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 − 1/x)3x. Give a decimal value.
f(x) = (1 − 1/x)3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 5x)/sin(2x), with x in radians.
f(x) = ln(1 + 5x)/sin(2x)x → 0
L =
3.DerivativesMedium
Let f(x) = x3·e−0.5x. Find f′(2).
f(x) = x3·e−0.5xx = 2
f′(2) =
4.DerivativesHard
Let f(x) = x3 + 10.5x2 + 36x. Find the x-coordinate x of the local minimum.
f(x) = x3 + 10.5x2 + 36x
x =
5.IntegralsMedium
Evaluate I = ∫₁2 x·ln x dx.
∫₁2 x ln x dx
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 0.5x2 and y = 4x.
y = 0.5x2y = 4x
A =
7.Complex numbersMedium
Given z₁ = 2.5∠30° and z₂ = 4∠−25°, find the real part of w = z₁/z₂.
z₁ = 2.5∠30°z₂ = 4∠−25°
Re(w) =
8.Complex numbersHard
One root of z³ + z² − 12z − 90 = 0 is z₁ = −3 + j3. The coefficients are real. Find the real root r.
z₁ = −3 + j3
r =
9.Differential equationsMedium
Solve y″ + 8y′ + 16y = 0 with y(0) = 2 and y′(0) = -1. Find y(1.5).
y″ + 8y′ + 16y = 0y(0) = 2y′(0) = -1
y(1.5) =
10.Differential equationsHard
A system obeys y″ + 3y′ + 11.25y = 22.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 3y′ + 11.25y = 22.5y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [4 −3 3; 1 −3 −4; −1 1 −3]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [6 3; −5 −2]
λ =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (2, 6, −3) and b = (4, 4, −6), in degrees.
a = (2, 6, −3)b = (4, 4, −6)
θ = °
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(−2, 2, −1), B(−2, 4, −4) and C(3, −1, 2). Coordinates are in metres.
A(−2, 2, −1)B(−2, 4, −4)C(3, −1, 2)
A = m²
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 10, 20, 30, 40p(x): 0.25, 0.2, 0.25, 0.3
σ =
16.ProbabilityHard
11 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?
n = 11p = 0.2
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 29, 27, 45, 18, 19, 25.
n = 6
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 15. Find the z-score z of a score of 35.
μ = 50σ = 15x = 35
z =
19.Normal distributionMedium
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 166.8 and 183.6 cm? Give P as a decimal.
μ = 178σ = 7x1 = 166.8x2 = 183.6
P =
20.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
3Solutions
1.Limits
Answer: L = 0.04979
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→∞ 3·x·ln(1 − 1/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 3·(−1) = −3
Undo the logarithm
L = e−3 = 0.04979
Answer
L = 0.04979
2.Limits
Answer: L = 2.5
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).
L'Hôpital's rule
L = limx→0 [5/(1 + 5x)]/[2·cos 2x]
Substitute x = 0
L = 5/2 = 2.5
Answer
L = 2.5
3.Derivatives
Answer: f′(2) = 2.943
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′(2) = 0.3679·(12 + (−4)) = 2.943
Answer
f′(2) = 2.943
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 + 21x + 36 = 3·(x2 + 7x + 12)
Solve f′(x) = 0
x = −4 or x = −3
Which one is which
f″(x) = 6x + 21: negative at a maximum, positive at a minimum.
f″(−4) = −3, f″(−3) = 3
x = −3
Answer
x = −3
5.Integrals
Answer: I = 0.6363
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
(2·ln 2 − 1) − (0 − 0.25) = 0.6363
Answer
I = 0.6363
6.Integrals
Answer: A = 42.67
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
0.5x2 = 4x → x = 0, x = 8
Integrate upper minus lower
A = [2x2 − 0.1667x3]08 = 42.67
Answer
A = 42.67
7.Complex numbers
Answer: Re(w) = 0.3585
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| = 2.5 / 4 = 0.625
θ = 30° − (−25)° = 55°
Rectangular form
Re(w) = 0.625·cos(55°) = 0.3585
Im(w) = 0.625·sin(55°) = 0.512
Re(w) = 0.3585
Answer
Re(w) = 0.3585
8.Complex numbers
Answer: r = 5
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.
Check with the sum of the roots: z₁ + z₂ + r = −6 + 5 = −1 = −(1).
9.Differential equations
Answer: y(1.5) = 0.03098
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.
Double root
(r + 4)² = 0 → r = −4
Constants
A = y(0) = 2
y′(0) = B − 4A → B = -1 + 8 = 7
Evaluate
y(1.5) = (2 + 7·1.5)·e−6 = 0.03098
Answer
y(1.5) = 0.03098
10.Differential equations
Answer: ymax = 2.416
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.
Steady value and roots
yss = 2, r = −1.5 ± 3j
Constants from the initial values
y(0) = 2 + A = 0 → A = −2
y′(0) = −1.5A + 3B = 0 → B = −1
Peak
t = π/3 = 1.047
ymax = 2·(1 + e−1.5·π/3) = 2.416
Answer
ymax = 2.416
11.Matrices & determinants
Answer: det A = 25
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 = 13, M₁₂ = −7, M₁₃ = −2
Cofactor expansion
det A = 4·13 − (−3)·(−7) + 3·(−2) = 25
Answer
det A = 25
12.Matrices & determinants
Answer: λ = 3
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 − (4)λ + (3) = 0
Solve
λ = (4 ± √(4))/2 → λ = 1, 3
λ = 3
Answer
λ = 3
13.Vectors: dot & cross product
Answer: θ = 29.98 °
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 = 2·4 + 6·4 + (−3)·(−6) = 50
Magnitudes
|a| = √(2² + 6² + (−3)²) = 7
|b| = √(4² + 4² + (−6)²) = 8.246
Angle
cos θ = 50/(7·8.246) = 0.8662
θ = arccos(0.8662) = 29.98°
Answer
θ = 29.98°
14.Vectors: dot & cross product
Answer: A = 9.138 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.
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.
Zero and one failure
P(X = 0) = 0.811 = 0.0859
P(X = 1) = 11·0.2·0.810 = 0.2362
Add them
P = 0.0859 + 0.2362 = 0.3221
Answer
P = 0.3221
17.Descriptive statistics
Answer: s = 9.766
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 = (35 − 50)/15 = −1
Answer
z = −1
19.Normal distribution
Answer: P = 0.7333
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.
z-scores
z₁ = (166.8 − 178)/7 = −1.6
z₂ = (183.6 − 178)/7 = 0.8
Area between
P = Φ(0.8) − Φ(−1.6) = 0.7881 − 0.0548 = 0.7333
Answer
P = 0.7333
20.Normal distribution
Answer: P = 0.8413
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).