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→∞ (−2x2 + 6x − 3)/(3x2 + 4x + 8).
f(x) = (−2x2 + 6x − 3)/(3x2 + 4x + 8)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 5x)/(3x²), with x in radians.
f(x) = (1 − cos 5x)/(3x²)x → 0
L =
3.DerivativesMedium
Let f(x) = (2x − 3)4. Find f′(1).
f(x) = (2x − 3)4x = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 − 12x2 + 36x − 6. Find the x-coordinate x of the local minimum.
f(x) = x3 − 12x2 + 36x − 6
x =
5.IntegralsMedium
Evaluate I = ∫₁5 x·ln x dx.
∫₁5 x ln x dx
I =
6.IntegralsHard
The region between the line y = 2x and the parabola y = 0.5x² is rotated about the x-axis. Find the volume V of the solid.
y = 2xy = 0.5x²
V =
7.Complex numbersMedium
Given z₁ = 3∠−60° and z₂ = 2∠45°, find the real part of w = z₁·z₂.
z₁ = 3∠−60°z₂ = 2∠45°
Re(w) =
8.Complex numbersHard
Use De Moivre's theorem to find the imaginary part of z⁶ for z = 1 − j.
z = 1 − jn = 6
Im(z⁶) =
9.Differential equationsMedium
A cup of coffee at 84 °C is left in a room at 24 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.12 min⁻¹, find its temperature T after 8 min.
T0 = 84 °CTa = 24 °Ck = 0.12 min⁻¹t = 8 min
T = °C
10.Differential equationsHard
A tank holds 1000 L of water in which 10 kg of salt is dissolved. 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. After how many minutes t does the tank contain 26 kg of salt?
V = 1000 Lm0 = 10 kgc = 0.05 kg/Lr = 5 L/min
t = min
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [0 −5 0; 4 0 3; −4 4 −4]
det A =
12.Matrices & determinantsHard
Solve the system −6x − 4y = −5 and x − 2y = 9. Find x.
−6x − 4y = −5x − 2y = 9
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (−4, 2, 0) onto b = (0, −2, −5), that is, the component of a in the direction of b.
a = (−4, 2, 0)b = (0, −2, −5)
compb a =
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(1, 1, −4), B(−4, 4, 2) and C(−1, 0, −4). Coordinates are in metres.
A(1, 1, −4)B(−4, 4, 2)C(−1, 0, −4)
A = m²
15.ProbabilityMedium
A bag holds 3 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
3 / 5
P =
16.ProbabilityHard
X has the distribution shown, and Y = 3X + 4. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 17, 29, 43, 18, 45.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is below 43.6?
μ = 50σ = 8x = 43.6
P =
19.Normal distributionMedium
X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 34.5 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 34.5 min
P =
20.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 97.5% of them lie (the 97.5th percentile).
μ = 250σ = 5P = 97.5%
x = g
3Solutions
1.Limits
Answer: L = −0.6667
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
(−2 + 6/x + (−3)/x2)/(3 + 4/x + 8/x2)
Let x → ∞
L = −2/3 = −0.6667
Answer
L = −0.6667
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 4.167
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/(6x) → 0/0
Second application
25·cos 5x/6
Substitute x = 0
L = 25/6 = 4.167
Answer
L = 4.167
3.Derivatives
Answer: f′(1) = −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.
Chain rule
f′(x) = 4·(2x − 3)3·2
Substitute
f′(1) = 4·(−1)3·2 = −8
Answer
f′(1) = −8
4.Derivatives
Answer: x = 6
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.
Derivative
f′(x) = 3x2 − 24x + 36 = 3·(x2 − 8x + 12)
Solve f′(x) = 0
x = 2 or x = 6
Which one is which
f″(x) = 6x − 24: negative at a maximum, positive at a minimum.
f″(2) = −12, f″(6) = 12
x = 6
Answer
x = 6
5.Integrals
Answer: I = 14.12
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
(12.5·ln 5 − 6.25) − (0 − 0.25) = 14.12
Answer
I = 14.12
6.Integrals
Answer: V = 107.2
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
2x = 0.5x² → x = 0, x = 4
Washer method
V = π∫₀4 ((2x)² − (0.5x²)²) dx
Integrate
V = π(4·4³/3 − 0.25·4⁵/5) = π·34.13 = 107.2
Answer
V = 107.2
7.Complex numbers
Answer: Re(w) = 5.796
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| = 3 · 2 = 6
θ = −60° + 45° = −15°
Rectangular form
Re(w) = 6·cos(−15°) = 5.796
Im(w) = 6·sin(−15°) = −1.553
Re(w) = 5.796
Answer
Re(w) = 5.796
8.Complex numbers
Answer: Im(z⁶) = 8
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
z⁶ = (√2)⁶∠−270° = 8∠90°
Rectangular form
z⁶ = 8·(cos 90° + j sin 90°) = j8
Im(z⁶) = 8
Answer
Im(z⁶) = 8
9.Differential equations
Answer: T = 46.97 °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) = 24 + 60·e−0.12t
Substitute t
T(8) = 24 + 60·0.3829 = 46.97 °C
Answer
T = 46.97 °C
10.Differential equations
Answer: t = 102.2 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) = 50 + (−40)·e−t/200
Solve for t
e−t/200 = (50 − 26)/(50 − 10) = 0.6
t = 200·ln(1.667) = 102.2 min
Answer
t = 102.2 min
11.Matrices & determinants
Answer: det A = −20
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 = −4, M₁₃ = 16
Cofactor expansion
det A = 0·(−12) − (−5)·(−4) + 0·16 = −20
Answer
det A = −20
12.Matrices & determinants
Answer: x = 2.875
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 = (−6)·(−2) − (−4)·1 = 16
Replace the x column
det Aₓ = (−5)·(−2) − (−4)·9 = 46
x = 46/16 = 2.875
Answer
x = 2.875
y = −3.063
13.Vectors: dot & cross product
Answer: compb a = −0.7428
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 = (−4)·0 + 2·(−2) + 0·(−5) = −4
Length of b
|b| = √(0² + (−2)² + (−5)²) = 5.385
Scalar projection
compb a = a·b/|b| = −4/5.385 = −0.7428
Answer
compb a = −0.7428
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: A = 8.675 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.
First and second draw
P = 3/8 · 2/7 = 6/56 = 0.1071
Answer
P = 0.1071
16.Probability
Answer: σY = 4.056
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.
Spread of X
μ = 1.65, Var(X) = 1.828, σX = 1.352
Linear change
σY = |3|·1.352 = 4.056
Answer
σY = 4.056
The mean does move: E(Y) = 3·1.65 + 4 = 8.95.
17.Descriptive statistics
Answer: s = 13.3
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 = −0.8
Standard normal table
Φ(−0.8) = 0.2119
P = 0.2119
Answer
P = 0.2119
19.Normal distribution
Answer: P = 0.1587
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 = 32 min, σ = √(1.5² + 2²) = 2.5 min
Standardise
z = (34.5 − 32)/2.5 = 1
Upper tail
P = 1 − Φ(1) = 1 − 0.8413 = 0.1587
Answer
P = 0.1587
20.Normal distribution
Answer: x = 259.8 g
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.