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→∞ (x3 − x + 8)/(5x3 + x + 3).
f(x) = (x3 − x + 8)/(5x3 + x + 3)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 2x)/(2x²), with x in radians.
f(x) = (1 − cos 2x)/(2x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(2x), with x in radians. Find f′(1.5).
f(x) = x·sin(2x)x = 1.5 rad
f′(1.5) =
4.DerivativesHard
Let f(x) = x3 + 1.5x2 − 6x + 6. Find the x-coordinate x of the local minimum.
f(x) = x3 + 1.5x2 − 6x + 6
x =
5.IntegralsMedium
Evaluate I = ∫01 ex dx.
f(x) = ex[0, 1]
I =
6.IntegralsHard
Evaluate I = ∫₀2.5 x²·eˣ dx.
∫₀2.5 x²eˣ dx
I =
7.Complex numbersMedium
Find the argument θ of z = −9 + j4 in degrees, as the principal value between −180° and 180°.
z = −9 + j4
θ = °
8.Complex numbersHard
One root of z³ + 5z² + 16z + 30 = 0 is z₁ = −1 + j3. The coefficients are real. Find the real root r.
z₁ = −1 + j3
r =
9.Differential equationsMedium
Solve y″ + 6y′ + 9y = 0 with y(0) = 2 and y′(0) = -2. Find y(1.5).
y″ + 6y′ + 9y = 0y(0) = 2y′(0) = -2
y(1.5) =
10.Differential equationsHard
A system obeys y″ + y′ + 4.25y = 4.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1); angles in radians.
y″ + y′ + 4.25y = 4.25y(0) = 0y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [0 −1 −5; 4 −3 −2; 2 1 −3]
det A =
12.Matrices & determinantsHard
Solve the system −2x + y = 8 and 6x + 4y = 1. Find x.
−2x + y = 86x + 4y = 1
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (−5, −6, −1) and b = (k, −4, 6) are perpendicular.
a = (−5, −6, −1)b = (k, −4, 6)
k =
14.Vectors: dot & cross productHard
For a = (2, −5, 3) and b = (4, 4, −2), find the x-component (a × b)x of the cross product a × b.
a = (2, −5, 3)b = (4, 4, −2)
(a × b)x =
15.ProbabilityMedium
In how many ways N can a committee of 2 people be chosen from a group of 8? The order does not matter.
n = 8k = 2
N =
16.ProbabilityHard
A game costs €5 to play. You win €25 with probability 0.2 and €250 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€25: p = 0.2€250: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 27, 55, 26, 25, 13, 49, 35.
n = 7
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 15. Find the z-score z of a score of 118.8.
μ = 100σ = 15x = 118.8
z =
19.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 442 and 534 h? Give P as a decimal.
μ = 500σ = 40x1 = 442x2 = 534
P =
20.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 80% of them lie (the 80th percentile).
μ = 32σ = 6P = 80%
x = min
3Solutions
1.Limits
Answer: L = 0.2
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).
Divide by the highest power
(1 + (−1)/x2 + 8/x3)/(5 + 1/x2 + 3/x3)
Let x → ∞
L = 1/5 = 0.2
Answer
L = 0.2
For equal degrees the limit is the ratio of the leading coefficients.
2.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).
First application
2·sin 2x/(4x) → 0/0
Second application
4·cos 2x/4
Substitute x = 0
L = 4/4 = 1
Answer
L = 1
3.Derivatives
Answer: f′(1.5) = −2.829
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 and chain rule
f′(x) = sin(2x) + 2x·cos(2x)
Substitute
f′(1.5) = 0.1411 + 3·(−0.99) = −2.829
Answer
f′(1.5) = −2.829
4.Derivatives
Answer: x = 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.
Derivative
f′(x) = 3x2 + 3x − 6 = 3·(x2 + x − 2)
Solve f′(x) = 0
x = −2 or x = 1
Which one is which
f″(x) = 6x + 3: negative at a maximum, positive at a minimum.
f″(−2) = −9, f″(1) = 9
x = 1
Answer
x = 1
5.Integrals
Answer: I = 1.718
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) = ex/1
Evaluate at the limits
I = (e1 − 1)/1 = (2.718 − 1)/1 = 1.718
Answer
I = 1.718
6.Integrals
Answer: I = 37.59
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.
First round
∫x²eˣ dx = x²eˣ − ∫2x eˣ dx
Second round
∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)
Limits
12.18·(3.25) − 1·2 = 37.59
Answer
I = 37.59
7.Complex numbers
Answer: θ = 156 °
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 = −9, b = 4 → II
Reference angle
α = atan(4/9) = 23.96°
Argument
θ = 180° − α = 156°
Answer
θ = 156°
In polar form z = 9.849∠156°.
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 = −2 + (−3) = −5 = −(5).
9.Differential equations
Answer: y(1.5) = 0.08887
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) = 2
y′(0) = B − 3A → B = -2 + 6 = 4
Evaluate
y(1.5) = (2 + 4·1.5)·e−4.5 = 0.08887
Answer
y(1.5) = 0.08887
10.Differential equations
Answer: y(1) = 1.115
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 = −0.5 ± 2j
Constants from the initial values
y(0) = 1 + A = 0 → A = −1
y′(0) = −0.5A + 2B = 0 → B = −0.25
Evaluate
y(1) = 1·(1 − e−0.5(cos 2 + 0.25·sin 2)) = 1.115
Answer
y(1) = 1.115
11.Matrices & determinants
Answer: det A = −58
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 = 11, M₁₂ = −8, M₁₃ = 10
Cofactor expansion
det A = 0·11 − (−1)·(−8) + (−5)·10 = −58
Answer
det A = −58
12.Matrices & determinants
Answer: x = −2.214
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 = (−2)·4 − 1·6 = −14
Replace the x column
det Aₓ = 8·4 − 1·1 = 31
x = 31/(−14) = −2.214
Answer
x = −2.214
y = 3.571
13.Vectors: dot & cross product
Answer: k = 3.6
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 equals zero
a·b = (−5)·k + (−6)·(−4) + (−1)·6 = 0
18 + (−5)·k = 0
Solve for k
k = −18/(−5) = 3.6
Answer
k = 3.6
14.Vectors: dot & cross product
Answer: (a × b)x = −2
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 = (−5)·(−2) − 3·4 = −2
(a × b)y = azbx − axbz = 3·4 − 2·(−2) = 16
(a × b)z = axby − aybx = 2·4 − (−5)·4 = 28
Requested component
(a × b)x = −2
Answer
(a × b)x = −2
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: N = 28
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.
Combinations
C(8, 2) = (8·7)/(2·1) = 28
Answer
N = 28
16.Probability
Answer: E(G) = 2.5 €
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 = 0.2), 245 (p = 0.01), −5 (p = 0.79)
Expected value
E(G) = 20·0.2 + 245·0.01 − 5·0.79 = 2.5 €
Answer
E(G) = 2.5 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 14.68
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 = (118.8 − 100)/15 = 1.25
Answer
z = 1.25
19.Normal distribution
Answer: P = 0.7288
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.
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.