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→∞ (√(4x2 + 3x) − 2x).
f(x) = √(4x2 + 3x) − 2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e−2x − 1 + 2x)/x².
f(x) = (e−2x − 1 + 2x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = (3x + 3)3. Find f′(2).
f(x) = (3x + 3)3x = 2
f′(2) =
4.DerivativesHard
The tangent line to f(x) = 2x2 + 5x − 5 at x = −2 crosses the y-axis at y = b. Find b.
f(x) = 2x2 + 5x − 5x = −2
b =
5.IntegralsMedium
Evaluate I = ∫₁2 x·ln x dx.
∫₁2 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
Find the argument θ of z = 1 + j8 in degrees, as the principal value between −180° and 180°.
z = 1 + j8
θ = °
8.Complex numbersHard
Use De Moivre's theorem to find the real part of z⁶ for z = −√3 + j.
z = −√3 + jn = 6
Re(z⁶) =
9.Differential equationsMedium
A cup of coffee at 82 °C is left in a room at 18 °C. After 19 min its temperature is 38 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 82 °CTa = 18 °CT(19) = 38 °C
k = min⁻¹
10.Differential equationsHard
A system obeys y″ + 4y′ + 29y = 145 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 4y′ + 29y = 145y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [6 0; 5 0]B = [−3 6; 0 5]
c21 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−1 2; 1 0]
λ =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−6, 6, −3) and b = (−2, −5, 6), in degrees.
a = (−6, 6, −3)b = (−2, −5, 6)
θ = °
14.Vectors: dot & cross productHard
For a = (−1, −4, 2) and b = (−2, 3, −6), find the y-component (a × b)y of the cross product a × b.
a = (−1, −4, 2)b = (−2, 3, −6)
(a × b)y =
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 0.25 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 = 0.25€40: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 11, 41, 31, 13, 16, 23, 15.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 10. What proportion P of values is below 185?
μ = 170σ = 10x = 185
P =
19.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 490 and 510 h? Give P as a decimal.
μ = 500σ = 40x1 = 490x2 = 510
P =
20.Normal distributionHard
The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
3Solutions
1.Limits
Answer: L = 0.75
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) = 3x/(√(4x2 + 3x) + 2x)
Divide by x
f(x) = 3/(√(4 + 3/x) + 2)
Let x → ∞
L = 3/(2 + 2) = 0.75
Answer
L = 0.75
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
(−2·e−2x + 2)/(2x) → 0/0
Second application
4·e−2x/2
Substitute x = 0
L = 4·1/2 = 2
Answer
L = 2
3.Derivatives
Answer: f′(2) = 729
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
f′(x) = 3·(3x + 3)2·3
Substitute
f′(2) = 3·92·3 = 729
Answer
f′(2) = 729
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) = −7
f′(x) = 4x + 5 → f′(−2) = −3
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −7 − (−3)·(−2) = −13
Answer
b = −13
5.Integrals
Answer: I = 0.6363
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
(2·ln 2 − 1) − (0 − 0.25) = 0.6363
Answer
I = 0.6363
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: θ = 82.87 °
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 = 1, b = 8 → I
Reference angle
α = atan(8/1) = 82.87°
Argument
θ = α = 82.87°
Answer
θ = 82.87°
In polar form z = 8.062∠82.87°.
8.Complex numbers
Answer: Re(z⁶) = −64
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.
Polar form
z = −√3 + j = 2∠150°
De Moivre
z⁶ = (2)⁶∠900° = 64∠180°
Rectangular form
z⁶ = 64·(cos 180° + j sin 180°) = −64
Re(z⁶) = −64
Answer
Re(z⁶) = −64
9.Differential equations
Answer: k = 0.06122 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−19k = 20/64
Solve for k
k = ln(64/20)/19 = 0.06122 min⁻¹
Answer
k = 0.06122 min⁻¹
10.Differential equations
Answer: ymax = 6.423
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 = −2 ± 5j
Constants from the initial values
y(0) = 5 + A = 0 → A = −5
y′(0) = −2A + 5B = 0 → B = −2
Peak
t = π/5 = 0.6283
ymax = 5·(1 + e−2·π/5) = 6.423
Answer
ymax = 6.423
11.Matrices & determinants
Answer: c21 = −15
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.
Row times column
c21 = 5·(−3) + 0·0 = −15
Answer
c21 = −15
12.Matrices & determinants
Answer: λ = 1
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 − (−1)λ + (−2) = 0
Solve
λ = (−1 ± √(9))/2 → λ = −2, 1
λ = 1
Answer
λ = 1
13.Vectors: dot & cross product
Answer: θ = 119.7 °
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 = (−6)·(−2) + 6·(−5) + (−3)·6 = −36
Magnitudes
|a| = √((−6)² + 6² + (−3)²) = 9
|b| = √((−2)² + (−5)² + 6²) = 8.062
Angle
cos θ = −36/(9·8.062) = −0.4961
θ = arccos(−0.4961) = 119.7°
Answer
θ = 119.7°
14.Vectors: dot & cross product
Answer: (a × b)y = −10
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 = (−4)·(−6) − 2·3 = 18
(a × b)y = azbx − axbz = 2·(−2) − (−1)·(−6) = −10
(a × b)z = axby − aybx = (−1)·3 − (−4)·(−2) = −11
Requested component
(a × b)y = −10
Answer
(a × b)y = −10
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
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) = 1.3 €
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 = 0.25), 38 (p = 0.02), −2 (p = 0.73)
Expected value
E(G) = 8·0.25 + 38·0.02 − 2·0.73 = 1.3 €
Answer
E(G) = 1.3 €
A positive value means you win 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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 0.9332
Answer
P = 0.9332
19.Normal distribution
Answer: P = 0.1974
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 = (490 − 500)/40 = −0.25
z₂ = (510 − 500)/40 = 0.25
Area between
P = Φ(0.25) − Φ(−0.25) = 0.5987 − 0.4013 = 0.1974
Answer
P = 0.1974
20.Normal distribution
Answer: P = 0.7389
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).