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→∞ (√(9x2 − 8x) − 3x).
f(x) = √(9x2 − 8x) − 3xx → ∞
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) = (3x + 3)4. Find f′(0).
f(x) = (3x + 3)4x = 0
f′(0) =
4.DerivativesHard
The tangent line to f(x) = −x2 + 3x + 5 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = −x2 + 3x + 5x = 3
b =
5.IntegralsMedium
Evaluate I = ∫01 2x·(x² + 1)2 dx.
f(x) = 2x·(x² + 1)2[0, 1]
I =
6.IntegralsHard
The region between the line y = 5x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.
y = 5xy = 3x²
V =
7.Complex numbersMedium
Given z₁ = 5∠55° and z₂ = 0.5∠110°, find the argument θ of w = z₁/z₂ in degrees, between −180° and 180°.
z₁ = 5∠55°z₂ = 0.5∠110°
θ = °
8.Complex numbersHard
Use De Moivre's theorem to find the real part of z⁶ for z = 1 + j√3.
z = 1 + j√3n = 6
Re(z⁶) =
9.Differential equationsMedium
Solve y″ + 4y′ + 3y = 0 with y(0) = 2 and y′(0) = -4. Find y(2).
y″ + 4y′ + 3y = 0y(0) = 2y′(0) = -4
y(2) =
10.Differential equationsHard
Solve y′ + 0.5y = 6e−2t with y(0) = 4. Find y(2).
y′ + 0.5y = 6e−2ty(0) = 4
y(2) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [4 −2 5; 5 1 0; −4 −4 2]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [5 3; −6 −6]
λ =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−3, 4, −2) and b = (6, 4, 4), in degrees.
a = (−3, 4, −2)b = (6, 4, 4)
θ = °
14.Vectors: dot & cross productHard
For a = (0, −4, 4) and b = (−5, −2, 5), find the x-component (a × b)x of the cross product a × b.
a = (0, −4, 4)b = (−5, −2, 5)
(a × b)x =
15.ProbabilityMedium
A fair die is rolled 5 times. What is the probability P of getting at least one six?
n = 5
P =
16.ProbabilityHard
A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 4 correct answers?
n = 12p = 0.5k = 4
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 52, 21, 47, 23, 34.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 40. What proportion P of values is above 516?
μ = 500σ = 40x = 516
P =
19.Normal distributionMedium
The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 10 of them is below 793.1 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 10< 793.1 kg
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 95% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points95%
x = points
3Solutions
1.Limits
Answer: L = −1.333
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) = −8x/(√(9x2 − 8x) + 3x)
Divide by x
f(x) = −8/(√(9 + (−8)/x) + 3)
Let x → ∞
L = −8/(3 + 3) = −1.333
Answer
L = −1.333
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) = 324
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) = 4·(3x + 3)3·3
Substitute
f′(0) = 4·33·3 = 324
Answer
f′(0) = 324
4.Derivatives
Answer: b = 14
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) = 5
f′(x) = −2x + 3 → f′(3) = −3
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 5 − (−3)·3 = 14
Answer
b = 14
5.Integrals
Answer: I = 2.333
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.
Substitution
u = x² + 1, du = 2x dx
I = ∫12 u2 du
Integrate and evaluate
I = [u3/3]12 = (8 − 1)/3 = 2.333
Answer
I = 2.333
6.Integrals
Answer: V = 48.48
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
5x = 3x² → x = 0, x = 1.667
Washer method
V = π∫₀1.667 ((5x)² − (3x²)²) dx
Integrate
V = π(25·1.667³/3 − 9·1.667⁵/5) = π·15.43 = 48.48
Answer
V = 48.48
7.Complex numbers
Answer: θ = −55 °
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| = 5 / 0.5 = 10
θ = 55° − 110° = −55°
Argument
θ = −55°
Answer
θ = −55°
8.Complex numbers
Answer: Re(z⁶) = 64
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 = 1 + j√3 = 2∠60°
De Moivre
z⁶ = (2)⁶∠360° = 64∠0°
Rectangular form
z⁶ = 64·(cos 0° + j sin 0°) = 64
Re(z⁶) = 64
Answer
Re(z⁶) = 64
9.Differential equations
Answer: y(2) = 0.1378
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 + 1)(r + 3) = 0 → r = −1, −3
Constants from the initial values
A + B = 2, −1A − 3B = -4
A = 1, B = 1
Evaluate
y(2) = 1·e−2 + 1·e−6 = 0.1378
Answer
y(2) = 0.1378
10.Differential equations
Answer: y(2) = 2.87
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 = e0.5t
(e0.5ty)′ = 6e−1.5t
General solution
y = 6/(−1.5)·e−2t + C·e−0.5t = −4·e−2t + C·e−0.5t
Initial condition
4 = −4 + C → C = 8
Evaluate
y(2) = −4·0.01832 + 8·0.3679 = 2.87
Answer
y(2) = 2.87
11.Matrices & determinants
Answer: det A = −52
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 = 10, M₁₃ = −16
Cofactor expansion
det A = 4·2 − (−2)·10 + 5·(−16) = −52
Answer
det A = −52
12.Matrices & determinants
Answer: λ = 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.
Characteristic equation
λ² − (−1)λ + (−12) = 0
Solve
λ = (−1 ± √(49))/2 → λ = −4, 3
λ = 3
Answer
λ = 3
13.Vectors: dot & cross product
Answer: θ = 103 °
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 = (−3)·6 + 4·4 + (−2)·4 = −10
Magnitudes
|a| = √((−3)² + 4² + (−2)²) = 5.385
|b| = √(6² + 4² + 4²) = 8.246
Angle
cos θ = −10/(5.385·8.246) = −0.2252
θ = arccos(−0.2252) = 103°
Answer
θ = 103°
14.Vectors: dot & cross product
Answer: (a × b)x = −12
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.
Cross product, component by component
(a × b)x = aybz − azby = (−4)·5 − 4·(−2) = −12
(a × b)y = azbx − axbz = 4·(−5) − 0·5 = −20
(a × b)z = axby − aybx = 0·(−2) − (−4)·(−5) = −20
Requested component
(a × b)x = −12
Answer
(a × b)x = −12
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: P = 0.5981
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.
Complement
P(no six) = (5/6)5 = 0.4019
P = 1 − 0.4019 = 0.5981
Answer
P = 0.5981
16.Probability
Answer: P = 0.1208
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 = 0.4
Standard normal table
Φ(0.4) = 0.6554
P = 1 − 0.6554 = 0.3446
Answer
P = 0.3446
19.Normal distribution
Answer: P = 0.06681
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 = 850 kg, σT = 12·√10 = 37.95 kg
Standardise
z = (793.1 − 850)/37.95 = −1.5
Probability
P = Φ(−1.5) = 0.06681
Answer
P = 0.06681
20.Normal distribution
Answer: x = 48.6 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.