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→∞ (−2x3 − x + 3)/(3x3 + 4x + 5).
f(x) = (−2x3 − x + 3)/(3x3 + 4x + 5)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 2x)/sin(x), with x in radians.
f(x) = ln(1 + 2x)/sin(x)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(2x), with x in radians. Find f′(0.5).
f(x) = x·sin(2x)x = 0.5 rad
f′(0.5) =
4.DerivativesHard
The tangent line to f(x) = 3x2 + 3x + 2 at x = −1 crosses the y-axis at y = b. Find b.
f(x) = 3x2 + 3x + 2x = −1
b =
5.IntegralsMedium
The region under y = √(2x) from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(2x)0 ≤ x ≤ 1
V =
6.IntegralsHard
Find the area A of the region between the parabola y = 1 − x² and the x-axis.
y = 1 − x²
A =
7.Complex numbersMedium
Given z₁ = 3∠30° and z₂ = 1.5∠80°, find the real part of w = z₁/z₂.
z₁ = 3∠30°z₂ = 1.5∠80°
Re(w) =
8.Complex numbersHard
Use De Moivre's theorem to find the imaginary part of z⁹ for z = −√3 + j.
z = −√3 + jn = 9
Im(z⁹) =
9.Differential equationsMedium
Solve y″ + 8y′ + 16y = 0 with y(0) = 1 and y′(0) = -1. Find y(1.5).
y″ + 8y′ + 16y = 0y(0) = 1y′(0) = -1
y(1.5) =
10.Differential equationsHard
A system obeys y″ + 2y′ + 26y = 130 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1.5); angles in radians.
y″ + 2y′ + 26y = 130y(0) = 0y′(0) = 0
y(1.5) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−2 3 0; 2 −2 −4; −1 0 2]
det A =
12.Matrices & determinantsHard
Solve the system −3x + 2y = −6 and −4x + y = −11. Find x.
−3x + 2y = −6−4x + y = −11
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−2, 4, 0) and b = (−6, −4, 1), in degrees.
a = (−2, 4, 0)b = (−6, −4, 1)
θ = °
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(−3, −1, 0), B(3, −3, −4) and C(−3, −4, 4). Coordinates are in metres.
A(−3, −1, 0)B(3, −3, −4)C(−3, −4, 4)
A = m²
15.ProbabilityMedium
A bag holds 5 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
5 / 4
P =
16.ProbabilityHard
X has the distribution shown, and Y = 2X + 1. Find the standard deviation σY of Y.
x: 0, 2, 4, 6p(x): 0.3, 0.15, 0.2, 0.35Y = 2X + 1
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 34, 32, 35, 19, 47.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 20. Find the z-score z of a score of 78.
μ = 70σ = 20x = 78
z =
19.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9564. Find z to two decimals.
A = 0.9564
z =
20.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 95% of them lie (the 95th percentile).
μ = 178σ = 7P = 95%
x = cm
3Solutions
1.Limits
Answer: L = −0.6667
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 + (−1)/x2 + 3/x3)/(3 + 4/x2 + 5/x3)
Let x → ∞
L = −2/3 = −0.6667
Answer
L = −0.6667
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 2
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 [2/(1 + 2x)]/[cos x]
Substitute x = 0
L = 2/1 = 2
Answer
L = 2
3.Derivatives
Answer: f′(0.5) = 1.382
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.
Product and chain rule
f′(x) = sin(2x) + 2x·cos(2x)
Substitute
f′(0.5) = 0.8415 + 1·0.5403 = 1.382
Answer
f′(0.5) = 1.382
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(−1) = 2
f′(x) = 6x + 3 → f′(−1) = −3
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 2 − (−3)·(−1) = −1
Answer
b = −1
5.Integrals
Answer: V = 3.142
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∫₀1 2x dx
Integrate
V = π·2·1²/2 = 3.142
Answer
V = 3.142
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.
Zeros
1 − x² = 0 → x = ±1
Integrate
A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333
Answer
A = 1.333
7.Complex numbers
Answer: Re(w) = 1.286
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 / 1.5 = 2
θ = 30° − 80° = −50°
Rectangular form
Re(w) = 2·cos(−50°) = 1.286
Im(w) = 2·sin(−50°) = −1.532
Re(w) = 1.286
Answer
Re(w) = 1.286
8.Complex numbers
Answer: Im(z⁹) = −512
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 = −√3 + j = 2∠150°
De Moivre
z⁹ = (2)⁹∠1350° = 512∠−90°
Rectangular form
z⁹ = 512·(cos −90° + j sin −90°) = −j512
Im(z⁹) = −512
Answer
Im(z⁹) = −512
9.Differential equations
Answer: y(1.5) = 0.01363
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) = 1
y′(0) = B − 4A → B = -1 + 4 = 3
Evaluate
y(1.5) = (1 + 3·1.5)·e−6 = 0.01363
Answer
y(1.5) = 0.01363
10.Differential equations
Answer: y(1.5) = 4.404
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.
2×2 minors of the first row
M₁₁ = −4, M₁₂ = 0, M₁₃ = −2
Cofactor expansion
det A = (−2)·(−4) − 3·0 + 0·(−2) = 8
Answer
det A = 8
12.Matrices & determinants
Answer: x = 3.2
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 = (−3)·1 − 2·(−4) = 5
Replace the x column
det Aₓ = (−6)·1 − 2·(−11) = 16
x = 16/5 = 3.2
Answer
x = 3.2
y = 1.8
13.Vectors: dot & cross product
Answer: θ = 97.06 °
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 = (−2)·(−6) + 4·(−4) + 0·1 = −4
Magnitudes
|a| = √((−2)² + 4² + 0²) = 4.472
|b| = √((−6)² + (−4)² + 1²) = 7.28
Angle
cos θ = −4/(4.472·7.28) = −0.1229
θ = arccos(−0.1229) = 97.06°
Answer
θ = 97.06°
14.Vectors: dot & cross product
Answer: A = 18.03 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.
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.
First and second draw
P = 5/9 · 4/8 = 20/72 = 0.2778
Answer
P = 0.2778
16.Probability
Answer: σY = 4.964
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.
Spread of X
μ = 3.2, Var(X) = 6.16, σX = 2.482
Linear change
σY = |2|·2.482 = 4.964
Answer
σY = 4.964
The mean does move: E(Y) = 2·3.2 + 1 = 7.4.
17.Descriptive statistics
Answer: s = 9.965
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 = (78 − 70)/20 = 0.4
Answer
z = 0.4
19.Normal distribution
Answer: z = −1.71
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.
Area to the left of z
Φ(z) = 1 − 0.9564 = 0.0436
Read the z-table backwards
Find 0.0436 inside the table and read off z.
z = −1.71
Answer
z = −1.71
20.Normal distribution
Answer: x = 189.5 cm
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.