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 − 2/x)x. Give a decimal value.
f(x) = (1 − 2/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 5x)/(4x²), with x in radians.
f(x) = (1 − cos 5x)/(4x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x3·e−x. Find f′(0.5).
f(x) = x3·e−xx = 0.5
f′(0.5) =
4.DerivativesHard
Let f(x) = x3 + 3x2 − 9x − 1. Find the x-coordinate x of the local maximum.
f(x) = x3 + 3x2 − 9x − 1
x =
5.IntegralsMedium
The region under y = x from x = 0 to x = 4 is rotated about the x-axis. Find the volume V of the solid (a cone).
y = x0 ≤ x ≤ 4
V =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = x.
y = x2y = x
A =
7.Complex numbersMedium
Given z₁ = 2∠125° and z₂ = 4∠20°, find the imaginary part of w = z₁·z₂.
z₁ = 2∠125°z₂ = 4∠20°
Im(w) =
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
Solve y″ + 2y′ + y = 0 with y(0) = 3 and y′(0) = 4. Find y(1.5).
y″ + 2y′ + y = 0y(0) = 3y′(0) = 4
y(1.5) =
10.Differential equationsHard
Solve y″ + 3y′ + 11.25y = 0 with y(0) = 10 and y′(0) = 0 (released from rest). Find y(0.5); angles in radians.
y″ + 3y′ + 11.25y = 0y(0) = 10y′(0) = 0
y(0.5) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−4 −2 1; 0 −5 −2; 0 5 −4]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−2 −3; 5 6]
λ =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−5, 2, −6) and b = (−2, 5, −1), in degrees.
a = (−5, 2, −6)b = (−2, 5, −1)
θ = °
14.Vectors: dot & cross productHard
For a = (−4, 6, −5) and b = (−4, 1, −2), find the z-component (a × b)z of the cross product a × b.
a = (−4, 6, −5)b = (−4, 1, −2)
(a × b)z =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 10, 20, 30, 40p(x): 0.25, 0.3, 0.15, 0.3
Var(X) =
16.ProbabilityHard
A game costs €1 to play. You win €5 with probability 0.1 and €10 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €1€5: p = 0.1€10: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 23, 58, 51, 37, 35, 56, 29.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is below 38?
μ = 50σ = 8x = 38
P =
19.Normal distributionMedium
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 242.25 and 256.25 g? Give P as a decimal.
μ = 250σ = 5x1 = 242.25x2 = 256.25
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 bottom 30% is to the left of it. Find the cut-off x.
μ = 500 hσ = 40 h30%
x = h
3Solutions
1.Limits
Answer: L = 0.1353
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→∞ x·ln(1 − 2/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·(−2) = −2
Undo the logarithm
L = e−2 = 0.1353
Answer
L = 0.1353
2.Limits
Answer: L = 3.125
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
5·sin 5x/(8x) → 0/0
Second application
25·cos 5x/8
Substitute x = 0
L = 25/8 = 3.125
Answer
L = 3.125
3.Derivatives
Answer: f′(0.5) = 0.3791
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′(0.5) = 0.6065·(0.75 + (−0.125)) = 0.3791
Answer
f′(0.5) = 0.3791
4.Derivatives
Answer: x = −3
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 − 9 = 3·(x2 + 2x − 3)
Solve f′(x) = 0
x = −3 or x = 1
Which one is which
f″(x) = 6x + 6: negative at a maximum, positive at a minimum.
f″(−3) = −12, f″(1) = 12
x = −3
Answer
x = −3
5.Integrals
Answer: V = 67.02
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∫₀4 1x² dx
Integrate
V = π·1·4³/3 = 67.02
Answer
V = 67.02
Check: a cone of radius 4 and height 4 has (1/3)πr²h = 67.02.
6.Integrals
Answer: A = 0.1667
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.
Intersection points
x2 = x → x = 0, x = 1
Integrate upper minus lower
A = [0.5x2 − 0.3333x3]01 = 0.1667
Answer
A = 0.1667
7.Complex numbers
Answer: Im(w) = 4.589
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.
Modulus and argument of w
|w| = 2 · 4 = 8
θ = 125° + 20° = 145°
Rectangular form
Re(w) = 8·cos(145°) = −6.553
Im(w) = 8·sin(145°) = 4.589
Im(w) = 4.589
Answer
Im(w) = 4.589
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∠45°
De Moivre
z⁵ = (2√2)⁵∠225° = 181∠−135°
Rectangular form
z⁵ = 181·(cos −135° + j sin −135°) = −128 − j128
Im(z⁵) = −128
Answer
Im(z⁵) = −128
9.Differential equations
Answer: y(1.5) = 3.012
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.
Double root
(r + 1)² = 0 → r = −1
Constants
A = y(0) = 3
y′(0) = B − 1A → B = 4 + 3 = 7
Evaluate
y(1.5) = (3 + 7·1.5)·e−1.5 = 3.012
Answer
y(1.5) = 3.012
10.Differential equations
Answer: y(0.5) = 2.69
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 = 10, B = 1.5·10/3 = 5
Evaluate
y(0.5) = e−0.75(10·cos 1.5 + 5·sin 1.5) = 2.69
Answer
y(0.5) = 2.69
11.Matrices & determinants
Answer: det A = −120
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 = 30, M₁₂ = 0, M₁₃ = 0
Cofactor expansion
det A = (−4)·30 − (−2)·0 + 1·0 = −120
Answer
det A = −120
12.Matrices & determinants
Answer: λ = 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.
Characteristic equation
λ² − (4)λ + (3) = 0
Solve
λ = (4 ± √(4))/2 → λ = 1, 3
λ = 3
Answer
λ = 3
13.Vectors: dot & cross product
Answer: θ = 53.93 °
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 = (−5)·(−2) + 2·5 + (−6)·(−1) = 26
Magnitudes
|a| = √((−5)² + 2² + (−6)²) = 8.062
|b| = √((−2)² + 5² + (−1)²) = 5.477
Angle
cos θ = 26/(8.062·5.477) = 0.5888
θ = arccos(0.5888) = 53.93°
Answer
θ = 53.93°
14.Vectors: dot & cross product
Answer: (a × b)z = 20
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.
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: Var(X) = 135
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.
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: 4 (p = 0.1), 9 (p = 0.02), −1 (p = 0.88)
Expected value
E(G) = 4·0.1 + 9·0.02 − 1·0.88 = −0.3 €
Answer
E(G) = −0.3 €
A negative value means you lose money on average.
17.Descriptive statistics
Answer: s = 13.74
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 = −1.5
Standard normal table
Φ(−1.5) = 0.06681
P = 0.06681
Answer
P = 0.06681
19.Normal distribution
Answer: P = 0.8338
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.
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.