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→∞ (2x2 + 6x − 1)/(5x2 + 4x + 5).
f(x) = (2x2 + 6x − 1)/(5x2 + 4x + 5)x → ∞
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) = ln(3x2 + 4). Find f′(−1).
f(x) = ln(3x2 + 4)x = −1
f′(−1) =
4.DerivativesHard
The tangent line to f(x) = x2 − 5 at x = −1 crosses the y-axis at y = b. Find b.
f(x) = x2 − 5x = −1
b =
5.IntegralsMedium
Evaluate I = ∫01 2x·(x² + 1)1 dx.
f(x) = 2x·(x² + 1)1[0, 1]
I =
6.IntegralsHard
The region between the line y = x and the parabola y = 0.5x² is rotated about the x-axis. Find the volume V of the solid.
y = xy = 0.5x²
V =
7.Complex numbersMedium
Given z₁ = 4∠100° and z₂ = 0.5∠−95°, find the argument θ of w = z₁/z₂ in degrees, between −180° and 180°.
z₁ = 4∠100°z₂ = 0.5∠−95°
θ = °
8.Complex numbersHard
Use De Moivre's theorem to find the imaginary part of z⁸ for z = 1 + j√3.
z = 1 + j√3n = 8
Im(z⁸) =
9.Differential equationsMedium
A can of soda at 4 °C is taken out of the fridge into a room at 28 °C. After 10 min its temperature is 19 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 12 °C?
T0 = 4 °CTa = 28 °CT(10) = 19 °CT = 12 °C
t = min
10.Differential equationsHard
A system obeys y″ + 3y′ + 27.25y = 27.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 3y′ + 27.25y = 27.25y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−1 −5 0; −2 −5 1; −4 0 −1]
det A =
12.Matrices & determinantsHard
Solve the system x + 3y = −5 and −3x + 2y = 0. Find x.
x + 3y = −5−3x + 2y = 0
x =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (−2, 4, −2) and b = (−7, −3, −5).
a = (−2, 4, −2)b = (−7, −3, −5)
a·b =
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(0, −1, −4), B(2, 4, −2) and C(−4, 0, 1). Coordinates are in metres.
A(0, −1, −4)B(2, 4, −2)C(−4, 0, 1)
A = m²
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 10, 20, 30, 40p(x): 0.3, 0.25, 0.3, 0.15
Var(X) =
16.ProbabilityHard
A game costs €2 to play. You win €10 with probability 0.1 and €100 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €2€10: p = 0.1€100: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 20, 49, 41, 43, 18, 56.
n = 6
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
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 26 and 44 min? Give P as a decimal.
μ = 32σ = 6x1 = 26x2 = 44
P =
20.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 90% of them lie (the 90th percentile).
μ = 500σ = 40P = 90%
x = h
3Solutions
1.Limits
Answer: L = 0.4
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
(2 + 6/x + (−1)/x2)/(5 + 4/x + 5/x2)
Let x → ∞
L = 2/5 = 0.4
Answer
L = 0.4
For equal degrees the limit is the ratio of the leading coefficients.
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′(−1) = −0.8571
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 + 4)
Substitute
f′(−1) = −6/7 = −0.8571
Answer
f′(−1) = −0.8571
4.Derivatives
Answer: b = −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.
Point and slope
f(−1) = −4
f′(x) = 2x → f′(−1) = −2
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −4 − (−2)·(−1) = −6
Answer
b = −6
5.Integrals
Answer: I = 1.5
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 u1 du
Integrate and evaluate
I = [u2/2]12 = (4 − 1)/2 = 1.5
Answer
I = 1.5
6.Integrals
Answer: V = 3.351
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
1x = 0.5x² → x = 0, x = 2
Washer method
V = π∫₀2 ((1x)² − (0.5x²)²) dx
Integrate
V = π(1·2³/3 − 0.25·2⁵/5) = π·1.067 = 3.351
Answer
V = 3.351
7.Complex numbers
Answer: θ = −165 °
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| = 4 / 0.5 = 8
θ = 100° − (−95)° = 195° → −165°
Argument
θ = −165°
Answer
θ = −165°
8.Complex numbers
Answer: Im(z⁸) = 221.7
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 = 1 + j√3 = 2∠60°
De Moivre
z⁸ = (2)⁸∠480° = 256∠120°
Rectangular form
z⁸ = 256·(cos 120° + j sin 120°) = −128 + j221.7
Im(z⁸) = 221.7
Answer
Im(z⁸) = 221.7
9.Differential equations
Answer: t = 4.134 min
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.
Cooling constant
k = ln(-24/-9)/10 = 0.09808 min⁻¹
Time to reach the target
t = ln(-24/(−16))/0.09808 = 4.134 min
Answer
t = 4.134 min
10.Differential equations
Answer: ymax = 1.39
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 ± 5j
Constants from the initial values
y(0) = 1 + A = 0 → A = −1
y′(0) = −1.5A + 5B = 0 → B = −0.3
Peak
t = π/5 = 0.6283
ymax = 1·(1 + e−1.5·π/5) = 1.39
Answer
ymax = 1.39
11.Matrices & determinants
Answer: det A = 25
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 = 5, M₁₂ = 6, M₁₃ = −20
Cofactor expansion
det A = (−1)·5 − (−5)·6 + 0·(−20) = 25
Answer
det A = 25
12.Matrices & determinants
Answer: x = −0.9091
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 = 1·2 − 3·(−3) = 11
Replace the x column
det Aₓ = (−5)·2 − 3·0 = −10
x = −10/11 = −0.9091
Answer
x = −0.9091
y = −1.364
13.Vectors: dot & cross product
Answer: a·b = 12
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.
Positive: the angle between the vectors is less than 90°.
14.Vectors: dot & cross product
Answer: A = 18.28 m²
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.
Edge vectors
AB = B − A = (2, 5, 2)
AC = C − A = (−4, 1, 5)
Cross product
(AB × AC)x = AByACz − ABzACy = 5·5 − 2·1 = 23
(AB × AC)y = ABzACx − ABxACz = 2·(−4) − 2·5 = −18
(AB × AC)z = ABxACy − AByACx = 2·1 − 5·(−4) = 22
|AB × AC| = √(23² + (−18)² + 22²) = 36.57
Area of the triangle
A = ½·36.57 = 18.28 m²
Answer
A = 18.28 m²
15.Probability
Answer: Var(X) = 111
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: 8 (p = 0.1), 98 (p = 0.05), −2 (p = 0.85)
Expected value
E(G) = 8·0.1 + 98·0.05 − 2·0.85 = 4 €
Answer
E(G) = 4 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 15.51
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 = 0.8186
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 = (26 − 32)/6 = −1
z₂ = (44 − 32)/6 = 2
Area between
P = Φ(2) − Φ(−1) = 0.9772 − 0.1587 = 0.8186
Answer
P = 0.8186
20.Normal distribution
Answer: x = 551.3 h
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.