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)3x. Give a decimal value.
f(x) = (1 − 2/x)3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e−0.5x − 1 + 0.5x)/x².
f(x) = (e−0.5x − 1 + 0.5x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = ln(3x2 + 1). Find f′(3).
f(x) = ln(3x2 + 1)x = 3
f′(3) =
4.DerivativesHard
The tangent line to f(x) = 2x2 − 4x − 5 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = 2x2 − 4x − 5x = 1
b =
5.IntegralsMedium
The region under y = √(x) from x = 0 to x = 3 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(x)0 ≤ x ≤ 3
V =
6.IntegralsHard
The region between the line y = 4x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.
y = 4xy = 3x²
V =
7.Complex numbersMedium
Write z = −4 − j7 in exponential form r·ejθ. Give θ in radians, between −π and π.
z = −4 − j7
θ = rad
8.Complex numbersHard
The roots of zⁿ = w with n = 3 and w = 64∠60° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2. Find the argument of z₀ in degrees, between −180° and 180°.
w = 64∠60°n = 3k = 0
θ₀ = °
9.Differential equationsMedium
Solve y″ + 8y′ + 16y = 0 with y(0) = 2 and y′(0) = 0. Find y(1).
y″ + 8y′ + 16y = 0y(0) = 2y′(0) = 0
y(1) =
10.Differential equationsHard
A tank holds 400 L of pure water. Brine with 0.2 kg of salt per litre flows in at 4 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 60 min?
V = 400 Lm0 = 0 kgc = 0.2 kg/Lr = 4 L/min
m = kg
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).
A = [−5 4; 0 −2]B = [−1 −3; 4 −5]
c12 =
12.Matrices & determinantsHard
Solve the system x − 3y = −9 and 3x − 6y = −9. Find x.
x − 3y = −93x − 6y = −9
x =
13.Vectors: dot & cross productMedium
A constant force F = (30, −5, 15) N acts on an object while it moves in a straight line from point A(4, 4, 3) to point B(1, 4, 2). Coordinates are in metres. Find the work W done by the force.
F = (30, −5, 15) NA(4, 4, 3) mB(1, 4, 2) m
W = J
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−2, −2, 4) and b = (−5, 3, 1) and has a positive z-component. Find its y-component n̂y.
a = (−2, −2, 4)b = (−5, 3, 1)n̂z > 0
n̂y =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2, 3p(x): 0.15, 0.3, 0.2, 0.35
σ =
16.ProbabilityHard
A game costs €2 to play. You win €10 with probability 0.3 and €100 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €2€10: p = 0.3€100: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 12, 37, 23, 48, 51.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 5. What proportion P of values is above 177.5?
μ = 170σ = 5x = 177.5
P =
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 99.9 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 99.9 s
P =
20.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 75% of them lie (the 75th percentile).
μ = 250σ = 5P = 75%
x = g
3Solutions
1.Limits
Answer: L = 0.002479
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 − 2/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 3·(−2) = −6
Undo the logarithm
L = e−6 = 0.002479
Answer
L = 0.002479
2.Limits
Answer: L = 0.125
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
(−0.5·e−0.5x + 0.5)/(2x) → 0/0
Second application
0.25·e−0.5x/2
Substitute x = 0
L = 0.25·1/2 = 0.125
Answer
L = 0.125
3.Derivatives
Answer: f′(3) = 0.6429
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.
Chain rule for the logarithm
f′(x) = 6x/(3x2 + 1)
Substitute
f′(3) = 18/28 = 0.6429
Answer
f′(3) = 0.6429
4.Derivatives
Answer: b = −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.
Point and slope
f(1) = −7
f′(x) = 4x − 4 → f′(1) = 0
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −7 − 0·1 = −7
Answer
b = −7
5.Integrals
Answer: V = 14.14
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∫₀3 x dx
Integrate
V = π·1·3²/2 = 14.14
Answer
V = 14.14
6.Integrals
Answer: V = 15.89
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
4x = 3x² → x = 0, x = 1.333
Washer method
V = π∫₀1.333 ((4x)² − (3x²)²) dx
Integrate
V = π(16·1.333³/3 − 9·1.333⁵/5) = π·5.057 = 15.89
Answer
V = 15.89
7.Complex numbers
Answer: θ = −2.09 rad
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
r = √(16 + 49) = 8.062
Argument in radians
θ = −119.7° = −119.7·π/180 = −2.09 rad
Exponential form
z = 8.062·e−j2.09
Answer
θ = −2.09 rad
8.Complex numbers
Answer: θ₀ = 20 °
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.
Modulus of the roots
641/3 = 4
Argument of the root
θ₀ = (60° + 360°·0)/3 = 20°
Argument
θ₀ = 20°
Answer
θ₀ = 20°
The 3 roots lie on the circle of radius 4, 120° apart.
9.Differential equations
Answer: y(1) = 0.1832
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 + 4)² = 0 → r = −4
Constants
A = y(0) = 2
y′(0) = B − 4A → B = 0 + 8 = 8
Evaluate
y(1) = (2 + 8·1)·e−4 = 0.1832
Answer
y(1) = 0.1832
10.Differential equations
Answer: m = 36.1 kg
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.
Set up the equation
dm/dt = 4·0.2 − (4/400)·m = 0.8 − 0.01·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 80 + (−80)·e−t/100
Substitute t
m(60) = 80 + (−80)·0.5488 = 36.1 kg
Answer
m = 36.1 kg
11.Matrices & determinants
Answer: c12 = −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.
Row times column
c12 = (−5)·(−3) + 4·(−5) = −5
Answer
c12 = −5
12.Matrices & determinants
Answer: x = 9
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·(−6) − (−3)·3 = 3
Replace the x column
det Aₓ = (−9)·(−6) − (−3)·(−9) = 27
x = 27/3 = 9
Answer
x = 9
y = 6
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 = (1, 4, 2) − (4, 4, 3) = (−3, 0, −1) m
Work
W = F·d = 30·(−3) + (−5)·0 + 15·(−1) = −105 J
Answer
W = −105 J
Negative work: the force acts mostly against the motion.
14.Vectors: dot & cross product
Answer: n̂y = 0.6462
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 = (−2)·1 − 4·3 = −14
(a × b)y = azbx − axbz = 4·(−5) − (−2)·1 = −18
(a × b)z = axby − aybx = (−2)·3 − (−2)·(−5) = −16
Normalise
The z-component of a × b is negative, so use −(a × b).
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 = 0·0.15 + 1·0.3 + 2·0.2 + 3·0.35 = 1.75
Mean of the squares
E(X²) = 0·0.15 + 1·0.3 + 4·0.2 + 9·0.35 = 4.25
Variance
Var(X) = 4.25 − 1.75² = 1.188
σ = √1.188 = 1.09
Answer
σ = 1.09
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: 8 (p = 0.3), 98 (p = 0.01), −2 (p = 0.69)
Expected value
E(G) = 8·0.3 + 98·0.01 − 2·0.69 = 2 €
Answer
E(G) = 2 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 16.57
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) = 0.9332
P = 1 − 0.9332 = 0.06681
Answer
P = 0.06681
19.Normal distribution
Answer: P = 0.7881
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 = (99.9 − 105)/6.403 = −0.8
Upper tail
P = 1 − Φ(−0.8) = 1 − 0.2119 = 0.7881
Answer
P = 0.7881
20.Normal distribution
Answer: x = 253.4 g
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.