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→∞ (√(x2 + 9x) − x).
f(x) = √(x2 + 9x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 2x)/(x²), with x in radians.
f(x) = (1 − cos 2x)/(x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·e2x. Find f′(2).
f(x) = x·e2xx = 2
f′(2) =
4.DerivativesHard
The tangent line to f(x) = 3x2 + 4x − 4 at x = −1 crosses the y-axis at y = b. Find b.
f(x) = 3x2 + 4x − 4x = −1
b =
5.IntegralsMedium
Evaluate I = ∫02 e−0.5x dx.
f(x) = e−0.5x[0, 2]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = 2x.
y = x2y = 2x
A =
7.Complex numbersMedium
The number z = 2.5·ej0.4 has its angle in radians. Find its real part a.
r = 2.5θ = 0.4 rad
a =
8.Complex numbersHard
The roots of zⁿ = w with n = 3 and w = 27∠0° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2. Find the real part of z₂.
w = 27∠0°n = 3k = 2
Re(z₂) =
9.Differential equationsMedium
A cup of coffee at 92 °C is left in a room at 23 °C. After 20 min its temperature is 61 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 92 °CTa = 23 °CT(20) = 61 °C
k = min⁻¹
10.Differential equationsHard
A system obeys y″ + y′ + 9.25y = 18.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + y′ + 9.25y = 18.5y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [−2 0; 5 3]B = [−2 2; 0 0]
c21 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−1 6; −2 6]
λ =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (6, 4, −4) and b = (k, −4, 4) are perpendicular.
a = (6, 4, −4)b = (k, −4, 4)
k =
14.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (0, −2, −6) and b = (4, −4, −3).
a = (0, −2, −6)b = (4, −4, −3)
A =
15.ProbabilityMedium
A fair die is rolled 6 times. What is the probability P of getting at least one six?
n = 6
P =
16.ProbabilityHard
X has the distribution shown, and Y = 3X + 5. Find the standard deviation σY of Y.
x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = 3X + 5
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 55, 22, 59, 52, 54.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 50. Find the z-score z of a score of 450.
μ = 500σ = 50x = 450
z =
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 4 of them is below 376 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 4< 376 kg
P =
20.Normal distributionHard
Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.03 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
3Solutions
1.Limits
Answer: L = 4.5
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).
Multiply by the conjugate
f(x) = 9x/(√(x2 + 9x) + x)
Divide by x
f(x) = 9/(√(1 + 9/x) + 1)
Let x → ∞
L = 9/(1 + 1) = 4.5
Answer
L = 4.5
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).
First application
2·sin 2x/(2x) → 0/0
Second application
4·cos 2x/2
Substitute x = 0
L = 4/2 = 2
Answer
L = 2
3.Derivatives
Answer: f′(2) = 273
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′(2) = 54.6·(1 + 4) = 273
Answer
f′(2) = 273
4.Derivatives
Answer: b = −7
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) = −5
f′(x) = 6x + 4 → f′(−1) = −2
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −5 − (−2)·(−1) = −7
Answer
b = −7
5.Integrals
Answer: I = 1.264
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.
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.
Intersection points
x2 = 2x → x = 0, x = 2
Integrate upper minus lower
A = [x2 − 0.3333x3]02 = 1.333
Answer
A = 1.333
7.Complex numbers
Answer: a = 2.303
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.
Euler's formula
z = 2.5·(cos 0.4 + j sin 0.4)
Components (radians)
a = 2.5·cos 0.4 = 2.303
b = 2.5·sin 0.4 = 0.9735
a = 2.303
Answer
a = 2.303
8.Complex numbers
Answer: Re(z₂) = −1.5
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.
Modulus of the roots
271/3 = 3
Argument of the root
θ₂ = (0° + 360°·2)/3 = 240° → −120°
Rectangular form
z₂ = 3·(cos −120° + j sin −120°) = −1.5 − j2.598
Re(z₂) = −1.5
Answer
Re(z₂) = −1.5
The 3 roots lie on the circle of radius 3, 120° apart.
9.Differential equations
Answer: k = 0.02983 min⁻¹
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.
Ratio of temperature differences
e−20k = 38/69
Solve for k
k = ln(69/38)/20 = 0.02983 min⁻¹
Answer
k = 0.02983 min⁻¹
10.Differential equations
Answer: ymax = 3.185
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.
Steady value and roots
yss = 2, r = −0.5 ± 3j
Constants from the initial values
y(0) = 2 + A = 0 → A = −2
y′(0) = −0.5A + 3B = 0 → B = −0.3333
Peak
t = π/3 = 1.047
ymax = 2·(1 + e−0.5·π/3) = 3.185
Answer
ymax = 3.185
11.Matrices & determinants
Answer: c21 = −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.
Row times column
c21 = 5·(−2) + 3·0 = −10
Answer
c21 = −10
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 − (5)λ + (6) = 0
Solve
λ = (5 ± √(1))/2 → λ = 2, 3
λ = 3
Answer
λ = 3
13.Vectors: dot & cross product
Answer: k = 5.333
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 equals zero
a·b = 6·k + 4·(−4) + (−4)·4 = 0
−32 + 6·k = 0
Solve for k
k = −(−32)/6 = 5.333
Answer
k = 5.333
14.Vectors: dot & cross product
Answer: A = 31.05
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.
Complement
P(no six) = (5/6)6 = 0.3349
P = 1 − 0.3349 = 0.6651
Answer
P = 0.6651
16.Probability
Answer: σY = 2.414
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 = 1.05, Var(X) = 0.6475, σX = 0.8047
Linear change
σY = |3|·0.8047 = 2.414
Answer
σY = 2.414
The mean does move: E(Y) = 3·1.05 + 5 = 8.15.
17.Descriptive statistics
Answer: s = 14.98
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 = (450 − 500)/50 = −1
Answer
z = −1
19.Normal distribution
Answer: P = 0.9332
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 = 340 kg, σT = 12·√4 = 24 kg
Standardise
z = (376 − 340)/24 = 1.5
Probability
P = Φ(1.5) = 0.9332
Answer
P = 0.9332
20.Normal distribution
Answer: P = 0.8849
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).