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→∞ (−3x2 + 2x − 1)/(4x2 + 6x + 6).
f(x) = (−3x2 + 2x − 1)/(4x2 + 6x + 6)x → ∞
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(3x2 + 4). Find f′(3).
f(x) = ln(3x2 + 4)x = 3
f′(3) =
4.DerivativesHard
The tangent line to f(x) = 0.5x2 − 5x − 6 at x = −2 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 − 5x − 6x = −2
b =
5.IntegralsMedium
Evaluate I = ∫₀π/3 x·cos x dx (x in radians).
∫₀π/3 x·cos x dx
I =
6.IntegralsHard
Evaluate I = ∫₀0.5 x²·eˣ dx.
∫₀0.5 x²eˣ dx
I =
7.Complex numbersMedium
Given z₁ = 2∠125° and z₂ = 2∠−25°, find the argument θ of w = z₁·z₂ in degrees, between −180° and 180°.
z₁ = 2∠125°z₂ = 2∠−25°
θ = °
8.Complex numbersHard
Solve z² − 5z + 10.25 = 0. The roots are a ± jb; find b > 0.
p = −5q = 10.25
b =
9.Differential equationsMedium
Solve y″ + 3y′ + 2y = 0 with y(0) = 2 and y′(0) = 4. Find y(2).
y″ + 3y′ + 2y = 0y(0) = 2y′(0) = 4
y(2) =
10.Differential equationsHard
Solve y″ + 3y′ + 6.25y = 0 with y(0) = 10 and y′(0) = 0 (released from rest). Find y(1.5); angles in radians.
y″ + 3y′ + 6.25y = 0y(0) = 10y′(0) = 0
y(1.5) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [2 −2; 0 2]B = [5 −5; −4 6]
c21 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [5 −1; 6 0]
λ =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (0, −2, 4) and b = (−3, 2, −2), in degrees.
a = (0, −2, 4)b = (−3, 2, −2)
θ = °
14.Vectors: dot & cross productHard
A force F = (−30, 10, 30) N acts at point A(4, 4, 0). Coordinates are in metres. Find the y-component My of the moment of the force about the origin O(0, 0, 0).
F = (−30, 10, 30) NA(4, 4, 0) mO(0, 0, 0) m
My = N·m
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 2, 4, 6p(x): 0.1, 0.35, 0.2, 0.35
Var(X) =
16.ProbabilityHard
A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 3 correct answers?
n = 11p = 0.1k = 3
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 11, 45, 51, 32, 45, 58.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 5. What proportion P of values is above 47.5?
μ = 50σ = 5x = 47.5
P =
19.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 92.5 and 103.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 92.5x2 = 103.75
P =
20.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 70% is to the right of it. Find the cut-off x.
μ = 178 cmσ = 7 cm70%
x = cm
3Solutions
1.Limits
Answer: L = −0.75
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
(−3 + 2/x + (−1)/x2)/(4 + 6/x + 6/x2)
Let x → ∞
L = −3/4 = −0.75
Answer
L = −0.75
For equal degrees the limit is the ratio of the leading coefficients.
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′(3) = 0.5806
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) = 6x/(3x2 + 4)
Substitute
f′(3) = 18/31 = 0.5806
Answer
f′(3) = 0.5806
4.Derivatives
Answer: b = −8
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(−2) = 6
f′(x) = x − 5 → f′(−2) = −7
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 6 − (−7)·(−2) = −8
Answer
b = −8
5.Integrals
Answer: I = 0.4069
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 = cos x dx, v = sin x
Antiderivative
∫x cos x dx = x sin x − ∫sin x dx = x sin x + cos x
Limits
(π/3·sin π/3 + cos π/3) − (0 + 1) = 0.4069
Answer
I = 0.4069
6.Integrals
Answer: I = 0.0609
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.
First round
∫x²eˣ dx = x²eˣ − ∫2x eˣ dx
Second round
∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)
Limits
1.649·(1.25) − 1·2 = 0.0609
Answer
I = 0.0609
7.Complex numbers
Answer: θ = 100 °
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 · 2 = 4
θ = 125° + (−25)° = 100°
Argument
θ = 100°
Answer
θ = 100°
8.Complex numbers
Answer: b = 2
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 = 25 − 41 = −16 < 0
Quadratic formula
z = (5 ± j√16)/2 = 2.5 + j2, 2.5 − j2
Answer
b = 2
Answer
b = 2
9.Differential equations
Answer: y(2) = 0.9728
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 + 2) = 0 → r = −1, −2
Constants from the initial values
A + B = 2, −1A − 2B = 4
A = 8, B = −6
Evaluate
y(2) = 8·e−2 + (−6)·e−4 = 0.9728
Answer
y(2) = 0.9728
10.Differential equations
Answer: y(1.5) = −0.9319
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 ± 2j
Constants
A = 10, B = 1.5·10/2 = 7.5
Evaluate
y(1.5) = e−2.25(10·cos 3 + 7.5·sin 3) = −0.9319
Answer
y(1.5) = −0.9319
11.Matrices & determinants
Answer: c21 = −8
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
c21 = 0·5 + 2·(−4) = −8
Answer
c21 = −8
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 − (5)λ + (6) = 0
Solve
λ = (5 ± √(1))/2 → λ = 2, 3
λ = 3
Answer
λ = 3
13.Vectors: dot & cross product
Answer: θ = 130.6 °
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 = 0·(−3) + (−2)·2 + 4·(−2) = −12
Magnitudes
|a| = √(0² + (−2)² + 4²) = 4.472
|b| = √((−3)² + 2² + (−2)²) = 4.123
Angle
cos θ = −12/(4.472·4.123) = −0.6508
θ = arccos(−0.6508) = 130.6°
Answer
θ = 130.6°
14.Vectors: dot & cross product
Answer: My = −120 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.
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 = 0·0.1 + 2·0.35 + 4·0.2 + 6·0.35 = 3.6
Mean of the squares
E(X²) = 0·0.1 + 4·0.35 + 16·0.2 + 36·0.35 = 17.2
Variance
Var(X) = 17.2 − 3.6² = 4.24
Answer
Var(X) = 4.24
16.Probability
Answer: P = 0.07103
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 = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915
Answer
P = 0.6915
19.Normal distribution
Answer: P = 0.2902
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 = (92.5 − 100)/15 = −0.5
z₂ = (103.75 − 100)/15 = 0.25
Area between
P = Φ(0.25) − Φ(−0.5) = 0.5987 − 0.3085 = 0.2902
Answer
P = 0.2902
20.Normal distribution
Answer: x = 174.4 cm
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.