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 + 1/x)x. Give a decimal value.
f(x) = (1 + 1/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 3x)/(3x²), with x in radians.
f(x) = (1 − cos 3x)/(3x²)x → 0
L =
3.DerivativesMedium
Let f(x) = (3x − 3)4. Find f′(2).
f(x) = (3x − 3)4x = 2
f′(2) =
4.DerivativesHard
Let f(x) = x3 + 4.5x2 + 6x + 4. Find the x-coordinate x of the local maximum.
f(x) = x3 + 4.5x2 + 6x + 4
x =
5.IntegralsMedium
The region under y = 0.5x from x = 0 to x = 3 is rotated about the x-axis. Find the volume V of the solid (a cone).
y = 0.5x0 ≤ x ≤ 3
V =
6.IntegralsHard
Find the area A of the region enclosed by y = 0.5x2 and y = x.
y = 0.5x2y = x
A =
7.Complex numbersMedium
Find the argument θ of z = −6 + j in degrees, as the principal value between −180° and 180°.
z = −6 + j
θ = °
8.Complex numbersHard
One root of z³ − 9z² + 43z − 75 = 0 is z₁ = 3 + j4. The coefficients are real. Find the real root r.
z₁ = 3 + j4
r =
9.Differential equationsMedium
Solve y″ + 6y′ + 5y = 0 with y(0) = 3 and y′(0) = -4. Find y(1.5).
y″ + 6y′ + 5y = 0y(0) = 3y′(0) = -4
y(1.5) =
10.Differential equationsHard
A system obeys y″ + y′ + 25.25y = 252.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + y′ + 25.25y = 252.5y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [5 −5 5; 1 −4 2; 3 4 3]
det A =
12.Matrices & determinantsHard
Solve the system 5x + y = 0 and −2x − 5y = −11. Find x.
5x + y = 0−2x − 5y = −11
x =
13.Vectors: dot & cross productMedium
A constant force F = (−25, 10, −5) N acts on an object while it moves in a straight line from point A(3, −2, −3) to point B(−5, 2, 1). Coordinates are in metres. Find the work W done by the force.
F = (−25, 10, −5) NA(3, −2, −3) mB(−5, 2, 1) m
W = J
14.Vectors: dot & cross productHard
For a = (6, −4, −1) and b = (−5, 6, −3), find the z-component (a × b)z of the cross product a × b.
a = (6, −4, −1)b = (−5, 6, −3)
(a × b)z =
15.ProbabilityMedium
A fair die is rolled 8 times. What is the probability P of getting at least one six?
n = 8
P =
16.ProbabilityHard
A game costs €5 to play. You win €10 with probability 0.25 and €100 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€10: p = 0.25€100: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 26, 15, 37, 41, 47.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 5. Find the z-score z of a score of 52.
μ = 50σ = 5x = 52
z =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.9660. Find z to two decimals.
A = 0.9660
z =
20.Normal distributionHard
Hole diameters D are normal with mean 25.02 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.02 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
3Solutions
1.Limits
Answer: L = 2.718
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 + 1/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·1 = 1
Undo the logarithm
L = e1 = 2.718
Answer
L = 2.718
2.Limits
Answer: L = 1.5
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
3·sin 3x/(6x) → 0/0
Second application
9·cos 3x/6
Substitute x = 0
L = 9/6 = 1.5
Answer
L = 1.5
3.Derivatives
Answer: f′(2) = 324
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 − 3)3·3
Substitute
f′(2) = 4·33·3 = 324
Answer
f′(2) = 324
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 + 9x + 6 = 3·(x2 + 3x + 2)
Solve f′(x) = 0
x = −2 or x = −1
Which one is which
f″(x) = 6x + 9: negative at a maximum, positive at a minimum.
f″(−2) = −3, f″(−1) = 3
x = −2
Answer
x = −2
5.Integrals
Answer: V = 7.069
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∫₀3 0.25x² dx
Integrate
V = π·0.25·3³/3 = 7.069
Answer
V = 7.069
Check: a cone of radius 1.5 and height 3 has (1/3)πr²h = 7.069.
6.Integrals
Answer: A = 0.6667
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.
Intersection points
0.5x2 = x → x = 0, x = 2
Integrate upper minus lower
A = [0.5x2 − 0.1667x3]02 = 0.6667
Answer
A = 0.6667
7.Complex numbers
Answer: θ = 170.5 °
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 = −6, b = 1 → II
Reference angle
α = atan(1/6) = 9.462°
Argument
θ = 180° − α = 170.5°
Answer
θ = 170.5°
In polar form z = 6.083∠170.5°.
8.Complex numbers
Answer: r = 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.
Check with the sum of the roots: z₁ + z₂ + r = 6 + 3 = 9 = −(−9).
9.Differential equations
Answer: y(1.5) = 0.6137
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 = 3, −1A − 5B = -4
A = 2.75, B = 0.25
Evaluate
y(1.5) = 2.75·e−1.5 + 0.25·e−7.5 = 0.6137
Answer
y(1.5) = 0.6137
10.Differential equations
Answer: ymax = 17.3
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 = 10, r = −0.5 ± 5j
Constants from the initial values
y(0) = 10 + A = 0 → A = −10
y′(0) = −0.5A + 5B = 0 → B = −1
Peak
t = π/5 = 0.6283
ymax = 10·(1 + e−0.5·π/5) = 17.3
Answer
ymax = 17.3
11.Matrices & determinants
Answer: det A = −35
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 = −20, M₁₂ = −3, M₁₃ = 16
Cofactor expansion
det A = 5·(−20) − (−5)·(−3) + 5·16 = −35
Answer
det A = −35
12.Matrices & determinants
Answer: x = −0.4783
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 = 5·(−5) − 1·(−2) = −23
Replace the x column
det Aₓ = 0·(−5) − 1·(−11) = 11
x = 11/(−23) = −0.4783
Answer
x = −0.4783
y = 2.391
13.Vectors: dot & cross product
Answer: W = 220 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, 2, 1) − (3, −2, −3) = (−8, 4, 4) m
Work
W = F·d = (−25)·(−8) + 10·4 + (−5)·4 = 220 J
Answer
W = 220 J
14.Vectors: dot & cross product
Answer: (a × b)z = 16
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, component by component
(a × b)x = aybz − azby = (−4)·(−3) − (−1)·6 = 18
(a × b)y = azbx − axbz = (−1)·(−5) − 6·(−3) = 23
(a × b)z = axby − aybx = 6·6 − (−4)·(−5) = 16
Requested component
(a × b)z = 16
Answer
(a × b)z = 16
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: P = 0.7674
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)8 = 0.2326
P = 1 − 0.2326 = 0.7674
Answer
P = 0.7674
16.Probability
Answer: E(G) = −0.5 €
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 = 0.25), 95 (p = 0.02), −5 (p = 0.73)
Expected value
E(G) = 5·0.25 + 95·0.02 − 5·0.73 = −0.5 €
Answer
E(G) = −0.5 €
A negative value means you lose money on average.
17.Descriptive statistics
Answer: s = 12.74
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 = (52 − 50)/5 = 0.4
Answer
z = 0.4
19.Normal distribution
Answer: z = 2.12
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.
Area to the left of z
Φ(z) = (1 + 0.9660)/2 = 0.9830
Read the z-table backwards
Find 0.9830 inside the table and read off z.
z = 2.12
Answer
z = 2.12
20.Normal distribution
Answer: P = 0.9207
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).