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)x. Give a decimal value.
f(x) = (1 − 2/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 3x)/(x²), with x in radians.
f(x) = (1 − cos 3x)/(x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·ex. Find f′(0.5).
f(x) = x·exx = 0.5
f′(0.5) =
4.DerivativesHard
Let f(x) = x3 + 4.5x2 − 5. Find the x-coordinate x of the local maximum.
f(x) = x3 + 4.5x2 − 5
x =
5.IntegralsMedium
Evaluate I = ∫₀1 x·sin x dx (x in radians).
∫₀1 x·sin 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
Write z = 2.5∠−20° in rectangular form a + jb and give the imaginary part b.
r = 2.5θ = −20°
b =
8.Complex numbersHard
Exactly one of the points A–D is a root of z³ = 27∠120°. Click it.
w = 27∠120°n = 3
1A (1.5∠40°)
2B (3∠40°)
3C (3∠−140°)
4D (3∠10°)
9.Differential equationsMedium
A forged steel part at 460 °C is left to cool in a workshop at 23 °C. After 18 min its temperature is 328 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 206 °C?
T0 = 460 °CTa = 23 °CT(18) = 328 °CT = 206 °C
t = min
10.Differential equationsHard
A system obeys y″ + 2y′ + 26y = 260 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1.5); angles in radians.
y″ + 2y′ + 26y = 260y(0) = 0y′(0) = 0
y(1.5) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [−5 4; 1 −1]B = [5 −1; −6 −1]
c11 =
12.Matrices & determinantsHard
Solve the system 4x + 6y = 4 and 6x − 2y = 8. Find x.
4x + 6y = 46x − 2y = 8
x =
13.Vectors: dot & cross productMedium
A constant force F = (30, 10, 25) N acts on an object while it moves in a straight line from point A(4, 1, 3) to point B(−1, −4, −4). Coordinates are in metres. Find the work W done by the force.
F = (30, 10, 25) NA(4, 1, 3) mB(−1, −4, −4) m
W = J
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−4, 4, −2) and b = (−2, −1, 1) and has a positive z-component. Find its x-component n̂x.
a = (−4, 4, −2)b = (−2, −1, 1)n̂z > 0
n̂x =
15.ProbabilityMedium
A fair die is rolled 4 times. What is the probability P of getting at least one six?
n = 4
P =
16.ProbabilityHard
11 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?
n = 11p = 0.7
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 37, 60, 10, 54, 24, 22, 37.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 12. What proportion P of values is below 84?
μ = 60σ = 12x = 84
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 8 of them exceeds 629.1 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 8> 629.1 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 bottom 90% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g90%
x = g
3Solutions
1.Limits
Answer: L = 0.1353
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→∞ x·ln(1 − 2/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·(−2) = −2
Undo the logarithm
L = e−2 = 0.1353
Answer
L = 0.1353
2.Limits
Answer: L = 4.5
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
3·sin 3x/(2x) → 0/0
Second application
9·cos 3x/2
Substitute x = 0
L = 9/2 = 4.5
Answer
L = 4.5
3.Derivatives
Answer: f′(0.5) = 2.473
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.
Product rule
f′(x) = u′v + uv′ = ex·(1 + 1·x)
Substitute
f′(0.5) = 1.649·(1 + 0.5) = 2.473
Answer
f′(0.5) = 2.473
4.Derivatives
Answer: x = −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.
Derivative
f′(x) = 3x2 + 9x = 3·(x2 + 3x)
Solve f′(x) = 0
x = −3 or x = 0
Which one is which
f″(x) = 6x + 9: negative at a maximum, positive at a minimum.
f″(−3) = −9, f″(0) = 9
x = −3
Answer
x = −3
5.Integrals
Answer: I = 0.3012
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 = x, du = dx; dv = sin x dx, v = −cos x
Antiderivative
∫x sin x dx = −x cos x + ∫cos x dx = −x cos x + sin x
Limits
(−1·cos 1 + sin 1) − (0 + 0) = 0.3012
Answer
I = 0.3012
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: b = −0.8551
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 = 2.5·cos(−20°) = 2.349
b = 2.5·sin(−20°) = −0.8551
Rectangular form
z = 2.349 − j0.8551
b = −0.8551
Answer
b = −0.8551
8.Complex numbers
Answer: B (3∠40°)
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 = 40°, 160°, −80°
The point
B = 3∠40°
Answer
B (3∠40°)
9.Differential equations
Answer: t = 43.57 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(437/305)/18 = 0.01998 min⁻¹
Time to reach the target
t = ln(437/183)/0.01998 = 43.57 min
Answer
t = 43.57 min
10.Differential equations
Answer: y(1.5) = 8.808
Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.
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 = (−5)·5 + 4·(−6) = −49
Answer
c11 = −49
12.Matrices & determinants
Answer: x = 1.273
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 = 4·(−2) − 6·6 = −44
Replace the x column
det Aₓ = 4·(−2) − 6·8 = −56
x = −56/(−44) = 1.273
Answer
x = 1.273
y = −0.1818
13.Vectors: dot & cross product
Answer: W = −375 J
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.
Displacement
d = B − A = (−1, −4, −4) − (4, 1, 3) = (−5, −5, −7) m
Work
W = F·d = 30·(−5) + 10·(−5) + 25·(−7) = −375 J
Answer
W = −375 J
Negative work: the force acts mostly against the motion.
14.Vectors: dot & cross product
Answer: n̂x = 0.1374
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 = 4·1 − (−2)·(−1) = 2
(a × b)y = azbx − axbz = (−2)·(−2) − (−4)·1 = 8
(a × b)z = axby − aybx = (−4)·(−1) − 4·(−2) = 12
Normalise
The z-component of a × b is already positive.
|a × b| = √(2² + 8² + 12²) = 14.56
n̂ = (2, 8, 12)/14.56 = (0.1374, 0.5494, 0.8242)
n̂x = 0.1374
Answer
n̂x = 0.1374
15.Probability
Answer: P = 0.5177
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)4 = 0.4823
P = 1 − 0.4823 = 0.5177
Answer
P = 0.5177
16.Probability
Answer: P = 4.724 × 10⁻⁵
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) = 0.311 = 1.771 × 10⁻⁶
P(X = 1) = 11·0.7·0.310 = 4.547 × 10⁻⁵
Add them
P = 1.771 × 10⁻⁶ + 4.547 × 10⁻⁵ = 4.724 × 10⁻⁵
Answer
P = 4.724 × 10⁻⁵
17.Descriptive statistics
Answer: s = 17.84
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 = 2
Standard normal table
Φ(2) = 0.9772
P = 0.9772
Answer
P = 0.9772
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 = 680 kg, σT = 12·√8 = 33.94 kg
Standardise
z = (629.1 − 680)/33.94 = −1.5
Probability
P = 1 − Φ(−1.5) = 0.9332
Answer
P = 0.9332
20.Normal distribution
Answer: x = 256.4 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.