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→∞ (√(9x2 + 7x) − 3x).
f(x) = √(9x2 + 7x) − 3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + x)/sin(5x), with x in radians.
f(x) = ln(1 + x)/sin(5x)x → 0
L =
3.DerivativesMedium
Let f(x) = x2·e2x. Find f′(0.5).
f(x) = x2·e2xx = 0.5
f′(0.5) =
4.DerivativesHard
Let f(x) = x3 + 1.5x2 − 6x + 6. Find the x-coordinate x of the local maximum.
f(x) = x3 + 1.5x2 − 6x + 6
x =
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
Find the area A of the region enclosed by y = 0.5x2 and y = 4x.
y = 0.5x2y = 4x
A =
7.Complex numbersMedium
Write z = 6∠−130° in rectangular form a + jb and give the imaginary part b.
r = 6θ = −130°
b =
8.Complex numbersHard
Exactly one of the points A–D is a root of z³ = 27∠0°. Click it.
w = 27∠0°n = 3
1A (3∠60°)
2B (3∠−60°)
3C (5.4∠−120°)
4D (3∠−120°)
9.Differential equationsMedium
Solve y″ + 5y′ + 6y = 0 with y(0) = 2 and y′(0) = 0. Find y(0.5).
y″ + 5y′ + 6y = 0y(0) = 2y′(0) = 0
y(0.5) =
10.Differential equationsHard
A tank holds 200 L of water in which 5 kg of salt is dissolved. Brine with 0.1 kg of salt per litre flows in at 4 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 25 min?
V = 200 Lm0 = 5 kgc = 0.1 kg/Lr = 4 L/min
m = kg
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−4 1 1; 4 5 −5; −1 −1 −5]
det A =
12.Matrices & determinantsHard
Solve the system 5x + 5y = −3 and −2x − 6y = 4. Find x.
5x + 5y = −3−2x − 6y = 4
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (1, 3, −1) and b = (−4, −3, k) are perpendicular.
a = (1, 3, −1)b = (−4, −3, k)
k =
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(−2, −3, 2), B(−3, 2, −1) and C(−3, 1, 4). Coordinates are in metres.
A(−2, −3, 2)B(−3, 2, −1)C(−3, 1, 4)
A = m²
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2, 3, 4p(x): 0.35, 0.05, 0.15, 0.2, 0.25
σ =
16.ProbabilityHard
A test has 8 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 4 correct answers?
n = 8p = 0.4k = 4
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 25, 17, 24, 24, 44.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is above 74.8?
μ = 70σ = 8x = 74.8
P =
19.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 50.5 and 80.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 50.5x2 = 80.5
P =
20.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 25% is to the left of it. Find the cut-off x.
μ = 178 cmσ = 7 cm25%
x = cm
3Solutions
1.Limits
Answer: L = 1.167
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).
Multiply by the conjugate
f(x) = 7x/(√(9x2 + 7x) + 3x)
Divide by x
f(x) = 7/(√(9 + 7/x) + 3)
Let x → ∞
L = 7/(3 + 3) = 1.167
Answer
L = 1.167
2.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).
L'Hôpital's rule
L = limx→0 [1/(1 + x)]/[5·cos 5x]
Substitute x = 0
L = 1/5 = 0.2
Answer
L = 0.2
3.Derivatives
Answer: f′(0.5) = 4.077
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·(2x + 2·x2)
Substitute
f′(0.5) = 2.718·(1 + 0.5) = 4.077
Answer
f′(0.5) = 4.077
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 + 3x − 6 = 3·(x2 + x − 2)
Solve f′(x) = 0
x = −2 or x = 1
Which one is which
f″(x) = 6x + 3: negative at a maximum, positive at a minimum.
f″(−2) = −9, f″(1) = 9
x = −2
Answer
x = −2
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: A = 42.67
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
0.5x2 = 4x → x = 0, x = 8
Integrate upper minus lower
A = [2x2 − 0.1667x3]08 = 42.67
Answer
A = 42.67
7.Complex numbers
Answer: b = −4.596
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 = 6·cos(−130°) = −3.857
b = 6·sin(−130°) = −4.596
Rectangular form
z = −3.857 − j4.596
b = −4.596
Answer
b = −4.596
8.Complex numbers
Answer: D (3∠−120°)
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 = 0°, 120°, −120°
The point
D = 3∠−120°
Answer
D (3∠−120°)
9.Differential equations
Answer: y(0.5) = 1.315
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 = 2, −2A − 3B = 0
A = 6, B = −4
Evaluate
y(0.5) = 6·e−1 + (−4)·e−1.5 = 1.315
Answer
y(0.5) = 1.315
10.Differential equations
Answer: m = 10.9 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 = 4·0.1 − (4/200)·m = 0.4 − 0.02·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 20 + (−15)·e−t/50
Substitute t
m(25) = 20 + (−15)·0.6065 = 10.9 kg
Answer
m = 10.9 kg
11.Matrices & determinants
Answer: det A = 146
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 = −30, M₁₂ = −25, M₁₃ = 1
Cofactor expansion
det A = (−4)·(−30) − 1·(−25) + 1·1 = 146
Answer
det A = 146
12.Matrices & determinants
Answer: x = 0.1
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 = 5·(−6) − 5·(−2) = −20
Replace the x column
det Aₓ = (−3)·(−6) − 5·4 = −2
x = −2/(−20) = 0.1
Answer
x = 0.1
y = −0.7
13.Vectors: dot & cross product
Answer: k = −13
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 = 1·(−4) + 3·(−3) + (−1)·k = 0
−13 + (−1)·k = 0
Solve for k
k = −(−13)/(−1) = −13
Answer
k = −13
14.Vectors: dot & cross product
Answer: A = 11.29 m²
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.
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.
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 = 0.6
Standard normal table
Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743
Answer
P = 0.2743
19.Normal distribution
Answer: P = 0.8659
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.
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.