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→∞ (4x3 + 7)/(5x3 + 2x + 1).
f(x) = (4x3 + 7)/(5x3 + 2x + 1)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e1.5x − 1 − 1.5x)/x².
f(x) = (e1.5x − 1 − 1.5x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x·e2x. Find f′(1).
f(x) = x·e2xx = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 − 15x2 + 63x − 5. Find the x-coordinate x of the local minimum.
f(x) = x3 − 15x2 + 63x − 5
x =
5.IntegralsMedium
The region under y = √(2x) from x = 0 to x = 2 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(2x)0 ≤ x ≤ 2
V =
6.IntegralsHard
Find the area A of the region between the parabola y = 4 − x² and the x-axis.
y = 4 − x²
A =
7.Complex numbersMedium
The diagram shows z = −4 + j4. With w = 2∠180°, click the point that represents z·w.
z = −4 + j4w = 2∠180°
1A (2 − j2)
2B (4 − j4)
3C (−8 + j8)
4D (8 − j8)
8.Complex numbersHard
Use De Moivre's theorem to find the real part of z⁷ for z = −√3 + j.
z = −√3 + jn = 7
Re(z⁷) =
9.Differential equationsMedium
A forged steel part at 300 °C is left to cool in a workshop at 22 °C. After 48 min its temperature is 63 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 300 °CTa = 22 °CT(48) = 63 °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 y(1); angles in radians.
y″ + 3y′ + 6.25y = 6.25y(0) = 0y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−3 5 −3; 3 −1 5; −3 2 −4]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [4 −1; −1 4]
λ =
13.Vectors: dot & cross productMedium
A constant force F = (−10, −20, −15) N acts on an object while it moves in a straight line from point A(0, −2, 0) to point B(5, −3, 5). Coordinates are in metres. Find the work W done by the force.
F = (−10, −20, −15) NA(0, −2, 0) mB(5, −3, 5) m
W = J
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−4, −1, −3) and b = (−3, 0, 3) and has a positive z-component. Find its x-component n̂x.
a = (−4, −1, −3)b = (−3, 0, 3)n̂z > 0
n̂x =
15.ProbabilityMedium
In how many ways N can a committee of 4 people be chosen from a group of 8? The order does not matter.
n = 8k = 4
N =
16.ProbabilityHard
A game costs €2 to play. You win €4 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€4: p = 0.2€20: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 32, 25, 38, 33, 30.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 10. Find the z-score z of a score of 70.
μ = 50σ = 10x = 70
z =
19.Normal distributionMedium
X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 111.4 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 111.4 s
P =
20.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
3Solutions
1.Limits
Answer: L = 0.8
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
(4 + 7/x3)/(5 + 2/x2 + 1/x3)
Let x → ∞
L = 4/5 = 0.8
Answer
L = 0.8
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 1.125
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
(1.5·e1.5x − 1.5)/(2x) → 0/0
Second application
2.25·e1.5x/2
Substitute x = 0
L = 2.25·1/2 = 1.125
Answer
L = 1.125
3.Derivatives
Answer: f′(1) = 22.17
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′ = e2x·(1 + 2·x)
Substitute
f′(1) = 7.389·(1 + 2) = 22.17
Answer
f′(1) = 22.17
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: V = 12.57
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.
Disc method
V = π∫₀2 2x dx
Integrate
V = π·2·2²/2 = 12.57
Answer
V = 12.57
6.Integrals
Answer: A = 10.67
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
4 − x² = 0 → x = ±2
Integrate
A = [4x − x³/3]−22 = 5.333 − (−5.333) = 10.67
Answer
A = 10.67
7.Complex numbers
Answer: D (8 − j8)
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.
Effect of w
|z·w| = 2·|z|, arg(z·w) = arg z + 180°
The point
z·w = 8 − j8 → D
Answer
D (8 − j8)
8.Complex numbers
Answer: Re(z⁷) = 110.9
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 = −√3 + j = 2∠150°
De Moivre
z⁷ = (2)⁷∠1050° = 128∠−30°
Rectangular form
z⁷ = 128·(cos −30° + j sin −30°) = 110.9 − j64
Re(z⁷) = 110.9
Answer
Re(z⁷) = 110.9
9.Differential equations
Answer: k = 0.03988 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−48k = 41/278
Solve for k
k = ln(278/41)/48 = 0.03988 min⁻¹
Answer
k = 0.03988 min⁻¹
10.Differential equations
Answer: y(1) = 0.9407
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
Evaluate
y(1) = 1·(1 − e−1.5(cos 2 + 0.75·sin 2)) = 0.9407
Answer
y(1) = 0.9407
11.Matrices & determinants
Answer: det A = −6
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 = −6, M₁₂ = 3, M₁₃ = 3
Cofactor expansion
det A = (−3)·(−6) − 5·3 + (−3)·3 = −6
Answer
det A = −6
12.Matrices & determinants
Answer: λ = 5
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.
Characteristic equation
λ² − (8)λ + (15) = 0
Solve
λ = (8 ± √(4))/2 → λ = 3, 5
λ = 5
Answer
λ = 5
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 = (5, −3, 5) − (0, −2, 0) = (5, −1, 5) m
Work
W = F·d = (−10)·5 + (−20)·(−1) + (−15)·5 = −105 J
Answer
W = −105 J
Negative work: the force acts mostly against the motion.
14.Vectors: dot & cross product
Answer: n̂x = 0.14
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 = (−1)·3 − (−3)·0 = −3
(a × b)y = azbx − axbz = (−3)·(−3) − (−4)·3 = 21
(a × b)z = axby − aybx = (−4)·0 − (−1)·(−3) = −3
Normalise
The z-component of a × b is negative, so use −(a × b).
|a × b| = √((−3)² + 21² + (−3)²) = 21.42
n̂ = −(−3, 21, −3)/21.42 = (0.14, −0.9802, 0.14)
n̂x = 0.14
Answer
n̂x = 0.14
15.Probability
Answer: N = 70
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.
Combinations
C(8, 4) = (8·7·6·5)/(4·3·2·1) = 70
Answer
N = 70
16.Probability
Answer: E(G) = −0.2 €
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: 2 (p = 0.2), 18 (p = 0.05), −2 (p = 0.75)
Expected value
E(G) = 2·0.2 + 18·0.05 − 2·0.75 = −0.2 €
Answer
E(G) = −0.2 €
A negative value means you lose money on average.
17.Descriptive statistics
Answer: s = 4.722
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 = (70 − 50)/10 = 2
Answer
z = 2
19.Normal distribution
Answer: P = 0.1587
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 sum
μ = 105 s, σ = √(5² + 4²) = 6.403 s
Standardise
z = (111.4 − 105)/6.403 = 1
Upper tail
P = 1 − Φ(1) = 1 − 0.8413 = 0.1587
Answer
P = 0.1587
20.Normal distribution
Answer: P = 0.8413
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).