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)3x. Give a decimal value.
f(x) = (1 − 1/x)3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 4x)/sin(3x), with x in radians.
f(x) = ln(1 + 4x)/sin(3x)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(3x), with x in radians. Find f′(1.5).
f(x) = x·sin(3x)x = 1.5 rad
f′(1.5) =
4.DerivativesHard
Let f(x) = x3 + 3x2 + 3. Find the x-coordinate x of the local maximum.
f(x) = x3 + 3x2 + 3
x =
5.IntegralsMedium
Evaluate I = ∫₁2.5 x·ln x dx.
∫₁2.5 x ln x dx
I =
6.IntegralsHard
The region between the line y = 5x and the parabola y = 2x² is rotated about the x-axis. Find the volume V of the solid.
y = 5xy = 2x²
V =
7.Complex numbersMedium
Write z = 8∠110° in rectangular form a + jb and give the real part a.
r = 8θ = 110°
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 340 °C is left to cool in a workshop at 30 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.05 min⁻¹, find its temperature T after 46 min.
T0 = 340 °CTa = 30 °Ck = 0.05 min⁻¹t = 46 min
T = °C
10.Differential equationsHard
Solve y′ + y = 3e−3t with y(0) = 1. Find y(1.5).
y′ + y = 3e−3ty(0) = 1
y(1.5) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [−3 −1; −1 6]B = [−1 −4; −4 0]
c11 =
12.Matrices & determinantsHard
Solve the system −x − 6y = −12 and 6x + 4y = −11. Find x.
−x − 6y = −126x + 4y = −11
x =
13.Vectors: dot & cross productMedium
A constant force F = (25, −30, 25) N acts on an object while it moves in a straight line from point A(3, −1, 3) to point B(4, 0, −1). Coordinates are in metres. Find the work W done by the force.
F = (25, −30, 25) NA(3, −1, 3) mB(4, 0, −1) m
W = J
14.Vectors: dot & cross productHard
A force F = (40, −50, −10) N acts at point A(−3, −3, 3). Coordinates are in metres. Find the y-component My of the moment of the force about the origin O(0, 0, 0).
F = (40, −50, −10) NA(−3, −3, 3) mO(0, 0, 0) m
My = N·m
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
A game costs €10 to play. You win €50 with probability 0.25 and €500 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €10€50: p = 0.25€500: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 35, 30, 55, 36, 52, 41, 49.
n = 7
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 25. Find the z-score z of a score of 475.
μ = 500σ = 25x = 475
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 164 and 179.75 cm? Give P as a decimal.
μ = 178σ = 7x1 = 164x2 = 179.75
P =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points95%
x = points
3Solutions
1.Limits
Answer: L = 0.04979
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→∞ 3·x·ln(1 − 1/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 3·(−1) = −3
Undo the logarithm
L = e−3 = 0.04979
Answer
L = 0.04979
2.Limits
Answer: L = 1.333
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 [4/(1 + 4x)]/[3·cos 3x]
Substitute x = 0
L = 4/3 = 1.333
Answer
L = 1.333
3.Derivatives
Answer: f′(1.5) = −1.926
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 and chain rule
f′(x) = sin(3x) + 3x·cos(3x)
Substitute
f′(1.5) = −0.9775 + 4.5·(−0.2108) = −1.926
Answer
f′(1.5) = −1.926
4.Derivatives
Answer: x = −2
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 + 6x = 3·(x2 + 2x)
Solve f′(x) = 0
x = −2 or x = 0
Which one is which
f″(x) = 6x + 6: negative at a maximum, positive at a minimum.
f″(−2) = −6, f″(0) = 6
x = −2
Answer
x = −2
5.Integrals
Answer: I = 1.551
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.
Choose u and dv
u = ln x, du = dx/x; dv = x dx, v = x²/2
Antiderivative
∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4
Limits
(3.125·ln 2.5 − 1.5625) − (0 − 0.25) = 1.551
Answer
I = 1.551
6.Integrals
Answer: V = 163.6
Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.
Intersection points
5x = 2x² → x = 0, x = 2.5
Washer method
V = π∫₀2.5 ((5x)² − (2x²)²) dx
Integrate
V = π(25·2.5³/3 − 4·2.5⁵/5) = π·52.08 = 163.6
Answer
V = 163.6
7.Complex numbers
Answer: a = −2.736
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 = 8·cos(110°) = −2.736
b = 8·sin(110°) = 7.518
Rectangular form
z = −2.736 + j7.518
a = −2.736
Answer
a = −2.736
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 = 61.08 °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) = 30 + 310·e−0.05t
Substitute t
T(46) = 30 + 310·0.1003 = 61.08 °C
Answer
T = 61.08 °C
10.Differential equations
Answer: y(1.5) = 0.5412
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.
Integrating factor
μ = et
(ety)′ = 3e−2t
General solution
y = 3/(−2)·e−3t + C·e−t = −1.5·e−3t + C·e−t
Initial condition
1 = −1.5 + C → C = 2.5
Evaluate
y(1.5) = −1.5·0.01111 + 2.5·0.2231 = 0.5412
Answer
y(1.5) = 0.5412
11.Matrices & determinants
Answer: c11 = 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.
Row times column
c11 = (−3)·(−1) + (−1)·(−4) = 7
Answer
c11 = 7
12.Matrices & determinants
Answer: x = −3.563
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)·4 − (−6)·6 = 32
Replace the x column
det Aₓ = (−12)·4 − (−6)·(−11) = −114
x = −114/32 = −3.563
Answer
x = −3.563
y = 2.594
13.Vectors: dot & cross product
Answer: W = −105 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 = (4, 0, −1) − (3, −1, 3) = (1, 1, −4) m
Work
W = F·d = 25·1 + (−30)·1 + 25·(−4) = −105 J
Answer
W = −105 J
Negative work: the force acts mostly against the motion.
14.Vectors: dot & cross product
Answer: My = 90 N·m
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.
Position vector from O to A
r = A − O = (−3, −3, 3) − (0, 0, 0) = (−3, −3, 3) m
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: E(G) = 27.5 €
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: 40 (p = 0.25), 490 (p = 0.05), −10 (p = 0.7)
Expected value
E(G) = 40·0.25 + 490·0.05 − 10·0.7 = 27.5 €
Answer
E(G) = 27.5 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 9.537
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 = (475 − 500)/25 = −1
Answer
z = −1
19.Normal distribution
Answer: P = 0.576
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 = (164 − 178)/7 = −2
z₂ = (179.75 − 178)/7 = 0.25
Area between
P = Φ(0.25) − Φ(−2) = 0.5987 − 0.02275 = 0.576
Answer
P = 0.576
20.Normal distribution
Answer: x = 48.6 points
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.