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 − 3x) − x).
f(x) = √(x2 − 3x) − xx → ∞
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(3x2 + 4). Find f′(−2).
f(x) = ln(3x2 + 4)x = −2
f′(−2) =
4.DerivativesHard
The tangent line to f(x) = 2x2 + 3x + 6 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = 2x2 + 3x + 6x = 3
b =
5.IntegralsMedium
The region under y = 0.5x from x = 0 to x = 2 is rotated about the x-axis. Find the volume V of the solid (a cone).
y = 0.5x0 ≤ x ≤ 2
V =
6.IntegralsHard
Evaluate I = ∫₀1.5 x²·eˣ dx.
∫₀1.5 x²eˣ dx
I =
7.Complex numbersMedium
Given z₁ = 5∠150° and z₂ = 2∠65°, find the real part of w = z₁/z₂.
z₁ = 5∠150°z₂ = 2∠65°
Re(w) =
8.Complex numbersHard
The roots of zⁿ = w with n = 2 and w = 36∠0° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1. Find the real part of z₁.
w = 36∠0°n = 2k = 1
Re(z₁) =
9.Differential equationsMedium
Solve y″ + 5y′ + 6y = 0 with y(0) = 5 and y′(0) = -4. Find y(0.5).
y″ + 5y′ + 6y = 0y(0) = 5y′(0) = -4
y(0.5) =
10.Differential equationsHard
A tank holds 1000 L of water in which 20 kg of salt is dissolved. Brine with 0.25 kg of salt per litre flows in at 5 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 85 min?
V = 1000 Lm0 = 20 kgc = 0.25 kg/Lr = 5 L/min
m = kg
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−3 −1 3; 1 −1 4; −4 −3 1]
det A =
12.Matrices & determinantsHard
Solve the system 6x − y = −1 and 4x − 2y = 4. Find x.
6x − y = −14x − 2y = 4
x =
13.Vectors: dot & cross productMedium
A constant force F = (30, −10, −30) N acts on an object while it moves in a straight line from point A(4, 2, 2) to point B(3, 1, −5). Coordinates are in metres. Find the work W done by the force.
F = (30, −10, −30) NA(4, 2, 2) mB(3, 1, −5) m
W = J
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (0, 2, −4) and b = (5, 5, 3) and has a positive z-component. Find its x-component n̂x.
a = (0, 2, −4)b = (5, 5, 3)n̂z > 0
n̂x =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3p(x): 0.15, 0.35, 0.15, 0.35
Var(X) =
16.ProbabilityHard
X has the distribution shown, and Y = −2X + 10. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 38, 37, 38, 28, 34, 56, 50.
n = 7
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 12. Find the z-score z of a score of 146.
μ = 170σ = 12x = 146
z =
19.Normal distributionMedium
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 9 of them is below 227.3 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 9< 227.3 kg
P =
20.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 97.5% of them lie (the 97.5th percentile).
μ = 500σ = 40P = 97.5%
x = h
3Solutions
1.Limits
Answer: L = −1.5
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) = −3x/(√(x2 − 3x) + x)
Divide by x
f(x) = −3/(√(1 + (−3)/x) + 1)
Let x → ∞
L = −3/(1 + 1) = −1.5
Answer
L = −1.5
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′(−2) = −0.75
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) = 6x/(3x2 + 4)
Substitute
f′(−2) = −12/16 = −0.75
Answer
f′(−2) = −0.75
4.Derivatives
Answer: b = −12
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) = 33
f′(x) = 4x + 3 → f′(3) = 15
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 33 − 15·3 = −12
Answer
b = −12
5.Integrals
Answer: V = 2.094
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 0.25x² dx
Integrate
V = π·0.25·2³/3 = 2.094
Answer
V = 2.094
Check: a cone of radius 1 and height 2 has (1/3)πr²h = 2.094.
6.Integrals
Answer: I = 3.602
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
4.482·(1.25) − 1·2 = 3.602
Answer
I = 3.602
7.Complex numbers
Answer: Re(w) = 0.2179
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.
Modulus and argument of w
|w| = 5 / 2 = 2.5
θ = 150° − 65° = 85°
Rectangular form
Re(w) = 2.5·cos(85°) = 0.2179
Im(w) = 2.5·sin(85°) = 2.49
Re(w) = 0.2179
Answer
Re(w) = 0.2179
8.Complex numbers
Answer: Re(z₁) = −6
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
361/2 = 6
Argument of the root
θ₁ = (0° + 360°·1)/2 = 180°
Rectangular form
z₁ = 6·(cos 180° + j sin 180°) = −6
Re(z₁) = −6
Answer
Re(z₁) = −6
The 2 roots lie on the circle of radius 6, 180° apart.
9.Differential equations
Answer: y(0.5) = 2.708
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.
Characteristic roots
(r + 2)(r + 3) = 0 → r = −2, −3
Constants from the initial values
A + B = 5, −2A − 3B = -4
A = 11, B = −6
Evaluate
y(0.5) = 11·e−1 + (−6)·e−1.5 = 2.708
Answer
y(0.5) = 2.708
10.Differential equations
Answer: m = 99.63 kg
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.
Set up the equation
dm/dt = 5·0.25 − (5/1000)·m = 1.25 − 0.005·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 250 + (−230)·e−t/200
Substitute t
m(85) = 250 + (−230)·0.6538 = 99.63 kg
Answer
m = 99.63 kg
11.Matrices & determinants
Answer: det A = −37
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 = 11, M₁₂ = 17, M₁₃ = −7
Cofactor expansion
det A = (−3)·11 − (−1)·17 + 3·(−7) = −37
Answer
det A = −37
12.Matrices & determinants
Answer: x = −0.75
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·(−2) − (−1)·4 = −8
Replace the x column
det Aₓ = (−1)·(−2) − (−1)·4 = 6
x = 6/(−8) = −0.75
Answer
x = −0.75
y = −3.5
13.Vectors: dot & cross product
Answer: W = 190 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 = (3, 1, −5) − (4, 2, 2) = (−1, −1, −7) m
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 = 2·3 − (−4)·5 = 26
(a × b)y = azbx − axbz = (−4)·5 − 0·3 = −20
(a × b)z = axby − aybx = 0·5 − 2·5 = −10
Normalise
The z-component of a × b is negative, so use −(a × b).
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 = 0·0.15 + 1·0.35 + 2·0.15 + 3·0.35 = 1.7
Mean of the squares
E(X²) = 0·0.15 + 1·0.35 + 4·0.15 + 9·0.35 = 4.1
Variance
Var(X) = 4.1 − 1.7² = 1.21
Answer
Var(X) = 1.21
16.Probability
Answer: σY = 2.683
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.
Spread of X
μ = 2, Var(X) = 1.8, σX = 1.342
Linear change
σY = |−2|·1.342 = 2.683
Answer
σY = 2.683
The mean does move: E(Y) = −2·2 + 10 = 6.
17.Descriptive statistics
Answer: s = 9.599
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 = (146 − 170)/12 = −2
Answer
z = −2
19.Normal distribution
Answer: P = 0.937
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 = 225 kg, σT = 0.5·√9 = 1.5 kg
Standardise
z = (227.3 − 225)/1.5 = 1.53
Probability
P = Φ(1.53) = 0.937
Answer
P = 0.937
20.Normal distribution
Answer: x = 578.4 h
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.