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 + 10x) − x).
f(x) = √(x2 + 10x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 5x)/sin(2x), with x in radians.
f(x) = ln(1 + 5x)/sin(2x)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(2x2 + 1). Find f′(3).
f(x) = ln(2x2 + 1)x = 3
f′(3) =
4.DerivativesHard
The tangent line to f(x) = x2 − 6 at x = −2 crosses the y-axis at y = b. Find b.
f(x) = x2 − 6x = −2
b =
5.IntegralsMedium
Evaluate I = ∫02 e−0.5x dx.
f(x) = e−0.5x[0, 2]
I =
6.IntegralsHard
Evaluate I = ∫₀1.5 x²·eˣ dx.
∫₀1.5 x²eˣ dx
I =
7.Complex numbersMedium
Find the argument θ of z = −8 − j8 in degrees, as the principal value between −180° and 180°.
z = −8 − j8
θ = °
8.Complex numbersHard
The equation z² − 3z + 27.25 = 0 has two complex roots. Find their modulus |z|.
p = −3q = 27.25
|z| =
9.Differential equationsMedium
A cup of coffee at 91 °C is left in a room at 21 °C. After 15 min its temperature is 59 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 91 °CTa = 21 °CT(15) = 59 °C
k = min⁻¹
10.Differential equationsHard
A system obeys y″ + 3y′ + 6.25y = 6.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 3y′ + 6.25y = 6.25y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−3 4 5; 1 1 −1; 0 3 −1]
det A =
12.Matrices & determinantsHard
Solve the system −x + 3y = 3 and −2x + 3y = −4. Find x.
−x + 3y = 3−2x + 3y = −4
x =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (1, 4, 1) and b = (−3, 7, 2).
a = (1, 4, 1)b = (−3, 7, 2)
a·b =
14.Vectors: dot & cross productHard
For a = (3, −1, −3) and b = (3, −1, 4), find the y-component (a × b)y of the cross product a × b.
a = (3, −1, −3)b = (3, −1, 4)
(a × b)y =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4, 5p(x): 0.15, 0.25, 0.15, 0.35, 0.1
σ =
16.ProbabilityHard
A game costs €2 to play. You win €8 with probability 0.2 and €20 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €2€8: p = 0.2€20: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 48, 15, 12, 30, 20, 55, 29.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is above 42?
μ = 50σ = 8x = 42
P =
19.Normal distributionMedium
The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 6 of them exceeds 74 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 6> 74 kg
P =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 10% of them lie (the 10th percentile).
μ = 65σ = 10P = 10%
x = points
3Solutions
1.Limits
Answer: L = 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) = 10x/(√(x2 + 10x) + x)
Divide by x
f(x) = 10/(√(1 + 10/x) + 1)
Let x → ∞
L = 10/(1 + 1) = 5
Answer
L = 5
2.Limits
Answer: L = 2.5
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).
L'Hôpital's rule
L = limx→0 [5/(1 + 5x)]/[2·cos 2x]
Substitute x = 0
L = 5/2 = 2.5
Answer
L = 2.5
3.Derivatives
Answer: f′(3) = 0.6316
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 + 1)
Substitute
f′(3) = 12/19 = 0.6316
Answer
f′(3) = 0.6316
4.Derivatives
Answer: b = −10
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) = −2
f′(x) = 2x → f′(−2) = −4
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −2 − (−4)·(−2) = −10
Answer
b = −10
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.
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
4.482·(1.25) − 1·2 = 3.602
Answer
I = 3.602
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 = −8, b = −8 → III
Reference angle
α = atan(8/8) = 45°
Argument
θ = −(180° − α) = −135°
Answer
θ = −135°
In polar form z = 11.31∠−135°.
8.Complex numbers
Answer: |z| = 5.22
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 − 109 = −100 < 0
Quadratic formula
z = (3 ± j√100)/2 = 1.5 + j5, 1.5 − j5
Answer
|z| = √(1.5² + 5²) = √27.25 = 5.22
Answer
|z| = 5.22
9.Differential equations
Answer: k = 0.04073 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−15k = 38/70
Solve for k
k = ln(70/38)/15 = 0.04073 min⁻¹
Answer
k = 0.04073 min⁻¹
10.Differential equations
Answer: ymax = 1.095
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 = 1, r = −1.5 ± 2j
Constants from the initial values
y(0) = 1 + A = 0 → A = −1
y′(0) = −1.5A + 2B = 0 → B = −0.75
Peak
t = π/2 = 1.571
ymax = 1·(1 + e−1.5·π/2) = 1.095
Answer
ymax = 1.095
11.Matrices & determinants
Answer: det A = 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.
2×2 minors of the first row
M₁₁ = 2, M₁₂ = −1, M₁₃ = 3
Cofactor expansion
det A = (−3)·2 − 4·(−1) + 5·3 = 13
Answer
det A = 13
12.Matrices & determinants
Answer: x = 7
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)·3 − 3·(−2) = 3
Replace the x column
det Aₓ = 3·3 − 3·(−4) = 21
x = 21/3 = 7
Answer
x = 7
y = 3.333
13.Vectors: dot & cross product
Answer: a·b = 27
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.
Multiply matching components and add
a·b = 1·(−3) + 4·7 + 1·2 = (−3) + 28 + 2 = 27
Answer
a·b = 27
Positive: the angle between the vectors is less than 90°.
14.Vectors: dot & cross product
Answer: (a × b)y = −21
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, component by component
(a × b)x = aybz − azby = (−1)·4 − (−3)·(−1) = −7
(a × b)y = azbx − axbz = (−3)·3 − 3·4 = −21
(a × b)z = axby − aybx = 3·(−1) − (−1)·3 = 0
Requested component
(a × b)y = −21
Answer
(a × b)y = −21
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: σ = 1.265
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.
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: 6 (p = 0.2), 18 (p = 0.05), −2 (p = 0.75)
Expected value
E(G) = 6·0.2 + 18·0.05 − 2·0.75 = 0.6 €
Answer
E(G) = 0.6 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 16.32
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
Standard normal table
Φ(−1) = 0.1587
P = 1 − 0.1587 = 0.8413
Answer
P = 0.8413
19.Normal distribution
Answer: P = 0.1539
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 = 72 kg, σT = 0.8·√6 = 1.96 kg
Standardise
z = (74 − 72)/1.96 = 1.02
Probability
P = 1 − Φ(1.02) = 0.1539
Answer
P = 0.1539
20.Normal distribution
Answer: x = 52.18 points
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.