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 − 5x − 6)/(5x2 + 5x + 9).
f(x) = (−x2 − 5x − 6)/(5x2 + 5x + 9)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 3x)/sin(x), with x in radians.
f(x) = ln(1 + 3x)/sin(x)x → 0
L =
3.DerivativesMedium
Let f(x) = x·e2x. Find f′(0.5).
f(x) = x·e2xx = 0.5
f′(0.5) =
4.DerivativesHard
Let f(x) = x3 − 4.5x2 + 6x + 3. Find the x-coordinate x of the local maximum.
f(x) = x3 − 4.5x2 + 6x + 3
x =
5.IntegralsMedium
The region under y = √(x) from x = 0 to x = 2 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(x)0 ≤ x ≤ 2
V =
6.IntegralsHard
The region between the line y = 3x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.
y = 3xy = 3x²
V =
7.Complex numbersMedium
Write z = 2∠−20° in rectangular form a + jb and give the real part a.
r = 2θ = −20°
a =
8.Complex numbersHard
The equation z² + 8z + 41 = 0 has two complex roots. Find their modulus |z|.
p = 8q = 41
|z| =
9.Differential equationsMedium
Solve y″ + 6y′ + 9y = 0 with y(0) = 5 and y′(0) = -1. Find y(0.5).
y″ + 6y′ + 9y = 0y(0) = 5y′(0) = -1
y(0.5) =
10.Differential equationsHard
Solve y″ + 3y′ + 6.25y = 0 with y(0) = 3 and y′(0) = 0 (released from rest). Find y(1); angles in radians.
y″ + 3y′ + 6.25y = 0y(0) = 3y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [2 1; 6 3]B = [3 0; 1 −4]
c21 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−1 2; 4 1]
λ =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (2, 4, −3) and b = (k, 6, 0) are perpendicular.
a = (2, 4, −3)b = (k, 6, 0)
k =
14.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (0, 4, −6) and b = (0, −3, 3).
a = (0, 4, −6)b = (0, −3, 3)
A =
15.ProbabilityMedium
In how many ways N can a committee of 5 people be chosen from a group of 8? The order does not matter.
n = 8k = 5
N =
16.ProbabilityHard
A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 3 correct answers?
n = 7p = 0.2k = 3
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 29, 32, 25, 31, 52, 24.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 20. What proportion P of values is above 80?
μ = 50σ = 20x = 80
P =
19.Normal distributionMedium
The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 4 of them is below 47.2 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 4< 47.2 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 97.5% of them lie (the 97.5th percentile).
μ = 65σ = 10P = 97.5%
x = points
3Solutions
1.Limits
Answer: L = −0.2
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
(−1 + (−5)/x + (−6)/x2)/(5 + 5/x + 9/x2)
Let x → ∞
L = −1/5 = −0.2
Answer
L = −0.2
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 3
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).
L'Hôpital's rule
L = limx→0 [3/(1 + 3x)]/[cos x]
Substitute x = 0
L = 3/1 = 3
Answer
L = 3
3.Derivatives
Answer: f′(0.5) = 5.437
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 rule
f′(x) = u′v + uv′ = e2x·(1 + 2·x)
Substitute
f′(0.5) = 2.718·(1 + 1) = 5.437
Answer
f′(0.5) = 5.437
4.Derivatives
Answer: x = 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.
Derivative
f′(x) = 3x2 − 9x + 6 = 3·(x2 − 3x + 2)
Solve f′(x) = 0
x = 1 or x = 2
Which one is which
f″(x) = 6x − 9: negative at a maximum, positive at a minimum.
f″(1) = −3, f″(2) = 3
x = 1
Answer
x = 1
5.Integrals
Answer: V = 6.283
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.
Disc method
V = π∫₀2 x dx
Integrate
V = π·1·2²/2 = 6.283
Answer
V = 6.283
6.Integrals
Answer: V = 3.77
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
3x = 3x² → x = 0, x = 1
Washer method
V = π∫₀1 ((3x)² − (3x²)²) dx
Integrate
V = π(9·1³/3 − 9·1⁵/5) = π·1.2 = 3.77
Answer
V = 3.77
7.Complex numbers
Answer: a = 1.879
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·cos(−20°) = 1.879
b = 2·sin(−20°) = −0.684
Rectangular form
z = 1.879 − j0.684
a = 1.879
Answer
a = 1.879
8.Complex numbers
Answer: |z| = 6.403
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 − 164 = −100 < 0
Quadratic formula
z = (−8 ± j√100)/2 = −4 + j5, −4 − j5
Answer
|z| = √((−4)² + 5²) = √41 = 6.403
Answer
|z| = 6.403
9.Differential equations
Answer: y(0.5) = 2.678
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) = 5
y′(0) = B − 3A → B = -1 + 15 = 14
Evaluate
y(0.5) = (5 + 14·0.5)·e−1.5 = 2.678
Answer
y(0.5) = 2.678
10.Differential equations
Answer: y(1) = 0.1779
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 = −1.5 ± 2j
Constants
A = 3, B = 1.5·3/2 = 2.25
Evaluate
y(1) = e−1.5(3·cos 2 + 2.25·sin 2) = 0.1779
Answer
y(1) = 0.1779
11.Matrices & determinants
Answer: c21 = 21
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
c21 = 6·3 + 3·1 = 21
Answer
c21 = 21
12.Matrices & determinants
Answer: λ = 3
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 − (0)λ + (−9) = 0
Solve
λ = (0 ± √(36))/2 → λ = −3, 3
λ = 3
Answer
λ = 3
13.Vectors: dot & cross product
Answer: k = −12
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 = 2·k + 4·6 + (−3)·0 = 0
24 + 2·k = 0
Solve for k
k = −24/2 = −12
Answer
k = −12
14.Vectors: dot & cross product
Answer: A = 6
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 = 4·3 − (−6)·(−3) = −6
(a × b)y = azbx − axbz = (−6)·0 − 0·3 = 0
(a × b)z = axby − aybx = 0·(−3) − 4·0 = 0
Area
A = |a × b| = √((−6)² + 0² + 0²) = 6
Answer
A = 6
15.Probability
Answer: N = 56
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.
Combinations
C(8, 5) = (8·7·6·5·4)/(5·4·3·2·1) = 56
Answer
N = 56
16.Probability
Answer: P = 0.1147
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 = 1 − 0.9332 = 0.06681
Answer
P = 0.06681
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 = 48 kg, σT = 0.8·√4 = 1.6 kg
Standardise
z = (47.2 − 48)/1.6 = −0.5
Probability
P = Φ(−0.5) = 0.3085
Answer
P = 0.3085
20.Normal distribution
Answer: x = 84.6 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.