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 − 3x − 8)/(5x2 + 3x + 5).
f(x) = (−x2 − 3x − 8)/(5x2 + 3x + 5)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e1.5x − 1 − 1.5x)/x².
f(x) = (e1.5x − 1 − 1.5x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = (4x + 3)4. Find f′(1).
f(x) = (4x + 3)4x = 1
f′(1) =
4.DerivativesHard
The tangent line to f(x) = 0.5x2 + 4x − 6 at x = −2 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 + 4x − 6x = −2
b =
5.IntegralsMedium
Evaluate I = ∫02 e2x dx.
f(x) = e2x[0, 2]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 2x2 and y = 2x.
y = 2x2y = 2x
A =
7.Complex numbersMedium
Given z₁ = 2.5∠55° and z₂ = 1.5∠20°, find the real part of w = z₁/z₂.
z₁ = 2.5∠55°z₂ = 1.5∠20°
Re(w) =
8.Complex numbersHard
Solve z² − 3z + 8.5 = 0. The roots are a ± jb; find b > 0.
p = −3q = 8.5
b =
9.Differential equationsMedium
A can of soda at 3 °C is taken out of the fridge into a room at 28 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.08 min⁻¹, find its temperature T after 25 min.
T0 = 3 °CTa = 28 °Ck = 0.08 min⁻¹t = 25 min
T = °C
10.Differential equationsHard
A tank holds 200 L of pure water. Brine with 0.2 kg of salt per litre flows in at 2 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 24 kg of salt?
V = 200 Lm0 = 0 kgc = 0.2 kg/Lr = 2 L/min
t = min
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [−1 −5; −4 6]B = [−6 −6; 4 4]
c11 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [3 2; 4 1]
λ =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (−5, 4, −1) and b = (2, −3, k) are perpendicular.
a = (−5, 4, −1)b = (2, −3, k)
k =
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(−2, 0, −3), B(−2, −2, 0) and C(−4, −2, −1). Coordinates are in metres.
A(−2, 0, −3)B(−2, −2, 0)C(−4, −2, −1)
A = m²
15.ProbabilityMedium
A fair die is rolled 2 times. What is the probability P of getting at least one six?
n = 2
P =
16.ProbabilityHard
A game costs €10 to play. You win €40 with probability 0.15 and €200 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €10€40: p = 0.15€200: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 47, 58, 46, 38, 32, 37, 28.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 20. What proportion P of values is above 36?
μ = 60σ = 20x = 36
P =
19.Normal distributionMedium
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 12 of them exceeds 300.9 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 12> 300.9 kg
P =
20.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 60% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g60%
x = g
3Solutions
1.Limits
Answer: L = −0.2
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
(−1 + (−3)/x + (−8)/x2)/(5 + 3/x + 5/x2)
Let x → ∞
L = −1/5 = −0.2
Answer
L = −0.2
For equal degrees the limit is the ratio of the leading coefficients.
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
(1.5·e1.5x − 1.5)/(2x) → 0/0
Second application
2.25·e1.5x/2
Substitute x = 0
L = 2.25·1/2 = 1.125
Answer
L = 1.125
3.Derivatives
Answer: f′(1) = 5488
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) = 4·(4x + 3)3·4
Substitute
f′(1) = 4·73·4 = 5488
Answer
f′(1) = 5488
4.Derivatives
Answer: b = −8
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) = −12
f′(x) = x + 4 → f′(−2) = 2
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −12 − 2·(−2) = −8
Answer
b = −8
5.Integrals
Answer: I = 26.8
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) = e2x/2
Evaluate at the limits
I = (e4 − 1)/2 = (54.6 − 1)/2 = 26.8
Answer
I = 26.8
6.Integrals
Answer: A = 0.3333
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
2x2 = 2x → x = 0, x = 1
Integrate upper minus lower
A = [x2 − 0.6667x3]01 = 0.3333
Answer
A = 0.3333
7.Complex numbers
Answer: Re(w) = 1.365
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 / 1.5 = 1.667
θ = 55° − 20° = 35°
Rectangular form
Re(w) = 1.667·cos(35°) = 1.365
Im(w) = 1.667·sin(35°) = 0.956
Re(w) = 1.365
Answer
Re(w) = 1.365
8.Complex numbers
Answer: b = 2.5
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.
Discriminant
p² − 4q = 9 − 34 = −25 < 0
Quadratic formula
z = (3 ± j√25)/2 = 1.5 + j2.5, 1.5 − j2.5
Answer
b = 2.5
Answer
b = 2.5
9.Differential equations
Answer: T = 24.62 °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) = 28 + (−25)·e−0.08t
Substitute t
T(25) = 28 + (−25)·0.1353 = 24.62 °C
Answer
T = 24.62 °C
10.Differential equations
Answer: t = 91.63 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.
Result
m(t) = 40 + (−40)·e−t/100
Solve for t
e−t/100 = (40 − 24)/(40 − 0) = 0.4
t = 100·ln(2.5) = 91.63 min
Answer
t = 91.63 min
11.Matrices & determinants
Answer: c11 = −14
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 = (−1)·(−6) + (−5)·4 = −14
Answer
c11 = −14
12.Matrices & determinants
Answer: λ = 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.
Characteristic equation
λ² − (4)λ + (−5) = 0
Solve
λ = (4 ± √(36))/2 → λ = −1, 5
λ = 5
Answer
λ = 5
13.Vectors: dot & cross product
Answer: k = −22
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 = (−5)·2 + 4·(−3) + (−1)·k = 0
−22 + (−1)·k = 0
Solve for k
k = −(−22)/(−1) = −22
Answer
k = −22
14.Vectors: dot & cross product
Answer: A = 3.742 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.
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
Complement
P(no six) = (5/6)2 = 0.6944
P = 1 − 0.6944 = 0.3056
Answer
P = 0.3056
16.Probability
Answer: E(G) = 6 €
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: 30 (p = 0.15), 190 (p = 0.05), −10 (p = 0.8)
Expected value
E(G) = 30·0.15 + 190·0.05 − 10·0.8 = 6 €
Answer
E(G) = 6 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 10.21
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 = −1.2
Standard normal table
Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849
Answer
P = 0.8849
19.Normal distribution
Answer: P = 0.3015
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 total
μT = 300 kg, σT = 0.5·√12 = 1.732 kg
Standardise
z = (300.9 − 300)/1.732 = 0.52
Probability
P = 1 − Φ(0.52) = 0.3015
Answer
P = 0.3015
20.Normal distribution
Answer: x = 248.8 g
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.