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 + 0.5/x)x. Give a decimal value.
f(x) = (1 + 0.5/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 5x)/(x²), with x in radians.
f(x) = (1 − cos 5x)/(x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(x), with x in radians. Find f′(2).
f(x) = x·sin(x)x = 2 rad
f′(2) =
4.DerivativesHard
The tangent line to f(x) = −x2 − 4x at x = 3 crosses the y-axis at y = b. Find b.
f(x) = −x2 − 4xx = 3
b =
5.IntegralsMedium
The region under y = √(x) from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(x)0 ≤ x ≤ 1
V =
6.IntegralsHard
The region between the line y = 5x and the parabola y = 2x² is rotated about the x-axis. Find the volume V of the solid.
y = 5xy = 2x²
V =
7.Complex numbersMedium
Write z = −4 − j9 in exponential form r·ejθ. Give θ in radians, between −π and π.
z = −4 − j9
θ = rad
8.Complex numbersHard
The roots of zⁿ = w with n = 6 and w = 64∠−60° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, …, 5. Find the argument of z₅ in degrees, between −180° and 180°.
w = 64∠−60°n = 6k = 5
θ₅ = °
9.Differential equationsMedium
Solve y″ + 8y′ + 16y = 0 with y(0) = 3 and y′(0) = 0. Find y(0.5).
y″ + 8y′ + 16y = 0y(0) = 3y′(0) = 0
y(0.5) =
10.Differential equationsHard
A system obeys y″ + 3y′ + 11.25y = 11.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1.5); angles in radians.
y″ + 3y′ + 11.25y = 11.25y(0) = 0y′(0) = 0
y(1.5) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−4 1 1; 5 −3 −4; −5 −2 −4]
det A =
12.Matrices & determinantsHard
Solve the system 6x + 6y = 5 and −3x − 2y = −9. Find x.
6x + 6y = 5−3x − 2y = −9
x =
13.Vectors: dot & cross productMedium
A constant force F = (30, 5, 5) N acts on an object while it moves in a straight line from point A(−4, −4, 3) to point B(3, 0, −3). Coordinates are in metres. Find the work W done by the force.
F = (30, 5, 5) NA(−4, −4, 3) mB(3, 0, −3) m
W = J
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (3, 5, −1), b = (0, 2, −1) and c = (0, −2, −3).
a = (3, 5, −1)b = (0, 2, −1)c = (0, −2, −3)
V =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3p(x): 0.35, 0.2, 0.3, 0.15
Var(X) =
16.ProbabilityHard
X has the distribution shown, and Y = 3X + 5. Find the standard deviation σY of Y.
x: 0, 1, 2p(x): 0.35, 0.3, 0.35Y = 3X + 5
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 28, 57, 11, 44, 19.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 60 and standard deviation 20. Find the z-score z of a score of 40.
μ = 60σ = 20x = 40
z =
19.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 100.75 and 103 points? Give P as a decimal.
μ = 100σ = 15x1 = 100.75x2 = 103
P =
20.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 5% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g5%
x = g
3Solutions
1.Limits
Answer: L = 1.649
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 + 0.5/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·0.5 = 0.5
Undo the logarithm
L = e0.5 = 1.649
Answer
L = 1.649
2.Limits
Answer: L = 12.5
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
5·sin 5x/(2x) → 0/0
Second application
25·cos 5x/2
Substitute x = 0
L = 25/2 = 12.5
Answer
L = 12.5
3.Derivatives
Answer: f′(2) = 0.077
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 and chain rule
f′(x) = sin(x) + x·cos(x)
Substitute
f′(2) = 0.9093 + 2·(−0.4161) = 0.077
Answer
f′(2) = 0.077
4.Derivatives
Answer: b = 9
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) = −21
f′(x) = −2x − 4 → f′(3) = −10
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −21 − (−10)·3 = 9
Answer
b = 9
5.Integrals
Answer: V = 1.571
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.
Disc method
V = π∫₀1 x dx
Integrate
V = π·1·1²/2 = 1.571
Answer
V = 1.571
6.Integrals
Answer: V = 163.6
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
5x = 2x² → x = 0, x = 2.5
Washer method
V = π∫₀2.5 ((5x)² − (2x²)²) dx
Integrate
V = π(25·2.5³/3 − 4·2.5⁵/5) = π·52.08 = 163.6
Answer
V = 163.6
7.Complex numbers
Answer: θ = −1.989 rad
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
r = √(16 + 81) = 9.849
Argument in radians
θ = −114° = −114·π/180 = −1.989 rad
Exponential form
z = 9.849·e−j1.989
Answer
θ = −1.989 rad
8.Complex numbers
Answer: θ₅ = −70 °
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
641/6 = 2
Argument of the root
θ₅ = (−60° + 360°·5)/6 = 290° → −70°
Argument
θ₅ = −70°
Answer
θ₅ = −70°
The 6 roots lie on the circle of radius 2, 60° apart.
9.Differential equations
Answer: y(0.5) = 1.218
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 + 4)² = 0 → r = −4
Constants
A = y(0) = 3
y′(0) = B − 4A → B = 0 + 12 = 12
Evaluate
y(0.5) = (3 + 12·0.5)·e−2 = 1.218
Answer
y(0.5) = 1.218
10.Differential equations
Answer: y(1.5) = 1.074
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.
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 = 4, M₁₂ = −40, M₁₃ = −25
Cofactor expansion
det A = (−4)·4 − 1·(−40) + 1·(−25) = −1
Answer
det A = −1
12.Matrices & determinants
Answer: x = 7.333
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·(−2) − 6·(−3) = 6
Replace the x column
det Aₓ = 5·(−2) − 6·(−9) = 44
x = 44/6 = 7.333
Answer
x = 7.333
y = −6.5
13.Vectors: dot & cross product
Answer: W = 200 J
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.
Displacement
d = B − A = (3, 0, −3) − (−4, −4, 3) = (7, 4, −6) m
Work
W = F·d = 30·7 + 5·4 + 5·(−6) = 200 J
Answer
W = 200 J
14.Vectors: dot & cross product
Answer: V = 24
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·(−3) − (−1)·(−2) = −8
(b × c)y = bzcx − bxcz = (−1)·0 − 0·(−3) = 0
(b × c)z = bxcy − bycx = 0·(−2) − 2·0 = 0
Dot product with a
a·(b × c) = 3·(−8) + 5·0 + (−1)·0 = −24
V = |−24| = 24
Answer
V = 24
15.Probability
Answer: Var(X) = 1.188
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 + 1·0.2 + 2·0.3 + 3·0.15 = 1.25
Mean of the squares
E(X²) = 0·0.35 + 1·0.2 + 4·0.3 + 9·0.15 = 2.75
Variance
Var(X) = 2.75 − 1.25² = 1.188
Answer
Var(X) = 1.188
16.Probability
Answer: σY = 2.51
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, Var(X) = 0.7, σX = 0.8367
Linear change
σY = |3|·0.8367 = 2.51
Answer
σY = 2.51
The mean does move: E(Y) = 3·1 + 5 = 8.
17.Descriptive statistics
Answer: s = 18.67
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 = (40 − 60)/20 = −1
Answer
z = −1
19.Normal distribution
Answer: P = 0.05932
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.
z-scores
z₁ = (100.75 − 100)/15 = 0.05
z₂ = (103 − 100)/15 = 0.2
Area between
P = Φ(0.2) − Φ(0.05) = 0.5793 − 0.5199 = 0.05932
Answer
P = 0.05932
20.Normal distribution
Answer: x = 258.2 g
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.