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 − 5x) − x).
f(x) = √(x2 − 5x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 3x)/(x²), with x in radians.
f(x) = (1 − cos 3x)/(x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(x), with x in radians. Find f′(1.5).
f(x) = x·sin(x)x = 1.5 rad
f′(1.5) =
4.DerivativesHard
Let f(x) = x3 − 10.5x2 + 36x − 1. Find the x-coordinate x of the local minimum.
f(x) = x3 − 10.5x2 + 36x − 1
x =
5.IntegralsMedium
Evaluate I = ∫₁2.5 x·ln x dx.
∫₁2.5 x ln x dx
I =
6.IntegralsHard
Evaluate I = ∫₀3 x²·eˣ dx.
∫₀3 x²eˣ dx
I =
7.Complex numbersMedium
Given z₁ = 1.5∠−60° and z₂ = 4∠−25°, find the imaginary part of w = z₁/z₂.
z₁ = 1.5∠−60°z₂ = 4∠−25°
Im(w) =
8.Complex numbersHard
The roots of zⁿ = w with n = 5 and w = 32∠−120° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, …, 4. Find the imaginary part of z₁.
w = 32∠−120°n = 5k = 1
Im(z₁) =
9.Differential equationsMedium
Solve y″ + 4y′ + 4y = 0 with y(0) = 1 and y′(0) = -1. Find y(2).
y″ + 4y′ + 4y = 0y(0) = 1y′(0) = -1
y(2) =
10.Differential equationsHard
A tank holds 100 L of pure water. Brine with 0.05 kg of salt per litre flows in at 4 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 3 kg of salt?
V = 100 Lm0 = 0 kgc = 0.05 kg/Lr = 4 L/min
t = min
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [2 0 −4; −1 5 −1; 5 −3 0]
det A =
12.Matrices & determinantsHard
Solve the system 5x − 3y = −9 and 6x − 2y = 9. Find x.
5x − 3y = −96x − 2y = 9
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (2, −3, −6) and b = (k, 6, 4) are perpendicular.
a = (2, −3, −6)b = (k, 6, 4)
k =
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−1, −2, 3) and b = (−5, 1, 3) and has a positive z-component. Find its y-component n̂y.
a = (−1, −2, 3)b = (−5, 1, 3)n̂z > 0
n̂y =
15.ProbabilityMedium
In how many ways N can a committee of 4 people be chosen from a group of 7? The order does not matter.
n = 7k = 4
N =
16.ProbabilityHard
A game costs €1 to play. You win €2 with probability 0.1 and €50 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €1€2: p = 0.1€50: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 59, 60, 42, 41, 31, 32.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 12. What proportion P of values is above 82?
μ = 70σ = 12x = 82
P =
19.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.6950. Find z to two decimals.
A = 0.6950
z =
20.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 95% of them lie (the 95th percentile).
μ = 250σ = 5P = 95%
x = g
3Solutions
1.Limits
Answer: L = −2.5
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).
Multiply by the conjugate
f(x) = −5x/(√(x2 − 5x) + x)
Divide by x
f(x) = −5/(√(1 + (−5)/x) + 1)
Let x → ∞
L = −5/(1 + 1) = −2.5
Answer
L = −2.5
2.Limits
Answer: L = 4.5
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/(2x) → 0/0
Second application
9·cos 3x/2
Substitute x = 0
L = 9/2 = 4.5
Answer
L = 4.5
3.Derivatives
Answer: f′(1.5) = 1.104
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(x) + x·cos(x)
Substitute
f′(1.5) = 0.9975 + 1.5·0.07074 = 1.104
Answer
f′(1.5) = 1.104
4.Derivatives
Answer: x = 4
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 − 21x + 36 = 3·(x2 − 7x + 12)
Solve f′(x) = 0
x = 3 or x = 4
Which one is which
f″(x) = 6x − 21: negative at a maximum, positive at a minimum.
f″(3) = −3, f″(4) = 3
x = 4
Answer
x = 4
5.Integrals
Answer: I = 1.551
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
(3.125·ln 2.5 − 1.5625) − (0 − 0.25) = 1.551
Answer
I = 1.551
6.Integrals
Answer: I = 98.43
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
20.09·(5) − 1·2 = 98.43
Answer
I = 98.43
7.Complex numbers
Answer: Im(w) = −0.2151
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 / 4 = 0.375
θ = −60° − (−25)° = −35°
Rectangular form
Re(w) = 0.375·cos(−35°) = 0.3072
Im(w) = 0.375·sin(−35°) = −0.2151
Im(w) = −0.2151
Answer
Im(w) = −0.2151
8.Complex numbers
Answer: Im(z₁) = 1.486
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 of the roots
321/5 = 2
Argument of the root
θ₁ = (−120° + 360°·1)/5 = 48°
Rectangular form
z₁ = 2·(cos 48° + j sin 48°) = 1.338 + j1.486
Im(z₁) = 1.486
Answer
Im(z₁) = 1.486
The 5 roots lie on the circle of radius 2, 72° apart.
9.Differential equations
Answer: y(2) = 0.05495
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 + 2)² = 0 → r = −2
Constants
A = y(0) = 1
y′(0) = B − 2A → B = -1 + 2 = 1
Evaluate
y(2) = (1 + 1·2)·e−4 = 0.05495
Answer
y(2) = 0.05495
10.Differential equations
Answer: t = 22.91 min
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.
Result
m(t) = 5 + (−5)·e−t/25
Solve for t
e−t/25 = (5 − 3)/(5 − 0) = 0.4
t = 25·ln(2.5) = 22.91 min
Answer
t = 22.91 min
11.Matrices & determinants
Answer: det A = 82
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 = −3, M₁₂ = 5, M₁₃ = −22
Cofactor expansion
det A = 2·(−3) − 0·5 + (−4)·(−22) = 82
Answer
det A = 82
12.Matrices & determinants
Answer: x = 5.625
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·(−2) − (−3)·6 = 8
Replace the x column
det Aₓ = (−9)·(−2) − (−3)·9 = 45
x = 45/8 = 5.625
Answer
x = 5.625
y = 12.38
13.Vectors: dot & cross product
Answer: k = 21
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 = 2·k + (−3)·6 + (−6)·4 = 0
−42 + 2·k = 0
Solve for k
k = −(−42)/2 = 21
Answer
k = 21
14.Vectors: dot & cross product
Answer: n̂y = 0.6451
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
(a × b)x = aybz − azby = (−2)·3 − 3·1 = −9
(a × b)y = azbx − axbz = 3·(−5) − (−1)·3 = −12
(a × b)z = axby − aybx = (−1)·1 − (−2)·(−5) = −11
Normalise
The z-component of a × b is negative, so use −(a × b).
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(7, 4) = (7·6·5·4)/(4·3·2·1) = 35
Answer
N = 35
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: 1 (p = 0.1), 49 (p = 0.02), −1 (p = 0.88)
Expected value
E(G) = 1·0.1 + 49·0.02 − 1·0.88 = 0.2 €
Answer
E(G) = 0.2 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 12.7
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) = 0.8413
P = 1 − 0.8413 = 0.1587
Answer
P = 0.1587
19.Normal distribution
Answer: z = −0.51
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 − 0.6950 = 0.3050
Read the z-table backwards
Find 0.3050 inside the table and read off z.
z = −0.51
Answer
z = −0.51
20.Normal distribution
Answer: x = 258.2 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.