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→∞ (−x3 + 6)/(5x3 + 5x + 7).
f(x) = (−x3 + 6)/(5x3 + 5x + 7)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(2x2 + 1). Find f′(1).
f(x) = ln(2x2 + 1)x = 1
f′(1) =
4.DerivativesHard
The tangent line to f(x) = 2x2 + 2x − 5 at x = −2 crosses the y-axis at y = b. Find b.
f(x) = 2x2 + 2x − 5x = −2
b =
5.IntegralsMedium
The region under y = 3x from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a cone).
y = 3x0 ≤ x ≤ 1
V =
6.IntegralsHard
Find the area A of the region between the parabola y = 1 − x² and the x-axis.
y = 1 − x²
A =
7.Complex numbersMedium
The number z = 3·e−j1 has its angle in radians. Find its real part a.
r = 3θ = −1 rad
a =
8.Complex numbersHard
Exactly one of the points A–D is a root of z³ = 27∠90°. Click it.
w = 27∠90°n = 3
1A (5.4∠150°)
2B (3∠−150°)
3C (3∠−30°)
4D (3∠150°)
9.Differential equationsMedium
A forged steel part at 500 °C is left to cool in a workshop at 28 °C. After 58 min its temperature is 54 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 37 °C?
T0 = 500 °CTa = 28 °CT(58) = 54 °CT = 37 °C
t = min
10.Differential equationsHard
Solve y′ + y = 9e−2t with y(0) = 5. Find y(0.5).
y′ + y = 9e−2ty(0) = 5
y(0.5) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [0 −1 −2; −3 −3 −3; −1 −1 3]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [0 6; −3 9]
λ =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (6, −2, −6) and b = (2, 2, 5), in degrees.
a = (6, −2, −6)b = (2, 2, 5)
θ = °
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(1, 1, −1), B(2, 4, −3) and C(1, −4, 1). Coordinates are in metres.
A(1, 1, −1)B(2, 4, −3)C(1, −4, 1)
A = m²
15.ProbabilityMedium
In how many ways N can a committee of 5 people be chosen from a group of 14? The order does not matter.
n = 14k = 5
N =
16.ProbabilityHard
A game costs €5 to play. You win €25 with probability 0.2 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 = €5€25: p = 0.2€100: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 11, 17, 48, 23, 42, 36, 50.
n = 7
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 10. Find the z-score z of a score of 85.
μ = 100σ = 10x = 85
z =
19.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9545. Find z to two decimals.
A = 0.9545
z =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 90% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points90%
x = points
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 + 6/x3)/(5 + 5/x2 + 7/x3)
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 = 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) = 1.333
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) = 4x/(2x2 + 1)
Substitute
f′(1) = 4/3 = 1.333
Answer
f′(1) = 1.333
4.Derivatives
Answer: b = −13
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) = −1
f′(x) = 4x + 2 → f′(−2) = −6
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −1 − (−6)·(−2) = −13
Answer
b = −13
5.Integrals
Answer: V = 9.425
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∫₀1 9x² dx
Integrate
V = π·9·1³/3 = 9.425
Answer
V = 9.425
Check: a cone of radius 3 and height 1 has (1/3)πr²h = 9.425.
6.Integrals
Answer: A = 1.333
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.
Zeros
1 − x² = 0 → x = ±1
Integrate
A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333
Answer
A = 1.333
7.Complex numbers
Answer: a = 1.621
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 = 3·(cos −1 + j sin −1)
Components (radians)
a = 3·cos −1 = 1.621
b = 3·sin −1 = −2.524
a = 1.621
Answer
a = 1.621
8.Complex numbers
Answer: D (3∠150°)
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.
Modulus and arguments of the roots
|z| = 3
θ = 30°, 150°, −90°
The point
D = 3∠150°
Answer
D (3∠150°)
9.Differential equations
Answer: t = 79.23 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(472/26)/58 = 0.04998 min⁻¹
Time to reach the target
t = ln(472/9)/0.04998 = 79.23 min
Answer
t = 79.23 min
10.Differential equations
Answer: y(0.5) = 5.181
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 = et
(ety)′ = 9e−t
General solution
y = 9/(−1)·e−2t + C·e−t = −9·e−2t + C·e−t
Initial condition
5 = −9 + C → C = 14
Evaluate
y(0.5) = −9·0.3679 + 14·0.6065 = 5.181
Answer
y(0.5) = 5.181
11.Matrices & determinants
Answer: det A = −12
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 = −12, M₁₂ = −12, M₁₃ = 0
Cofactor expansion
det A = 0·(−12) − (−1)·(−12) + (−2)·0 = −12
Answer
det A = −12
12.Matrices & determinants
Answer: λ = 6
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 − (9)λ + (18) = 0
Solve
λ = (9 ± √(9))/2 → λ = 3, 6
λ = 6
Answer
λ = 6
13.Vectors: dot & cross product
Answer: θ = 116.1 °
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 = 6·2 + (−2)·2 + (−6)·5 = −22
Magnitudes
|a| = √(6² + (−2)² + (−6)²) = 8.718
|b| = √(2² + 2² + 5²) = 5.745
Angle
cos θ = −22/(8.718·5.745) = −0.4393
θ = arccos(−0.4393) = 116.1°
Answer
θ = 116.1°
14.Vectors: dot & cross product
Answer: A = 3.354 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.
Combinations
C(14, 5) = (14·13·12·11·10)/(5·4·3·2·1) = 2002
Answer
N = 2002
16.Probability
Answer: E(G) = 5 €
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: 20 (p = 0.2), 95 (p = 0.05), −5 (p = 0.75)
Expected value
E(G) = 20·0.2 + 95·0.05 − 5·0.75 = 5 €
Answer
E(G) = 5 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 15.5
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 = (85 − 100)/10 = −1.5
Answer
z = −1.5
19.Normal distribution
Answer: z = −1.69
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.
Area to the left of z
Φ(z) = 1 − 0.9545 = 0.0455
Read the z-table backwards
Find 0.0455 inside the table and read off z.
z = −1.69
Answer
z = −1.69
20.Normal distribution
Answer: x = 52.2 points
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.