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→∞ (3x3 + 4x − 7)/(5x3 + x + 1).
f(x) = (3x3 + 4x − 7)/(5x3 + x + 1)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (ex − 1 − x)/x².
f(x) = (ex − 1 − x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = ln(3x2 + 1). Find f′(2).
f(x) = ln(3x2 + 1)x = 2
f′(2) =
4.DerivativesHard
Let f(x) = x3 + 10.5x2 + 36x + 6. Find the x-coordinate x of the local maximum.
f(x) = x3 + 10.5x2 + 36x + 6
x =
5.IntegralsMedium
Evaluate I = ∫₀π x·cos x dx (x in radians).
∫₀π x·cos x dx
I =
6.IntegralsHard
Evaluate I = ∫₀1 x²·eˣ dx.
∫₀1 x²eˣ dx
I =
7.Complex numbersMedium
Write z = 5∠140° in rectangular form a + jb and give the imaginary part b.
r = 5θ = 140°
b =
8.Complex numbersHard
The equation z² + 8z + 18.25 = 0 has two complex roots. Find their modulus |z|.
p = 8q = 18.25
|z| =
9.Differential equationsMedium
Solve y″ + 6y′ + 5y = 0 with y(0) = 1 and y′(0) = 1. Find y(2).
y″ + 6y′ + 5y = 0y(0) = 1y′(0) = 1
y(2) =
10.Differential equationsHard
Solve y″ + y′ + 4.25y = 0 with y(0) = 1 and y′(0) = 0 (released from rest). Find y(1.5); angles in radians.
y″ + y′ + 4.25y = 0y(0) = 1y′(0) = 0
y(1.5) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [6 −3; 4 −1]B = [−2 5; 6 2]
c11 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [4 6; −7 −9]
λ =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (3, 5, −3) and b = (4, −6, k) are perpendicular.
a = (3, 5, −3)b = (4, −6, k)
k =
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−3, 1, 0) and b = (4, 3, 3) and has a positive z-component. Find its y-component n̂y.
a = (−3, 1, 0)b = (4, 3, 3)n̂z > 0
n̂y =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 10, 20, 30, 40p(x): 0.25, 0.3, 0.3, 0.15
Var(X) =
16.ProbabilityHard
A test has 8 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 5 correct answers?
n = 8p = 0.5k = 5
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 32, 28, 59, 40, 23, 36.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 15. What proportion P of values is below 72.5?
μ = 50σ = 15x = 72.5
P =
19.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 462 and 582 h? Give P as a decimal.
μ = 500σ = 40x1 = 462x2 = 582
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 90% of them lie (the 90th percentile).
μ = 100σ = 15P = 90%
x = points
3Solutions
1.Limits
Answer: L = 0.6
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 + 4/x2 + (−7)/x3)/(5 + 1/x2 + 1/x3)
Let x → ∞
L = 3/5 = 0.6
Answer
L = 0.6
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·ex − 1)/(2x) → 0/0
Second application
1·ex/2
Substitute x = 0
L = 1·1/2 = 0.5
Answer
L = 0.5
3.Derivatives
Answer: f′(2) = 0.9231
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 + 1)
Substitute
f′(2) = 12/13 = 0.9231
Answer
f′(2) = 0.9231
4.Derivatives
Answer: x = −4
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 = −4
Answer
x = −4
5.Integrals
Answer: I = −2
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·sin π + cos π) − (0 + 1) = −2
Answer
I = −2
6.Integrals
Answer: I = 0.7183
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
2.718·(1) − 1·2 = 0.7183
Answer
I = 0.7183
7.Complex numbers
Answer: b = 3.214
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.
Components
a = 5·cos(140°) = −3.83
b = 5·sin(140°) = 3.214
Rectangular form
z = −3.83 + j3.214
b = 3.214
Answer
b = 3.214
8.Complex numbers
Answer: |z| = 4.272
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 = 64 − 73 = −9 < 0
Quadratic formula
z = (−8 ± j√9)/2 = −4 + j1.5, −4 − j1.5
Answer
|z| = √((−4)² + 1.5²) = √18.25 = 4.272
Answer
|z| = 4.272
9.Differential equations
Answer: y(2) = 0.203
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 + 5) = 0 → r = −1, −5
Constants from the initial values
A + B = 1, −1A − 5B = 1
A = 1.5, B = −0.5
Evaluate
y(2) = 1.5·e−2 + (−0.5)·e−10 = 0.203
Answer
y(2) = 0.203
10.Differential equations
Answer: y(1.5) = −0.451
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 = −0.5 ± 2j
Constants
A = 1, B = 0.5·1/2 = 0.25
Evaluate
y(1.5) = e−0.75(1·cos 3 + 0.25·sin 3) = −0.451
Answer
y(1.5) = −0.451
11.Matrices & determinants
Answer: c11 = −30
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 = 6·(−2) + (−3)·6 = −30
Answer
c11 = −30
12.Matrices & determinants
Answer: λ = −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.
Characteristic equation
λ² − (−5)λ + (6) = 0
Solve
λ = (−5 ± √(1))/2 → λ = −3, −2
λ = −2
Answer
λ = −2
13.Vectors: dot & cross product
Answer: k = −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 equals zero
a·b = 3·4 + 5·(−6) + (−3)·k = 0
−18 + (−3)·k = 0
Solve for k
k = −(−18)/(−3) = −6
Answer
k = −6
14.Vectors: dot & cross product
Answer: n̂y = −0.5592
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 = 1·3 − 0·3 = 3
(a × b)y = azbx − axbz = 0·4 − (−3)·3 = 9
(a × b)z = axby − aybx = (−3)·3 − 1·4 = −13
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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 0.9332
Answer
P = 0.9332
19.Normal distribution
Answer: P = 0.8088
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 = (462 − 500)/40 = −0.95
z₂ = (582 − 500)/40 = 2.05
Area between
P = Φ(2.05) − Φ(−0.95) = 0.9798 − 0.1711 = 0.8088
Answer
P = 0.8088
20.Normal distribution
Answer: x = 119.2 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.