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 + 3/x)x. Give a decimal value.
f(x) = (1 + 3/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e1.5x − 1 − 1.5x)/x².
f(x) = (e1.5x − 1 − 1.5x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = ln(2x2 + 1). Find f′(2).
f(x) = ln(2x2 + 1)x = 2
f′(2) =
4.DerivativesHard
The tangent line to f(x) = 0.5x2 − 3x at x = −1 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 − 3xx = −1
b =
5.IntegralsMedium
Evaluate I = ∫₁5 x·ln x dx.
∫₁5 x ln x dx
I =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = 4x.
y = x2y = 4x
A =
7.Complex numbersMedium
Find the argument θ of z = −3 + j4 in degrees, as the principal value between −180° and 180°.
z = −3 + j4
θ = °
8.Complex numbersHard
The equation z² + 3z + 3.25 = 0 has two complex roots. Find their modulus |z|.
p = 3q = 3.25
|z| =
9.Differential equationsMedium
Solve y″ + 4y′ + 4y = 0 with y(0) = 3 and y′(0) = -2. Find y(2).
y″ + 4y′ + 4y = 0y(0) = 3y′(0) = -2
y(2) =
10.Differential equationsHard
Solve y″ + 3y′ + 18.25y = 0 with y(0) = 5 and y′(0) = 0 (released from rest). Find y(0.75); angles in radians.
y″ + 3y′ + 18.25y = 0y(0) = 5y′(0) = 0
y(0.75) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [4 −4; −1 −4]B = [5 −2; −3 3]
c11 =
12.Matrices & determinantsHard
Solve the system 4x − 2y = −3 and −x + 4y = −12. Find x.
4x − 2y = −3−x + 4y = −12
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−6, −1, −6) and b = (−4, −1, 2), in degrees.
a = (−6, −1, −6)b = (−4, −1, 2)
θ = °
14.Vectors: dot & cross productHard
A force F = (10, 20, −20) N acts at point A(3, −4, −3). Coordinates are in metres. Find the z-component Mz of the moment of the force about point O(3, 3, 1).
F = (10, 20, −20) NA(3, −4, −3) mO(3, 3, 1) m
Mz = N·m
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4, 5p(x): 0.2, 0.05, 0.3, 0.15, 0.3
σ =
16.ProbabilityHard
9 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?
n = 9p = 0.4
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 40, 14, 19, 56, 10, 47, 23.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is above 120?
μ = 100σ = 20x = 120
P =
19.Normal distributionMedium
X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 113 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 113 s
P =
20.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 97.5% of them lie (the 97.5th percentile).
μ = 100σ = 15P = 97.5%
x = points
3Solutions
1.Limits
Answer: L = 20.09
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 + 3/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·3 = 3
Undo the logarithm
L = e3 = 20.09
Answer
L = 20.09
2.Limits
Answer: L = 1.125
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
(1.5·e1.5x − 1.5)/(2x) → 0/0
Second application
2.25·e1.5x/2
Substitute x = 0
L = 2.25·1/2 = 1.125
Answer
L = 1.125
3.Derivatives
Answer: f′(2) = 0.8889
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 for the logarithm
f′(x) = 4x/(2x2 + 1)
Substitute
f′(2) = 8/9 = 0.8889
Answer
f′(2) = 0.8889
4.Derivatives
Answer: b = −0.5
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) = 3.5
f′(x) = x − 3 → f′(−1) = −4
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 3.5 − (−4)·(−1) = −0.5
Answer
b = −0.5
5.Integrals
Answer: I = 14.12
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
(12.5·ln 5 − 6.25) − (0 − 0.25) = 14.12
Answer
I = 14.12
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.
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 = 126.9 °
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 = −3, b = 4 → II
Reference angle
α = atan(4/3) = 53.13°
Argument
θ = 180° − α = 126.9°
Answer
θ = 126.9°
In polar form z = 5∠126.9°.
8.Complex numbers
Answer: |z| = 1.803
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.
Discriminant
p² − 4q = 9 − 13 = −4 < 0
Quadratic formula
z = (−3 ± j√4)/2 = −1.5 + j, −1.5 − j
Answer
|z| = √((−1.5)² + 1²) = √3.25 = 1.803
Answer
|z| = 1.803
9.Differential equations
Answer: y(2) = 0.2015
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 + 2)² = 0 → r = −2
Constants
A = y(0) = 3
y′(0) = B − 2A → B = -2 + 6 = 4
Evaluate
y(2) = (3 + 4·2)·e−4 = 0.2015
Answer
y(2) = 0.2015
10.Differential equations
Answer: y(0.75) = −1.521
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.
Complex roots
r = −1.5 ± 4j
Constants
A = 5, B = 1.5·5/4 = 1.875
Evaluate
y(0.75) = e−1.125(5·cos 3 + 1.875·sin 3) = −1.521
Answer
y(0.75) = −1.521
11.Matrices & determinants
Answer: c11 = 32
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 = 4·5 + (−4)·(−3) = 32
Answer
c11 = 32
12.Matrices & determinants
Answer: x = −2.571
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 = 4·4 − (−2)·(−1) = 14
Replace the x column
det Aₓ = (−3)·4 − (−2)·(−12) = −36
x = −36/14 = −2.571
Answer
x = −2.571
y = −3.643
13.Vectors: dot & cross product
Answer: θ = 70.61 °
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)·(−4) + (−1)·(−1) + (−6)·2 = 13
Magnitudes
|a| = √((−6)² + (−1)² + (−6)²) = 8.544
|b| = √((−4)² + (−1)² + 2²) = 4.583
Angle
cos θ = 13/(8.544·4.583) = 0.332
θ = arccos(0.332) = 70.61°
Answer
θ = 70.61°
14.Vectors: dot & cross product
Answer: Mz = 70 N·m
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.
Position vector from O to A
r = A − O = (3, −4, −3) − (3, 3, 1) = (0, −7, −4) m
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) = 0.69 = 0.01008
P(X = 1) = 9·0.4·0.68 = 0.06047
Add them
P = 0.01008 + 0.06047 = 0.07054
Answer
P = 0.07054
17.Descriptive statistics
Answer: s = 17.75
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 = 1
Standard normal table
Φ(1) = 0.8413
P = 1 − 0.8413 = 0.1587
Answer
P = 0.1587
19.Normal distribution
Answer: P = 0.1056
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 = 105 s, σ = √(5² + 4²) = 6.403 s
Standardise
z = (113 − 105)/6.403 = 1.25
Upper tail
P = 1 − Φ(1.25) = 1 − 0.8944 = 0.1056
Answer
P = 0.1056
20.Normal distribution
Answer: x = 129.4 points
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.