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→∞ (−4x2 − 6x − 7)/(4x2 + 5x + 8).
f(x) = (−4x2 − 6x − 7)/(4x2 + 5x + 8)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 3x)/(4x²), with x in radians.
f(x) = (1 − cos 3x)/(4x²)x → 0
L =
3.DerivativesMedium
Let f(x) = (2x + 1)4. Find f′(0).
f(x) = (2x + 1)4x = 0
f′(0) =
4.DerivativesHard
The tangent line to f(x) = 3x2 − 2x − 5 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = 3x2 − 2x − 5x = 3
b =
5.IntegralsMedium
The region under y = √(2x) from x = 0 to x = 4 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(2x)0 ≤ x ≤ 4
V =
6.IntegralsHard
Evaluate I = ∫₀1 x²·eˣ dx.
∫₀1 x²eˣ dx
I =
7.Complex numbersMedium
The number z = 2·ej0.8 has its angle in radians. Find its imaginary part b.
r = 2θ = 0.8 rad
b =
8.Complex numbersHard
Find the imaginary part of (−1 − j3)³.
z = −1 − j3n = 3
Im(z³) =
9.Differential equationsMedium
A cup of coffee at 94 °C is left in a room at 23 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.1 min⁻¹, find its temperature T after 11 min.
T0 = 94 °CTa = 23 °Ck = 0.1 min⁻¹t = 11 min
T = °C
10.Differential equationsHard
A tank holds 100 L of water in which 10 kg of salt is dissolved. Brine with 0.2 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 10 min?
V = 100 Lm0 = 10 kgc = 0.2 kg/Lr = 4 L/min
m = kg
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [−4 3; −3 1]B = [2 −5; −3 −3]
c21 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [0 −1; 6 7]
λ =
13.Vectors: dot & cross productMedium
A constant force F = (5, −15, 10) N acts on an object while it moves in a straight line from point A(2, 3, 5) to point B(5, −4, 3). Coordinates are in metres. Find the work W done by the force.
F = (5, −15, 10) NA(2, 3, 5) mB(5, −4, 3) m
W = J
14.Vectors: dot & cross productHard
For a = (6, −2, −5) and b = (2, −4, 1), find the z-component (a × b)z of the cross product a × b.
a = (6, −2, −5)b = (2, −4, 1)
(a × b)z =
15.ProbabilityMedium
A fair die is rolled 5 times. What is the probability P of getting at least one six?
n = 5
P =
16.ProbabilityHard
X has the distribution shown, and Y = 3X + 1. Find the standard deviation σY of Y.
Find the sample standard deviation s of the data: 55, 24, 22, 31, 12, 24.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 100. What proportion P of values is above 300?
μ = 500σ = 100x = 300
P =
19.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9099. Find z to two decimals.
A = 0.9099
z =
20.Normal distributionHard
The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
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 + (−6)/x + (−7)/x2)/(4 + 5/x + 8/x2)
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 = 1.125
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
3·sin 3x/(8x) → 0/0
Second application
9·cos 3x/8
Substitute x = 0
L = 9/8 = 1.125
Answer
L = 1.125
3.Derivatives
Answer: f′(0) = 8
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
f′(x) = 4·(2x + 1)3·2
Substitute
f′(0) = 4·13·2 = 8
Answer
f′(0) = 8
4.Derivatives
Answer: b = −32
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) = 16
f′(x) = 6x − 2 → f′(3) = 16
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 16 − 16·3 = −32
Answer
b = −32
5.Integrals
Answer: V = 50.27
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∫₀4 2x dx
Integrate
V = π·2·4²/2 = 50.27
Answer
V = 50.27
6.Integrals
Answer: I = 0.7183
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
2.718·(1) − 1·2 = 0.7183
Answer
I = 0.7183
7.Complex numbers
Answer: b = 1.435
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.
Euler's formula
z = 2·(cos 0.8 + j sin 0.8)
Components (radians)
a = 2·cos 0.8 = 1.393
b = 2·sin 0.8 = 1.435
b = 1.435
Answer
b = 1.435
8.Complex numbers
Answer: Im(z³) = 18
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.
Powers step by step
z² = (−1 − j3)(−1 − j3) = −8 + j6
z³ = (−8 + j6)(−1 − j3) = 26 + j18
Check with De Moivre
z = 3.162∠−108.4° → z³ = 31.62∠34.7°
Im(z³) = 18
Answer
Im(z³) = 18
9.Differential equations
Answer: T = 46.63 °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) = 23 + 71·e−0.1t
Substitute t
T(11) = 23 + 71·0.3329 = 46.63 °C
Answer
T = 46.63 °C
10.Differential equations
Answer: m = 13.3 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 = 4·0.2 − (4/100)·m = 0.8 − 0.04·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 20 + (−10)·e−t/25
Substitute t
m(10) = 20 + (−10)·0.6703 = 13.3 kg
Answer
m = 13.3 kg
11.Matrices & determinants
Answer: c21 = −9
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 = (−3)·2 + 1·(−3) = −9
Answer
c21 = −9
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 − (7)λ + (6) = 0
Solve
λ = (7 ± √(25))/2 → λ = 1, 6
λ = 6
Answer
λ = 6
13.Vectors: dot & cross product
Answer: W = 100 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 = (5, −4, 3) − (2, 3, 5) = (3, −7, −2) m
Work
W = F·d = 5·3 + (−15)·(−7) + 10·(−2) = 100 J
Answer
W = 100 J
14.Vectors: dot & cross product
Answer: (a × b)z = −20
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, component by component
(a × b)x = aybz − azby = (−2)·1 − (−5)·(−4) = −22
(a × b)y = azbx − axbz = (−5)·2 − 6·1 = −16
(a × b)z = axby − aybx = 6·(−4) − (−2)·2 = −20
Requested component
(a × b)z = −20
Answer
(a × b)z = −20
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: P = 0.5981
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)5 = 0.4019
P = 1 − 0.4019 = 0.5981
Answer
P = 0.5981
16.Probability
Answer: σY = 36.12
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 = 25, Var(X) = 145, σX = 12.04
Linear change
σY = |3|·12.04 = 36.12
Answer
σY = 36.12
The mean does move: E(Y) = 3·25 + 1 = 76.
17.Descriptive statistics
Answer: s = 14.57
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 = −2
Standard normal table
Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772
Answer
P = 0.9772
19.Normal distribution
Answer: z = −1.34
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 − 0.9099 = 0.0901
Read the z-table backwards
Find 0.0901 inside the table and read off z.
z = −1.34
Answer
z = −1.34
20.Normal distribution
Answer: P = 0.7389
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).