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 − 2/x)x. Give a decimal value.
f(x) = (1 − 2/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 5x)/sin(2x), with x in radians.
f(x) = ln(1 + 5x)/sin(2x)x → 0
L =
3.DerivativesMedium
Let f(x) = x·e2x. Find f′(1).
f(x) = x·e2xx = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 − 13.5x2 + 54x − 2. Find the x-coordinate x of the local minimum.
f(x) = x3 − 13.5x2 + 54x − 2
x =
5.IntegralsMedium
Evaluate I = ∫₁2 x·ln x dx.
∫₁2 x ln x dx
I =
6.IntegralsHard
Evaluate I = ∫₀0.5 x²·eˣ dx.
∫₀0.5 x²eˣ dx
I =
7.Complex numbersMedium
Find the argument θ of z = −6 + j3 in degrees, as the principal value between −180° and 180°.
z = −6 + j3
θ = °
8.Complex numbersHard
Use De Moivre's theorem to find the real part of z³ for z = −2 − j2.
z = −2 − j2n = 3
Re(z³) =
9.Differential equationsMedium
Solve y″ + 6y′ + 9y = 0 with y(0) = 5 and y′(0) = 2. Find y(1).
y″ + 6y′ + 9y = 0y(0) = 5y′(0) = 2
y(1) =
10.Differential equationsHard
Solve the initial value problem y′ + 2y/x = x2 for x > 0, with y(1) = 2. Find y(2).
y′ + 2y/x = x2y(1) = 2
y(2) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c22 (row 2, column 2).
A = [0 2; 2 −5]B = [2 1; −5 −1]
c22 =
12.Matrices & determinantsHard
Solve the system 6x − 3y = 9 and −4x − 4y = 2. Find x.
6x − 3y = 9−4x − 4y = 2
x =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (2, 5, 2) and b = (3, 7, −6).
a = (2, 5, 2)b = (3, 7, −6)
a·b =
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (0, −4, 2), b = (−2, 2, −2) and c = (−5, 0, 1).
a = (0, −4, 2)b = (−2, 2, −2)c = (−5, 0, 1)
V =
15.ProbabilityMedium
In how many ways N can a committee of 2 people be chosen from a group of 10? The order does not matter.
n = 10k = 2
N =
16.ProbabilityHard
6 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?
n = 6p = 0.25
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 36, 59, 40, 30, 46, 35.
n = 6
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 15. Find the z-score z of a score of 152.
μ = 170σ = 15x = 152
z =
19.Normal distributionMedium
The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 12 of them is below 486.2 g. Round z to two decimals.
μ = 40 gσ = 1.2 gn = 12< 486.2 g
P =
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 80% of them lie (the 80th percentile).
μ = 178σ = 7P = 80%
x = cm
3Solutions
1.Limits
Answer: L = 0.1353
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 − 2/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·(−2) = −2
Undo the logarithm
L = e−2 = 0.1353
Answer
L = 0.1353
2.Limits
Answer: L = 2.5
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)]/[2·cos 2x]
Substitute x = 0
L = 5/2 = 2.5
Answer
L = 2.5
3.Derivatives
Answer: f′(1) = 22.17
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′ = e2x·(1 + 2·x)
Substitute
f′(1) = 7.389·(1 + 2) = 22.17
Answer
f′(1) = 22.17
4.Derivatives
Answer: x = 6
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 − 27x + 54 = 3·(x2 − 9x + 18)
Solve f′(x) = 0
x = 3 or x = 6
Which one is which
f″(x) = 6x − 27: negative at a maximum, positive at a minimum.
f″(3) = −9, f″(6) = 9
x = 6
Answer
x = 6
5.Integrals
Answer: I = 0.6363
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
(2·ln 2 − 1) − (0 − 0.25) = 0.6363
Answer
I = 0.6363
6.Integrals
Answer: I = 0.0609
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
1.649·(1.25) − 1·2 = 0.0609
Answer
I = 0.0609
7.Complex numbers
Answer: θ = 153.4 °
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 = 3 → II
Reference angle
α = atan(3/6) = 26.57°
Argument
θ = 180° − α = 153.4°
Answer
θ = 153.4°
In polar form z = 6.708∠153.4°.
8.Complex numbers
Answer: Re(z³) = 16
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 = −2 − j2 = 2√2∠−135°
De Moivre
z³ = (2√2)³∠−405° = 22.63∠−45°
Rectangular form
z³ = 22.63·(cos −45° + j sin −45°) = 16 − j16
Re(z³) = 16
Answer
Re(z³) = 16
9.Differential equations
Answer: y(1) = 1.095
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 + 3)² = 0 → r = −3
Constants
A = y(0) = 5
y′(0) = B − 3A → B = 2 + 15 = 17
Evaluate
y(1) = (5 + 17·1)·e−3 = 1.095
Answer
y(1) = 1.095
10.Differential equations
Answer: y(2) = 2.05
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∫2/x dx = x2
Integrate
(x2y)′ = x4 → x2y = 1x5/5 + C
Initial condition
2 = 1/5 + C → C = 1.8
Evaluate
y(2) = (1·25/5 + 1.8)/22 = 2.05
Answer
y(2) = 2.05
11.Matrices & determinants
Answer: c22 = 7
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
c22 = 2·1 + (−5)·(−1) = 7
Answer
c22 = 7
12.Matrices & determinants
Answer: x = 0.8333
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 = 6·(−4) − (−3)·(−4) = −36
Replace the x column
det Aₓ = 9·(−4) − (−3)·2 = −30
x = −30/(−36) = 0.8333
Answer
x = 0.8333
y = −1.333
13.Vectors: dot & cross product
Answer: a·b = 29
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.
Multiply matching components and add
a·b = 2·3 + 5·7 + 2·(−6) = 6 + 35 + (−12) = 29
Answer
a·b = 29
Positive: the angle between the vectors is less than 90°.
14.Vectors: dot & cross product
Answer: V = 28
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 = 2·1 − (−2)·0 = 2
(b × c)y = bzcx − bxcz = (−2)·(−5) − (−2)·1 = 12
(b × c)z = bxcy − bycx = (−2)·0 − 2·(−5) = 10
Dot product with a
a·(b × c) = 0·2 + (−4)·12 + 2·10 = −28
V = |−28| = 28
Answer
V = 28
15.Probability
Answer: N = 45
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(10, 2) = (10·9)/(2·1) = 45
Answer
N = 45
16.Probability
Answer: P = 0.5339
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) = 0.756 = 0.178
P(X = 1) = 6·0.25·0.755 = 0.356
Add them
P = 0.178 + 0.356 = 0.5339
Answer
P = 0.5339
17.Descriptive statistics
Answer: s = 10.32
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 = (152 − 170)/15 = −1.2
Answer
z = −1.2
19.Normal distribution
Answer: P = 0.9319
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 = 480 g, σT = 1.2·√12 = 4.157 g
Standardise
z = (486.2 − 480)/4.157 = 1.49
Probability
P = Φ(1.49) = 0.9319
Answer
P = 0.9319
20.Normal distribution
Answer: x = 183.9 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.