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 − 2x) − x).
f(x) = √(x2 − 2x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 5x)/(3x²), with x in radians.
f(x) = (1 − cos 5x)/(3x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(3x), with x in radians. Find f′(2).
f(x) = x·sin(3x)x = 2 rad
f′(2) =
4.DerivativesHard
Let f(x) = x3 + 4.5x2 + 3. Find the x-coordinate x of the local maximum.
f(x) = x3 + 4.5x2 + 3
x =
5.IntegralsMedium
Evaluate I = ∫02 2x·(x² + 1)3 dx.
f(x) = 2x·(x² + 1)3[0, 2]
I =
6.IntegralsHard
The region between the line y = 2x and the parabola y = 0.5x² is rotated about the x-axis. Find the volume V of the solid.
y = 2xy = 0.5x²
V =
7.Complex numbersMedium
Write z = 2.5∠−130° in rectangular form a + jb and give the real part a.
r = 2.5θ = −130°
a =
8.Complex numbersHard
Exactly one of the points A–D is a root of z⁶ = 729∠120°. Click it.
w = 729∠120°n = 6
1A (3∠80°)
2B (1.5∠80°)
3C (3∠110°)
4D (3∠−80°)
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′ + y/x = 6x for x > 0, with y(1) = 1. Find y(3).
y′ + y/x = 6xy(1) = 1
y(3) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [3 −2 5; 4 0 2; 1 −4 −5]
det A =
12.Matrices & determinantsHard
Solve the system −6x + 5y = 9 and −6x − y = −10. Find x.
−6x + 5y = 9−6x − y = −10
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (5, 4, −1) and b = (−1, 3, 5), in degrees.
a = (5, 4, −1)b = (−1, 3, 5)
θ = °
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (−2, 5, 3), b = (4, 1, 4) and c = (4, 3, −4).
a = (−2, 5, 3)b = (4, 1, 4)c = (4, 3, −4)
V =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4p(x): 0.3, 0.35, 0.2, 0.15
Var(X) =
16.ProbabilityHard
12 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?
n = 12p = 0.6
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 31, 57, 32, 18, 37.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 25. What proportion P of values is above 487.5?
μ = 500σ = 25x = 487.5
P =
19.Normal distributionMedium
The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 9 of them is below 747 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 9< 747 kg
P =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 5% of them lie (the 5th percentile).
μ = 65σ = 10P = 5%
x = points
3Solutions
1.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).
Multiply by the conjugate
f(x) = −2x/(√(x2 − 2x) + x)
Divide by x
f(x) = −2/(√(1 + (−2)/x) + 1)
Let x → ∞
L = −2/(1 + 1) = −1
Answer
L = −1
2.Limits
Answer: L = 4.167
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
5·sin 5x/(6x) → 0/0
Second application
25·cos 5x/6
Substitute x = 0
L = 25/6 = 4.167
Answer
L = 4.167
3.Derivatives
Answer: f′(2) = 5.482
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.
Product and chain rule
f′(x) = sin(3x) + 3x·cos(3x)
Substitute
f′(2) = −0.2794 + 6·0.9602 = 5.482
Answer
f′(2) = 5.482
4.Derivatives
Answer: x = −3
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 + 9x = 3·(x2 + 3x)
Solve f′(x) = 0
x = −3 or x = 0
Which one is which
f″(x) = 6x + 9: negative at a maximum, positive at a minimum.
f″(−3) = −9, f″(0) = 9
x = −3
Answer
x = −3
5.Integrals
Answer: I = 156
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.
Substitution
u = x² + 1, du = 2x dx
I = ∫15 u3 du
Integrate and evaluate
I = [u4/4]15 = (625 − 1)/4 = 156
Answer
I = 156
6.Integrals
Answer: V = 107.2
Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.
Intersection points
2x = 0.5x² → x = 0, x = 4
Washer method
V = π∫₀4 ((2x)² − (0.5x²)²) dx
Integrate
V = π(4·4³/3 − 0.25·4⁵/5) = π·34.13 = 107.2
Answer
V = 107.2
7.Complex numbers
Answer: a = −1.607
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 = 2.5·cos(−130°) = −1.607
b = 2.5·sin(−130°) = −1.915
Rectangular form
z = −1.607 − j1.915
a = −1.607
Answer
a = −1.607
8.Complex numbers
Answer: A (3∠80°)
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 and arguments of the roots
|z| = 3
θ = 20°, 80°, 140°, −160°, −100°, −40°
The point
A = 3∠80°
Answer
A (3∠80°)
9.Differential equations
Answer: y(1) = 1.104
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 = 2
Evaluate
y(1) = (1 + 2·1)·e−1 = 1.104
Answer
y(1) = 1.104
10.Differential equations
Answer: y(3) = 17.67
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∫1/x dx = x1
Integrate
(x1y)′ = 6x2 → x1y = 6x3/3 + C
Initial condition
1 = 6/3 + C → C = −1
Evaluate
y(3) = (6·33/3 + (−1))/31 = 17.67
Answer
y(3) = 17.67
11.Matrices & determinants
Answer: det A = −100
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 = 8, M₁₂ = −22, M₁₃ = −16
Cofactor expansion
det A = 3·8 − (−2)·(−22) + 5·(−16) = −100
Answer
det A = −100
12.Matrices & determinants
Answer: x = 1.139
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 = (−6)·(−1) − 5·(−6) = 36
Replace the x column
det Aₓ = 9·(−1) − 5·(−10) = 41
x = 41/36 = 1.139
Answer
x = 1.139
y = 3.167
13.Vectors: dot & cross product
Answer: θ = 87.01 °
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 = 5·(−1) + 4·3 + (−1)·5 = 2
Magnitudes
|a| = √(5² + 4² + (−1)²) = 6.481
|b| = √((−1)² + 3² + 5²) = 5.916
Angle
cos θ = 2/(6.481·5.916) = 0.05216
θ = arccos(0.05216) = 87.01°
Answer
θ = 87.01°
14.Vectors: dot & cross product
Answer: V = 216
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 = 1·(−4) − 4·3 = −16
(b × c)y = bzcx − bxcz = 4·4 − 4·(−4) = 32
(b × c)z = bxcy − bycx = 4·3 − 1·4 = 8
Dot product with a
a·(b × c) = (−2)·(−16) + 5·32 + 3·8 = 216
V = |216| = 216
Answer
V = 216
15.Probability
Answer: Var(X) = 1.06
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 = 1·0.3 + 2·0.35 + 3·0.2 + 4·0.15 = 2.2
Mean of the squares
E(X²) = 1·0.3 + 4·0.35 + 9·0.2 + 16·0.15 = 5.9
Variance
Var(X) = 5.9 − 2.2² = 1.06
Answer
Var(X) = 1.06
16.Probability
Answer: P = 0.0003188
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.
Zero and one failure
P(X = 0) = 0.412 = 1.678 × 10⁻⁵
P(X = 1) = 12·0.6·0.411 = 0.000302
Add them
P = 1.678 × 10⁻⁵ + 0.000302 = 0.0003188
Answer
P = 0.0003188
17.Descriptive statistics
Answer: s = 14.16
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 = 0.3085
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 total
μT = 765 kg, σT = 12·√9 = 36 kg
Standardise
z = (747 − 765)/36 = −0.5
Probability
P = Φ(−0.5) = 0.3085
Answer
P = 0.3085
20.Normal distribution
Answer: x = 48.55 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.