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→∞ (√(9x2 + 2x) − 3x).
f(x) = √(9x2 + 2x) − 3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 3x)/(4x²), with x in radians.
f(x) = (1 − cos 3x)/(4x²)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(2x2 + 2). Find f′(−1).
f(x) = ln(2x2 + 2)x = −1
f′(−1) =
4.DerivativesHard
The tangent line to f(x) = −x2 + 4x + 4 at x = 2 crosses the y-axis at y = b. Find b.
f(x) = −x2 + 4x + 4x = 2
b =
5.IntegralsMedium
Evaluate I = ∫₁6 x·ln x dx.
∫₁6 x ln x dx
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 9 − x² and the x-axis.
y = 9 − x²
A =
7.Complex numbersMedium
Write z = 2.5∠125° in rectangular form a + jb and give the real part a.
r = 2.5θ = 125°
a =
8.Complex numbersHard
Exactly one of the points A–D is a root of z³ = 8∠0°. Click it.
w = 8∠0°n = 3
1A (2∠−60°)
2B (2∠−120°)
3C (2∠−150°)
4D (1∠−120°)
9.Differential equationsMedium
A can of soda at 6 °C is taken out of the fridge into a room at 24 °C. After 10 min its temperature is 17 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 6 °CTa = 24 °CT(10) = 17 °C
k = min⁻¹
10.Differential equationsHard
A system obeys y″ + 2y′ + 26y = 130 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 2y′ + 26y = 130y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−4 2 0; 5 2 3; 1 2 0]
det A =
12.Matrices & determinantsHard
Solve the system −x + 5y = −6 and −3x + 3y = −7. Find x.
−x + 5y = −6−3x + 3y = −7
x =
13.Vectors: dot & cross productMedium
A constant force F = (−15, 15, 15) N acts on an object while it moves in a straight line from point A(0, 4, 2) to point B(0, −5, −1). Coordinates are in metres. Find the work W done by the force.
F = (−15, 15, 15) NA(0, 4, 2) mB(0, −5, −1) m
W = J
14.Vectors: dot & cross productHard
For a = (−4, 2, 6) and b = (5, −5, −6), find the x-component (a × b)x of the cross product a × b.
a = (−4, 2, 6)b = (5, −5, −6)
(a × b)x =
15.ProbabilityMedium
A fair die is rolled 2 times. What is the probability P of getting at least one six?
n = 2
P =
16.ProbabilityHard
A game costs €10 to play. You win €50 with probability 0.2 and €100 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €10€50: p = 0.2€100: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 52, 24, 35, 53, 26, 39.
n = 6
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 50. Find the z-score z of a score of 425.
μ = 500σ = 50x = 425
z =
19.Normal distributionMedium
X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 759.8 g. Round z to two decimals.
μX = 250, σX = 4μY = 500, σY = 9X + Y > 759.8 g
P =
20.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 25% is to the left of it. Find the cut-off x.
μ = 500 hσ = 40 h25%
x = h
3Solutions
1.Limits
Answer: L = 0.3333
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/(√(9x2 + 2x) + 3x)
Divide by x
f(x) = 2/(√(9 + 2/x) + 3)
Let x → ∞
L = 2/(3 + 3) = 0.3333
Answer
L = 0.3333
2.Limits
Answer: L = 1.125
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
3·sin 3x/(8x) → 0/0
Second application
9·cos 3x/8
Substitute x = 0
L = 9/8 = 1.125
Answer
L = 1.125
3.Derivatives
Answer: f′(−1) = −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.
Chain rule for the logarithm
f′(x) = 4x/(2x2 + 2)
Substitute
f′(−1) = −4/4 = −1
Answer
f′(−1) = −1
4.Derivatives
Answer: b = 8
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) = 8
f′(x) = −2x + 4 → f′(2) = 0
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 8 − 0·2 = 8
Answer
b = 8
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: A = 36
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
9 − x² = 0 → x = ±3
Integrate
A = [9x − x³/3]−33 = 18 − (−18) = 36
Answer
A = 36
7.Complex numbers
Answer: a = −1.434
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(125°) = −1.434
b = 2.5·sin(125°) = 2.048
Rectangular form
z = −1.434 + j2.048
a = −1.434
Answer
a = −1.434
8.Complex numbers
Answer: B (2∠−120°)
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 and arguments of the roots
|z| = 2
θ = 0°, 120°, −120°
The point
B = 2∠−120°
Answer
B (2∠−120°)
9.Differential equations
Answer: k = 0.09445 min⁻¹
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.
Ratio of temperature differences
e−10k = -7/-18
Solve for k
k = ln(-18/-7)/10 = 0.09445 min⁻¹
Answer
k = 0.09445 min⁻¹
10.Differential equations
Answer: ymax = 7.667
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 = 5, r = −1 ± 5j
Constants from the initial values
y(0) = 5 + A = 0 → A = −5
y′(0) = −1A + 5B = 0 → B = −1
Peak
t = π/5 = 0.6283
ymax = 5·(1 + e−1·π/5) = 7.667
Answer
ymax = 7.667
11.Matrices & determinants
Answer: det A = 30
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 = −6, M₁₂ = −3, M₁₃ = 8
Cofactor expansion
det A = (−4)·(−6) − 2·(−3) + 0·8 = 30
Answer
det A = 30
12.Matrices & determinants
Answer: x = 1.417
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)·3 − 5·(−3) = 12
Replace the x column
det Aₓ = (−6)·3 − 5·(−7) = 17
x = 17/12 = 1.417
Answer
x = 1.417
y = −0.9167
13.Vectors: dot & cross product
Answer: W = −180 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 = (0, −5, −1) − (0, 4, 2) = (0, −9, −3) m
Work
W = F·d = (−15)·0 + 15·(−9) + 15·(−3) = −180 J
Answer
W = −180 J
Negative work: the force acts mostly against the motion.
14.Vectors: dot & cross product
Answer: (a × b)x = 18
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 = 2·(−6) − 6·(−5) = 18
(a × b)y = azbx − axbz = 6·5 − (−4)·(−6) = 6
(a × b)z = axby − aybx = (−4)·(−5) − 2·5 = 10
Requested component
(a × b)x = 18
Answer
(a × b)x = 18
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: P = 0.3056
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)2 = 0.6944
P = 1 − 0.6944 = 0.3056
Answer
P = 0.3056
16.Probability
Answer: E(G) = 1 €
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: 40 (p = 0.2), 90 (p = 0.01), −10 (p = 0.79)
Expected value
E(G) = 40·0.2 + 90·0.01 − 10·0.79 = 1 €
Answer
E(G) = 1 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 12.42
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 = (425 − 500)/50 = −1.5
Answer
z = −1.5
19.Normal distribution
Answer: P = 0.1587
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 = 750 g, σ = √(4² + 9²) = 9.849 g
Standardise
z = (759.8 − 750)/9.849 = 1
Upper tail
P = 1 − Φ(1) = 1 − 0.8413 = 0.1587
Answer
P = 0.1587
20.Normal distribution
Answer: x = 473 h
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.