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 − 2x) − x).
f(x) = √(x2 − 2x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 5x)/(x²), with x in radians.
f(x) = (1 − cos 5x)/(x²)x → 0
L =
3.DerivativesMedium
Let f(x) = (3x − 1)4. Find f′(0).
f(x) = (3x − 1)4x = 0
f′(0) =
4.DerivativesHard
Let f(x) = x3 − 1.5x2 − 6x − 1. Find the x-coordinate x of the local minimum.
f(x) = x3 − 1.5x2 − 6x − 1
x =
5.IntegralsMedium
Evaluate I = ∫28 2/x dx.
f(x) = 2/x[2, 8]
I =
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 = 2.5∠65° in rectangular form a + jb and give the imaginary part b.
r = 2.5θ = 65°
b =
8.Complex numbersHard
The roots of zⁿ = w with n = 3 and w = 27∠0° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2. Find the real part of z₀.
w = 27∠0°n = 3k = 0
Re(z₀) =
9.Differential equationsMedium
Solve y″ + 6y′ + 9y = 0 with y(0) = 3 and y′(0) = -1. Find y(1).
y″ + 6y′ + 9y = 0y(0) = 3y′(0) = -1
y(1) =
10.Differential equationsHard
A system obeys y″ + 3y′ + 6.25y = 6.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 3y′ + 6.25y = 6.25y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [−1 4; 2 3]B = [−3 5; −3 −6]
c21 =
12.Matrices & determinantsHard
Solve the system 3x + 6y = 7 and −3x + 6y = 2. Find x.
3x + 6y = 7−3x + 6y = 2
x =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (5, 6, 4) and b = (−7, −7, 1).
a = (5, 6, 4)b = (−7, −7, 1)
a·b =
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−1, 3, −5) and b = (2, −2, 4) and has a positive z-component. Find its y-component n̂y.
a = (−1, 3, −5)b = (2, −2, 4)n̂z > 0
n̂y =
15.ProbabilityMedium
A fair die is rolled 6 times. What is the probability P of getting at least one six?
n = 6
P =
16.ProbabilityHard
A game costs €1 to play. You win €2 with probability 0.15 and €10 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €1€2: p = 0.15€10: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 17, 11, 54, 12, 10, 38, 41.
n = 7
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 75. Find the z-score z of a score of 650.
μ = 500σ = 75x = 650
z =
19.Normal distributionMedium
The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 16 of them exceeds 1312 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 16> 1312 kg
P =
20.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.
μ = 100 pointsσ = 15 points95%
x = points
3Solutions
1.Limits
Answer: L = −1
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) = −2x/(√(x2 − 2x) + x)
Divide by x
f(x) = −2/(√(1 + (−2)/x) + 1)
Let x → ∞
L = −2/(1 + 1) = −1
Answer
L = −1
2.Limits
Answer: L = 12.5
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).
First application
5·sin 5x/(2x) → 0/0
Second application
25·cos 5x/2
Substitute x = 0
L = 25/2 = 12.5
Answer
L = 12.5
3.Derivatives
Answer: f′(0) = −12
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.
Chain rule
f′(x) = 4·(3x − 1)3·3
Substitute
f′(0) = 4·(−1)3·3 = −12
Answer
f′(0) = −12
4.Derivatives
Answer: x = 2
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 − 6 = 3·(x2 − x − 2)
Solve f′(x) = 0
x = −1 or x = 2
Which one is which
f″(x) = 6x − 3: negative at a maximum, positive at a minimum.
f″(−1) = −9, f″(2) = 9
x = 2
Answer
x = 2
5.Integrals
Answer: I = 2.773
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 8 − ln 2) = 2·ln 4 = 2.773
Answer
I = 2.773
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 = 2.266
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 = 2.5·cos(65°) = 1.057
b = 2.5·sin(65°) = 2.266
Rectangular form
z = 1.057 + j2.266
b = 2.266
Answer
b = 2.266
8.Complex numbers
Answer: Re(z₀) = 3
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 of the roots
271/3 = 3
Argument of the root
θ₀ = (0° + 360°·0)/3 = 0°
Rectangular form
z₀ = 3·(cos 0° + j sin 0°) = 3
Re(z₀) = 3
Answer
Re(z₀) = 3
The 3 roots lie on the circle of radius 3, 120° apart.
9.Differential equations
Answer: y(1) = 0.5477
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.
Double root
(r + 3)² = 0 → r = −3
Constants
A = y(0) = 3
y′(0) = B − 3A → B = -1 + 9 = 8
Evaluate
y(1) = (3 + 8·1)·e−3 = 0.5477
Answer
y(1) = 0.5477
10.Differential equations
Answer: ymax = 1.095
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 = 1, r = −1.5 ± 2j
Constants from the initial values
y(0) = 1 + A = 0 → A = −1
y′(0) = −1.5A + 2B = 0 → B = −0.75
Peak
t = π/2 = 1.571
ymax = 1·(1 + e−1.5·π/2) = 1.095
Answer
ymax = 1.095
11.Matrices & determinants
Answer: c21 = −15
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
c21 = 2·(−3) + 3·(−3) = −15
Answer
c21 = −15
12.Matrices & determinants
Answer: x = 0.8333
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 = 3·6 − 6·(−3) = 36
Replace the x column
det Aₓ = 7·6 − 6·2 = 30
x = 30/36 = 0.8333
Answer
x = 0.8333
y = 0.75
13.Vectors: dot & cross product
Answer: a·b = −73
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.
Negative: the angle between the vectors is more than 90°.
14.Vectors: dot & cross product
Answer: n̂y = 0.8018
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 = 3·4 − (−5)·(−2) = 2
(a × b)y = azbx − axbz = (−5)·2 − (−1)·4 = −6
(a × b)z = axby − aybx = (−1)·(−2) − 3·2 = −4
Normalise
The z-component of a × b is negative, so use −(a × b).
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.
Complement
P(no six) = (5/6)6 = 0.3349
P = 1 − 0.3349 = 0.6651
Answer
P = 0.6651
16.Probability
Answer: E(G) = −0.2 €
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: 1 (p = 0.15), 9 (p = 0.05), −1 (p = 0.8)
Expected value
E(G) = 1·0.15 + 9·0.05 − 1·0.8 = −0.2 €
Answer
E(G) = −0.2 €
A negative value means you lose money on average.
17.Descriptive statistics
Answer: s = 17.85
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 = (650 − 500)/75 = 2
Answer
z = 2
19.Normal distribution
Answer: P = 0.8413
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 = 1360 kg, σT = 12·√16 = 48 kg
Standardise
z = (1312 − 1360)/48 = −1
Probability
P = 1 − Φ(−1) = 0.8413
Answer
P = 0.8413
20.Normal distribution
Answer: x = 75.4 points
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.