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→∞ (1 + 1/x)x. Give a decimal value.
f(x) = (1 + 1/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 2x)/(4x²), with x in radians.
f(x) = (1 − cos 2x)/(4x²)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(3x2 + 5). Find f′(−1).
f(x) = ln(3x2 + 5)x = −1
f′(−1) =
4.DerivativesHard
Let f(x) = x3 − 4.5x2 + 6x. Find the x-coordinate x of the local minimum.
f(x) = x3 − 4.5x2 + 6x
x =
5.IntegralsMedium
Evaluate I = ∫210 2/x dx.
f(x) = 2/x[2, 10]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = 4x.
y = x2y = 4x
A =
7.Complex numbersMedium
Given z₁ = 5∠−35° and z₂ = 4∠−50°, find the argument θ of w = z₁/z₂ in degrees, between −180° and 180°.
z₁ = 5∠−35°z₂ = 4∠−50°
θ = °
8.Complex numbersHard
One root of z³ + 5z² + 4z − 10 = 0 is z₁ = −3 + j. The coefficients are real. Find the real root r.
z₁ = −3 + j
r =
9.Differential equationsMedium
A forged steel part at 480 °C is left to cool in a workshop at 24 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.02 min⁻¹, find its temperature T after 18 min.
T0 = 480 °CTa = 24 °Ck = 0.02 min⁻¹t = 18 min
T = °C
10.Differential equationsHard
Solve y′ + 3y = 10e−t with y(0) = 3. Find y(0.5).
y′ + 3y = 10e−ty(0) = 3
y(0.5) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [2 −5; 6 5]B = [2 −3; 2 −1]
c21 =
12.Matrices & determinantsHard
Solve the system 3x − 3y = 8 and −6x − 6y = −10. Find x.
3x − 3y = 8−6x − 6y = −10
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (3, −1, −4) and b = (k, 4, 4) are perpendicular.
a = (3, −1, −4)b = (k, 4, 4)
k =
14.Vectors: dot & cross productHard
A force F = (0, −60, −50) N acts at point A(−1, 1, −3). Coordinates are in metres. Find the y-component My of the moment of the force about point O(−3, 3, −3).
F = (0, −60, −50) NA(−1, 1, −3) mO(−3, 3, −3) m
My = N·m
15.ProbabilityMedium
A bag holds 7 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
7 / 5
P =
16.ProbabilityHard
X has the distribution shown, and Y = 5X + 10. Find the standard deviation σY of Y.
x: 0, 1, 2p(x): 0.35, 0.35, 0.3Y = 5X + 10
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 45, 27, 13, 20, 43.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 60 and standard deviation 20. Find the z-score z of a score of 40.
μ = 60σ = 20x = 40
z =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.6579. Find z to two decimals.
A = 0.6579
z =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 99% of them lie (the 99th percentile).
μ = 65σ = 10P = 99%
x = points
3Solutions
1.Limits
Answer: L = 2.718
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).
Take the logarithm
ln L = limx→∞ x·ln(1 + 1/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·1 = 1
Undo the logarithm
L = e1 = 2.718
Answer
L = 2.718
2.Limits
Answer: L = 0.5
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).
First application
2·sin 2x/(8x) → 0/0
Second application
4·cos 2x/8
Substitute x = 0
L = 4/8 = 0.5
Answer
L = 0.5
3.Derivatives
Answer: f′(−1) = −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 + 5)
Substitute
f′(−1) = −6/8 = −0.75
Answer
f′(−1) = −0.75
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 − 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 = 2
Answer
x = 2
5.Integrals
Answer: I = 3.219
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.
Antiderivative
F(x) = 2·ln x
Evaluate at the limits
I = 2·(ln 10 − ln 2) = 2·ln 5 = 3.219
Answer
I = 3.219
6.Integrals
Answer: A = 10.67
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
x2 = 4x → x = 0, x = 4
Integrate upper minus lower
A = [2x2 − 0.3333x3]04 = 10.67
Answer
A = 10.67
7.Complex numbers
Answer: θ = 15 °
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 / 4 = 1.25
θ = −35° − (−50)° = 15°
Argument
θ = 15°
Answer
θ = 15°
8.Complex numbers
Answer: r = 1
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.
Check with the sum of the roots: z₁ + z₂ + r = −6 + 1 = −5 = −(5).
9.Differential equations
Answer: T = 342.1 °C
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.
Solution of the cooling law
T(t) = 24 + 456·e−0.02t
Substitute t
T(18) = 24 + 456·0.6977 = 342.1 °C
Answer
T = 342.1 °C
10.Differential equations
Answer: y(0.5) = 2.586
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 = e3t
(e3ty)′ = 10e2t
General solution
y = 10/2·e−t + C·e−3t = 5·e−t + C·e−3t
Initial condition
3 = 5 + C → C = −2
Evaluate
y(0.5) = 5·0.6065 + (−2)·0.2231 = 2.586
Answer
y(0.5) = 2.586
11.Matrices & determinants
Answer: c21 = 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.
Row times column
c21 = 6·2 + 5·2 = 22
Answer
c21 = 22
12.Matrices & determinants
Answer: x = 2.167
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·(−6) − (−3)·(−6) = −36
Replace the x column
det Aₓ = 8·(−6) − (−3)·(−10) = −78
x = −78/(−36) = 2.167
Answer
x = 2.167
y = −0.5
13.Vectors: dot & cross product
Answer: k = 6.667
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 = 3·k + (−1)·4 + (−4)·4 = 0
−20 + 3·k = 0
Solve for k
k = −(−20)/3 = 6.667
Answer
k = 6.667
14.Vectors: dot & cross product
Answer: My = 100 N·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.
Position vector from O to A
r = A − O = (−1, 1, −3) − (−3, 3, −3) = (2, −2, 0) m
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.
First and second draw
P = 7/12 · 6/11 = 42/132 = 0.3182
Answer
P = 0.3182
16.Probability
Answer: σY = 4.023
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 = 0.95, Var(X) = 0.6475, σX = 0.8047
Linear change
σY = |5|·0.8047 = 4.023
Answer
σY = 4.023
The mean does move: E(Y) = 5·0.95 + 10 = 14.75.
17.Descriptive statistics
Answer: s = 14.06
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 = (40 − 60)/20 = −1
Answer
z = −1
19.Normal distribution
Answer: z = 0.95
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.
Area to the left of z
Φ(z) = (1 + 0.6579)/2 = 0.8289
Read the z-table backwards
Find 0.8289 inside the table and read off z.
z = 0.95
Answer
z = 0.95
20.Normal distribution
Answer: x = 88.26 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.