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 (1 − cos 5x)/(2x²), with x in radians.
f(x) = (1 − cos 5x)/(2x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·e−x. Find f′(2).
f(x) = x·e−xx = 2
f′(2) =
4.DerivativesHard
Let f(x) = x3 − 12x2 + 36x + 4. Find the x-coordinate x of the local minimum.
f(x) = x3 − 12x2 + 36x + 4
x =
5.IntegralsMedium
Evaluate I = ∫01 2x·(x² + 1)2 dx.
f(x) = 2x·(x² + 1)2[0, 1]
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
Find the argument θ of z = 4 − j7 in degrees, as the principal value between −180° and 180°.
z = 4 − j7
θ = °
8.Complex numbersHard
Solve z² + 2z + 3.25 = 0. The roots are a ± jb; find b > 0.
p = 2q = 3.25
b =
9.Differential equationsMedium
Solve y″ + 8y′ + 15y = 0 with y(0) = 3 and y′(0) = 1. Find y(1).
y″ + 8y′ + 15y = 0y(0) = 3y′(0) = 1
y(1) =
10.Differential equationsHard
A system obeys y″ + y′ + 25.25y = 126.3 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + y′ + 25.25y = 126.3y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−2 2 3; −3 4 0; −2 1 −2]
det A =
12.Matrices & determinantsHard
Solve the system 5x + 2y = 12 and 5x + 4y = 9. Find x.
5x + 2y = 125x + 4y = 9
x =
13.Vectors: dot & cross productMedium
A constant force F = (−20, 10, 10) N acts on an object while it moves in a straight line from point A(3, −4, 4) to point B(−4, −2, 5). Coordinates are in metres. Find the work W done by the force.
F = (−20, 10, 10) NA(3, −4, 4) mB(−4, −2, 5) m
W = J
14.Vectors: dot & cross productHard
A force F = (−40, −50, 30) N acts at point A(−1, 2, −3). Coordinates are in metres. Find the magnitude |M| of the moment of the force about the origin O(0, 0, 0).
F = (−40, −50, 30) NA(−1, 2, −3) mO(0, 0, 0) m
|M| = N·m
15.ProbabilityMedium
A bag holds 5 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
5 / 6
P =
16.ProbabilityHard
6 components are tested, and each one fails with probability 0.5, independently. What is the probability P that at most one fails?
n = 6p = 0.5
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 16, 20, 35, 52, 58.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is below 40?
μ = 50σ = 10x = 40
P =
19.Normal distributionMedium
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 167.5 and 188.85 cm? Give P as a decimal.
μ = 178σ = 7x1 = 167.5x2 = 188.85
P =
20.Normal distributionHard
Hole diameters D are normal with mean 25.03 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 = 25.03 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
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 = 6.25
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
5·sin 5x/(4x) → 0/0
Second application
25·cos 5x/4
Substitute x = 0
L = 25/4 = 6.25
Answer
L = 6.25
3.Derivatives
Answer: f′(2) = −0.1353
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·(1 + (−1)·x)
Substitute
f′(2) = 0.1353·(1 + (−2)) = −0.1353
Answer
f′(2) = −0.1353
4.Derivatives
Answer: x = 6
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 − 24x + 36 = 3·(x2 − 8x + 12)
Solve f′(x) = 0
x = 2 or x = 6
Which one is which
f″(x) = 6x − 24: negative at a maximum, positive at a minimum.
f″(2) = −12, f″(6) = 12
x = 6
Answer
x = 6
5.Integrals
Answer: I = 2.333
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.
Substitution
u = x² + 1, du = 2x dx
I = ∫12 u2 du
Integrate and evaluate
I = [u3/3]12 = (8 − 1)/3 = 2.333
Answer
I = 2.333
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: θ = −60.26 °
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 = 4, b = −7 → IV
Reference angle
α = atan(7/4) = 60.26°
Argument
θ = −α = −60.26°
Answer
θ = −60.26°
In polar form z = 8.062∠−60.26°.
8.Complex numbers
Answer: b = 1.5
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 = 4 − 13 = −9 < 0
Quadratic formula
z = (−2 ± j√9)/2 = −1 + j1.5, −1 − j1.5
Answer
b = 1.5
Answer
b = 1.5
9.Differential equations
Answer: y(1) = 0.3646
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.
Characteristic roots
(r + 3)(r + 5) = 0 → r = −3, −5
Constants from the initial values
A + B = 3, −3A − 5B = 1
A = 8, B = −5
Evaluate
y(1) = 8·e−3 + (−5)·e−5 = 0.3646
Answer
y(1) = 0.3646
10.Differential equations
Answer: ymax = 8.652
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 = 5, r = −0.5 ± 5j
Constants from the initial values
y(0) = 5 + A = 0 → A = −5
y′(0) = −0.5A + 5B = 0 → B = −0.5
Peak
t = π/5 = 0.6283
ymax = 5·(1 + e−0.5·π/5) = 8.652
Answer
ymax = 8.652
11.Matrices & determinants
Answer: det A = 19
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 = −8, M₁₂ = 6, M₁₃ = 5
Cofactor expansion
det A = (−2)·(−8) − 2·6 + 3·5 = 19
Answer
det A = 19
12.Matrices & determinants
Answer: x = 3
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 = 5·4 − 2·5 = 10
Replace the x column
det Aₓ = 12·4 − 2·9 = 30
x = 30/10 = 3
Answer
x = 3
y = −1.5
13.Vectors: dot & cross product
Answer: W = 170 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, −2, 5) − (3, −4, 4) = (−7, 2, 1) m
Work
W = F·d = (−20)·(−7) + 10·2 + 10·1 = 170 J
Answer
W = 170 J
14.Vectors: dot & cross product
Answer: |M| = 217.9 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 = (−1, 2, −3) − (0, 0, 0) = (−1, 2, −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.
First and second draw
P = 5/11 · 4/10 = 20/110 = 0.1818
Answer
P = 0.1818
16.Probability
Answer: P = 0.1094
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.
Zero and one failure
P(X = 0) = 0.56 = 0.01563
P(X = 1) = 6·0.5·0.55 = 0.09375
Add them
P = 0.01563 + 0.09375 = 0.1094
Answer
P = 0.1094
17.Descriptive statistics
Answer: s = 18.69
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 = 0.1587
Answer
P = 0.1587
19.Normal distribution
Answer: P = 0.8726
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 = (167.5 − 178)/7 = −1.5
z₂ = (188.85 − 178)/7 = 1.55
Area between
P = Φ(1.55) − Φ(−1.5) = 0.9394 − 0.06681 = 0.8726
Answer
P = 0.8726
20.Normal distribution
Answer: P = 0.9515
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).