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 + 0.5/x)x. Give a decimal value.
f(x) = (1 + 0.5/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 4x)/(2x²), with x in radians.
f(x) = (1 − cos 4x)/(2x²)x → 0
L =
3.DerivativesMedium
Let f(x) = (3x + 1)5. Find f′(0).
f(x) = (3x + 1)5x = 0
f′(0) =
4.DerivativesHard
The tangent line to f(x) = x2 + 4x − 3 at x = −1 crosses the y-axis at y = b. Find b.
f(x) = x2 + 4x − 3x = −1
b =
5.IntegralsMedium
Evaluate I = ∫₁1.5 x·ln x dx.
∫₁1.5 x ln x dx
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 0.5x2 and y = x.
y = 0.5x2y = x
A =
7.Complex numbersMedium
Given z₁ = 3∠−60° and z₂ = 4∠−25°, find the imaginary part of w = z₁/z₂.
z₁ = 3∠−60°z₂ = 4∠−25°
Im(w) =
8.Complex numbersHard
Solve z² + 6z + 18 = 0. The roots are a ± jb; find b > 0.
p = 6q = 18
b =
9.Differential equationsMedium
Solve y″ + 8y′ + 16y = 0 with y(0) = 3 and y′(0) = 1. Find y(1).
y″ + 8y′ + 16y = 0y(0) = 3y′(0) = 1
y(1) =
10.Differential equationsHard
Solve the initial value problem y′ + 3y/x = 5x2 for x > 0, with y(1) = 2. Find y(3).
y′ + 3y/x = 5x2y(1) = 2
y(3) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [1 0; −1 2]B = [−4 2; 2 2]
c11 =
12.Matrices & determinantsHard
Solve the system −5x + 5y = −6 and 4x − 6y = 6. Find x.
−5x + 5y = −64x − 6y = 6
x =
13.Vectors: dot & cross productMedium
A constant force F = (0, 10, 15) N acts on an object while it moves in a straight line from point A(3, 5, −5) to point B(4, 1, 3). Coordinates are in metres. Find the work W done by the force.
F = (0, 10, 15) NA(3, 5, −5) mB(4, 1, 3) m
W = J
14.Vectors: dot & cross productHard
A force F = (30, −50, 60) N acts at point A(−1, −3, 0). Coordinates are in metres. Find the x-component Mx of the moment of the force about the origin O(0, 0, 0).
F = (30, −50, 60) NA(−1, −3, 0) mO(0, 0, 0) m
Mx = N·m
15.ProbabilityMedium
In how many ways N can a committee of 2 people be chosen from a group of 7? The order does not matter.
n = 7k = 2
N =
16.ProbabilityHard
A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 4 correct answers?
n = 6p = 0.3k = 4
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 52, 23, 24, 53, 51, 40, 26.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 50. What proportion P of values is below 475?
μ = 500σ = 50x = 475
P =
19.Normal distributionMedium
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 16 of them is below 404 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 16< 404 kg
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 90% of them lie (the 90th percentile).
μ = 32σ = 6P = 90%
x = min
3Solutions
1.Limits
Answer: L = 1.649
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→∞ x·ln(1 + 0.5/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·0.5 = 0.5
Undo the logarithm
L = e0.5 = 1.649
Answer
L = 1.649
2.Limits
Answer: L = 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).
First application
4·sin 4x/(4x) → 0/0
Second application
16·cos 4x/4
Substitute x = 0
L = 16/4 = 4
Answer
L = 4
3.Derivatives
Answer: f′(0) = 15
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) = 5·(3x + 1)4·3
Substitute
f′(0) = 5·14·3 = 15
Answer
f′(0) = 15
4.Derivatives
Answer: b = −4
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) = −6
f′(x) = 2x + 4 → f′(−1) = 2
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −6 − 2·(−1) = −4
Answer
b = −4
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: A = 0.6667
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
0.5x2 = x → x = 0, x = 2
Integrate upper minus lower
A = [0.5x2 − 0.1667x3]02 = 0.6667
Answer
A = 0.6667
7.Complex numbers
Answer: Im(w) = −0.4302
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| = 3 / 4 = 0.75
θ = −60° − (−25)° = −35°
Rectangular form
Re(w) = 0.75·cos(−35°) = 0.6144
Im(w) = 0.75·sin(−35°) = −0.4302
Im(w) = −0.4302
Answer
Im(w) = −0.4302
8.Complex numbers
Answer: b = 3
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 = 36 − 72 = −36 < 0
Quadratic formula
z = (−6 ± j√36)/2 = −3 + j3, −3 − j3
Answer
b = 3
Answer
b = 3
9.Differential equations
Answer: y(1) = 0.2931
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 + 4)² = 0 → r = −4
Constants
A = y(0) = 3
y′(0) = B − 4A → B = 1 + 12 = 13
Evaluate
y(1) = (3 + 13·1)·e−4 = 0.2931
Answer
y(1) = 0.2931
10.Differential equations
Answer: y(3) = 22.54
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)′ = 5x5 → x3y = 5x6/6 + C
Initial condition
2 = 5/6 + C → C = 1.167
Evaluate
y(3) = (5·36/6 + 1.167)/33 = 22.54
Answer
y(3) = 22.54
11.Matrices & determinants
Answer: c11 = −4
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 = 1·(−4) + 0·2 = −4
Answer
c11 = −4
12.Matrices & determinants
Answer: x = 0.6
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.
Determinant of the coefficient matrix
det A = (−5)·(−6) − 5·4 = 10
Replace the x column
det Aₓ = (−6)·(−6) − 5·6 = 6
x = 6/10 = 0.6
Answer
x = 0.6
y = −0.6
13.Vectors: dot & cross product
Answer: W = 80 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 = (4, 1, 3) − (3, 5, −5) = (1, −4, 8) m
Work
W = F·d = 0·1 + 10·(−4) + 15·8 = 80 J
Answer
W = 80 J
14.Vectors: dot & cross product
Answer: Mx = −180 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, 0) − (0, 0, 0) = (−1, −3, 0) m
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.
Combinations
C(7, 2) = (7·6)/(2·1) = 21
Answer
N = 21
16.Probability
Answer: P = 0.05953
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 = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 0.3085
Answer
P = 0.3085
19.Normal distribution
Answer: P = 0.9772
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 total
μT = 400 kg, σT = 0.5·√16 = 2 kg
Standardise
z = (404 − 400)/2 = 2
Probability
P = Φ(2) = 0.9772
Answer
P = 0.9772
20.Normal distribution
Answer: x = 39.69 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.