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 ln(1 + 5x)/sin(4x), with x in radians.
f(x) = ln(1 + 5x)/sin(4x)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(2x2 + 4). Find f′(2).
f(x) = ln(2x2 + 4)x = 2
f′(2) =
4.DerivativesHard
The tangent line to f(x) = x2 + 3x + 2 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = x2 + 3x + 2x = 1
b =
5.IntegralsMedium
The region under y = √(x) from x = 0 to x = 4 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(x)0 ≤ x ≤ 4
V =
6.IntegralsHard
Find the area A of the region enclosed by y = 2x2 and y = 4x.
y = 2x2y = 4x
A =
7.Complex numbersMedium
Given z₁ = 1.5∠−60° and z₂ = 2∠−140°, find the imaginary part of w = z₁/z₂.
z₁ = 1.5∠−60°z₂ = 2∠−140°
Im(w) =
8.Complex numbersHard
Exactly one of the points A–D is a root of z⁴ = 16∠90°. Click it.
w = 16∠90°n = 4
1A (2∠90°)
2B (2∠157.5°)
3C (3.6∠−157.5°)
4D (2∠−157.5°)
9.Differential equationsMedium
Solve y″ + 4y′ + 3y = 0 with y(0) = 4 and y′(0) = 6. Find y(0.5).
y″ + 4y′ + 3y = 0y(0) = 4y′(0) = 6
y(0.5) =
10.Differential equationsHard
A system obeys y″ + 3y′ + 27.25y = 54.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 3y′ + 27.25y = 54.5y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [1 −5 4; −1 0 0; −3 0 5]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [0 −2; 3 −5]
λ =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (−2, −6, −4) onto b = (3, −2, 2), that is, the component of a in the direction of b.
a = (−2, −6, −4)b = (3, −2, 2)
compb a =
14.Vectors: dot & cross productHard
For a = (−1, 6, 5) and b = (−1, 5, 1), find the x-component (a × b)x of the cross product a × b.
a = (−1, 6, 5)b = (−1, 5, 1)
(a × b)x =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3p(x): 0.35, 0.3, 0.3, 0.05
Var(X) =
16.ProbabilityHard
A game costs €5 to play. You win €25 with probability 0.3 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€25: p = 0.3€100: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 15, 18, 41, 12, 41, 54.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 50. What proportion P of values is above 450?
μ = 500σ = 50x = 450
P =
19.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 70.75 and 122.5 points? Give P as a decimal.
μ = 100σ = 15x1 = 70.75x2 = 122.5
P =
20.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 5% of them lie (the 5th percentile).
μ = 500σ = 40P = 5%
x = h
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 = 1.25
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).
L'Hôpital's rule
L = limx→0 [5/(1 + 5x)]/[4·cos 4x]
Substitute x = 0
L = 5/4 = 1.25
Answer
L = 1.25
3.Derivatives
Answer: f′(2) = 0.6667
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 + 4)
Substitute
f′(2) = 8/12 = 0.6667
Answer
f′(2) = 0.6667
4.Derivatives
Answer: b = 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.
Point and slope
f(1) = 6
f′(x) = 2x + 3 → f′(1) = 5
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 6 − 5·1 = 1
Answer
b = 1
5.Integrals
Answer: V = 25.13
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∫₀4 x dx
Integrate
V = π·1·4²/2 = 25.13
Answer
V = 25.13
6.Integrals
Answer: A = 2.667
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
2x2 = 4x → x = 0, x = 2
Integrate upper minus lower
A = [2x2 − 0.6667x3]02 = 2.667
Answer
A = 2.667
7.Complex numbers
Answer: Im(w) = 0.7386
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.
Modulus and argument of w
|w| = 1.5 / 2 = 0.75
θ = −60° − (−140)° = 80°
Rectangular form
Re(w) = 0.75·cos(80°) = 0.1302
Im(w) = 0.75·sin(80°) = 0.7386
Im(w) = 0.7386
Answer
Im(w) = 0.7386
8.Complex numbers
Answer: D (2∠−157.5°)
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 = 22.5°, 112.5°, −157.5°, −67.5°
The point
D = 2∠−157.5°
Answer
D (2∠−157.5°)
9.Differential equations
Answer: y(0.5) = 4.343
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 + 3) = 0 → r = −1, −3
Constants from the initial values
A + B = 4, −1A − 3B = 6
A = 9, B = −5
Evaluate
y(0.5) = 9·e−0.5 + (−5)·e−1.5 = 4.343
Answer
y(0.5) = 4.343
10.Differential equations
Answer: ymax = 2.779
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 = 2, r = −1.5 ± 5j
Constants from the initial values
y(0) = 2 + A = 0 → A = −2
y′(0) = −1.5A + 5B = 0 → B = −0.6
Peak
t = π/5 = 0.6283
ymax = 2·(1 + e−1.5·π/5) = 2.779
Answer
ymax = 2.779
11.Matrices & determinants
Answer: det A = −25
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 = 0, M₁₂ = −5, M₁₃ = 0
Cofactor expansion
det A = 1·0 − (−5)·(−5) + 4·0 = −25
Answer
det A = −25
12.Matrices & determinants
Answer: λ = −2
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 − (−5)λ + (6) = 0
Solve
λ = (−5 ± √(1))/2 → λ = −3, −2
λ = −2
Answer
λ = −2
13.Vectors: dot & cross product
Answer: compb a = −0.4851
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.
Dot product
a·b = (−2)·3 + (−6)·(−2) + (−4)·2 = −2
Length of b
|b| = √(3² + (−2)² + 2²) = 4.123
Scalar projection
compb a = a·b/|b| = −2/4.123 = −0.4851
Answer
compb a = −0.4851
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: (a × b)x = −19
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 = 6·1 − 5·5 = −19
(a × b)y = azbx − axbz = 5·(−1) − (−1)·1 = −4
(a × b)z = axby − aybx = (−1)·5 − 6·(−1) = 1
Requested component
(a × b)x = −19
Answer
(a × b)x = −19
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: Var(X) = 0.8475
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 = 0·0.35 + 1·0.3 + 2·0.3 + 3·0.05 = 1.05
Mean of the squares
E(X²) = 0·0.35 + 1·0.3 + 4·0.3 + 9·0.05 = 1.95
Variance
Var(X) = 1.95 − 1.05² = 0.8475
Answer
Var(X) = 0.8475
16.Probability
Answer: E(G) = 4.5 €
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: 20 (p = 0.3), 95 (p = 0.02), −5 (p = 0.68)
Expected value
E(G) = 20·0.3 + 95·0.02 − 5·0.68 = 4.5 €
Answer
E(G) = 4.5 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 17.38
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 = −1
Standard normal table
Φ(−1) = 0.1587
P = 1 − 0.1587 = 0.8413
Answer
P = 0.8413
19.Normal distribution
Answer: P = 0.9076
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.
z-scores
z₁ = (70.75 − 100)/15 = −1.95
z₂ = (122.5 − 100)/15 = 1.5
Area between
P = Φ(1.5) − Φ(−1.95) = 0.9332 − 0.02559 = 0.9076
Answer
P = 0.9076
20.Normal distribution
Answer: x = 434.2 h
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.