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 − x + 8)/(3x3 + 4x + 2).
f(x) = (−x3 − x + 8)/(3x3 + 4x + 2)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e−x − 1 + x)/x².
f(x) = (e−x − 1 + x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x3·e−0.5x. Find f′(1).
f(x) = x3·e−0.5xx = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 − 9x2 + 24x − 3. Find the x-coordinate x of the local maximum.
f(x) = x3 − 9x2 + 24x − 3
x =
5.IntegralsMedium
Evaluate I = ∫03 e−0.5x dx.
f(x) = e−0.5x[0, 3]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = 4x.
y = x2y = 4x
A =
7.Complex numbersMedium
Write z = 4 + j2 in exponential form r·ejθ. Give θ in radians, between −π and π.
z = 4 + j2
θ = rad
8.Complex numbersHard
Find the real part of (2 + j2)⁴.
z = 2 + j2n = 4
Re(z⁴) =
9.Differential equationsMedium
A forged steel part at 480 °C is left to cool in a workshop at 22 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.04 min⁻¹, find its temperature T after 20 min.
T0 = 480 °CTa = 22 °Ck = 0.04 min⁻¹t = 20 min
T = °C
10.Differential equationsHard
A system obeys y″ + y′ + 25.25y = 25.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1.5); angles in radians.
y″ + y′ + 25.25y = 25.25y(0) = 0y′(0) = 0
y(1.5) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−1 5 −5; −2 −2 4; 3 3 −4]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−1 2; 9 2]
λ =
13.Vectors: dot & cross productMedium
A constant force F = (20, 25, −30) N acts on an object while it moves in a straight line from point A(0, −5, −3) to point B(4, 1, −2). Coordinates are in metres. Find the work W done by the force.
F = (20, 25, −30) NA(0, −5, −3) mB(4, 1, −2) m
W = J
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−1, −2, 0) and b = (−2, 3, −4) and has a positive z-component. Find its x-component n̂x.
a = (−1, −2, 0)b = (−2, 3, −4)n̂z > 0
n̂x =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2, 3, 4p(x): 0.25, 0.2, 0.05, 0.15, 0.35
σ =
16.ProbabilityHard
A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 5 correct answers?
n = 7p = 0.4k = 5
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 48, 60, 52, 12, 34.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 12. Find the z-score z of a score of 40.4.
μ = 50σ = 12x = 40.4
z =
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 6 of them exceeds 495.3 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 6> 495.3 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 85% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g85%
x = g
3Solutions
1.Limits
Answer: L = −0.3333
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 + (−1)/x2 + 8/x3)/(3 + 4/x2 + 2/x3)
Let x → ∞
L = −1/3 = −0.3333
Answer
L = −0.3333
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 0.5
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
(−1·e−x + 1)/(2x) → 0/0
Second application
1·e−x/2
Substitute x = 0
L = 1·1/2 = 0.5
Answer
L = 0.5
3.Derivatives
Answer: f′(1) = 1.516
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′ = e−0.5x·(3x2 + (−0.5)·x3)
Substitute
f′(1) = 0.6065·(3 + (−0.5)) = 1.516
Answer
f′(1) = 1.516
4.Derivatives
Answer: x = 2
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 − 18x + 24 = 3·(x2 − 6x + 8)
Solve f′(x) = 0
x = 2 or x = 4
Which one is which
f″(x) = 6x − 18: negative at a maximum, positive at a minimum.
f″(2) = −6, f″(4) = 6
x = 2
Answer
x = 2
5.Integrals
Answer: I = 1.554
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.
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
x2 = 4x → x = 0, x = 4
Integrate upper minus lower
A = [2x2 − 0.3333x3]04 = 10.67
Answer
A = 10.67
7.Complex numbers
Answer: θ = 0.4636 rad
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
r = √(16 + 4) = 4.472
Argument in radians
θ = 26.57° = 26.57·π/180 = 0.4636 rad
Exponential form
z = 4.472·ej0.4636
Answer
θ = 0.4636 rad
8.Complex numbers
Answer: Re(z⁴) = −64
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 = (2 + j2)(2 + j2) = j8
z³ = (j8)(2 + j2) = −16 + j16
z⁴ = (−16 + j16)(2 + j2) = −64
Check with De Moivre
z = 2.828∠45° → z⁴ = 64∠180°
Re(z⁴) = −64
Answer
Re(z⁴) = −64
9.Differential equations
Answer: T = 227.8 °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) = 22 + 458·e−0.04t
Substitute t
T(20) = 22 + 458·0.4493 = 227.8 °C
Answer
T = 227.8 °C
10.Differential equations
Answer: y(1.5) = 0.792
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.
2×2 minors of the first row
M₁₁ = −4, M₁₂ = −4, M₁₃ = 0
Cofactor expansion
det A = (−1)·(−4) − 5·(−4) + (−5)·0 = 24
Answer
det A = 24
12.Matrices & determinants
Answer: λ = 5
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 − (1)λ + (−20) = 0
Solve
λ = (1 ± √(81))/2 → λ = −4, 5
λ = 5
Answer
λ = 5
13.Vectors: dot & cross product
Answer: W = 200 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 = (4, 1, −2) − (0, −5, −3) = (4, 6, 1) m
Work
W = F·d = 20·4 + 25·6 + (−30)·1 = 200 J
Answer
W = 200 J
14.Vectors: dot & cross product
Answer: n̂x = −0.7044
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 = (−2)·(−4) − 0·3 = 8
(a × b)y = azbx − axbz = 0·(−2) − (−1)·(−4) = −4
(a × b)z = axby − aybx = (−1)·3 − (−2)·(−2) = −7
Normalise
The z-component of a × b is negative, so use −(a × b).
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.
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.
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 = (40.4 − 50)/12 = −0.8
Answer
z = −0.8
19.Normal distribution
Answer: P = 0.6915
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 = 510 kg, σT = 12·√6 = 29.39 kg
Standardise
z = (495.3 − 510)/29.39 = −0.5
Probability
P = 1 − Φ(−0.5) = 0.6915
Answer
P = 0.6915
20.Normal distribution
Answer: x = 244.8 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.