x3 − 1.5x2
x =
5.IntegralsMedium
Evaluate I = ∫₁6 x·ln x dx.
∫₁6 x ln x dx
I =
6.IntegralsHard
Evaluate I = ∫₀1 x²·eˣ dx.
∫₀1 x²eˣ dx
I =
7.Complex numbersMedium
Find the argument θ of z = −1 + j3 in degrees, as the principal value between −180° and 180°.
z = −1 + j3
θ = °
8.Complex numbersHard
One root of z³ − 5z² + 9z − 5 = 0 is z₁ = 2 + j. The coefficients are real. Find the real root r.
z₁ = 2 + j
r =
9.Differential equationsMedium
A can of soda at 5 °C is taken out of the fridge into a room at 26 °C. After 5 min its temperature is 10 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 18 °C?
T0 = 5 °CTa = 26 °CT(5) = 10 °CT = 18 °C
t = min
10.Differential equationsHard
Solve y″ + 4y′ + 29y = 0 with y(0) = 1 and y′(0) = 0 (released from rest). Find y(0.5); angles in radians.
y″ + 4y′ + 29y = 0y(0) = 1y′(0) = 0
y(0.5) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−3 5 5; 5 −3 −4; −3 −3 1]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [0 −1; 5 6]
λ =
13.Vectors: dot & cross productMedium
A constant force F = (−10, −5, 15) N acts on an object while it moves in a straight line from point A(−4, 3, −1) to point B(−5, 5, 5). Coordinates are in metres. Find the work W done by the force.
F = (−10, −5, 15) NA(−4, 3, −1) mB(−5, 5, 5) m
W = J
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (2, −1, 4), b = (−4, 0, 1) and c = (−5, 1, 2).
a = (2, −1, 4)b = (−4, 0, 1)c = (−5, 1, 2)
V =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3p(x): 0.3, 0.3, 0.15, 0.25
Var(X) =
16.ProbabilityHard
A game costs €5 to play. You win €10 with probability 0.2 and €250 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€10: p = 0.2€250: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 28, 15, 20, 26, 13, 52.
n = 6
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 10. Find the z-score z of a score of 65.
μ = 50σ = 10x = 65
z =
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 95.4 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 95.4 s
P =
20.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 80% of them lie (the 80th percentile).
μ = 250σ = 5P = 80%
x = g
3Solutions
1.Limits
Answer: L = 1.649
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 + 0.5/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·0.5 = 0.5
Undo the logarithm
L = e0.5 = 1.649
Answer
L = 1.649
2.Limits
Answer: L = 2
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
4·sin 4x/(8x) → 0/0
Second application
16·cos 4x/8
Substitute x = 0
L = 16/8 = 2
Answer
L = 2
3.Derivatives
Answer: f′(0.5) = 0.9182
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(x) + x·cos(x)
Substitute
f′(0.5) = 0.4794 + 0.5·0.8776 = 0.9182
Answer
f′(0.5) = 0.9182
4.Derivatives
Answer: x = 1
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 − 3x = 3·(x2 − x)
Solve f′(x) = 0
x = 0 or x = 1
Which one is which
f″(x) = 6x − 3: negative at a maximum, positive at a minimum.
f″(0) = −3, f″(1) = 3
x = 1
Answer
x = 1
5.Integrals
Answer: I = 23.5
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
(18·ln 6 − 9) − (0 − 0.25) = 23.5
Answer
I = 23.5
6.Integrals
Answer: I = 0.7183
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
2.718·(1) − 1·2 = 0.7183
Answer
I = 0.7183
7.Complex numbers
Answer: θ = 108.4 °
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 = −1, b = 3 → II
Reference angle
α = atan(3/1) = 71.57°
Argument
θ = 180° − α = 108.4°
Answer
θ = 108.4°
In polar form z = 3.162∠108.4°.
8.Complex numbers
Answer: r = 1
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.
Check with the sum of the roots: z₁ + z₂ + r = 4 + 1 = 5 = −(−5).
9.Differential equations
Answer: t = 17.74 min
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.
Cooling constant
k = ln(-21/-16)/5 = 0.05439 min⁻¹
Time to reach the target
t = ln(-21/(−8))/0.05439 = 17.74 min
Answer
t = 17.74 min
10.Differential equations
Answer: y(0.5) = −0.2067
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.
Complex roots
r = −2 ± 5j
Constants
A = 1, B = 2·1/5 = 0.4
Evaluate
y(0.5) = e−1(1·cos 2.5 + 0.4·sin 2.5) = −0.2067
Answer
y(0.5) = −0.2067
11.Matrices & determinants
Answer: det A = −40
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 = −15, M₁₂ = −7, M₁₃ = −24
Cofactor expansion
det A = (−3)·(−15) − 5·(−7) + 5·(−24) = −40
Answer
det A = −40
12.Matrices & determinants
Answer: λ = 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.
Characteristic equation
λ² − (6)λ + (5) = 0
Solve
λ = (6 ± √(16))/2 → λ = 1, 5
λ = 5
Answer
λ = 5
13.Vectors: dot & cross product
Answer: W = 90 J
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.
Displacement
d = B − A = (−5, 5, 5) − (−4, 3, −1) = (−1, 2, 6) m
Work
W = F·d = (−10)·(−1) + (−5)·2 + 15·6 = 90 J
Answer
W = 90 J
14.Vectors: dot & cross product
Answer: V = 21
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 = 0·2 − 1·1 = −1
(b × c)y = bzcx − bxcz = 1·(−5) − (−4)·2 = 3
(b × c)z = bxcy − bycx = (−4)·1 − 0·(−5) = −4
Dot product with a
a·(b × c) = 2·(−1) + (−1)·3 + 4·(−4) = −21
V = |−21| = 21
Answer
V = 21
15.Probability
Answer: Var(X) = 1.327
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 = 0·0.3 + 1·0.3 + 2·0.15 + 3·0.25 = 1.35
Mean of the squares
E(X²) = 0·0.3 + 1·0.3 + 4·0.15 + 9·0.25 = 3.15
Variance
Var(X) = 3.15 − 1.35² = 1.327
Answer
Var(X) = 1.327
16.Probability
Answer: E(G) = 9.5 €
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: 5 (p = 0.2), 245 (p = 0.05), −5 (p = 0.75)
Expected value
E(G) = 5·0.2 + 245·0.05 − 5·0.75 = 9.5 €
Answer
E(G) = 9.5 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 14.18
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 = (65 − 50)/10 = 1.5
Answer
z = 1.5
19.Normal distribution
Answer: P = 0.9332
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 = (95.4 − 105)/6.403 = −1.5
Upper tail
P = 1 − Φ(−1.5) = 1 − 0.06681 = 0.9332
Answer
P = 0.9332
20.Normal distribution
Answer: x = 254.2 g
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.