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→∞ (x3 − x − 7)/(2x3 + 5x + 4).
f(x) = (x3 − x − 7)/(2x3 + 5x + 4)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 3x)/sin(4x), with x in radians.
f(x) = ln(1 + 3x)/sin(4x)x → 0
L =
3.DerivativesMedium
Let f(x) = (2x − 3)4. Find f′(0).
f(x) = (2x − 3)4x = 0
f′(0) =
4.DerivativesHard
The tangent line to f(x) = 0.5x2 + 2x + 5 at x = −1 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 + 2x + 5x = −1
b =
5.IntegralsMedium
Evaluate I = ∫01 2x·(x² + 1)1 dx.
f(x) = 2x·(x² + 1)1[0, 1]
I =
6.IntegralsHard
Evaluate I = ∫₀3 x²·eˣ dx.
∫₀3 x²eˣ dx
I =
7.Complex numbersMedium
Write z = 4∠50° in rectangular form a + jb and give the real part a.
r = 4θ = 50°
a =
8.Complex numbersHard
Exactly one of the points A–D is a root of z³ = 27∠90°. Click it.
w = 27∠90°n = 3
1A (3∠−30°)
2B (3∠180°)
3C (3∠150°)
4D (5.4∠150°)
9.Differential equationsMedium
Solve y″ + 2y′ + y = 0 with y(0) = 2 and y′(0) = 4. Find y(1.5).
y″ + 2y′ + y = 0y(0) = 2y′(0) = 4
y(1.5) =
10.Differential equationsHard
A system obeys y″ + 4y′ + 8y = 8 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.75); angles in radians.
y″ + 4y′ + 8y = 8y(0) = 0y′(0) = 0
y(0.75) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−4 1 2; 5 −4 2; 3 4 1]
det A =
12.Matrices & determinantsHard
Solve the system −x − 6y = −2 and 3x + y = 1. Find x.
−x − 6y = −23x + y = 1
x =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (−2, 3, 4) and b = (1, −2, 0).
a = (−2, 3, 4)b = (1, −2, 0)
a·b =
14.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (5, 0, −6) and b = (−3, 2, −5).
a = (5, 0, −6)b = (−3, 2, −5)
A =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4p(x): 0.15, 0.35, 0.2, 0.3
Var(X) =
16.ProbabilityHard
A game costs €1 to play. You win €4 with probability 0.1 and €10 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €1€4: p = 0.1€10: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 57, 22, 29, 43, 42, 47, 15.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 12. What proportion P of values is below 74.8?
μ = 70σ = 12x = 74.8
P =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.3688. Find z to two decimals.
A = 0.3688
z =
20.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 10% of them lie (the 10th percentile).
μ = 500σ = 40P = 10%
x = h
3Solutions
1.Limits
Answer: L = 0.5
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).
Divide by the highest power
(1 + (−1)/x2 + (−7)/x3)/(2 + 5/x2 + 4/x3)
Let x → ∞
L = 1/2 = 0.5
Answer
L = 0.5
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 0.75
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 [3/(1 + 3x)]/[4·cos 4x]
Substitute x = 0
L = 3/4 = 0.75
Answer
L = 0.75
3.Derivatives
Answer: f′(0) = −216
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) = 4·(2x − 3)3·2
Substitute
f′(0) = 4·(−3)3·2 = −216
Answer
f′(0) = −216
4.Derivatives
Answer: b = 4.5
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) = 3.5
f′(x) = x + 2 → f′(−1) = 1
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 3.5 − 1·(−1) = 4.5
Answer
b = 4.5
5.Integrals
Answer: I = 1.5
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.
Substitution
u = x² + 1, du = 2x dx
I = ∫12 u1 du
Integrate and evaluate
I = [u2/2]12 = (4 − 1)/2 = 1.5
Answer
I = 1.5
6.Integrals
Answer: I = 98.43
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
20.09·(5) − 1·2 = 98.43
Answer
I = 98.43
7.Complex numbers
Answer: a = 2.571
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(50°) = 2.571
b = 4·sin(50°) = 3.064
Rectangular form
z = 2.571 + j3.064
a = 2.571
Answer
a = 2.571
8.Complex numbers
Answer: C (3∠150°)
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 and arguments of the roots
|z| = 3
θ = 30°, 150°, −90°
The point
C = 3∠150°
Answer
C (3∠150°)
9.Differential equations
Answer: y(1.5) = 2.454
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 + 1)² = 0 → r = −1
Constants
A = y(0) = 2
y′(0) = B − 1A → B = 4 + 2 = 6
Evaluate
y(1.5) = (2 + 6·1.5)·e−1.5 = 2.454
Answer
y(1.5) = 2.454
10.Differential equations
Answer: y(0.75) = 0.7616
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 = −12, M₁₂ = −1, M₁₃ = 32
Cofactor expansion
det A = (−4)·(−12) − 1·(−1) + 2·32 = 113
Answer
det A = 113
12.Matrices & determinants
Answer: x = 0.2353
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 = (−1)·1 − (−6)·3 = 17
Replace the x column
det Aₓ = (−2)·1 − (−6)·1 = 4
x = 4/17 = 0.2353
Answer
x = 0.2353
y = 0.2941
13.Vectors: dot & cross product
Answer: a·b = −8
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.
Negative: the angle between the vectors is more than 90°.
14.Vectors: dot & cross product
Answer: A = 45.75
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
(a × b)x = aybz − azby = 0·(−5) − (−6)·2 = 12
(a × b)y = azbx − axbz = (−6)·(−3) − 5·(−5) = 43
(a × b)z = axby − aybx = 5·2 − 0·(−3) = 10
Area
A = |a × b| = √(12² + 43² + 10²) = 45.75
Answer
A = 45.75
15.Probability
Answer: Var(X) = 1.127
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 = 1·0.15 + 2·0.35 + 3·0.2 + 4·0.3 = 2.65
Mean of the squares
E(X²) = 1·0.15 + 4·0.35 + 9·0.2 + 16·0.3 = 8.15
Variance
Var(X) = 8.15 − 2.65² = 1.127
Answer
Var(X) = 1.127
16.Probability
Answer: E(G) = −0.5 €
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.
Net outcomes
G: 3 (p = 0.1), 9 (p = 0.01), −1 (p = 0.89)
Expected value
E(G) = 3·0.1 + 9·0.01 − 1·0.89 = −0.5 €
Answer
E(G) = −0.5 €
A negative value means you lose money on average.
17.Descriptive statistics
Answer: s = 14.9
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 = 0.4
Standard normal table
Φ(0.4) = 0.6554
P = 0.6554
Answer
P = 0.6554
19.Normal distribution
Answer: z = 0.48
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.
Area to the left of z
Φ(z) = (1 + 0.3688)/2 = 0.6844
Read the z-table backwards
Find 0.6844 inside the table and read off z.
z = 0.48
Answer
z = 0.48
20.Normal distribution
Answer: x = 448.7 h
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.