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 − 2/x)3x. Give a decimal value.
f(x) = (1 − 2/x)3xx → ∞
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) = x2·ex. Find f′(0.5).
f(x) = x2·exx = 0.5
f′(0.5) =
4.DerivativesHard
The tangent line to f(x) = 2x2 − 4x + 1 at x = −2 crosses the y-axis at y = b. Find b.
f(x) = 2x2 − 4x + 1x = −2
b =
5.IntegralsMedium
The region under y = 3x from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a cone).
y = 3x0 ≤ x ≤ 1
V =
6.IntegralsHard
Find the area A of the region between the parabola y = 4 − x² and the x-axis.
y = 4 − x²
A =
7.Complex numbersMedium
Given z₁ = 2∠55° and z₂ = 0.5∠−25°, find the imaginary part of w = z₁·z₂.
z₁ = 2∠55°z₂ = 0.5∠−25°
Im(w) =
8.Complex numbersHard
Find the real part of (−2 − j3)⁵.
z = −2 − j3n = 5
Re(z⁵) =
9.Differential equationsMedium
Solve y″ + 5y′ + 6y = 0 with y(0) = 1 and y′(0) = 0. Find y(1).
y″ + 5y′ + 6y = 0y(0) = 1y′(0) = 0
y(1) =
10.Differential equationsHard
A system obeys y″ + y′ + 16.25y = 16.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + y′ + 16.25y = 16.25y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [0 −3 0; 0 3 3; −3 −4 4]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−1 −2; −2 −1]
λ =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (3, 3, 0) and b = (k, −5, −5) are perpendicular.
a = (3, 3, 0)b = (k, −5, −5)
k =
14.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (3, −3, 6) and b = (3, −2, 6).
a = (3, −3, 6)b = (3, −2, 6)
A =
15.ProbabilityMedium
A fair die is rolled 2 times. What is the probability P of getting at least one six?
n = 2
P =
16.ProbabilityHard
10 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?
n = 10p = 0.4
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 34, 52, 58, 30, 54, 34.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 5. What proportion P of values is above 76.25?
μ = 70σ = 5x = 76.25
P =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.9676. Find z to two decimals.
A = 0.9676
z =
20.Normal distributionHard
Hole diameters D are normal with mean 25.02 mm and standard deviation 0.015 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 = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
3Solutions
1.Limits
Answer: L = 0.002479
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→∞ 3·x·ln(1 − 2/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 3·(−2) = −6
Undo the logarithm
L = e−6 = 0.002479
Answer
L = 0.002479
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′(0.5) = 2.061
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 rule
f′(x) = u′v + uv′ = ex·(2x + 1·x2)
Substitute
f′(0.5) = 1.649·(1 + 0.25) = 2.061
Answer
f′(0.5) = 2.061
4.Derivatives
Answer: b = −7
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) = 17
f′(x) = 4x − 4 → f′(−2) = −12
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 17 − (−12)·(−2) = −7
Answer
b = −7
5.Integrals
Answer: V = 9.425
Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.
Disc method
V = π∫₀1 9x² dx
Integrate
V = π·9·1³/3 = 9.425
Answer
V = 9.425
Check: a cone of radius 3 and height 1 has (1/3)πr²h = 9.425.
6.Integrals
Answer: A = 10.67
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
4 − x² = 0 → x = ±2
Integrate
A = [4x − x³/3]−22 = 5.333 − (−5.333) = 10.67
Answer
A = 10.67
7.Complex numbers
Answer: Im(w) = 0.5
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| = 2 · 0.5 = 1
θ = 55° + (−25)° = 30°
Rectangular form
Re(w) = 1·cos(30°) = 0.866
Im(w) = 1·sin(30°) = 0.5
Im(w) = 0.5
Answer
Im(w) = 0.5
8.Complex numbers
Answer: Re(z⁵) = −122
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.
Powers step by step
z² = (−2 − j3)(−2 − j3) = −5 + j12
z³ = (−5 + j12)(−2 − j3) = 46 − j9
z⁴ = (46 − j9)(−2 − j3) = −119 − j120
z⁵ = (−119 − j120)(−2 − j3) = −122 + j597
Check with De Moivre
z = 3.606∠−123.7° → z⁵ = 609.3∠101.5°
Re(z⁵) = −122
Answer
Re(z⁵) = −122
9.Differential equations
Answer: y(1) = 0.3064
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 + 2)(r + 3) = 0 → r = −2, −3
Constants from the initial values
A + B = 1, −2A − 3B = 0
A = 3, B = −2
Evaluate
y(1) = 3·e−2 + (−2)·e−3 = 0.3064
Answer
y(1) = 0.3064
10.Differential equations
Answer: ymax = 1.675
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 ± 4j
Constants from the initial values
y(0) = 1 + A = 0 → A = −1
y′(0) = −0.5A + 4B = 0 → B = −0.125
Peak
t = π/4 = 0.7854
ymax = 1·(1 + e−0.5·π/4) = 1.675
Answer
ymax = 1.675
11.Matrices & determinants
Answer: det A = 27
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 = 24, M₁₂ = 9, M₁₃ = 9
Cofactor expansion
det A = 0·24 − (−3)·9 + 0·9 = 27
Answer
det A = 27
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 − (−2)λ + (−3) = 0
Solve
λ = (−2 ± √(16))/2 → λ = −3, 1
λ = 1
Answer
λ = 1
13.Vectors: dot & cross product
Answer: k = 5
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 = 3·k + 3·(−5) + 0·(−5) = 0
−15 + 3·k = 0
Solve for k
k = −(−15)/3 = 5
Answer
k = 5
14.Vectors: dot & cross product
Answer: A = 6.708
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 = (−3)·6 − 6·(−2) = −6
(a × b)y = azbx − axbz = 6·3 − 3·6 = 0
(a × b)z = axby − aybx = 3·(−2) − (−3)·3 = 3
Area
A = |a × b| = √((−6)² + 0² + 3²) = 6.708
Answer
A = 6.708
15.Probability
Answer: P = 0.3056
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.
Complement
P(no six) = (5/6)2 = 0.6944
P = 1 − 0.6944 = 0.3056
Answer
P = 0.3056
16.Probability
Answer: P = 0.04636
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.
Zero and one failure
P(X = 0) = 0.610 = 0.006047
P(X = 1) = 10·0.4·0.69 = 0.04031
Add them
P = 0.006047 + 0.04031 = 0.04636
Answer
P = 0.04636
17.Descriptive statistics
Answer: s = 12.29
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 = 1.25
Standard normal table
Φ(1.25) = 0.8944
P = 1 − 0.8944 = 0.1056
Answer
P = 0.1056
19.Normal distribution
Answer: z = 2.14
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 + 0.9676)/2 = 0.9838
Read the z-table backwards
Find 0.9838 inside the table and read off z.
z = 2.14
Answer
z = 2.14
20.Normal distribution
Answer: P = 0.8665
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).