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 (1 − cos 3x)/(3x²), with x in radians.
f(x) = (1 − cos 3x)/(3x²)x → 0
L =
3.DerivativesMedium
Let f(x) = (2x + 3)3. Find f′(0).
f(x) = (2x + 3)3x = 0
f′(0) =
4.DerivativesHard
The tangent line to f(x) = 3x2 + 2x at x = −2 crosses the y-axis at y = b. Find b.
f(x) = 3x2 + 2xx = −2
b =
5.IntegralsMedium
Evaluate I = ∫14 2/x dx.
f(x) = 2/x[1, 4]
I =
6.IntegralsHard
The region between the line y = 6x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.
y = 6xy = 3x²
V =
7.Complex numbersMedium
Find the argument θ of z = −1 − j in degrees, as the principal value between −180° and 180°.
z = −1 − j
θ = °
8.Complex numbersHard
Solve z² + 5z + 7.25 = 0. The roots are a ± jb; find b > 0.
p = 5q = 7.25
b =
9.Differential equationsMedium
Solve y″ + 4y′ + 4y = 0 with y(0) = 5 and y′(0) = 1. Find y(0.5).
y″ + 4y′ + 4y = 0y(0) = 5y′(0) = 1
y(0.5) =
10.Differential equationsHard
A tank holds 100 L of pure water. Brine with 0.05 kg of salt per litre flows in at 4 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 4 kg of salt?
V = 100 Lm0 = 0 kgc = 0.05 kg/Lr = 4 L/min
t = min
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [0 6; 3 2]B = [−1 −2; −5 3]
c21 =
12.Matrices & determinantsHard
Solve the system −3x + y = −3 and −2x + 5y = 7. Find x.
−3x + y = −3−2x + 5y = 7
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (−5, 2, 3) onto b = (0, −6, 6), that is, the component of a in the direction of b.
a = (−5, 2, 3)b = (0, −6, 6)
compb a =
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (2, −2, −5), b = (−4, −1, −1) and c = (5, −4, −1).
a = (2, −2, −5)b = (−4, −1, −1)c = (5, −4, −1)
V =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4, 5p(x): 0.3, 0.25, 0.1, 0.25, 0.1
σ =
16.ProbabilityHard
A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 2 correct answers?
n = 12p = 0.7k = 2
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 29, 25, 53, 41, 11, 31, 53.
n = 7
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 5. Find the z-score z of a score of 55.
μ = 50σ = 5x = 55
z =
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 6 of them exceeds 148.2 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 6> 148.2 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 95% of them lie (the 95th percentile).
μ = 32σ = 6P = 95%
x = min
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 = 1.5
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).
First application
3·sin 3x/(6x) → 0/0
Second application
9·cos 3x/6
Substitute x = 0
L = 9/6 = 1.5
Answer
L = 1.5
3.Derivatives
Answer: f′(0) = 54
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) = 3·(2x + 3)2·2
Substitute
f′(0) = 3·32·2 = 54
Answer
f′(0) = 54
4.Derivatives
Answer: b = −12
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(−2) = 8
f′(x) = 6x + 2 → f′(−2) = −10
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 8 − (−10)·(−2) = −12
Answer
b = −12
5.Integrals
Answer: I = 2.773
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.
Antiderivative
F(x) = 2·ln x
Evaluate at the limits
I = 2·(ln 4 − ln 1) = 2·ln 4 = 2.773
Answer
I = 2.773
6.Integrals
Answer: V = 120.6
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
6x = 3x² → x = 0, x = 2
Washer method
V = π∫₀2 ((6x)² − (3x²)²) dx
Integrate
V = π(36·2³/3 − 9·2⁵/5) = π·38.4 = 120.6
Answer
V = 120.6
7.Complex numbers
Answer: θ = −135 °
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 = −1, b = −1 → III
Reference angle
α = atan(1/1) = 45°
Argument
θ = −(180° − α) = −135°
Answer
θ = −135°
In polar form z = 1.414∠−135°.
8.Complex numbers
Answer: b = 1
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 = 25 − 29 = −4 < 0
Quadratic formula
z = (−5 ± j√4)/2 = −2.5 + j, −2.5 − j
Answer
b = 1
Answer
b = 1
9.Differential equations
Answer: y(0.5) = 3.863
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 + 2)² = 0 → r = −2
Constants
A = y(0) = 5
y′(0) = B − 2A → B = 1 + 10 = 11
Evaluate
y(0.5) = (5 + 11·0.5)·e−1 = 3.863
Answer
y(0.5) = 3.863
10.Differential equations
Answer: t = 40.24 min
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.
Result
m(t) = 5 + (−5)·e−t/25
Solve for t
e−t/25 = (5 − 4)/(5 − 0) = 0.2
t = 25·ln(5) = 40.24 min
Answer
t = 40.24 min
11.Matrices & determinants
Answer: c21 = −13
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
c21 = 3·(−1) + 2·(−5) = −13
Answer
c21 = −13
12.Matrices & determinants
Answer: x = 1.692
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)·5 − 1·(−2) = −13
Replace the x column
det Aₓ = (−3)·5 − 1·7 = −22
x = −22/(−13) = 1.692
Answer
x = 1.692
y = 2.077
13.Vectors: dot & cross product
Answer: compb a = 0.7071
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)·0 + 2·(−6) + 3·6 = 6
Length of b
|b| = √(0² + (−6)² + 6²) = 8.485
Scalar projection
compb a = a·b/|b| = 6/8.485 = 0.7071
Answer
compb a = 0.7071
14.Vectors: dot & cross product
Answer: V = 93
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.
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.
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 = (55 − 50)/5 = 1
Answer
z = 1
19.Normal distribution
Answer: P = 0.9292
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 = 150 kg, σT = 0.5·√6 = 1.225 kg
Standardise
z = (148.2 − 150)/1.225 = −1.47
Probability
P = 1 − Φ(−1.47) = 0.9292
Answer
P = 0.9292
20.Normal distribution
Answer: x = 41.87 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.