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 (e−x − 1 + x)/x².
f(x) = (e−x − 1 + x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(2x), with x in radians. Find f′(1).
f(x) = x·sin(2x)x = 1 rad
f′(1) =
4.DerivativesHard
Let f(x) = x3 + 9x2 + 24x − 4. Find the x-coordinate x of the local maximum.
f(x) = x3 + 9x2 + 24x − 4
x =
5.IntegralsMedium
Evaluate I = ∫03 e0.5x dx.
f(x) = e0.5x[0, 3]
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 + j6 in degrees, as the principal value between −180° and 180°.
z = −1 + j6
θ = °
8.Complex numbersHard
Solve z² − 6z + 25 = 0. The roots are a ± jb; find the real part a.
p = −6q = 25
a =
9.Differential equationsMedium
Solve y″ + 8y′ + 15y = 0 with y(0) = 1 and y′(0) = 2. Find y(0.5).
y″ + 8y′ + 15y = 0y(0) = 1y′(0) = 2
y(0.5) =
10.Differential equationsHard
A system obeys y″ + 4y′ + 13y = 130 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1); angles in radians.
y″ + 4y′ + 13y = 130y(0) = 0y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [5 1 −5; 2 2 5; 4 −2 1]
det A =
12.Matrices & determinantsHard
Solve the system 4x + y = −2 and −4x − 4y = −6. Find x.
4x + y = −2−4x − 4y = −6
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (−6, −5, 6) and b = (k, 1, −3) are perpendicular.
a = (−6, −5, 6)b = (k, 1, −3)
k =
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (3, −3, −4), b = (−2, 3, −3) and c = (−3, −4, 2).
a = (3, −3, −4)b = (−2, 3, −3)c = (−3, −4, 2)
V =
15.ProbabilityMedium
In how many ways N can a committee of 2 people be chosen from a group of 7? The order does not matter.
n = 7k = 2
N =
16.ProbabilityHard
A game costs €3 to play. You win €12 with probability 0.25 and €30 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€12: p = 0.25€30: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 25, 31, 13, 20, 18, 25, 37.
n = 7
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 5. Find the z-score z of a score of 172.
μ = 170σ = 5x = 172
z =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.3401. Find z to two decimals.
A = 0.3401
z =
20.Normal distributionHard
Hole diameters D are normal with mean 25.02 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.02 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
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 = 0.5
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
(−1·e−x + 1)/(2x) → 0/0
Second application
1·e−x/2
Substitute x = 0
L = 1·1/2 = 0.5
Answer
L = 0.5
3.Derivatives
Answer: f′(1) = 0.077
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) = 0.9093 + 2·(−0.4161) = 0.077
Answer
f′(1) = 0.077
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 + 18x + 24 = 3·(x2 + 6x + 8)
Solve f′(x) = 0
x = −4 or x = −2
Which one is which
f″(x) = 6x + 18: negative at a maximum, positive at a minimum.
f″(−4) = −6, f″(−2) = 6
x = −4
Answer
x = −4
5.Integrals
Answer: I = 6.963
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) = e0.5x/0.5
Evaluate at the limits
I = (e1.5 − 1)/0.5 = (4.482 − 1)/0.5 = 6.963
Answer
I = 6.963
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 = 99.46 °
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 = 6 → II
Reference angle
α = atan(6/1) = 80.54°
Argument
θ = 180° − α = 99.46°
Answer
θ = 99.46°
In polar form z = 6.083∠99.46°.
8.Complex numbers
Answer: a = 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.
Discriminant
p² − 4q = 36 − 100 = −64 < 0
Quadratic formula
z = (6 ± j√64)/2 = 3 + j4, 3 − j4
Answer
a = 3
Answer
a = 3
9.Differential equations
Answer: y(0.5) = 0.5757
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 + 3)(r + 5) = 0 → r = −3, −5
Constants from the initial values
A + B = 1, −3A − 5B = 2
A = 3.5, B = −2.5
Evaluate
y(0.5) = 3.5·e−1.5 + (−2.5)·e−2.5 = 0.5757
Answer
y(0.5) = 0.5757
10.Differential equations
Answer: y(1) = 11.21
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 = 10, r = −2 ± 3j
Constants from the initial values
y(0) = 10 + A = 0 → A = −10
y′(0) = −2A + 3B = 0 → B = −6.667
Evaluate
y(1) = 10·(1 − e−2(cos 3 + 0.6667·sin 3)) = 11.21
Answer
y(1) = 11.21
11.Matrices & determinants
Answer: det A = 138
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 = 12, M₁₂ = −18, M₁₃ = −12
Cofactor expansion
det A = 5·12 − 1·(−18) + (−5)·(−12) = 138
Answer
det A = 138
12.Matrices & determinants
Answer: x = −1.167
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 = 4·(−4) − 1·(−4) = −12
Replace the x column
det Aₓ = (−2)·(−4) − 1·(−6) = 14
x = 14/(−12) = −1.167
Answer
x = −1.167
y = 2.667
13.Vectors: dot & cross product
Answer: k = −3.833
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 = (−6)·k + (−5)·1 + 6·(−3) = 0
−23 + (−6)·k = 0
Solve for k
k = −(−23)/(−6) = −3.833
Answer
k = −3.833
14.Vectors: dot & cross product
Answer: V = 125
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 = 3·2 − (−3)·(−4) = −6
(b × c)y = bzcx − bxcz = (−3)·(−3) − (−2)·2 = 13
(b × c)z = bxcy − bycx = (−2)·(−4) − 3·(−3) = 17
Dot product with a
a·(b × c) = 3·(−6) + (−3)·13 + (−4)·17 = −125
V = |−125| = 125
Answer
V = 125
15.Probability
Answer: N = 21
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, 2) = (7·6)/(2·1) = 21
Answer
N = 21
16.Probability
Answer: E(G) = 1.5 €
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: 9 (p = 0.25), 27 (p = 0.05), −3 (p = 0.7)
Expected value
E(G) = 9·0.25 + 27·0.05 − 3·0.7 = 1.5 €
Answer
E(G) = 1.5 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 8.092
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 = (172 − 170)/5 = 0.4
Answer
z = 0.4
19.Normal distribution
Answer: z = 0.44
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 + 0.3401)/2 = 0.6700
Read the z-table backwards
Find 0.6700 inside the table and read off z.
z = 0.44
Answer
z = 0.44
20.Normal distribution
Answer: P = 0.7611
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).