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→∞ (1 + 0.5/x)2x. Give a decimal value.
f(x) = (1 + 0.5/x)2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 4x)/(3x²), with x in radians.
f(x) = (1 − cos 4x)/(3x²)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(3x2 + 4). Find f′(−2).
f(x) = ln(3x2 + 4)x = −2
f′(−2) =
4.DerivativesHard
The tangent line to f(x) = 2x2 − 4x − 6 at x = −1 crosses the y-axis at y = b. Find b.
f(x) = 2x2 − 4x − 6x = −1
b =
5.IntegralsMedium
Evaluate I = ∫02 ex dx.
f(x) = ex[0, 2]
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 1 − x² and the x-axis.
y = 1 − x²
A =
7.Complex numbersMedium
The diagram shows z = −1 + j3. With w = 2∠−90°, click the point that represents z·w.
z = −1 + j3w = 2∠−90°
1A (3 + j)
2B (−6 − j2)
3C (6 + j2)
4D (−2 + j6)
8.Complex numbersHard
Exactly one of the points A–D is a root of z⁶ = 64∠−90°. Click it.
w = 64∠−90°n = 6
1A (2∠−45°)
2B (3.6∠−45°)
3C (2∠−75°)
4D (2∠75°)
9.Differential equationsMedium
A forged steel part at 480 °C is left to cool in a workshop at 29 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.01 min⁻¹, find its temperature T after 54 min.
T0 = 480 °CTa = 29 °Ck = 0.01 min⁻¹t = 54 min
T = °C
10.Differential equationsHard
Solve y′ + 2y = 6e−t with y(0) = 3. Find y(0.5).
y′ + 2y = 6e−ty(0) = 3
y(0.5) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [3 −3 −2; 1 1 −5; −5 −4 1]
det A =
12.Matrices & determinantsHard
Solve the system −3x − 2y = −7 and −3x + 2y = 12. Find x.
−3x − 2y = −7−3x + 2y = 12
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−1, 3, −3) and b = (−3, 3, −6), in degrees.
a = (−1, 3, −3)b = (−3, 3, −6)
θ = °
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (5, −4, 2) and b = (−1, 4, −4) and has a positive z-component. Find its y-component n̂y.
a = (5, −4, 2)b = (−1, 4, −4)n̂z > 0
n̂y =
15.ProbabilityMedium
A bag holds 5 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
5 / 6
P =
16.ProbabilityHard
A game costs €1 to play. You win €2 with probability 0.3 and €10 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 = 0.3€10: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 16, 37, 33, 24, 57, 52, 58.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 5. What proportion P of values is above 50?
μ = 60σ = 5x = 50
P =
19.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 62.5 and 80.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 62.5x2 = 80.5
P =
20.Normal distributionHard
Hole diameters D are normal with mean 25.02 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 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 = 0.01 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
3Solutions
1.Limits
Answer: L = 2.718
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).
Take the logarithm
ln L = limx→∞ 2·x·ln(1 + 0.5/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 2·0.5 = 1
Undo the logarithm
L = e1 = 2.718
Answer
L = 2.718
2.Limits
Answer: L = 2.667
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
4·sin 4x/(6x) → 0/0
Second application
16·cos 4x/6
Substitute x = 0
L = 16/6 = 2.667
Answer
L = 2.667
3.Derivatives
Answer: f′(−2) = −0.75
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) = 6x/(3x2 + 4)
Substitute
f′(−2) = −12/16 = −0.75
Answer
f′(−2) = −0.75
4.Derivatives
Answer: b = −8
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) = 0
f′(x) = 4x − 4 → f′(−1) = −8
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 0 − (−8)·(−1) = −8
Answer
b = −8
5.Integrals
Answer: I = 6.389
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 = (e2 − 1)/1 = (7.389 − 1)/1 = 6.389
Answer
I = 6.389
6.Integrals
Answer: A = 1.333
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
1 − x² = 0 → x = ±1
Integrate
A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333
Answer
A = 1.333
7.Complex numbers
Answer: C (6 + j2)
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.
Effect of w
|z·w| = 2·|z|, arg(z·w) = arg z + (−90)°
The point
z·w = 6 + j2 → C
Answer
C (6 + j2)
8.Complex numbers
Answer: C (2∠−75°)
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 = −15°, 45°, 105°, 165°, −135°, −75°
The point
C = 2∠−75°
Answer
C (2∠−75°)
9.Differential equations
Answer: T = 291.8 °C
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.
Solution of the cooling law
T(t) = 29 + 451·e−0.01t
Substitute t
T(54) = 29 + 451·0.5827 = 291.8 °C
Answer
T = 291.8 °C
10.Differential equations
Answer: y(0.5) = 2.536
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.
Integrating factor
μ = e2t
(e2ty)′ = 6et
General solution
y = 6/1·e−t + C·e−2t = 6·e−t + C·e−2t
Initial condition
3 = 6 + C → C = −3
Evaluate
y(0.5) = 6·0.6065 + (−3)·0.3679 = 2.536
Answer
y(0.5) = 2.536
11.Matrices & determinants
Answer: det A = −131
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 = −19, M₁₂ = −24, M₁₃ = 1
Cofactor expansion
det A = 3·(−19) − (−3)·(−24) + (−2)·1 = −131
Answer
det A = −131
12.Matrices & determinants
Answer: x = −0.8333
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 = (−3)·2 − (−2)·(−3) = −12
Replace the x column
det Aₓ = (−7)·2 − (−2)·12 = 10
x = 10/(−12) = −0.8333
Answer
x = −0.8333
y = 4.75
13.Vectors: dot & cross product
Answer: θ = 20.51 °
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 = (−1)·(−3) + 3·3 + (−3)·(−6) = 30
Magnitudes
|a| = √((−1)² + 3² + (−3)²) = 4.359
|b| = √((−3)² + 3² + (−6)²) = 7.348
Angle
cos θ = 30/(4.359·7.348) = 0.9366
θ = arccos(0.9366) = 20.51°
Answer
θ = 20.51°
14.Vectors: dot & cross product
Answer: n̂y = 0.7093
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 = (−4)·(−4) − 2·4 = 8
(a × b)y = azbx − axbz = 2·(−1) − 5·(−4) = 18
(a × b)z = axby − aybx = 5·4 − (−4)·(−1) = 16
Normalise
The z-component of a × b is already positive.
|a × b| = √(8² + 18² + 16²) = 25.38
n̂ = (8, 18, 16)/25.38 = (0.3152, 0.7093, 0.6305)
n̂y = 0.7093
Answer
n̂y = 0.7093
15.Probability
Answer: P = 0.1818
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 = 5/11 · 4/10 = 20/110 = 0.1818
Answer
P = 0.1818
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 = 0.3), 9 (p = 0.02), −1 (p = 0.68)
Expected value
E(G) = 1·0.3 + 9·0.02 − 1·0.68 = −0.2 €
Answer
E(G) = −0.2 €
A negative value means you lose money on average.
17.Descriptive statistics
Answer: s = 16.56
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 = −2
Standard normal table
Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772
Answer
P = 0.9772
19.Normal distribution
Answer: P = 0.5381
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 = (62.5 − 65)/10 = −0.25
z₂ = (80.5 − 65)/10 = 1.55
Area between
P = Φ(1.55) − Φ(−0.25) = 0.9394 − 0.4013 = 0.5381
Answer
P = 0.5381
20.Normal distribution
Answer: P = 0.9207
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).