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 + 3/x)2x. Give a decimal value.
f(x) = (1 + 3/x)2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 5x)/(3x²), with x in radians.
f(x) = (1 − cos 5x)/(3x²)x → 0
L =
3.DerivativesMedium
Let f(x) = (−2x + 1)3. Find f′(0).
f(x) = (−2x + 1)3x = 0
f′(0) =
4.DerivativesHard
The tangent line to f(x) = x2 + 3x − 4 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = x2 + 3x − 4x = 1
b =
5.IntegralsMedium
The region under y = √(2x) from x = 0 to x = 2 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(2x)0 ≤ x ≤ 2
V =
6.IntegralsHard
The region between the line y = x and the parabola y = x² is rotated about the x-axis. Find the volume V of the solid.
y = xy = x²
V =
7.Complex numbersMedium
Find the argument θ of z = −8 + j6 in degrees, as the principal value between −180° and 180°.
z = −8 + j6
θ = °
8.Complex numbersHard
Use De Moivre's theorem to find the real part of z⁷ for z = −1 + j.
z = −1 + jn = 7
Re(z⁷) =
9.Differential equationsMedium
Solve y″ + 5y′ + 4y = 0 with y(0) = 5 and y′(0) = 4. Find y(2).
y″ + 5y′ + 4y = 0y(0) = 5y′(0) = 4
y(2) =
10.Differential equationsHard
A system obeys y″ + 4y′ + 8y = 8 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 4y′ + 8y = 8y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−4 −1 −4; 4 −1 2; 0 3 −5]
det A =
12.Matrices & determinantsHard
Solve the system −6x + 6y = −2 and −6x + 2y = −2. Find x.
−6x + 6y = −2−6x + 2y = −2
x =
13.Vectors: dot & cross productMedium
A constant force F = (10, 10, 10) N acts on an object while it moves in a straight line from point A(4, −5, −5) to point B(−1, −1, 0). Coordinates are in metres. Find the work W done by the force.
F = (10, 10, 10) NA(4, −5, −5) mB(−1, −1, 0) m
W = J
14.Vectors: dot & cross productHard
A force F = (20, 50, −60) N acts at point A(−3, −1, 0). Coordinates are in metres. Find the magnitude |M| of the moment of the force about the origin O(0, 0, 0).
F = (20, 50, −60) NA(−3, −1, 0) mO(0, 0, 0) m
|M| = N·m
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3p(x): 0.3, 0.15, 0.2, 0.35
Var(X) =
16.ProbabilityHard
9 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?
n = 9p = 0.25
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 39, 29, 49, 29, 40.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 8. Find the z-score z of a score of 73.2.
μ = 70σ = 8x = 73.2
z =
19.Normal distributionMedium
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 172.75 and 192 cm? Give P as a decimal.
μ = 178σ = 7x1 = 172.75x2 = 192
P =
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 0.02 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 = 0.015 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
3Solutions
1.Limits
Answer: L = 403.4
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 + 3/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 2·3 = 6
Undo the logarithm
L = e6 = 403.4
Answer
L = 403.4
2.Limits
Answer: L = 4.167
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
5·sin 5x/(6x) → 0/0
Second application
25·cos 5x/6
Substitute x = 0
L = 25/6 = 4.167
Answer
L = 4.167
3.Derivatives
Answer: f′(0) = −6
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·(−2x + 1)2·(−2)
Substitute
f′(0) = 3·12·(−2) = −6
Answer
f′(0) = −6
4.Derivatives
Answer: b = −5
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) = 2x + 3 → f′(1) = 5
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 0 − 5·1 = −5
Answer
b = −5
5.Integrals
Answer: V = 12.57
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∫₀2 2x dx
Integrate
V = π·2·2²/2 = 12.57
Answer
V = 12.57
6.Integrals
Answer: V = 0.4189
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.
Intersection points
1x = 1x² → x = 0, x = 1
Washer method
V = π∫₀1 ((1x)² − (1x²)²) dx
Integrate
V = π(1·1³/3 − 1·1⁵/5) = π·0.1333 = 0.4189
Answer
V = 0.4189
7.Complex numbers
Answer: θ = 143.1 °
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 = −8, b = 6 → II
Reference angle
α = atan(6/8) = 36.87°
Argument
θ = 180° − α = 143.1°
Answer
θ = 143.1°
In polar form z = 10∠143.1°.
8.Complex numbers
Answer: Re(z⁷) = −8
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 = −1 + j = √2∠135°
De Moivre
z⁷ = (√2)⁷∠945° = 11.31∠−135°
Rectangular form
z⁷ = 11.31·(cos −135° + j sin −135°) = −8 − j8
Re(z⁷) = −8
Answer
Re(z⁷) = −8
9.Differential equations
Answer: y(2) = 1.082
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 + 1)(r + 4) = 0 → r = −1, −4
Constants from the initial values
A + B = 5, −1A − 4B = 4
A = 8, B = −3
Evaluate
y(2) = 8·e−2 + (−3)·e−8 = 1.082
Answer
y(2) = 1.082
10.Differential equations
Answer: ymax = 1.043
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 = −2 ± 2j
Constants from the initial values
y(0) = 1 + A = 0 → A = −1
y′(0) = −2A + 2B = 0 → B = −1
Peak
t = π/2 = 1.571
ymax = 1·(1 + e−2·π/2) = 1.043
Answer
ymax = 1.043
11.Matrices & determinants
Answer: det A = −64
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 = −1, M₁₂ = −20, M₁₃ = 12
Cofactor expansion
det A = (−4)·(−1) − (−1)·(−20) + (−4)·12 = −64
Answer
det A = −64
12.Matrices & determinants
Answer: x = 0.3333
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 = (−6)·2 − 6·(−6) = 24
Replace the x column
det Aₓ = (−2)·2 − 6·(−2) = 8
x = 8/24 = 0.3333
Answer
x = 0.3333
y = 0
13.Vectors: dot & cross product
Answer: W = 40 J
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.
Displacement
d = B − A = (−1, −1, 0) − (4, −5, −5) = (−5, 4, 5) m
Work
W = F·d = 10·(−5) + 10·4 + 10·5 = 40 J
Answer
W = 40 J
14.Vectors: dot & cross product
Answer: |M| = 230 N·m
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.
Position vector from O to A
r = A − O = (−3, −1, 0) − (0, 0, 0) = (−3, −1, 0) m
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.
Mean
μ = 0·0.3 + 1·0.15 + 2·0.2 + 3·0.35 = 1.6
Mean of the squares
E(X²) = 0·0.3 + 1·0.15 + 4·0.2 + 9·0.35 = 4.1
Variance
Var(X) = 4.1 − 1.6² = 1.54
Answer
Var(X) = 1.54
16.Probability
Answer: P = 0.3003
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) = 0.759 = 0.07508
P(X = 1) = 9·0.25·0.758 = 0.2253
Add them
P = 0.07508 + 0.2253 = 0.3003
Answer
P = 0.3003
17.Descriptive statistics
Answer: s = 8.438
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 = (73.2 − 70)/8 = 0.4
Answer
z = 0.4
19.Normal distribution
Answer: P = 0.7506
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 = (172.75 − 178)/7 = −0.75
z₂ = (192 − 178)/7 = 2
Area between
P = Φ(2) − Φ(−0.75) = 0.9772 − 0.2266 = 0.7506
Answer
P = 0.7506
20.Normal distribution
Answer: P = 0.7881
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).