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→∞ (−2x2 + 3x − 4)/(4x2 + 4x + 4).
f(x) = (−2x2 + 3x − 4)/(4x2 + 4x + 4)x → ∞
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) = (3x + 1)3. Find f′(1).
f(x) = (3x + 1)3x = 1
f′(1) =
4.DerivativesHard
The tangent line to f(x) = −x2 + 3x − 5 at x = −2 crosses the y-axis at y = b. Find b.
f(x) = −x2 + 3x − 5x = −2
b =
5.IntegralsMedium
Evaluate I = ∫13 1/x dx.
f(x) = 1/x[1, 3]
I =
6.IntegralsHard
The region between the line y = 4x and the parabola y = x² is rotated about the x-axis. Find the volume V of the solid.
y = 4xy = x²
V =
7.Complex numbersMedium
Write z = 12∠25° in rectangular form a + jb and give the imaginary part b.
r = 12θ = 25°
b =
8.Complex numbersHard
Find the imaginary part of (−3 − j2)⁵.
z = −3 − j2n = 5
Im(z⁵) =
9.Differential equationsMedium
Solve y″ + 4y′ + 4y = 0 with y(0) = 5 and y′(0) = -2. Find y(2).
y″ + 4y′ + 4y = 0y(0) = 5y′(0) = -2
y(2) =
10.Differential equationsHard
Solve the initial value problem y′ + 3y/x = 2 for x > 0, with y(1) = 2. Find y(2).
y′ + 3y/x = 2y(1) = 2
y(2) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [3 −2; −6 −1]B = [−3 −5; −3 −1]
c11 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [5 1; 3 3]
λ =
13.Vectors: dot & cross productMedium
A constant force F = (−30, −10, −15) N acts on an object while it moves in a straight line from point A(−2, −4, 0) to point B(−2, −5, 2). Coordinates are in metres. Find the work W done by the force.
F = (−30, −10, −15) NA(−2, −4, 0) mB(−2, −5, 2) m
W = J
14.Vectors: dot & cross productHard
A force F = (10, 0, 50) N acts at point A(−1, −3, −4). Coordinates are in metres. Find the z-component Mz of the moment of the force about point O(2, 2, 3).
F = (10, 0, 50) NA(−1, −3, −4) mO(2, 2, 3) m
Mz = N·m
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2p(x): 0.35, 0.35, 0.3
σ =
16.ProbabilityHard
A game costs €5 to play. You win €25 with probability 0.2 and €50 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€25: p = 0.2€50: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 34, 53, 52, 60, 11, 32, 18.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 12. What proportion P of values is below 90.4?
μ = 100σ = 12x = 90.4
P =
19.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 60 and 85.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 60x2 = 85.5
P =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points40%
x = points
3Solutions
1.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).
Divide by the highest power
(−2 + 3/x + (−4)/x2)/(4 + 4/x + 4/x2)
Let x → ∞
L = −2/4 = −0.5
Answer
L = −0.5
For equal degrees the limit is the ratio of the leading coefficients.
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′(1) = 144
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 + 1)2·3
Substitute
f′(1) = 3·42·3 = 144
Answer
f′(1) = 144
4.Derivatives
Answer: b = −1
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) = −15
f′(x) = −2x + 3 → f′(−2) = 7
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −15 − 7·(−2) = −1
Answer
b = −1
5.Integrals
Answer: I = 1.099
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) = ln x
Evaluate at the limits
I = (ln 3 − ln 1) = ln 3 = 1.099
Answer
I = 1.099
6.Integrals
Answer: V = 428.9
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
4x = 1x² → x = 0, x = 4
Washer method
V = π∫₀4 ((4x)² − (1x²)²) dx
Integrate
V = π(16·4³/3 − 1·4⁵/5) = π·136.5 = 428.9
Answer
V = 428.9
7.Complex numbers
Answer: b = 5.071
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.
Components
a = 12·cos(25°) = 10.88
b = 12·sin(25°) = 5.071
Rectangular form
z = 10.88 + j5.071
b = 5.071
Answer
b = 5.071
8.Complex numbers
Answer: Im(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 = (−3 − j2)(−3 − j2) = 5 + j12
z³ = (5 + j12)(−3 − j2) = 9 − j46
z⁴ = (9 − j46)(−3 − j2) = −119 + j120
z⁵ = (−119 + j120)(−3 − j2) = 597 − j122
Check with De Moivre
z = 3.606∠−146.3° → z⁵ = 609.3∠−11.55°
Im(z⁵) = −122
Answer
Im(z⁵) = −122
9.Differential equations
Answer: y(2) = 0.3846
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.
Double root
(r + 2)² = 0 → r = −2
Constants
A = y(0) = 5
y′(0) = B − 2A → B = -2 + 10 = 8
Evaluate
y(2) = (5 + 8·2)·e−4 = 0.3846
Answer
y(2) = 0.3846
10.Differential equations
Answer: y(2) = 1.188
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 = e∫3/x dx = x3
Integrate
(x3y)′ = 2x3 → x3y = 2x4/4 + C
Initial condition
2 = 2/4 + C → C = 1.5
Evaluate
y(2) = (2·24/4 + 1.5)/23 = 1.188
Answer
y(2) = 1.188
11.Matrices & determinants
Answer: c11 = −3
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
c11 = 3·(−3) + (−2)·(−3) = −3
Answer
c11 = −3
12.Matrices & determinants
Answer: λ = 6
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 − (8)λ + (12) = 0
Solve
λ = (8 ± √(16))/2 → λ = 2, 6
λ = 6
Answer
λ = 6
13.Vectors: dot & cross product
Answer: W = −20 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 = (−2, −5, 2) − (−2, −4, 0) = (0, −1, 2) m
Work
W = F·d = (−30)·0 + (−10)·(−1) + (−15)·2 = −20 J
Answer
W = −20 J
Negative work: the force acts mostly against the motion.
14.Vectors: dot & cross product
Answer: Mz = 50 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 = (−1, −3, −4) − (2, 2, 3) = (−3, −5, −7) 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 = 0·0.35 + 1·0.35 + 2·0.3 = 0.95
Mean of the squares
E(X²) = 0·0.35 + 1·0.35 + 4·0.3 = 1.55
Variance
Var(X) = 1.55 − 0.95² = 0.6475
σ = √0.6475 = 0.8047
Answer
σ = 0.8047
16.Probability
Answer: E(G) = 2.5 €
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: 20 (p = 0.2), 45 (p = 0.05), −5 (p = 0.75)
Expected value
E(G) = 20·0.2 + 45·0.05 − 5·0.75 = 2.5 €
Answer
E(G) = 2.5 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 18.62
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 = −0.8
Standard normal table
Φ(−0.8) = 0.2119
P = 0.2119
Answer
P = 0.2119
19.Normal distribution
Answer: P = 0.6713
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 = (60 − 65)/10 = −0.5
z₂ = (85.5 − 65)/10 = 2.05
Area between
P = Φ(2.05) − Φ(−0.5) = 0.9798 − 0.3085 = 0.6713
Answer
P = 0.6713
20.Normal distribution
Answer: x = 67.5 points
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.