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→∞ (6x2 − 4x − 9)/(x2 + 2x + 7).
f(x) = (6x2 − 4x − 9)/(x2 + 2x + 7)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e3x − 1 − 3x)/x².
f(x) = (e3x − 1 − 3x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(3x), with x in radians. Find f′(1.5).
f(x) = x·sin(3x)x = 1.5 rad
f′(1.5) =
4.DerivativesHard
The tangent line to f(x) = −x2 + 2x + 4 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = −x2 + 2x + 4x = 1
b =
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
Evaluate I = ∫₀1 x²·eˣ dx.
∫₀1 x²eˣ dx
I =
7.Complex numbersMedium
Given z₁ = 2.5∠40° and z₂ = 0.5∠65°, find the argument θ of w = z₁·z₂ in degrees, between −180° and 180°.
z₁ = 2.5∠40°z₂ = 0.5∠65°
θ = °
8.Complex numbersHard
One root of z³ − 5z² + 17z − 13 = 0 is z₁ = 2 + j3. The coefficients are real. Find the real root r.
z₁ = 2 + j3
r =
9.Differential equationsMedium
A forged steel part at 300 °C is left to cool in a workshop at 21 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.05 min⁻¹, find its temperature T after 13 min.
T0 = 300 °CTa = 21 °Ck = 0.05 min⁻¹t = 13 min
T = °C
10.Differential equationsHard
A tank holds 400 L of pure water. Brine with 0.05 kg of salt per litre flows in at 5 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 90 min?
V = 400 Lm0 = 0 kgc = 0.05 kg/Lr = 5 L/min
m = kg
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [0 4 1; −2 2 −3; −3 −5 2]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [5 −3; 6 −6]
λ =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (−4, 1, 6) and b = (−3, 2, 7).
a = (−4, 1, 6)b = (−3, 2, 7)
a·b =
14.Vectors: dot & cross productHard
For a = (−1, −6, −5) and b = (−3, 6, 6), find the x-component (a × b)x of the cross product a × b.
a = (−1, −6, −5)b = (−3, 6, 6)
(a × b)x =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2p(x): 0.35, 0.3, 0.35
Var(X) =
16.ProbabilityHard
A game costs €10 to play. You win €20 with probability 0.1 and €500 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €10€20: p = 0.1€500: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 36, 48, 14, 24, 32.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 12. Find the z-score z of a score of 76.
μ = 100σ = 12x = 76
z =
19.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 30.5 and 35.3 min? Give P as a decimal.
μ = 32σ = 6x1 = 30.5x2 = 35.3
P =
20.Normal distributionHard
The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
3Solutions
1.Limits
Answer: L = 6
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
(6 + (−4)/x + (−9)/x2)/(1 + 2/x + 7/x2)
Let x → ∞
L = 6/1 = 6
Answer
L = 6
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 4.5
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·e3x − 3)/(2x) → 0/0
Second application
9·e3x/2
Substitute x = 0
L = 9·1/2 = 4.5
Answer
L = 4.5
3.Derivatives
Answer: f′(1.5) = −1.926
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 and chain rule
f′(x) = sin(3x) + 3x·cos(3x)
Substitute
f′(1.5) = −0.9775 + 4.5·(−0.2108) = −1.926
Answer
f′(1.5) = −1.926
4.Derivatives
Answer: b = 5
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) = 5
f′(x) = −2x + 2 → f′(1) = 0
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 5 − 0·1 = 5
Answer
b = 5
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: I = 0.7183
Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.
First round
∫x²eˣ dx = x²eˣ − ∫2x eˣ dx
Second round
∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)
Limits
2.718·(1) − 1·2 = 0.7183
Answer
I = 0.7183
7.Complex numbers
Answer: θ = 105 °
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| = 2.5 · 0.5 = 1.25
θ = 40° + 65° = 105°
Argument
θ = 105°
Answer
θ = 105°
8.Complex numbers
Answer: r = 1
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.
Check with the sum of the roots: z₁ + z₂ + r = 4 + 1 = 5 = −(−5).
9.Differential equations
Answer: T = 166.7 °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) = 21 + 279·e−0.05t
Substitute t
T(13) = 21 + 279·0.522 = 166.7 °C
Answer
T = 166.7 °C
10.Differential equations
Answer: m = 13.51 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 = 5·0.05 − (5/400)·m = 0.25 − 0.0125·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 20 + (−20)·e−t/80
Substitute t
m(90) = 20 + (−20)·0.3247 = 13.51 kg
Answer
m = 13.51 kg
11.Matrices & determinants
Answer: det A = 68
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 = −11, M₁₂ = −13, M₁₃ = 16
Cofactor expansion
det A = 0·(−11) − 4·(−13) + 1·16 = 68
Answer
det A = 68
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 − (−1)λ + (−12) = 0
Solve
λ = (−1 ± √(49))/2 → λ = −4, 3
λ = 3
Answer
λ = 3
13.Vectors: dot & cross product
Answer: a·b = 56
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.
Multiply matching components and add
a·b = (−4)·(−3) + 1·2 + 6·7 = 12 + 2 + 42 = 56
Answer
a·b = 56
Positive: the angle between the vectors is less than 90°.
14.Vectors: dot & cross product
Answer: (a × b)x = −6
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, component by component
(a × b)x = aybz − azby = (−6)·6 − (−5)·6 = −6
(a × b)y = azbx − axbz = (−5)·(−3) − (−1)·6 = 21
(a × b)z = axby − aybx = (−1)·6 − (−6)·(−3) = −24
Requested component
(a × b)x = −6
Answer
(a × b)x = −6
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: Var(X) = 0.7
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.
Mean
μ = 0·0.35 + 1·0.3 + 2·0.35 = 1
Mean of the squares
E(X²) = 0·0.35 + 1·0.3 + 4·0.35 = 1.7
Variance
Var(X) = 1.7 − 1² = 0.7
Answer
Var(X) = 0.7
16.Probability
Answer: E(G) = −3 €
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: 10 (p = 0.1), 490 (p = 0.01), −10 (p = 0.89)
Expected value
E(G) = 10·0.1 + 490·0.01 − 10·0.89 = −3 €
Answer
E(G) = −3 €
A negative value means you lose money on average.
17.Descriptive statistics
Answer: s = 12.77
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 = (76 − 100)/12 = −2
Answer
z = −2
19.Normal distribution
Answer: P = 0.3075
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 = (30.5 − 32)/6 = −0.25
z₂ = (35.3 − 32)/6 = 0.55
Area between
P = Φ(0.55) − Φ(−0.25) = 0.7088 − 0.4013 = 0.3075
Answer
P = 0.3075
20.Normal distribution
Answer: P = 0.7389
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).