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)2x. Give a decimal value.
f(x) = (1 + 2/x)2xx → ∞
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) = (−2x − 1)3. Find f′(1).
f(x) = (−2x − 1)3x = 1
f′(1) =
4.DerivativesHard
The tangent line to f(x) = 0.5x2 + 4x + 2 at x = −1 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 + 4x + 2x = −1
b =
5.IntegralsMedium
Evaluate I = ∫₁3 x·ln x dx.
∫₁3 x ln x dx
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 0.5x2 and y = 3x.
y = 0.5x2y = 3x
A =
7.Complex numbersMedium
Write z = 12∠−75° in rectangular form a + jb and give the imaginary part b.
r = 12θ = −75°
b =
8.Complex numbersHard
The equation z² − 3z + 11.25 = 0 has two complex roots. Find their modulus |z|.
p = −3q = 11.25
|z| =
9.Differential equationsMedium
A cup of coffee at 82 °C is left in a room at 19 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.08 min⁻¹, find its temperature T after 22 min.
T0 = 82 °CTa = 19 °Ck = 0.08 min⁻¹t = 22 min
T = °C
10.Differential equationsHard
Solve y′ + 0.5y = 9e−2t with y(0) = 0. Find y(1).
y′ + 0.5y = 9e−2ty(0) = 0
y(1) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [3 −4; 4 −5]B = [−6 −1; 3 6]
c21 =
12.Matrices & determinantsHard
Solve the system −x − 4y = −10 and −4x + 2y = −5. Find x.
−x − 4y = −10−4x + 2y = −5
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−1, −2, −6) and b = (6, 6, 0), in degrees.
a = (−1, −2, −6)b = (6, 6, 0)
θ = °
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−2, 0, −5) and b = (4, 2, 4) and has a positive z-component. Find its x-component n̂x.
a = (−2, 0, −5)b = (4, 2, 4)n̂z > 0
n̂x =
15.ProbabilityMedium
A fair die is rolled 7 times. What is the probability P of getting at least one six?
n = 7
P =
16.ProbabilityHard
A game costs €5 to play. You win €10 with probability 0.1 and €100 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€10: p = 0.1€100: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 13, 24, 56, 13, 33.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 60. Find the z-score z of a score of 452.
μ = 500σ = 60x = 452
z =
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 117.8 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 117.8 s
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 = 54.6
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).
Take the logarithm
ln L = limx→∞ 2·x·ln(1 + 2/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 2·2 = 4
Undo the logarithm
L = e4 = 54.6
Answer
L = 54.6
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′(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 = 1.5
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) = −1.5
f′(x) = x + 4 → f′(−1) = 3
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −1.5 − 3·(−1) = 1.5
Answer
b = 1.5
5.Integrals
Answer: I = 2.944
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.
Choose u and dv
u = ln x, du = dx/x; dv = x dx, v = x²/2
Antiderivative
∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4
Limits
(4.5·ln 3 − 2.25) − (0 − 0.25) = 2.944
Answer
I = 2.944
6.Integrals
Answer: A = 18
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
0.5x2 = 3x → x = 0, x = 6
Integrate upper minus lower
A = [1.5x2 − 0.1667x3]06 = 18
Answer
A = 18
7.Complex numbers
Answer: b = −11.59
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.
Components
a = 12·cos(−75°) = 3.106
b = 12·sin(−75°) = −11.59
Rectangular form
z = 3.106 − j11.59
b = −11.59
Answer
b = −11.59
8.Complex numbers
Answer: |z| = 3.354
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 − 45 = −36 < 0
Quadratic formula
z = (3 ± j√36)/2 = 1.5 + j3, 1.5 − j3
Answer
|z| = √(1.5² + 3²) = √11.25 = 3.354
Answer
|z| = 3.354
9.Differential equations
Answer: T = 29.84 °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) = 19 + 63·e−0.08t
Substitute t
T(22) = 19 + 63·0.172 = 29.84 °C
Answer
T = 29.84 °C
10.Differential equations
Answer: y(1) = 2.827
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.
Integrating factor
μ = e0.5t
(e0.5ty)′ = 9e−1.5t
General solution
y = 9/(−1.5)·e−2t + C·e−0.5t = −6·e−2t + C·e−0.5t
Initial condition
0 = −6 + C → C = 6
Evaluate
y(1) = −6·0.1353 + 6·0.6065 = 2.827
Answer
y(1) = 2.827
11.Matrices & determinants
Answer: c21 = −39
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
c21 = 4·(−6) + (−5)·3 = −39
Answer
c21 = −39
12.Matrices & determinants
Answer: x = 2.222
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 − (−4)·(−4) = −18
Replace the x column
det Aₓ = (−10)·2 − (−4)·(−5) = −40
x = −40/(−18) = 2.222
Answer
x = 2.222
y = 1.944
13.Vectors: dot & cross product
Answer: θ = 109.3 °
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
a·b = (−1)·6 + (−2)·6 + (−6)·0 = −18
Magnitudes
|a| = √((−1)² + (−2)² + (−6)²) = 6.403
|b| = √(6² + 6² + 0²) = 8.485
Angle
cos θ = −18/(6.403·8.485) = −0.3313
θ = arccos(−0.3313) = 109.3°
Answer
θ = 109.3°
14.Vectors: dot & cross product
Answer: n̂x = −0.6202
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 = 0·4 − (−5)·2 = 10
(a × b)y = azbx − axbz = (−5)·4 − (−2)·4 = −12
(a × b)z = axby − aybx = (−2)·2 − 0·4 = −4
Normalise
The z-component of a × b is negative, so use −(a × b).
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)7 = 0.2791
P = 1 − 0.2791 = 0.7209
Answer
P = 0.7209
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: 5 (p = 0.1), 95 (p = 0.02), −5 (p = 0.88)
Expected value
E(G) = 5·0.1 + 95·0.02 − 5·0.88 = −2 €
Answer
E(G) = −2 €
A negative value means you lose money on average.
17.Descriptive statistics
Answer: s = 17.85
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 = (452 − 500)/60 = −0.8
Answer
z = −0.8
19.Normal distribution
Answer: P = 0.02275
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 = (117.8 − 105)/6.403 = 2
Upper tail
P = 1 − Φ(2) = 1 − 0.9772 = 0.02275
Answer
P = 0.02275
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).