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 − x + 5)/(4x2 + 5x + 9).
f(x) = (−x2 − x + 5)/(4x2 + 5x + 9)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 3x)/(2x²), with x in radians.
f(x) = (1 − cos 3x)/(2x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x2·e0.5x. Find f′(2).
f(x) = x2·e0.5xx = 2
f′(2) =
4.DerivativesHard
The tangent line to f(x) = x2 − 5x + 5 at x = −2 crosses the y-axis at y = b. Find b.
f(x) = x2 − 5x + 5x = −2
b =
5.IntegralsMedium
Evaluate I = ∫12 3/x dx.
f(x) = 3/x[1, 2]
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
Write z = 2∠50° in rectangular form a + jb and give the real part a.
r = 2θ = 50°
a =
8.Complex numbersHard
Find the imaginary part of (2 − j3)⁴.
z = 2 − j3n = 4
Im(z⁴) =
9.Differential equationsMedium
A forged steel part at 480 °C is left to cool in a workshop at 24 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.03 min⁻¹, find its temperature T after 26 min.
T0 = 480 °CTa = 24 °Ck = 0.03 min⁻¹t = 26 min
T = °C
10.Differential equationsHard
Solve y″ + 2y′ + 26y = 0 with y(0) = 3 and y′(0) = 0 (released from rest). Find y(1); angles in radians.
y″ + 2y′ + 26y = 0y(0) = 3y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c22 (row 2, column 2).
A = [−2 1; 5 −1]B = [0 −2; −5 0]
c22 =
12.Matrices & determinantsHard
Solve the system −2x + 4y = 12 and −5x + y = −7. Find x.
−2x + 4y = 12−5x + y = −7
x =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (0, −2, 5) and b = (0, −1, 3).
a = (0, −2, 5)b = (0, −1, 3)
a·b =
14.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (−4, 4, 6) and b = (−1, −2, −6).
a = (−4, 4, 6)b = (−1, −2, −6)
A =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3p(x): 0.2, 0.3, 0.35, 0.15
Var(X) =
16.ProbabilityHard
A game costs €10 to play. You win €20 with probability 0.3 and €100 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €10€20: p = 0.3€100: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 11, 14, 23, 15, 37.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 5. Find the z-score z of a score of 67.5.
μ = 70σ = 5x = 67.5
z =
19.Normal distributionMedium
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 171.35 and 188.85 cm? Give P as a decimal.
μ = 178σ = 7x1 = 171.35x2 = 188.85
P =
20.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.
μ = 500 hσ = 40 h95%
x = h
3Solutions
1.Limits
Answer: L = −0.25
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)/x + 5/x2)/(4 + 5/x + 9/x2)
Let x → ∞
L = −1/4 = −0.25
Answer
L = −0.25
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 2.25
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
3·sin 3x/(4x) → 0/0
Second application
9·cos 3x/4
Substitute x = 0
L = 9/4 = 2.25
Answer
L = 2.25
3.Derivatives
Answer: f′(2) = 16.31
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′ = e0.5x·(2x + 0.5·x2)
Substitute
f′(2) = 2.718·(4 + 2) = 16.31
Answer
f′(2) = 16.31
4.Derivatives
Answer: b = 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.
Point and slope
f(−2) = 19
f′(x) = 2x − 5 → f′(−2) = −9
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 19 − (−9)·(−2) = 1
Answer
b = 1
5.Integrals
Answer: I = 2.079
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) = 3·ln x
Evaluate at the limits
I = 3·(ln 2 − ln 1) = 3·ln 2 = 2.079
Answer
I = 2.079
6.Integrals
Answer: A = 1.333
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: a = 1.286
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 = 2·cos(50°) = 1.286
b = 2·sin(50°) = 1.532
Rectangular form
z = 1.286 + j1.532
a = 1.286
Answer
a = 1.286
8.Complex numbers
Answer: Im(z⁴) = 120
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.
Powers step by step
z² = (2 − j3)(2 − j3) = −5 − j12
z³ = (−5 − j12)(2 − j3) = −46 − j9
z⁴ = (−46 − j9)(2 − j3) = −119 + j120
Check with De Moivre
z = 3.606∠−56.31° → z⁴ = 169∠134.8°
Im(z⁴) = 120
Answer
Im(z⁴) = 120
9.Differential equations
Answer: T = 233 °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) = 24 + 456·e−0.03t
Substitute t
T(26) = 24 + 456·0.4584 = 233 °C
Answer
T = 233 °C
10.Differential equations
Answer: y(1) = 0.1014
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 = 3, B = 1·3/5 = 0.6
Evaluate
y(1) = e−1(3·cos 5 + 0.6·sin 5) = 0.1014
Answer
y(1) = 0.1014
11.Matrices & determinants
Answer: c22 = −10
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
c22 = 5·(−2) + (−1)·0 = −10
Answer
c22 = −10
12.Matrices & determinants
Answer: x = 2.222
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 = (−2)·1 − 4·(−5) = 18
Replace the x column
det Aₓ = 12·1 − 4·(−7) = 40
x = 40/18 = 2.222
Answer
x = 2.222
y = 4.111
13.Vectors: dot & cross product
Answer: a·b = 17
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 = 0·0 + (−2)·(−1) + 5·3 = 0 + 2 + 15 = 17
Answer
a·b = 17
Positive: the angle between the vectors is less than 90°.
14.Vectors: dot & cross product
Answer: A = 34.47
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 = 4·(−6) − 6·(−2) = −12
(a × b)y = azbx − axbz = 6·(−1) − (−4)·(−6) = −30
(a × b)z = axby − aybx = (−4)·(−2) − 4·(−1) = 12
Area
A = |a × b| = √((−12)² + (−30)² + 12²) = 34.47
Answer
A = 34.47
15.Probability
Answer: Var(X) = 0.9475
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 = 0·0.2 + 1·0.3 + 2·0.35 + 3·0.15 = 1.45
Mean of the squares
E(X²) = 0·0.2 + 1·0.3 + 4·0.35 + 9·0.15 = 3.05
Variance
Var(X) = 3.05 − 1.45² = 0.9475
Answer
Var(X) = 0.9475
16.Probability
Answer: E(G) = −2 €
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: 10 (p = 0.3), 90 (p = 0.02), −10 (p = 0.68)
Expected value
E(G) = 10·0.3 + 90·0.02 − 10·0.68 = −2 €
Answer
E(G) = −2 €
A negative value means you lose money on average.
17.Descriptive statistics
Answer: s = 10.49
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 = (67.5 − 70)/5 = −0.5
Answer
z = −0.5
19.Normal distribution
Answer: P = 0.7684
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 = (171.35 − 178)/7 = −0.95
z₂ = (188.85 − 178)/7 = 1.55
Area between
P = Φ(1.55) − Φ(−0.95) = 0.9394 − 0.1711 = 0.7684
Answer
P = 0.7684
20.Normal distribution
Answer: x = 434 h
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.