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→∞ (−4x3 − x + 5)/(4x3 + 3x + 3).
f(x) = (−4x3 − x + 5)/(4x3 + 3x + 3)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e3x − 1 − 3x)/x².
f(x) = (e3x − 1 − 3x)/x²x → 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 − 9x2 + 24x − 4. Find the x-coordinate x of the local maximum.
f(x) = x3 − 9x2 + 24x − 4
x =
5.IntegralsMedium
Evaluate I = ∫13 5/x dx.
f(x) = 5/x[1, 3]
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 1 − x² and the x-axis.
y = 1 − x²
A =
7.Complex numbersMedium
Write z = 9 − j8 in exponential form r·ejθ. Give θ in radians, between −π and π.
z = 9 − j8
θ = rad
8.Complex numbersHard
Exactly one of the points A–D is a root of z⁶ = 729∠180°. Click it.
w = 729∠180°n = 6
1A (3∠90°)
2B (3∠120°)
3C (3∠180°)
4D (1.5∠90°)
9.Differential equationsMedium
A cup of coffee at 85 °C is left in a room at 22 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.05 min⁻¹, find its temperature T after 22 min.
T0 = 85 °CTa = 22 °Ck = 0.05 min⁻¹t = 22 min
T = °C
10.Differential equationsHard
A tank holds 400 L of water in which 5 kg of salt is dissolved. Brine with 0.2 kg of salt per litre flows in at 5 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 50 kg of salt?
V = 400 Lm0 = 5 kgc = 0.2 kg/Lr = 5 L/min
t = min
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).
A = [6 2; −6 −5]B = [5 −4; 4 4]
c12 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [3 3; 3 3]
λ =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (−1, −6, −7) and b = (2, −6, −5).
a = (−1, −6, −7)b = (2, −6, −5)
a·b =
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(3, 1, 2), B(−2, 0, −3) and C(−1, 3, −1). Coordinates are in metres.
A(3, 1, 2)B(−2, 0, −3)C(−1, 3, −1)
A = m²
15.ProbabilityMedium
A fair die is rolled 2 times. What is the probability P of getting at least one six?
n = 2
P =
16.ProbabilityHard
A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 2 correct answers?
n = 12p = 0.6k = 2
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 19, 60, 11, 33, 38.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 8. What proportion P of values is below 162?
μ = 170σ = 8x = 162
P =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.7995. Find z to two decimals.
A = 0.7995
z =
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 top 40% is to the right of it. Find the cut-off x.
μ = 178 cmσ = 7 cm40%
x = cm
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).
Divide by the highest power
(−4 + (−1)/x2 + 5/x3)/(4 + 3/x2 + 3/x3)
Let x → ∞
L = −4/4 = −1
Answer
L = −1
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 4.5
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
(3·e3x − 3)/(2x) → 0/0
Second application
9·e3x/2
Substitute x = 0
L = 9·1/2 = 4.5
Answer
L = 4.5
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 − 18x + 24 = 3·(x2 − 6x + 8)
Solve f′(x) = 0
x = 2 or x = 4
Which one is which
f″(x) = 6x − 18: negative at a maximum, positive at a minimum.
f″(2) = −6, f″(4) = 6
x = 2
Answer
x = 2
5.Integrals
Answer: I = 5.493
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) = 5·ln x
Evaluate at the limits
I = 5·(ln 3 − ln 1) = 5·ln 3 = 5.493
Answer
I = 5.493
6.Integrals
Answer: A = 1.333
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.
Zeros
1 − x² = 0 → x = ±1
Integrate
A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333
Answer
A = 1.333
7.Complex numbers
Answer: θ = −0.7266 rad
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
r = √(81 + 64) = 12.04
Argument in radians
θ = −41.63° = −41.63·π/180 = −0.7266 rad
Exponential form
z = 12.04·e−j0.7266
Answer
θ = −0.7266 rad
8.Complex numbers
Answer: A (3∠90°)
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 = 30°, 90°, 150°, −150°, −90°, −30°
The point
A = 3∠90°
Answer
A (3∠90°)
9.Differential equations
Answer: T = 42.97 °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) = 22 + 63·e−0.05t
Substitute t
T(22) = 22 + 63·0.3329 = 42.97 °C
Answer
T = 42.97 °C
10.Differential equations
Answer: t = 73.3 min
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.
Result
m(t) = 80 + (−75)·e−t/80
Solve for t
e−t/80 = (80 − 50)/(80 − 5) = 0.4
t = 80·ln(2.5) = 73.3 min
Answer
t = 73.3 min
11.Matrices & determinants
Answer: c12 = −16
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
c12 = 6·(−4) + 2·4 = −16
Answer
c12 = −16
12.Matrices & determinants
Answer: λ = 6
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 − (6)λ + (0) = 0
Solve
λ = (6 ± √(36))/2 → λ = 0, 6
λ = 6
Answer
λ = 6
13.Vectors: dot & cross product
Answer: a·b = 69
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.
Positive: the angle between the vectors is less than 90°.
14.Vectors: dot & cross product
Answer: A = 9.874 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.
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.
Complement
P(no six) = (5/6)2 = 0.6944
P = 1 − 0.6944 = 0.3056
Answer
P = 0.3056
16.Probability
Answer: P = 0.002491
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
Standard normal table
Φ(−1) = 0.1587
P = 0.1587
Answer
P = 0.1587
19.Normal distribution
Answer: z = 1.28
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 + 0.7995)/2 = 0.8997
Read the z-table backwards
Find 0.8997 inside the table and read off z.
z = 1.28
Answer
z = 1.28
20.Normal distribution
Answer: x = 179.8 cm
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.