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→∞ (√(x2 − 8x) − x).
f(x) = √(x2 − 8x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 3x)/(4x²), with x in radians.
f(x) = (1 − cos 3x)/(4x²)x → 0
L =
3.DerivativesMedium
Let f(x) = (2x − 1)3. Find f′(−1).
f(x) = (2x − 1)3x = −1
f′(−1) =
4.DerivativesHard
The tangent line to f(x) = 0.5x2 − x + 4 at x = −2 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 − x + 4x = −2
b =
5.IntegralsMedium
Evaluate I = ∫14 2/x dx.
f(x) = 2/x[1, 4]
I =
6.IntegralsHard
The region between the line y = 6x and the parabola y = 2x² is rotated about the x-axis. Find the volume V of the solid.
y = 6xy = 2x²
V =
7.Complex numbersMedium
The number z = 1.5·ej2 has its angle in radians. Find its imaginary part b.
r = 1.5θ = 2 rad
b =
8.Complex numbersHard
Use De Moivre's theorem to find the real part of z⁴ for z = 1 − j.
z = 1 − jn = 4
Re(z⁴) =
9.Differential equationsMedium
A forged steel part at 420 °C is left to cool in a workshop at 26 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.04 min⁻¹, find its temperature T after 11 min.
T0 = 420 °CTa = 26 °Ck = 0.04 min⁻¹t = 11 min
T = °C
10.Differential equationsHard
A system obeys y″ + y′ + 16.25y = 16.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + y′ + 16.25y = 16.25y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [2 6; 2 3]B = [−5 −5; 6 2]
c11 =
12.Matrices & determinantsHard
Solve the system 5x − y = 5 and 4x + 2y = −9. Find x.
5x − y = 54x + 2y = −9
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (2, −3, 5) and b = (k, −3, 5) are perpendicular.
a = (2, −3, 5)b = (k, −3, 5)
k =
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (−3, −5, −1), b = (5, 2, 0) and c = (−2, −1, 2).
a = (−3, −5, −1)b = (5, 2, 0)c = (−2, −1, 2)
V =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2p(x): 0.3, 0.35, 0.35
σ =
16.ProbabilityHard
A game costs €10 to play. You win €50 with probability 0.3 and €200 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €10€50: p = 0.3€200: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 42, 13, 33, 39, 35, 17, 20.
n = 7
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 8. Find the z-score z of a score of 93.6.
μ = 100σ = 8x = 93.6
z =
19.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 26.3 and 42.8 min? Give P as a decimal.
μ = 32σ = 6x1 = 26.3x2 = 42.8
P =
20.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.
μ = 178 cmσ = 7 cm40%
x = cm
3Solutions
1.Limits
Answer: L = −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).
Multiply by the conjugate
f(x) = −8x/(√(x2 − 8x) + x)
Divide by x
f(x) = −8/(√(1 + (−8)/x) + 1)
Let x → ∞
L = −8/(1 + 1) = −4
Answer
L = −4
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
3·sin 3x/(8x) → 0/0
Second application
9·cos 3x/8
Substitute x = 0
L = 9/8 = 1.125
Answer
L = 1.125
3.Derivatives
Answer: f′(−1) = 54
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
f′(x) = 3·(2x − 1)2·2
Substitute
f′(−1) = 3·(−3)2·2 = 54
Answer
f′(−1) = 54
4.Derivatives
Answer: b = 2
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(−2) = 8
f′(x) = x − 1 → f′(−2) = −3
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 8 − (−3)·(−2) = 2
Answer
b = 2
5.Integrals
Answer: I = 2.773
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.
Antiderivative
F(x) = 2·ln x
Evaluate at the limits
I = 2·(ln 4 − ln 1) = 2·ln 4 = 2.773
Answer
I = 2.773
6.Integrals
Answer: V = 407.2
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
6x = 2x² → x = 0, x = 3
Washer method
V = π∫₀3 ((6x)² − (2x²)²) dx
Integrate
V = π(36·3³/3 − 4·3⁵/5) = π·129.6 = 407.2
Answer
V = 407.2
7.Complex numbers
Answer: b = 1.364
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.
Euler's formula
z = 1.5·(cos 2 + j sin 2)
Components (radians)
a = 1.5·cos 2 = −0.6242
b = 1.5·sin 2 = 1.364
b = 1.364
Answer
b = 1.364
8.Complex numbers
Answer: Re(z⁴) = −4
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 = √2∠−45°
De Moivre
z⁴ = (√2)⁴∠−180° = 4∠180°
Rectangular form
z⁴ = 4·(cos 180° + j sin 180°) = −4
Re(z⁴) = −4
Answer
Re(z⁴) = −4
9.Differential equations
Answer: T = 279.8 °C
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.
Solution of the cooling law
T(t) = 26 + 394·e−0.04t
Substitute t
T(11) = 26 + 394·0.644 = 279.8 °C
Answer
T = 279.8 °C
10.Differential equations
Answer: ymax = 1.675
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 = −0.5 ± 4j
Constants from the initial values
y(0) = 1 + A = 0 → A = −1
y′(0) = −0.5A + 4B = 0 → B = −0.125
Peak
t = π/4 = 0.7854
ymax = 1·(1 + e−0.5·π/4) = 1.675
Answer
ymax = 1.675
11.Matrices & determinants
Answer: c11 = 26
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
c11 = 2·(−5) + 6·6 = 26
Answer
c11 = 26
12.Matrices & determinants
Answer: x = 0.07143
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 = 5·2 − (−1)·4 = 14
Replace the x column
det Aₓ = 5·2 − (−1)·(−9) = 1
x = 1/14 = 0.07143
Answer
x = 0.07143
y = −4.643
13.Vectors: dot & cross product
Answer: k = −17
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 equals zero
a·b = 2·k + (−3)·(−3) + 5·5 = 0
34 + 2·k = 0
Solve for k
k = −34/2 = −17
Answer
k = −17
14.Vectors: dot & cross product
Answer: V = 39
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 b × c
(b × c)x = bycz − bzcy = 2·2 − 0·(−1) = 4
(b × c)y = bzcx − bxcz = 0·(−2) − 5·2 = −10
(b × c)z = bxcy − bycx = 5·(−1) − 2·(−2) = −1
Dot product with a
a·(b × c) = (−3)·4 + (−5)·(−10) + (−1)·(−1) = 39
V = |39| = 39
Answer
V = 39
15.Probability
Answer: σ = 0.8047
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 = 0·0.3 + 1·0.35 + 2·0.35 = 1.05
Mean of the squares
E(X²) = 0·0.3 + 1·0.35 + 4·0.35 = 1.75
Variance
Var(X) = 1.75 − 1.05² = 0.6475
σ = √0.6475 = 0.8047
Answer
σ = 0.8047
16.Probability
Answer: E(G) = 7 €
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: 40 (p = 0.3), 190 (p = 0.01), −10 (p = 0.69)
Expected value
E(G) = 40·0.3 + 190·0.01 − 10·0.69 = 7 €
Answer
E(G) = 7 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 11.54
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 = (93.6 − 100)/8 = −0.8
Answer
z = −0.8
19.Normal distribution
Answer: P = 0.793
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 = (26.3 − 32)/6 = −0.95
z₂ = (42.8 − 32)/6 = 1.8
Area between
P = Φ(1.8) − Φ(−0.95) = 0.9641 − 0.1711 = 0.793
Answer
P = 0.793
20.Normal distribution
Answer: x = 179.8 cm
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.