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)x. Give a decimal value.
f(x) = (1 + 1/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e−2x − 1 + 2x)/x².
f(x) = (e−2x − 1 + 2x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x·e2x. Find f′(1).
f(x) = x·e2xx = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 + 1.5x2 + 5. Find the x-coordinate x of the local maximum.
f(x) = x3 + 1.5x2 + 5
x =
5.IntegralsMedium
Evaluate I = ∫₁2.5 x·ln x dx.
∫₁2.5 x ln x dx
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 4 − x² and the x-axis.
y = 4 − x²
A =
7.Complex numbersMedium
Find the argument θ of z = 4 − j6 in degrees, as the principal value between −180° and 180°.
z = 4 − j6
θ = °
8.Complex numbersHard
Use De Moivre's theorem to find the imaginary part of z³ for z = −√3 + j.
z = −√3 + jn = 3
Im(z³) =
9.Differential equationsMedium
Solve y″ + 4y′ + 3y = 0 with y(0) = 4 and y′(0) = -2. Find y(1.5).
y″ + 4y′ + 3y = 0y(0) = 4y′(0) = -2
y(1.5) =
10.Differential equationsHard
A system obeys y″ + 2y′ + 10y = 50 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 2y′ + 10y = 50y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [4 3 −2; 5 −5 0; 5 −4 −2]
det A =
12.Matrices & determinantsHard
Solve the system 5x + 6y = 11 and −5x − 2y = 7. Find x.
5x + 6y = 11−5x − 2y = 7
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (−4, −2, −4) and b = (−2, 3, k) are perpendicular.
a = (−4, −2, −4)b = (−2, 3, k)
k =
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−3, 3, 5) and b = (4, 1, 5) and has a positive z-component. Find its y-component n̂y.
a = (−3, 3, 5)b = (4, 1, 5)n̂z > 0
n̂y =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4p(x): 0.25, 0.15, 0.35, 0.25
σ =
16.ProbabilityHard
A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 2 correct answers?
n = 9p = 0.3k = 2
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 10, 44, 56, 15, 33, 46.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 5. What proportion P of values is below 60?
μ = 50σ = 5x = 60
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 29.5 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 29.5 min
P =
20.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 75% of them lie (the 75th percentile).
μ = 178σ = 7P = 75%
x = cm
3Solutions
1.Limits
Answer: L = 2.718
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 + 1/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·1 = 1
Undo the logarithm
L = e1 = 2.718
Answer
L = 2.718
2.Limits
Answer: L = 2
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
(−2·e−2x + 2)/(2x) → 0/0
Second application
4·e−2x/2
Substitute x = 0
L = 4·1/2 = 2
Answer
L = 2
3.Derivatives
Answer: f′(1) = 22.17
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′ = e2x·(1 + 2·x)
Substitute
f′(1) = 7.389·(1 + 2) = 22.17
Answer
f′(1) = 22.17
4.Derivatives
Answer: x = −1
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 + 3x = 3·(x2 + x)
Solve f′(x) = 0
x = −1 or x = 0
Which one is which
f″(x) = 6x + 3: negative at a maximum, positive at a minimum.
f″(−1) = −3, f″(0) = 3
x = −1
Answer
x = −1
5.Integrals
Answer: I = 1.551
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
(3.125·ln 2.5 − 1.5625) − (0 − 0.25) = 1.551
Answer
I = 1.551
6.Integrals
Answer: A = 10.67
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.
Zeros
4 − x² = 0 → x = ±2
Integrate
A = [4x − x³/3]−22 = 5.333 − (−5.333) = 10.67
Answer
A = 10.67
7.Complex numbers
Answer: θ = −56.31 °
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 = 4, b = −6 → IV
Reference angle
α = atan(6/4) = 56.31°
Argument
θ = −α = −56.31°
Answer
θ = −56.31°
In polar form z = 7.211∠−56.31°.
8.Complex numbers
Answer: Im(z³) = 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 = −√3 + j = 2∠150°
De Moivre
z³ = (2)³∠450° = 8∠90°
Rectangular form
z³ = 8·(cos 90° + j sin 90°) = j8
Im(z³) = 8
Answer
Im(z³) = 8
9.Differential equations
Answer: y(1.5) = 1.105
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.
Characteristic roots
(r + 1)(r + 3) = 0 → r = −1, −3
Constants from the initial values
A + B = 4, −1A − 3B = -2
A = 5, B = −1
Evaluate
y(1.5) = 5·e−1.5 + (−1)·e−4.5 = 1.105
Answer
y(1.5) = 1.105
10.Differential equations
Answer: ymax = 6.755
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 = 5, r = −1 ± 3j
Constants from the initial values
y(0) = 5 + A = 0 → A = −5
y′(0) = −1A + 3B = 0 → B = −1.667
Peak
t = π/3 = 1.047
ymax = 5·(1 + e−1·π/3) = 6.755
Answer
ymax = 6.755
11.Matrices & determinants
Answer: det A = 60
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 = 10, M₁₂ = −10, M₁₃ = 5
Cofactor expansion
det A = 4·10 − 3·(−10) + (−2)·5 = 60
Answer
det A = 60
12.Matrices & determinants
Answer: x = −3.2
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 = 5·(−2) − 6·(−5) = 20
Replace the x column
det Aₓ = 11·(−2) − 6·7 = −64
x = −64/20 = −3.2
Answer
x = −3.2
y = 4.5
13.Vectors: dot & cross product
Answer: k = 0.5
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 equals zero
a·b = (−4)·(−2) + (−2)·3 + (−4)·k = 0
2 + (−4)·k = 0
Solve for k
k = −2/(−4) = 0.5
Answer
k = 0.5
14.Vectors: dot & cross product
Answer: n̂y = −0.889
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 = 3·5 − 5·1 = 10
(a × b)y = azbx − axbz = 5·4 − (−3)·5 = 35
(a × b)z = axby − aybx = (−3)·1 − 3·4 = −15
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.
Mean
μ = 1·0.25 + 2·0.15 + 3·0.35 + 4·0.25 = 2.6
Mean of the squares
E(X²) = 1·0.25 + 4·0.15 + 9·0.35 + 16·0.25 = 8
Variance
Var(X) = 8 − 2.6² = 1.24
σ = √1.24 = 1.114
Answer
σ = 1.114
16.Probability
Answer: P = 0.2668
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 = 2
Standard normal table
Φ(2) = 0.9772
P = 0.9772
Answer
P = 0.9772
19.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).
Distribution of the sum
μ = 32 min, σ = √(1.5² + 2²) = 2.5 min
Standardise
z = (29.5 − 32)/2.5 = −1
Upper tail
P = 1 − Φ(−1) = 1 − 0.1587 = 0.8413
Answer
P = 0.8413
20.Normal distribution
Answer: x = 182.7 cm
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.