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→∞ (√(4x2 + 9x) − 2x).
f(x) = √(4x2 + 9x) − 2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 2x)/(2x²), with x in radians.
f(x) = (1 − cos 2x)/(2x²)x → 0
L =
3.DerivativesMedium
Let f(x) = (4x − 3)5. Find f′(1).
f(x) = (4x − 3)5x = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 − 15x2 + 63x − 1. Find the x-coordinate x of the local minimum.
f(x) = x3 − 15x2 + 63x − 1
x =
5.IntegralsMedium
Evaluate I = ∫14 4/x dx.
f(x) = 4/x[1, 4]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 2x2 and y = 3x.
y = 2x2y = 3x
A =
7.Complex numbersMedium
The diagram shows z = 4 − j2. With w = 2∠180°, click the point that represents z·w.
z = 4 − j2w = 2∠180°
1A (8 − j4)
2B (−8 + j4)
3C (−4 + j2)
4D (−2 + j)
8.Complex numbersHard
The roots of zⁿ = w with n = 3 and w = 8∠60° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2. Find the imaginary part of z₀.
w = 8∠60°n = 3k = 0
Im(z₀) =
9.Differential equationsMedium
Solve y″ + 2y′ + y = 0 with y(0) = 2 and y′(0) = 2. Find y(0.5).
y″ + 2y′ + y = 0y(0) = 2y′(0) = 2
y(0.5) =
10.Differential equationsHard
Solve the initial value problem y′ + 3y/x = 2x2 for x > 0, with y(1) = −3. Find y(2).
y′ + 3y/x = 2x2y(1) = −3
y(2) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [0 5 3; 0 1 4; 3 1 4]
det A =
12.Matrices & determinantsHard
Solve the system 4x − 3y = −3 and −4x − y = −7. Find x.
4x − 3y = −3−4x − y = −7
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (4, 5, 7) onto b = (5, 6, −1), that is, the component of a in the direction of b.
a = (4, 5, 7)b = (5, 6, −1)
compb a =
14.Vectors: dot & cross productHard
For a = (3, −4, −5) and b = (6, −2, 4), find the x-component (a × b)x of the cross product a × b.
a = (3, −4, −5)b = (6, −2, 4)
(a × b)x =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3, 4p(x): 0.2, 0.1, 0.25, 0.3, 0.15
Var(X) =
16.ProbabilityHard
A game costs €1 to play. You win €2 with probability 0.2 and €50 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €1€2: p = 0.2€50: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 17, 60, 33, 37, 41, 46, 48.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 100. What proportion P of values is above 300?
μ = 500σ = 100x = 300
P =
19.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 62.5 and 80.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 62.5x2 = 80.5
P =
20.Normal distributionHard
Hole diameters D are normal with mean 25.05 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.05 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
3Solutions
1.Limits
Answer: L = 2.25
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).
Multiply by the conjugate
f(x) = 9x/(√(4x2 + 9x) + 2x)
Divide by x
f(x) = 9/(√(4 + 9/x) + 2)
Let x → ∞
L = 9/(2 + 2) = 2.25
Answer
L = 2.25
2.Limits
Answer: L = 1
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·sin 2x/(4x) → 0/0
Second application
4·cos 2x/4
Substitute x = 0
L = 4/4 = 1
Answer
L = 1
3.Derivatives
Answer: f′(1) = 20
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
f′(x) = 5·(4x − 3)4·4
Substitute
f′(1) = 5·14·4 = 20
Answer
f′(1) = 20
4.Derivatives
Answer: x = 7
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 − 30x + 63 = 3·(x2 − 10x + 21)
Solve f′(x) = 0
x = 3 or x = 7
Which one is which
f″(x) = 6x − 30: negative at a maximum, positive at a minimum.
f″(3) = −12, f″(7) = 12
x = 7
Answer
x = 7
5.Integrals
Answer: I = 5.545
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) = 4·ln x
Evaluate at the limits
I = 4·(ln 4 − ln 1) = 4·ln 4 = 5.545
Answer
I = 5.545
6.Integrals
Answer: A = 1.125
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
2x2 = 3x → x = 0, x = 1.5
Integrate upper minus lower
A = [1.5x2 − 0.6667x3]01.5 = 1.125
Answer
A = 1.125
7.Complex numbers
Answer: B (−8 + j4)
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 + 180°
The point
z·w = −8 + j4 → B
Answer
B (−8 + j4)
8.Complex numbers
Answer: Im(z₀) = 0.684
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
81/3 = 2
Argument of the root
θ₀ = (60° + 360°·0)/3 = 20°
Rectangular form
z₀ = 2·(cos 20° + j sin 20°) = 1.879 + j0.684
Im(z₀) = 0.684
Answer
Im(z₀) = 0.684
The 3 roots lie on the circle of radius 2, 120° apart.
9.Differential equations
Answer: y(0.5) = 2.426
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) = 2
y′(0) = B − 1A → B = 2 + 2 = 4
Evaluate
y(0.5) = (2 + 4·0.5)·e−0.5 = 2.426
Answer
y(0.5) = 2.426
10.Differential equations
Answer: y(2) = 2.25
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∫3/x dx = x3
Integrate
(x3y)′ = 2x5 → x3y = 2x6/6 + C
Initial condition
−3 = 2/6 + C → C = −3.333
Evaluate
y(2) = (2·26/6 + (−3.333))/23 = 2.25
Answer
y(2) = 2.25
11.Matrices & determinants
Answer: det A = 51
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 = 0, M₁₂ = −12, M₁₃ = −3
Cofactor expansion
det A = 0·0 − 5·(−12) + 3·(−3) = 51
Answer
det A = 51
12.Matrices & determinants
Answer: x = 1.125
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·(−1) − (−3)·(−4) = −16
Replace the x column
det Aₓ = (−3)·(−1) − (−3)·(−7) = −18
x = −18/(−16) = 1.125
Answer
x = 1.125
y = 2.5
13.Vectors: dot & cross product
Answer: compb a = 5.461
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 = 4·5 + 5·6 + 7·(−1) = 43
Length of b
|b| = √(5² + 6² + (−1)²) = 7.874
Scalar projection
compb a = a·b/|b| = 43/7.874 = 5.461
Answer
compb a = 5.461
14.Vectors: dot & cross product
Answer: (a × b)x = −26
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, component by component
(a × b)x = aybz − azby = (−4)·4 − (−5)·(−2) = −26
(a × b)y = azbx − axbz = (−5)·6 − 3·4 = −42
(a × b)z = axby − aybx = 3·(−2) − (−4)·6 = 18
Requested component
(a × b)x = −26
Answer
(a × b)x = −26
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: Var(X) = 1.79
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.
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.
Net outcomes
G: 1 (p = 0.2), 49 (p = 0.01), −1 (p = 0.79)
Expected value
E(G) = 1·0.2 + 49·0.01 − 1·0.79 = −0.1 €
Answer
E(G) = −0.1 €
A negative value means you lose money on average.
17.Descriptive statistics
Answer: s = 13.46
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) = 0.02275
P = 1 − 0.02275 = 0.9772
Answer
P = 0.9772
19.Normal distribution
Answer: P = 0.5381
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 = (62.5 − 65)/10 = −0.25
z₂ = (80.5 − 65)/10 = 1.55
Area between
P = Φ(1.55) − Φ(−0.25) = 0.9394 − 0.4013 = 0.5381
Answer
P = 0.5381
20.Normal distribution
Answer: P = 0.9772
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).