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→∞ (3x3 + 4x − 5)/(4x3 + x + 9).
f(x) = (3x3 + 4x − 5)/(4x3 + x + 9)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 4x)/sin(3x), with x in radians.
f(x) = ln(1 + 4x)/sin(3x)x → 0
L =
3.DerivativesMedium
Let f(x) = (3x − 1)4. Find f′(0).
f(x) = (3x − 1)4x = 0
f′(0) =
4.DerivativesHard
The tangent line to f(x) = x2 − x − 2 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = x2 − x − 2x = 1
b =
5.IntegralsMedium
Evaluate I = ∫₁6 x·ln x dx.
∫₁6 x ln x dx
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 9 − x² and the x-axis.
y = 9 − x²
A =
7.Complex numbersMedium
Find the argument θ of z = −9 + j5 in degrees, as the principal value between −180° and 180°.
z = −9 + j5
θ = °
8.Complex numbersHard
Find the imaginary part of (−1 + j2)³.
z = −1 + j2n = 3
Im(z³) =
9.Differential equationsMedium
A forged steel part at 460 °C is left to cool in a workshop at 29 °C. After 29 min its temperature is 164 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 460 °CTa = 29 °CT(29) = 164 °C
k = min⁻¹
10.Differential equationsHard
A tank holds 100 L of water in which 5 kg of salt is dissolved. Brine with 0.25 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 19 kg of salt?
V = 100 Lm0 = 5 kgc = 0.25 kg/Lr = 5 L/min
t = min
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [5 −1 −1; 1 −2 5; −4 5 −4]
det A =
12.Matrices & determinantsHard
Solve the system 6x − 3y = −10 and −x − 2y = −3. Find x.
6x − 3y = −10−x − 2y = −3
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (6, −3, −1) onto b = (−2, −5, 1), that is, the component of a in the direction of b.
a = (6, −3, −1)b = (−2, −5, 1)
compb a =
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−4, 0, 3) and b = (3, 4, 3) and has a positive z-component. Find its y-component n̂y.
a = (−4, 0, 3)b = (3, 4, 3)n̂z > 0
n̂y =
15.ProbabilityMedium
In how many ways N can a committee of 4 people be chosen from a group of 13? The order does not matter.
n = 13k = 4
N =
16.ProbabilityHard
8 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?
n = 8p = 0.25
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 20, 13, 47, 56, 13.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is below 90?
μ = 100σ = 20x = 90
P =
19.Normal distributionMedium
The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 16 of them exceeds 1288 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 16> 1288 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 90% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g90%
x = g
3Solutions
1.Limits
Answer: L = 0.75
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
(3 + 4/x2 + (−5)/x3)/(4 + 1/x2 + 9/x3)
Let x → ∞
L = 3/4 = 0.75
Answer
L = 0.75
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 1.333
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).
L'Hôpital's rule
L = limx→0 [4/(1 + 4x)]/[3·cos 3x]
Substitute x = 0
L = 4/3 = 1.333
Answer
L = 1.333
3.Derivatives
Answer: f′(0) = −12
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·(3x − 1)3·3
Substitute
f′(0) = 4·(−1)3·3 = −12
Answer
f′(0) = −12
4.Derivatives
Answer: b = −3
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) = −2
f′(x) = 2x − 1 → f′(1) = 1
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −2 − 1·1 = −3
Answer
b = −3
5.Integrals
Answer: I = 23.5
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
(18·ln 6 − 9) − (0 − 0.25) = 23.5
Answer
I = 23.5
6.Integrals
Answer: A = 36
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
9 − x² = 0 → x = ±3
Integrate
A = [9x − x³/3]−33 = 18 − (−18) = 36
Answer
A = 36
7.Complex numbers
Answer: θ = 150.9 °
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.
Quadrant
a = −9, b = 5 → II
Reference angle
α = atan(5/9) = 29.05°
Argument
θ = 180° − α = 150.9°
Answer
θ = 150.9°
In polar form z = 10.3∠150.9°.
8.Complex numbers
Answer: Im(z³) = −2
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.
Powers step by step
z² = (−1 + j2)(−1 + j2) = −3 − j4
z³ = (−3 − j4)(−1 + j2) = 11 − j2
Check with De Moivre
z = 2.236∠116.6° → z³ = 11.18∠−10.3°
Im(z³) = −2
Answer
Im(z³) = −2
9.Differential equations
Answer: k = 0.04003 min⁻¹
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.
Ratio of temperature differences
e−29k = 135/431
Solve for k
k = ln(431/135)/29 = 0.04003 min⁻¹
Answer
k = 0.04003 min⁻¹
10.Differential equations
Answer: t = 24.08 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) = 25 + (−20)·e−t/20
Solve for t
e−t/20 = (25 − 19)/(25 − 5) = 0.3
t = 20·ln(3.333) = 24.08 min
Answer
t = 24.08 min
11.Matrices & determinants
Answer: det A = −66
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 = −17, M₁₂ = 16, M₁₃ = −3
Cofactor expansion
det A = 5·(−17) − (−1)·16 + (−1)·(−3) = −66
Answer
det A = −66
12.Matrices & determinants
Answer: x = −0.7333
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) − (−3)·(−1) = −15
Replace the x column
det Aₓ = (−10)·(−2) − (−3)·(−3) = 11
x = 11/(−15) = −0.7333
Answer
x = −0.7333
y = 1.867
13.Vectors: dot & cross product
Answer: compb a = 0.3651
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) + (−3)·(−5) + (−1)·1 = 2
Length of b
|b| = √((−2)² + (−5)² + 1²) = 5.477
Scalar projection
compb a = a·b/|b| = 2/5.477 = 0.3651
Answer
compb a = 0.3651
14.Vectors: dot & cross product
Answer: n̂y = −0.7241
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 = 0·3 − 3·4 = −12
(a × b)y = azbx − axbz = 3·3 − (−4)·3 = 21
(a × b)z = axby − aybx = (−4)·4 − 0·3 = −16
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.
Combinations
C(13, 4) = (13·12·11·10)/(4·3·2·1) = 715
Answer
N = 715
16.Probability
Answer: P = 0.3671
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.
Zero and one failure
P(X = 0) = 0.758 = 0.1001
P(X = 1) = 8·0.25·0.757 = 0.267
Add them
P = 0.1001 + 0.267 = 0.3671
Answer
P = 0.3671
17.Descriptive statistics
Answer: s = 20.27
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 = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 0.3085
Answer
P = 0.3085
19.Normal distribution
Answer: P = 0.9332
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 = 1360 kg, σT = 12·√16 = 48 kg
Standardise
z = (1288 − 1360)/48 = −1.5
Probability
P = 1 − Φ(−1.5) = 0.9332
Answer
P = 0.9332
20.Normal distribution
Answer: x = 243.6 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.