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→∞ (−6x3 − 5x + 5)/(3x3 + 6x + 2).
f(x) = (−6x3 − 5x + 5)/(3x3 + 6x + 2)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 3x)/sin(5x), with x in radians.
f(x) = ln(1 + 3x)/sin(5x)x → 0
L =
3.DerivativesMedium
Let f(x) = x3·e−x. Find f′(2).
f(x) = x3·e−xx = 2
f′(2) =
4.DerivativesHard
Let f(x) = x3 − 15x2 + 63x − 2. Find the x-coordinate x of the local minimum.
f(x) = x3 − 15x2 + 63x − 2
x =
5.IntegralsMedium
Evaluate I = ∫01 e−0.5x dx.
f(x) = e−0.5x[0, 1]
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 1 − x² and the x-axis.
y = 1 − x²
A =
7.Complex numbersMedium
Find the argument θ of z = 9 + j3 in degrees, as the principal value between −180° and 180°.
z = 9 + j3
θ = °
8.Complex numbersHard
Use De Moivre's theorem to find the imaginary part of z⁵ for z = −2 − j2.
z = −2 − j2n = 5
Im(z⁵) =
9.Differential equationsMedium
A cup of coffee at 94 °C is left in a room at 23 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.11 min⁻¹, find its temperature T after 18 min.
T0 = 94 °CTa = 23 °Ck = 0.11 min⁻¹t = 18 min
T = °C
10.Differential equationsHard
Solve y″ + 3y′ + 11.25y = 0 with y(0) = 5 and y′(0) = 0 (released from rest). Find y(1); angles in radians.
y″ + 3y′ + 11.25y = 0y(0) = 5y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [−2 −1; 5 5]B = [2 2; 5 5]
c21 =
12.Matrices & determinantsHard
Solve the system −x + 3y = 5 and −6x − 2y = 3. Find x.
−x + 3y = 5−6x − 2y = 3
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−6, 3, −4) and b = (0, −5, −4), in degrees.
a = (−6, 3, −4)b = (0, −5, −4)
θ = °
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (−3, 0, −2), b = (3, 4, 3) and c = (0, −4, −4).
a = (−3, 0, −2)b = (3, 4, 3)c = (0, −4, −4)
V =
15.ProbabilityMedium
A fair die is rolled 4 times. What is the probability P of getting at least one six?
n = 4
P =
16.ProbabilityHard
A game costs €10 to play. You win €40 with probability 0.1 and €200 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €10€40: p = 0.1€200: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 47, 19, 46, 47, 48, 22.
n = 6
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 12. Find the z-score z of a score of 158.
μ = 170σ = 12x = 158
z =
19.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 85.75 and 130.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 85.75x2 = 130.75
P =
20.Normal distributionHard
Hole diameters D are normal with mean 25.04 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.04 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
3Solutions
1.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).
Divide by the highest power
(−6 + (−5)/x2 + 5/x3)/(3 + 6/x2 + 2/x3)
Let x → ∞
L = −6/3 = −2
Answer
L = −2
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 0.6
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)]/[5·cos 5x]
Substitute x = 0
L = 3/5 = 0.6
Answer
L = 0.6
3.Derivatives
Answer: f′(2) = 0.5413
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′ = e−x·(3x2 + (−1)·x3)
Substitute
f′(2) = 0.1353·(12 + (−8)) = 0.5413
Answer
f′(2) = 0.5413
4.Derivatives
Answer: x = 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.
Derivative
f′(x) = 3x2 − 30x + 63 = 3·(x2 − 10x + 21)
Solve f′(x) = 0
x = 3 or x = 7
Which one is which
f″(x) = 6x − 30: negative at a maximum, positive at a minimum.
f″(3) = −12, f″(7) = 12
x = 7
Answer
x = 7
5.Integrals
Answer: I = 0.7869
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.
Zeros
1 − x² = 0 → x = ±1
Integrate
A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333
Answer
A = 1.333
7.Complex numbers
Answer: θ = 18.43 °
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 = 9, b = 3 → I
Reference angle
α = atan(3/9) = 18.43°
Argument
θ = α = 18.43°
Answer
θ = 18.43°
In polar form z = 9.487∠18.43°.
8.Complex numbers
Answer: Im(z⁵) = 128
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.
Polar form
z = −2 − j2 = 2√2∠−135°
De Moivre
z⁵ = (2√2)⁵∠−675° = 181∠45°
Rectangular form
z⁵ = 181·(cos 45° + j sin 45°) = 128 + j128
Im(z⁵) = 128
Answer
Im(z⁵) = 128
9.Differential equations
Answer: T = 32.8 °C
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.
Solution of the cooling law
T(t) = 23 + 71·e−0.11t
Substitute t
T(18) = 23 + 71·0.1381 = 32.8 °C
Answer
T = 32.8 °C
10.Differential equations
Answer: y(1) = −1.026
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 = −1.5 ± 3j
Constants
A = 5, B = 1.5·5/3 = 2.5
Evaluate
y(1) = e−1.5(5·cos 3 + 2.5·sin 3) = −1.026
Answer
y(1) = −1.026
11.Matrices & determinants
Answer: c21 = 35
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 + 5·5 = 35
Answer
c21 = 35
12.Matrices & determinants
Answer: x = −0.95
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)·(−2) − 3·(−6) = 20
Replace the x column
det Aₓ = 5·(−2) − 3·3 = −19
x = −19/20 = −0.95
Answer
x = −0.95
y = 1.35
13.Vectors: dot & cross product
Answer: θ = 88.85 °
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
a·b = (−6)·0 + 3·(−5) + (−4)·(−4) = 1
Magnitudes
|a| = √((−6)² + 3² + (−4)²) = 7.81
|b| = √(0² + (−5)² + (−4)²) = 6.403
Angle
cos θ = 1/(7.81·6.403) = 0.02
θ = arccos(0.02) = 88.85°
Answer
θ = 88.85°
14.Vectors: dot & cross product
Answer: V = 36
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 = 4·(−4) − 3·(−4) = −4
(b × c)y = bzcx − bxcz = 3·0 − 3·(−4) = 12
(b × c)z = bxcy − bycx = 3·(−4) − 4·0 = −12
Dot product with a
a·(b × c) = (−3)·(−4) + 0·12 + (−2)·(−12) = 36
V = |36| = 36
Answer
V = 36
15.Probability
Answer: P = 0.5177
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)4 = 0.4823
P = 1 − 0.4823 = 0.5177
Answer
P = 0.5177
16.Probability
Answer: E(G) = 4 €
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: 30 (p = 0.1), 190 (p = 0.05), −10 (p = 0.85)
Expected value
E(G) = 30·0.1 + 190·0.05 − 10·0.85 = 4 €
Answer
E(G) = 4 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 13.73
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 = (158 − 170)/12 = −1
Answer
z = −1
19.Normal distribution
Answer: P = 0.8088
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 = (85.75 − 100)/15 = −0.95
z₂ = (130.75 − 100)/15 = 2.05
Area between
P = Φ(2.05) − Φ(−0.95) = 0.9798 − 0.1711 = 0.8088
Answer
P = 0.8088
20.Normal distribution
Answer: P = 0.9868
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).