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→∞ (3x2 − 2x + 6)/(2x2 + x + 3).
f(x) = (3x2 − 2x + 6)/(2x2 + x + 3)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 4x)/(x²), with x in radians.
f(x) = (1 − cos 4x)/(x²)x → 0
L =
3.DerivativesMedium
Let f(x) = (3x + 1)3. Find f′(0).
f(x) = (3x + 1)3x = 0
f′(0) =
4.DerivativesHard
The tangent line to f(x) = 2x2 + 4x − 2 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = 2x2 + 4x − 2x = 3
b =
5.IntegralsMedium
Evaluate I = ∫₁1.5 x·ln x dx.
∫₁1.5 x ln x dx
I =
6.IntegralsHard
The region between the line y = 3x and the parabola y = 2x² is rotated about the x-axis. Find the volume V of the solid.
y = 3xy = 2x²
V =
7.Complex numbersMedium
Write z = −8 + j8 in exponential form r·ejθ. Give θ in radians, between −π and π.
z = −8 + j8
θ = rad
8.Complex numbersHard
Solve z² + 5z + 31.25 = 0. The roots are a ± jb; find the real part a.
p = 5q = 31.25
a =
9.Differential equationsMedium
Solve y″ + 7y′ + 12y = 0 with y(0) = 5 and y′(0) = -2. Find y(0.5).
y″ + 7y′ + 12y = 0y(0) = 5y′(0) = -2
y(0.5) =
10.Differential equationsHard
Solve y″ + 3y′ + 11.25y = 0 with y(0) = 2 and y′(0) = 0 (released from rest). Find y(0.25); angles in radians.
y″ + 3y′ + 11.25y = 0y(0) = 2y′(0) = 0
y(0.25) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [3 1; −3 1]B = [−4 2; −1 −3]
c21 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [0 −2; 5 7]
λ =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−1, 4, −2) and b = (2, 6, −5), in degrees.
a = (−1, 4, −2)b = (2, 6, −5)
θ = °
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(−3, −4, −1), B(1, −2, 2) and C(1, −3, −2). Coordinates are in metres.
A(−3, −4, −1)B(1, −2, 2)C(1, −3, −2)
A = m²
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2, 3p(x): 0.3, 0.1, 0.3, 0.3
σ =
16.ProbabilityHard
A game costs €5 to play. You win €25 with probability 0.25 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 = 0.25€50: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 42, 37, 10, 29, 57.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 40. Find the z-score z of a score of 560.
μ = 500σ = 40x = 560
z =
19.Normal distributionMedium
X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 35.1 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 35.1 min
P =
20.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
3Solutions
1.Limits
Answer: L = 1.5
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
(3 + (−2)/x + 6/x2)/(2 + 1/x + 3/x2)
Let x → ∞
L = 3/2 = 1.5
Answer
L = 1.5
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 8
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/(2x) → 0/0
Second application
16·cos 4x/2
Substitute x = 0
L = 16/2 = 8
Answer
L = 8
3.Derivatives
Answer: f′(0) = 9
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′(0) = 3·12·3 = 9
Answer
f′(0) = 9
4.Derivatives
Answer: b = −20
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(3) = 28
f′(x) = 4x + 4 → f′(3) = 16
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 28 − 16·3 = −20
Answer
b = −20
5.Integrals
Answer: I = 0.1436
Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.
Choose u and dv
u = ln x, du = dx/x; dv = x dx, v = x²/2
Antiderivative
∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4
Limits
(1.125·ln 1.5 − 0.5625) − (0 − 0.25) = 0.1436
Answer
I = 0.1436
6.Integrals
Answer: V = 12.72
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
3x = 2x² → x = 0, x = 1.5
Washer method
V = π∫₀1.5 ((3x)² − (2x²)²) dx
Integrate
V = π(9·1.5³/3 − 4·1.5⁵/5) = π·4.05 = 12.72
Answer
V = 12.72
7.Complex numbers
Answer: θ = 2.356 rad
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
r = √(64 + 64) = 11.31
Argument in radians
θ = 135° = 135·π/180 = 2.356 rad
Exponential form
z = 11.31·ej2.356
Answer
θ = 2.356 rad
8.Complex numbers
Answer: a = −2.5
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 = 25 − 125 = −100 < 0
Quadratic formula
z = (−5 ± j√100)/2 = −2.5 + j5, −2.5 − j5
Answer
a = −2.5
Answer
a = −2.5
9.Differential equations
Answer: y(0.5) = 2.257
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 + 4) = 0 → r = −3, −4
Constants from the initial values
A + B = 5, −3A − 4B = -2
A = 18, B = −13
Evaluate
y(0.5) = 18·e−1.5 + (−13)·e−2 = 2.257
Answer
y(0.5) = 2.257
10.Differential equations
Answer: y(0.25) = 1.474
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.
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
c21 = (−3)·(−4) + 1·(−1) = 11
Answer
c21 = 11
12.Matrices & determinants
Answer: λ = 5
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 − (7)λ + (10) = 0
Solve
λ = (7 ± √(9))/2 → λ = 2, 5
λ = 5
Answer
λ = 5
13.Vectors: dot & cross product
Answer: θ = 29.99 °
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)·2 + 4·6 + (−2)·(−5) = 32
Magnitudes
|a| = √((−1)² + 4² + (−2)²) = 4.583
|b| = √(2² + 6² + (−5)²) = 8.062
Angle
cos θ = 32/(4.583·8.062) = 0.8661
θ = arccos(0.8661) = 29.99°
Answer
θ = 29.99°
14.Vectors: dot & cross product
Answer: A = 8.617 m²
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.
Edge vectors
AB = B − A = (4, 2, 3)
AC = C − A = (4, 1, −1)
Cross product
(AB × AC)x = AByACz − ABzACy = 2·(−1) − 3·1 = −5
(AB × AC)y = ABzACx − ABxACz = 3·4 − 4·(−1) = 16
(AB × AC)z = ABxACy − AByACx = 4·1 − 2·4 = −4
|AB × AC| = √((−5)² + 16² + (−4)²) = 17.23
Area of the triangle
A = ½·17.23 = 8.617 m²
Answer
A = 8.617 m²
15.Probability
Answer: σ = 1.2
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.
Mean
μ = 0·0.3 + 1·0.1 + 2·0.3 + 3·0.3 = 1.6
Mean of the squares
E(X²) = 0·0.3 + 1·0.1 + 4·0.3 + 9·0.3 = 4
Variance
Var(X) = 4 − 1.6² = 1.44
σ = √1.44 = 1.2
Answer
σ = 1.2
16.Probability
Answer: E(G) = 3.75 €
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 = 0.25), 45 (p = 0.05), −5 (p = 0.7)
Expected value
E(G) = 20·0.25 + 45·0.05 − 5·0.7 = 3.75 €
Answer
E(G) = 3.75 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 17.31
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 = (560 − 500)/40 = 1.5
Answer
z = 1.5
19.Normal distribution
Answer: P = 0.1075
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).
Distribution of the sum
μ = 32 min, σ = √(1.5² + 2²) = 2.5 min
Standardise
z = (35.1 − 32)/2.5 = 1.24
Upper tail
P = 1 − Φ(1.24) = 1 − 0.8925 = 0.1075
Answer
P = 0.1075
20.Normal distribution
Answer: P = 0.8413
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).