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→∞ (√(x2 + 6x) − x).
f(x) = √(x2 + 6x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 5x)/sin(3x), with x in radians.
f(x) = ln(1 + 5x)/sin(3x)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(2x2 + 4). Find f′(3).
f(x) = ln(2x2 + 4)x = 3
f′(3) =
4.DerivativesHard
Let f(x) = x3 − 9x2 + 15x − 5. Find the x-coordinate x of the local maximum.
f(x) = x3 − 9x2 + 15x − 5
x =
5.IntegralsMedium
Evaluate I = ∫01 2x·(x² + 1)3 dx.
f(x) = 2x·(x² + 1)3[0, 1]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = 2x.
y = x2y = 2x
A =
7.Complex numbersMedium
Find the argument θ of z = −6 − j7 in degrees, as the principal value between −180° and 180°.
z = −6 − j7
θ = °
8.Complex numbersHard
Exactly one of the points A–D is a root of z³ = 27∠0°. Click it.
w = 27∠0°n = 3
1A (3∠60°)
2B (3∠−150°)
3C (5.4∠−120°)
4D (3∠−120°)
9.Differential equationsMedium
Solve y″ + 5y′ + 6y = 0 with y(0) = 1 and y′(0) = 4. Find y(1.5).
y″ + 5y′ + 6y = 0y(0) = 1y′(0) = 4
y(1.5) =
10.Differential equationsHard
A system obeys y″ + 2y′ + 10y = 100 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 2y′ + 10y = 100y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [2 −4 0; 0 −1 1; −4 −4 2]
det A =
12.Matrices & determinantsHard
Solve the system x + y = 1 and −2x − 5y = 6. Find x.
x + y = 1−2x − 5y = 6
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (−3, 5, −1) and b = (k, −3, 4) are perpendicular.
a = (−3, 5, −1)b = (k, −3, 4)
k =
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(1, 3, −4), B(2, 1, −3) and C(−1, 4, −2). Coordinates are in metres.
A(1, 3, −4)B(2, 1, −3)C(−1, 4, −2)
A = m²
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 10, 20, 30, 40p(x): 0.25, 0.15, 0.3, 0.3
Var(X) =
16.ProbabilityHard
A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 5 correct answers?
n = 11p = 0.6k = 5
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 18, 24, 20, 40, 26.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 15. Find the z-score z of a score of 72.5.
μ = 50σ = 15x = 72.5
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 171.35 and 179.4 cm? Give P as a decimal.
μ = 178σ = 7x1 = 171.35x2 = 179.4
P =
20.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 5% of them lie (the 5th percentile).
μ = 32σ = 6P = 5%
x = min
3Solutions
1.Limits
Answer: L = 3
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).
Multiply by the conjugate
f(x) = 6x/(√(x2 + 6x) + x)
Divide by x
f(x) = 6/(√(1 + 6/x) + 1)
Let x → ∞
L = 6/(1 + 1) = 3
Answer
L = 3
2.Limits
Answer: L = 1.667
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).
L'Hôpital's rule
L = limx→0 [5/(1 + 5x)]/[3·cos 3x]
Substitute x = 0
L = 5/3 = 1.667
Answer
L = 1.667
3.Derivatives
Answer: f′(3) = 0.5455
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) = 4x/(2x2 + 4)
Substitute
f′(3) = 12/22 = 0.5455
Answer
f′(3) = 0.5455
4.Derivatives
Answer: x = 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.
Derivative
f′(x) = 3x2 − 18x + 15 = 3·(x2 − 6x + 5)
Solve f′(x) = 0
x = 1 or x = 5
Which one is which
f″(x) = 6x − 18: negative at a maximum, positive at a minimum.
f″(1) = −12, f″(5) = 12
x = 1
Answer
x = 1
5.Integrals
Answer: I = 3.75
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.
Substitution
u = x² + 1, du = 2x dx
I = ∫12 u3 du
Integrate and evaluate
I = [u4/4]12 = (16 − 1)/4 = 3.75
Answer
I = 3.75
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.
Intersection points
x2 = 2x → x = 0, x = 2
Integrate upper minus lower
A = [x2 − 0.3333x3]02 = 1.333
Answer
A = 1.333
7.Complex numbers
Answer: θ = −130.6 °
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 = −6, b = −7 → III
Reference angle
α = atan(7/6) = 49.4°
Argument
θ = −(180° − α) = −130.6°
Answer
θ = −130.6°
In polar form z = 9.22∠−130.6°.
8.Complex numbers
Answer: D (3∠−120°)
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| = 3
θ = 0°, 120°, −120°
The point
D = 3∠−120°
Answer
D (3∠−120°)
9.Differential equations
Answer: y(1.5) = 0.2819
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 = 4
A = 7, B = −6
Evaluate
y(1.5) = 7·e−3 + (−6)·e−4.5 = 0.2819
Answer
y(1.5) = 0.2819
10.Differential equations
Answer: ymax = 13.51
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 = 10, r = −1 ± 3j
Constants from the initial values
y(0) = 10 + A = 0 → A = −10
y′(0) = −1A + 3B = 0 → B = −3.333
Peak
t = π/3 = 1.047
ymax = 10·(1 + e−1·π/3) = 13.51
Answer
ymax = 13.51
11.Matrices & determinants
Answer: det A = 20
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 = 2, M₁₂ = 4, M₁₃ = −4
Cofactor expansion
det A = 2·2 − (−4)·4 + 0·(−4) = 20
Answer
det A = 20
12.Matrices & determinants
Answer: x = 3.667
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 = 1·(−5) − 1·(−2) = −3
Replace the x column
det Aₓ = 1·(−5) − 1·6 = −11
x = −11/(−3) = 3.667
Answer
x = 3.667
y = −2.667
13.Vectors: dot & cross product
Answer: k = −6.333
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 + 5·(−3) + (−1)·4 = 0
−19 + (−3)·k = 0
Solve for k
k = −(−19)/(−3) = −6.333
Answer
k = −6.333
14.Vectors: dot & cross product
Answer: A = 3.536 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.
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.
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.
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 = (72.5 − 50)/15 = 1.5
Answer
z = 1.5
19.Normal distribution
Answer: P = 0.4082
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 = (171.35 − 178)/7 = −0.95
z₂ = (179.4 − 178)/7 = 0.2
Area between
P = Φ(0.2) − Φ(−0.95) = 0.5793 − 0.1711 = 0.4082
Answer
P = 0.4082
20.Normal distribution
Answer: x = 22.13 min
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.