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→∞ (√(x2 − 9x) − x).
f(x) = √(x2 − 9x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 3x)/(4x²), with x in radians.
f(x) = (1 − cos 3x)/(4x²)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(x2 + 1). Find f′(1).
f(x) = ln(x2 + 1)x = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 − 10.5x2 + 30x − 6. Find the x-coordinate x of the local maximum.
f(x) = x3 − 10.5x2 + 30x − 6
x =
5.IntegralsMedium
Evaluate I = ∫₀π/2 x·cos x dx (x in radians).
∫₀π/2 x·cos x dx
I =
6.IntegralsHard
Evaluate I = ∫₀1 x²·eˣ dx.
∫₀1 x²eˣ dx
I =
7.Complex numbersMedium
The diagram shows z = −3 − j4. With w = 2∠−90°, click the point that represents z·w.
z = −3 − j4w = 2∠−90°
1A (−8 + j6)
2B (8 − j6)
3C (−6 − j8)
4D (−4 + j3)
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 argument of z₁ in degrees, between −180° and 180°.
w = 8∠−60°n = 3k = 1
θ₁ = °
9.Differential equationsMedium
Solve y″ + 6y′ + 9y = 0 with y(0) = 2 and y′(0) = 0. Find y(2).
y″ + 6y′ + 9y = 0y(0) = 2y′(0) = 0
y(2) =
10.Differential equationsHard
Solve y″ + 2y′ + 10y = 0 with y(0) = 3 and y′(0) = 0 (released from rest). Find y(0.5); angles in radians.
y″ + 2y′ + 10y = 0y(0) = 3y′(0) = 0
y(0.5) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−4 −1 −1; 0 5 3; −1 0 1]
det A =
12.Matrices & determinantsHard
Solve the system 3x − 2y = −1 and −4x − 3y = −4. Find x.
3x − 2y = −1−4x − 3y = −4
x =
13.Vectors: dot & cross productMedium
A constant force F = (25, −15, 20) N acts on an object while it moves in a straight line from point A(2, −4, 4) to point B(−4, −4, 3). Coordinates are in metres. Find the work W done by the force.
F = (25, −15, 20) NA(2, −4, 4) mB(−4, −4, 3) m
W = J
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (−1, 2, −1), b = (4, −2, −4) and c = (5, 4, 5).
a = (−1, 2, −1)b = (4, −2, −4)c = (5, 4, 5)
V =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4, 5p(x): 0.3, 0.15, 0.05, 0.3, 0.2
Var(X) =
16.ProbabilityHard
A game costs €5 to play. You win €25 with probability 0.1 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 = €5€25: p = 0.1€50: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 10, 53, 27, 53, 22.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is below 60?
μ = 50σ = 8x = 60
P =
19.Normal distributionMedium
X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 745.1 g. Round z to two decimals.
μX = 250, σX = 4μY = 500, σY = 9X + Y > 745.1 g
P =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 70% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points70%
x = points
3Solutions
1.Limits
Answer: L = −4.5
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).
Multiply by the conjugate
f(x) = −9x/(√(x2 − 9x) + x)
Divide by x
f(x) = −9/(√(1 + (−9)/x) + 1)
Let x → ∞
L = −9/(1 + 1) = −4.5
Answer
L = −4.5
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
3·sin 3x/(8x) → 0/0
Second application
9·cos 3x/8
Substitute x = 0
L = 9/8 = 1.125
Answer
L = 1.125
3.Derivatives
Answer: f′(1) = 1
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) = 2x/(x2 + 1)
Substitute
f′(1) = 2/2 = 1
Answer
f′(1) = 1
4.Derivatives
Answer: x = 2
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 + 30 = 3·(x2 − 7x + 10)
Solve f′(x) = 0
x = 2 or x = 5
Which one is which
f″(x) = 6x − 21: negative at a maximum, positive at a minimum.
f″(2) = −9, f″(5) = 9
x = 2
Answer
x = 2
5.Integrals
Answer: I = 0.5708
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/2·sin π/2 + cos π/2) − (0 + 1) = 0.5708
Answer
I = 0.5708
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: A (−8 + j6)
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 = −8 + j6 → A
Answer
A (−8 + j6)
8.Complex numbers
Answer: θ₁ = 100 °
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
θ₁ = (−60° + 360°·1)/3 = 100°
Argument
θ₁ = 100°
Answer
θ₁ = 100°
The 3 roots lie on the circle of radius 2, 120° apart.
9.Differential equations
Answer: y(2) = 0.0347
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 + 3)² = 0 → r = −3
Constants
A = y(0) = 2
y′(0) = B − 3A → B = 0 + 6 = 6
Evaluate
y(2) = (2 + 6·2)·e−6 = 0.0347
Answer
y(2) = 0.0347
10.Differential equations
Answer: y(0.5) = 0.7337
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 ± 3j
Constants
A = 3, B = 1·3/3 = 1
Evaluate
y(0.5) = e−0.5(3·cos 1.5 + 1·sin 1.5) = 0.7337
Answer
y(0.5) = 0.7337
11.Matrices & determinants
Answer: det A = −22
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 = 5, M₁₂ = 3, M₁₃ = 5
Cofactor expansion
det A = (−4)·5 − (−1)·3 + (−1)·5 = −22
Answer
det A = −22
12.Matrices & determinants
Answer: x = 0.2941
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 = 3·(−3) − (−2)·(−4) = −17
Replace the x column
det Aₓ = (−1)·(−3) − (−2)·(−4) = −5
x = −5/(−17) = 0.2941
Answer
x = 0.2941
y = 0.9412
13.Vectors: dot & cross product
Answer: W = −170 J
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.
Displacement
d = B − A = (−4, −4, 3) − (2, −4, 4) = (−6, 0, −1) m
Work
W = F·d = 25·(−6) + (−15)·0 + 20·(−1) = −170 J
Answer
W = −170 J
Negative work: the force acts mostly against the motion.
14.Vectors: dot & cross product
Answer: V = 112
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 b × c
(b × c)x = bycz − bzcy = (−2)·5 − (−4)·4 = 6
(b × c)y = bzcx − bxcz = (−4)·5 − 4·5 = −40
(b × c)z = bxcy − bycx = 4·4 − (−2)·5 = 26
Dot product with a
a·(b × c) = (−1)·6 + 2·(−40) + (−1)·26 = −112
V = |−112| = 112
Answer
V = 112
15.Probability
Answer: Var(X) = 2.447
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: 20 (p = 0.1), 45 (p = 0.01), −5 (p = 0.89)
Expected value
E(G) = 20·0.1 + 45·0.01 − 5·0.89 = −2 €
Answer
E(G) = −2 €
A negative value means you lose money on average.
17.Descriptive statistics
Answer: s = 19.27
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 = 1.25
Standard normal table
Φ(1.25) = 0.8944
P = 0.8944
Answer
P = 0.8944
19.Normal distribution
Answer: P = 0.6915
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 = 750 g, σ = √(4² + 9²) = 9.849 g
Standardise
z = (745.1 − 750)/9.849 = −0.5
Upper tail
P = 1 − Φ(−0.5) = 1 − 0.3085 = 0.6915
Answer
P = 0.6915
20.Normal distribution
Answer: x = 59.8 points
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.