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 + 5x) − 3x).
f(x) = √(9x2 + 5x) − 3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 4x)/(3x²), with x in radians.
f(x) = (1 − cos 4x)/(3x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x3·e0.5x. Find f′(1.5).
f(x) = x3·e0.5xx = 1.5
f′(1.5) =
4.DerivativesHard
The tangent line to f(x) = 2x2 − 2x + 4 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = 2x2 − 2x + 4x = 1
b =
5.IntegralsMedium
Evaluate I = ∫01 2x·(x² + 1)2 dx.
f(x) = 2x·(x² + 1)2[0, 1]
I =
6.IntegralsHard
Evaluate I = ∫₀2.5 x²·eˣ dx.
∫₀2.5 x²eˣ dx
I =
7.Complex numbersMedium
Given z₁ = 2.5∠55° and z₂ = 4∠65°, find the argument θ of w = z₁·z₂ in degrees, between −180° and 180°.
z₁ = 2.5∠55°z₂ = 4∠65°
θ = °
8.Complex numbersHard
Find the imaginary part of (−2 − j3)³.
z = −2 − j3n = 3
Im(z³) =
9.Differential equationsMedium
Solve y″ + 6y′ + 8y = 0 with y(0) = 5 and y′(0) = 0. Find y(2).
y″ + 6y′ + 8y = 0y(0) = 5y′(0) = 0
y(2) =
10.Differential equationsHard
Solve y′ + 2y = 9e−3t with y(0) = 3. Find y(1).
y′ + 2y = 9e−3ty(0) = 3
y(1) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−2 −2 −1; −5 −5 −5; −5 −3 2]
det A =
12.Matrices & determinantsHard
Solve the system 5x − 2y = −11 and −3x − 3y = 2. Find x.
5x − 2y = −11−3x − 3y = 2
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (5, −5, −3) and b = (−2, 6, 4), in degrees.
a = (5, −5, −3)b = (−2, 6, 4)
θ = °
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (1, 0, 4), b = (−3, −5, −4) and c = (4, −5, 1).
a = (1, 0, 4)b = (−3, −5, −4)c = (4, −5, 1)
V =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2p(x): 0.35, 0.35, 0.3
Var(X) =
16.ProbabilityHard
6 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?
n = 6p = 0.1
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 32, 27, 12, 35, 19, 15.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 20. What proportion P of values is below 200?
μ = 170σ = 20x = 200
P =
19.Normal distributionMedium
X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 200.3 mm. Round z to two decimals.
μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 200.3 mm
P =
20.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 99% of them lie (the 99th percentile).
μ = 500σ = 40P = 99%
x = h
3Solutions
1.Limits
Answer: L = 0.8333
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) = 5x/(√(9x2 + 5x) + 3x)
Divide by x
f(x) = 5/(√(9 + 5/x) + 3)
Let x → ∞
L = 5/(3 + 3) = 0.8333
Answer
L = 0.8333
2.Limits
Answer: L = 2.667
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).
First application
4·sin 4x/(6x) → 0/0
Second application
16·cos 4x/6
Substitute x = 0
L = 16/6 = 2.667
Answer
L = 2.667
3.Derivatives
Answer: f′(1.5) = 17.86
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 rule
f′(x) = u′v + uv′ = e0.5x·(3x2 + 0.5·x3)
Substitute
f′(1.5) = 2.117·(6.75 + 1.688) = 17.86
Answer
f′(1.5) = 17.86
4.Derivatives
Answer: b = 2
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) = 4
f′(x) = 4x − 2 → f′(1) = 2
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 4 − 2·1 = 2
Answer
b = 2
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: I = 37.59
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.
First round
∫x²eˣ dx = x²eˣ − ∫2x eˣ dx
Second round
∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)
Limits
12.18·(3.25) − 1·2 = 37.59
Answer
I = 37.59
7.Complex numbers
Answer: θ = 120 °
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| = 2.5 · 4 = 10
θ = 55° + 65° = 120°
Argument
θ = 120°
Answer
θ = 120°
8.Complex numbers
Answer: Im(z³) = −9
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.
Powers step by step
z² = (−2 − j3)(−2 − j3) = −5 + j12
z³ = (−5 + j12)(−2 − j3) = 46 − j9
Check with De Moivre
z = 3.606∠−123.7° → z³ = 46.87∠−11.07°
Im(z³) = −9
Answer
Im(z³) = −9
9.Differential equations
Answer: y(2) = 0.1815
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 + 4) = 0 → r = −2, −4
Constants from the initial values
A + B = 5, −2A − 4B = 0
A = 10, B = −5
Evaluate
y(2) = 10·e−4 + (−5)·e−8 = 0.1815
Answer
y(2) = 0.1815
10.Differential equations
Answer: y(1) = 1.176
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 = e2t
(e2ty)′ = 9e−t
General solution
y = 9/(−1)·e−3t + C·e−2t = −9·e−3t + C·e−2t
Initial condition
3 = −9 + C → C = 12
Evaluate
y(1) = −9·0.04979 + 12·0.1353 = 1.176
Answer
y(1) = 1.176
11.Matrices & determinants
Answer: det A = −10
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 = −25, M₁₂ = −35, M₁₃ = −10
Cofactor expansion
det A = (−2)·(−25) − (−2)·(−35) + (−1)·(−10) = −10
Answer
det A = −10
12.Matrices & determinants
Answer: x = −1.762
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·(−3) − (−2)·(−3) = −21
Replace the x column
det Aₓ = (−11)·(−3) − (−2)·2 = 37
x = 37/(−21) = −1.762
Answer
x = −1.762
y = 1.095
13.Vectors: dot & cross product
Answer: θ = 154.8 °
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 = 5·(−2) + (−5)·6 + (−3)·4 = −52
Magnitudes
|a| = √(5² + (−5)² + (−3)²) = 7.681
|b| = √((−2)² + 6² + 4²) = 7.483
Angle
cos θ = −52/(7.681·7.483) = −0.9047
θ = arccos(−0.9047) = 154.8°
Answer
θ = 154.8°
14.Vectors: dot & cross product
Answer: V = 115
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 b × c
(b × c)x = bycz − bzcy = (−5)·1 − (−4)·(−5) = −25
(b × c)y = bzcx − bxcz = (−4)·4 − (−3)·1 = −13
(b × c)z = bxcy − bycx = (−3)·(−5) − (−5)·4 = 35
Dot product with a
a·(b × c) = 1·(−25) + 0·(−13) + 4·35 = 115
V = |115| = 115
Answer
V = 115
15.Probability
Answer: Var(X) = 0.6475
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 = 0·0.35 + 1·0.35 + 2·0.3 = 0.95
Mean of the squares
E(X²) = 0·0.35 + 1·0.35 + 4·0.3 = 1.55
Variance
Var(X) = 1.55 − 0.95² = 0.6475
Answer
Var(X) = 0.6475
16.Probability
Answer: P = 0.8857
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.
Zero and one failure
P(X = 0) = 0.96 = 0.5314
P(X = 1) = 6·0.1·0.95 = 0.3543
Add them
P = 0.5314 + 0.3543 = 0.8857
Answer
P = 0.8857
17.Descriptive statistics
Answer: s = 9.395
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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 0.9332
Answer
P = 0.9332
19.Normal distribution
Answer: P = 0.2743
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 = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm
Standardise
z = (200.3 − 200)/0.5 = 0.6
Upper tail
P = 1 − Φ(0.6) = 1 − 0.7257 = 0.2743
Answer
P = 0.2743
20.Normal distribution
Answer: x = 593.1 h
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.