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 − 1/x)x. Give a decimal value.
f(x) = (1 − 1/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 2x)/sin(3x), with x in radians.
f(x) = ln(1 + 2x)/sin(3x)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(2x2 + 5). Find f′(1).
f(x) = ln(2x2 + 5)x = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 + 1.5x2 − 2. Find the x-coordinate x of the local maximum.
f(x) = x3 + 1.5x2 − 2
x =
5.IntegralsMedium
Evaluate I = ∫14 4/x dx.
f(x) = 4/x[1, 4]
I =
6.IntegralsHard
Evaluate I = ∫₀2.5 x²·eˣ dx.
∫₀2.5 x²eˣ dx
I =
7.Complex numbersMedium
Write z = 4∠65° in rectangular form a + jb and give the imaginary part b.
r = 4θ = 65°
b =
8.Complex numbersHard
Solve z² + 3z + 11.25 = 0. The roots are a ± jb; find the real part a.
p = 3q = 11.25
a =
9.Differential equationsMedium
Solve y″ + 4y′ + 4y = 0 with y(0) = 1 and y′(0) = 4. Find y(1).
y″ + 4y′ + 4y = 0y(0) = 1y′(0) = 4
y(1) =
10.Differential equationsHard
Solve y″ + 2y′ + 26y = 0 with y(0) = 2 and y′(0) = 0 (released from rest). Find y(0.5); angles in radians.
y″ + 2y′ + 26y = 0y(0) = 2y′(0) = 0
y(0.5) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [2 −2 −2; −2 −1 5; 1 −4 −5]
det A =
12.Matrices & determinantsHard
Solve the system 3x − 2y = 4 and −5x − 6y = −2. Find x.
3x − 2y = 4−5x − 6y = −2
x =
13.Vectors: dot & cross productMedium
A constant force F = (20, −25, 30) N acts on an object while it moves in a straight line from point A(−3, 5, −1) to point B(−1, 0, 0). Coordinates are in metres. Find the work W done by the force.
F = (20, −25, 30) NA(−3, 5, −1) mB(−1, 0, 0) m
W = J
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(4, 0, 4), B(−2, −2, 4) and C(−3, −4, −4). Coordinates are in metres.
A(4, 0, 4)B(−2, −2, 4)C(−3, −4, −4)
A = m²
15.ProbabilityMedium
A fair die is rolled 3 times. What is the probability P of getting at least one six?
n = 3
P =
16.ProbabilityHard
9 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?
n = 9p = 0.25
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 23, 28, 22, 49, 14, 45.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 20. What proportion P of values is above 30?
μ = 60σ = 20x = 30
P =
19.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 96.25 and 115.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 96.25x2 = 115.75
P =
20.Normal distributionHard
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
3Solutions
1.Limits
Answer: L = 0.3679
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 − 1/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·(−1) = −1
Undo the logarithm
L = e−1 = 0.3679
Answer
L = 0.3679
2.Limits
Answer: L = 0.6667
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 [2/(1 + 2x)]/[3·cos 3x]
Substitute x = 0
L = 2/3 = 0.6667
Answer
L = 0.6667
3.Derivatives
Answer: f′(1) = 0.5714
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 for the logarithm
f′(x) = 4x/(2x2 + 5)
Substitute
f′(1) = 4/7 = 0.5714
Answer
f′(1) = 0.5714
4.Derivatives
Answer: x = −1
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 + 3x = 3·(x2 + x)
Solve f′(x) = 0
x = −1 or x = 0
Which one is which
f″(x) = 6x + 3: negative at a maximum, positive at a minimum.
f″(−1) = −3, f″(0) = 3
x = −1
Answer
x = −1
5.Integrals
Answer: I = 5.545
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) = 4·ln x
Evaluate at the limits
I = 4·(ln 4 − ln 1) = 4·ln 4 = 5.545
Answer
I = 5.545
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: b = 3.625
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.
Components
a = 4·cos(65°) = 1.69
b = 4·sin(65°) = 3.625
Rectangular form
z = 1.69 + j3.625
b = 3.625
Answer
b = 3.625
8.Complex numbers
Answer: a = −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.
Discriminant
p² − 4q = 9 − 45 = −36 < 0
Quadratic formula
z = (−3 ± j√36)/2 = −1.5 + j3, −1.5 − j3
Answer
a = −1.5
Answer
a = −1.5
9.Differential equations
Answer: y(1) = 0.9473
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) = 1
y′(0) = B − 2A → B = 4 + 2 = 6
Evaluate
y(1) = (1 + 6·1)·e−2 = 0.9473
Answer
y(1) = 0.9473
10.Differential equations
Answer: y(0.5) = −0.8266
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.
Complex roots
r = −1 ± 5j
Constants
A = 2, B = 1·2/5 = 0.4
Evaluate
y(0.5) = e−0.5(2·cos 2.5 + 0.4·sin 2.5) = −0.8266
Answer
y(0.5) = −0.8266
11.Matrices & determinants
Answer: det A = 42
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 = 5, M₁₃ = 9
Cofactor expansion
det A = 2·25 − (−2)·5 + (−2)·9 = 42
Answer
det A = 42
12.Matrices & determinants
Answer: x = 1
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·(−6) − (−2)·(−5) = −28
Replace the x column
det Aₓ = 4·(−6) − (−2)·(−2) = −28
x = −28/(−28) = 1
Answer
x = 1
y = −0.5
13.Vectors: dot & cross product
Answer: W = 195 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 = (−1, 0, 0) − (−3, 5, −1) = (2, −5, 1) m
Work
W = F·d = 20·2 + (−25)·(−5) + 30·1 = 195 J
Answer
W = 195 J
14.Vectors: dot & cross product
Answer: A = 25.79 m²
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)3 = 0.5787
P = 1 − 0.5787 = 0.4213
Answer
P = 0.4213
16.Probability
Answer: P = 0.3003
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) = 0.759 = 0.07508
P(X = 1) = 9·0.25·0.758 = 0.2253
Add them
P = 0.07508 + 0.2253 = 0.3003
Answer
P = 0.3003
17.Descriptive statistics
Answer: s = 13.85
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) = 0.06681
P = 1 − 0.06681 = 0.9332
Answer
P = 0.9332
19.Normal distribution
Answer: P = 0.4518
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 = (96.25 − 100)/15 = −0.25
z₂ = (115.75 − 100)/15 = 1.05
Area between
P = Φ(1.05) − Φ(−0.25) = 0.8531 − 0.4013 = 0.4518
Answer
P = 0.4518
20.Normal distribution
Answer: P = 0.7881
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).