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→∞ (5x3 − x + 2)/(3x3 + 6x + 9).
f(x) = (5x3 − x + 2)/(3x3 + 6x + 9)x → ∞
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) = x2·e−x. Find f′(0.5).
f(x) = x2·e−xx = 0.5
f′(0.5) =
4.DerivativesHard
The tangent line to f(x) = 3x2 + 2x − 1 at x = 2 crosses the y-axis at y = b. Find b.
f(x) = 3x2 + 2x − 1x = 2
b =
5.IntegralsMedium
Evaluate I = ∫28 5/x dx.
f(x) = 5/x[2, 8]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = x.
y = x2y = x
A =
7.Complex numbersMedium
The diagram shows z = −2 + j. With w = 2∠90°, click the point that represents z·w.
z = −2 + jw = 2∠90°
1A (2 + j4)
2B (−4 + j2)
3C (−2 − j4)
4D (−1 − j2)
8.Complex numbersHard
Exactly one of the points A–D is a root of z³ = 27∠−90°. Click it.
w = 27∠−90°n = 3
1A (3∠180°)
2B (5.4∠−90°)
3C (3∠−150°)
4D (3∠−120°)
9.Differential equationsMedium
Solve y″ + 2y′ + y = 0 with y(0) = 3 and y′(0) = 2. Find y(1).
y″ + 2y′ + y = 0y(0) = 3y′(0) = 2
y(1) =
10.Differential equationsHard
Solve y″ + y′ + 4.25y = 0 with y(0) = 3 and y′(0) = 0 (released from rest). Find y(0.75); angles in radians.
y″ + y′ + 4.25y = 0y(0) = 3y′(0) = 0
y(0.75) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [0 0 1; 1 −5 3; 0 3 2]
det A =
12.Matrices & determinantsHard
Solve the system x − 5y = −8 and −x + 2y = −12. Find x.
x − 5y = −8−x + 2y = −12
x =
13.Vectors: dot & cross productMedium
A constant force F = (−15, 10, −25) N acts on an object while it moves in a straight line from point A(3, 1, −3) to point B(4, −2, −2). Coordinates are in metres. Find the work W done by the force.
F = (−15, 10, −25) NA(3, 1, −3) mB(4, −2, −2) m
W = J
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (0, 1, −3), b = (−5, 0, 3) and c = (2, 4, 4).
a = (0, 1, −3)b = (−5, 0, 3)c = (2, 4, 4)
V =
15.ProbabilityMedium
A bag holds 5 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
5 / 5
P =
16.ProbabilityHard
A game costs €2 to play. You win €10 with probability 0.1 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€10: p = 0.1€40: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 27, 21, 25, 49, 27.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 5. What proportion P of values is below 180?
μ = 170σ = 5x = 180
P =
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 764.8 g. Round z to two decimals.
μX = 250, σX = 4μY = 500, σY = 9X + Y > 764.8 g
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 75% of them lie (the 75th percentile).
μ = 250σ = 5P = 75%
x = g
3Solutions
1.Limits
Answer: L = 1.667
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).
Divide by the highest power
(5 + (−1)/x2 + 2/x3)/(3 + 6/x2 + 9/x3)
Let x → ∞
L = 5/3 = 1.667
Answer
L = 1.667
For equal degrees the limit is the ratio of the leading coefficients.
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) = 0.4549
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′ = e−x·(2x + (−1)·x2)
Substitute
f′(0.5) = 0.6065·(1 + (−0.25)) = 0.4549
Answer
f′(0.5) = 0.4549
4.Derivatives
Answer: b = −13
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) = 15
f′(x) = 6x + 2 → f′(2) = 14
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 15 − 14·2 = −13
Answer
b = −13
5.Integrals
Answer: I = 6.931
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) = 5·ln x
Evaluate at the limits
I = 5·(ln 8 − ln 2) = 5·ln 4 = 6.931
Answer
I = 6.931
6.Integrals
Answer: A = 0.1667
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
x2 = x → x = 0, x = 1
Integrate upper minus lower
A = [0.5x2 − 0.3333x3]01 = 0.1667
Answer
A = 0.1667
7.Complex numbers
Answer: C (−2 − j4)
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.
Effect of w
|z·w| = 2·|z|, arg(z·w) = arg z + 90°
The point
z·w = −2 − j4 → C
Answer
C (−2 − j4)
8.Complex numbers
Answer: C (3∠−150°)
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| = 3
θ = −30°, 90°, −150°
The point
C = 3∠−150°
Answer
C (3∠−150°)
9.Differential equations
Answer: y(1) = 2.943
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 + 1)² = 0 → r = −1
Constants
A = y(0) = 3
y′(0) = B − 1A → B = 2 + 3 = 5
Evaluate
y(1) = (3 + 5·1)·e−1 = 2.943
Answer
y(1) = 2.943
10.Differential equations
Answer: y(0.75) = 0.66
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.
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 = −19, M₁₂ = 2, M₁₃ = 3
Cofactor expansion
det A = 0·(−19) − 0·2 + 1·3 = 3
Answer
det A = 3
12.Matrices & determinants
Answer: x = 25.33
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 = 1·2 − (−5)·(−1) = −3
Replace the x column
det Aₓ = (−8)·2 − (−5)·(−12) = −76
x = −76/(−3) = 25.33
Answer
x = 25.33
y = 6.667
13.Vectors: dot & cross product
Answer: W = −70 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 = (4, −2, −2) − (3, 1, −3) = (1, −3, 1) m
Work
W = F·d = (−15)·1 + 10·(−3) + (−25)·1 = −70 J
Answer
W = −70 J
Negative work: the force acts mostly against the motion.
14.Vectors: dot & cross product
Answer: V = 86
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 = 0·4 − 3·4 = −12
(b × c)y = bzcx − bxcz = 3·2 − (−5)·4 = 26
(b × c)z = bxcy − bycx = (−5)·4 − 0·2 = −20
Dot product with a
a·(b × c) = 0·(−12) + 1·26 + (−3)·(−20) = 86
V = |86| = 86
Answer
V = 86
15.Probability
Answer: P = 0.2222
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.
First and second draw
P = 5/10 · 4/9 = 20/90 = 0.2222
Answer
P = 0.2222
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: 8 (p = 0.1), 38 (p = 0.02), −2 (p = 0.88)
Expected value
E(G) = 8·0.1 + 38·0.02 − 2·0.88 = −0.2 €
Answer
E(G) = −0.2 €
A negative value means you lose money on average.
17.Descriptive statistics
Answer: s = 11.01
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) = 0.9772
P = 0.9772
Answer
P = 0.9772
19.Normal distribution
Answer: P = 0.06681
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 = (764.8 − 750)/9.849 = 1.5
Upper tail
P = 1 − Φ(1.5) = 1 − 0.9332 = 0.06681
Answer
P = 0.06681
20.Normal distribution
Answer: x = 253.4 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.