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)2x. Give a decimal value.
f(x) = (1 − 1/x)2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e0.5x − 1 − 0.5x)/x².
f(x) = (e0.5x − 1 − 0.5x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = ln(2x2 + 1). Find f′(1).
f(x) = ln(2x2 + 1)x = 1
f′(1) =
4.DerivativesHard
The tangent line to f(x) = −x2 + 3x − 5 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = −x2 + 3x − 5x = 3
b =
5.IntegralsMedium
Evaluate I = ∫14 5/x dx.
f(x) = 5/x[1, 4]
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
The diagram shows z = 1 − j2. With w = 2∠90°, click the point that represents z·w.
z = 1 − j2w = 2∠90°
1A (2 − j4)
2B (−4 − j2)
3C (4 + j2)
4D (2 + j)
8.Complex numbersHard
The roots of zⁿ = w with n = 3 and w = 27∠120° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2. Find the real part of z₀.
w = 27∠120°n = 3k = 0
Re(z₀) =
9.Differential equationsMedium
Solve y″ + 2y′ + y = 0 with y(0) = 1 and y′(0) = -1. Find y(1).
y″ + 2y′ + y = 0y(0) = 1y′(0) = -1
y(1) =
10.Differential equationsHard
Solve the initial value problem y′ + 2y/x = 3x2 for x > 0, with y(1) = 2. Find y(2).
y′ + 2y/x = 3x2y(1) = 2
y(2) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [0 −6; −5 3]B = [−6 −5; −2 6]
c11 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [3 3; 5 1]
λ =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (6, −3, −4) and b = (−6, 5, −5).
a = (6, −3, −4)b = (−6, 5, −5)
a·b =
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(−4, 0, −3), B(−4, 3, −1) and C(−3, 4, 1). Coordinates are in metres.
A(−4, 0, −3)B(−4, 3, −1)C(−3, 4, 1)
A = m²
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 2, 4, 6p(x): 0.25, 0.2, 0.2, 0.35
σ =
16.ProbabilityHard
A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 2 correct answers?
n = 6p = 0.2k = 2
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 41, 26, 35, 57, 52, 18.
n = 6
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 15. Find the z-score z of a score of 92.5.
μ = 100σ = 15x = 92.5
z =
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 IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 80% of them lie (the 80th percentile).
μ = 100σ = 15P = 80%
x = points
3Solutions
1.Limits
Answer: L = 0.1353
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→∞ 2·x·ln(1 − 1/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 2·(−1) = −2
Undo the logarithm
L = e−2 = 0.1353
Answer
L = 0.1353
2.Limits
Answer: L = 0.125
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
(0.5·e0.5x − 0.5)/(2x) → 0/0
Second application
0.25·e0.5x/2
Substitute x = 0
L = 0.25·1/2 = 0.125
Answer
L = 0.125
3.Derivatives
Answer: f′(1) = 1.333
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′(1) = 4/3 = 1.333
Answer
f′(1) = 1.333
4.Derivatives
Answer: b = 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.
Point and slope
f(3) = −5
f′(x) = −2x + 3 → f′(3) = −3
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −5 − (−3)·3 = 4
Answer
b = 4
5.Integrals
Answer: I = 6.931
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.
Antiderivative
F(x) = 5·ln x
Evaluate at the limits
I = 5·(ln 4 − ln 1) = 5·ln 4 = 6.931
Answer
I = 6.931
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: C (4 + j2)
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.
Effect of w
|z·w| = 2·|z|, arg(z·w) = arg z + 90°
The point
z·w = 4 + j2 → C
Answer
C (4 + j2)
8.Complex numbers
Answer: Re(z₀) = 2.298
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.
Modulus of the roots
271/3 = 3
Argument of the root
θ₀ = (120° + 360°·0)/3 = 40°
Rectangular form
z₀ = 3·(cos 40° + j sin 40°) = 2.298 + j1.928
Re(z₀) = 2.298
Answer
Re(z₀) = 2.298
The 3 roots lie on the circle of radius 3, 120° apart.
9.Differential equations
Answer: y(1) = 0.3679
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 + 1)² = 0 → r = −1
Constants
A = y(0) = 1
y′(0) = B − 1A → B = -1 + 1 = 0
Evaluate
y(1) = (1 + 0·1)·e−1 = 0.3679
Answer
y(1) = 0.3679
10.Differential equations
Answer: y(2) = 5.15
First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e∫p dx: (μy)′ = μq, so μy = ∫μq dx + C.
Integrating factor
μ = e∫2/x dx = x2
Integrate
(x2y)′ = 3x4 → x2y = 3x5/5 + C
Initial condition
2 = 3/5 + C → C = 1.4
Evaluate
y(2) = (3·25/5 + 1.4)/22 = 5.15
Answer
y(2) = 5.15
11.Matrices & determinants
Answer: c11 = 12
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 = 0·(−6) + (−6)·(−2) = 12
Answer
c11 = 12
12.Matrices & determinants
Answer: λ = 6
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 + (−12) = 0
Solve
λ = (4 ± √(64))/2 → λ = −2, 6
λ = 6
Answer
λ = 6
13.Vectors: dot & cross product
Answer: a·b = −31
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.
Negative: the angle between the vectors is more than 90°.
14.Vectors: dot & cross product
Answer: A = 2.693 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.
Edge vectors
AB = B − A = (0, 3, 2)
AC = C − A = (1, 4, 4)
Cross product
(AB × AC)x = AByACz − ABzACy = 3·4 − 2·4 = 4
(AB × AC)y = ABzACx − ABxACz = 2·1 − 0·4 = 2
(AB × AC)z = ABxACy − AByACx = 0·4 − 3·1 = −3
|AB × AC| = √(4² + 2² + (−3)²) = 5.385
Area of the triangle
A = ½·5.385 = 2.693 m²
Answer
A = 2.693 m²
15.Probability
Answer: σ = 2.39
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 = 0·0.25 + 2·0.2 + 4·0.2 + 6·0.35 = 3.3
Mean of the squares
E(X²) = 0·0.25 + 4·0.2 + 16·0.2 + 36·0.35 = 16.6
Variance
Var(X) = 16.6 − 3.3² = 5.71
σ = √5.71 = 2.39
Answer
σ = 2.39
16.Probability
Answer: P = 0.2458
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.
Binomial formula
P(X = 2) = 15·0.22·0.84 = 15·0.04·0.4096 = 0.2458
Answer
P = 0.2458
17.Descriptive statistics
Answer: s = 14.96
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 = (92.5 − 100)/15 = −0.5
Answer
z = −0.5
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 = 112.6 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.