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 + 0.5/x)2x. Give a decimal value.
f(x) = (1 + 0.5/x)2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 5x)/sin(3x), with x in radians.
f(x) = ln(1 + 5x)/sin(3x)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(x2 + 4). Find f′(3).
f(x) = ln(x2 + 4)x = 3
f′(3) =
4.DerivativesHard
The tangent line to f(x) = 2x2 + 5x + 4 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = 2x2 + 5x + 4x = 1
b =
5.IntegralsMedium
Evaluate I = ∫02 2x·(x² + 1)2 dx.
f(x) = 2x·(x² + 1)2[0, 2]
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 9 − x² and the x-axis.
y = 9 − x²
A =
7.Complex numbersMedium
Find the argument θ of z = −2 − j4 in degrees, as the principal value between −180° and 180°.
z = −2 − j4
θ = °
8.Complex numbersHard
Find the real part of (3 + j3)⁵.
z = 3 + j3n = 5
Re(z⁵) =
9.Differential equationsMedium
A can of soda at 6 °C is taken out of the fridge into a room at 23 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.06 min⁻¹, find its temperature T after 5 min.
T0 = 6 °CTa = 23 °Ck = 0.06 min⁻¹t = 5 min
T = °C
10.Differential equationsHard
A tank holds 400 L of pure water. Brine with 0.1 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 70 min?
V = 400 Lm0 = 0 kgc = 0.1 kg/Lr = 5 L/min
m = kg
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [0 −1 −2; 0 −5 4; 1 −1 1]
det A =
12.Matrices & determinantsHard
Solve the system 6x − 4y = −1 and −4x − 5y = −9. Find x.
6x − 4y = −1−4x − 5y = −9
x =
13.Vectors: dot & cross productMedium
A constant force F = (−25, −25, 0) N acts on an object while it moves in a straight line from point A(−4, 1, −3) to point B(−5, 4, −5). Coordinates are in metres. Find the work W done by the force.
F = (−25, −25, 0) NA(−4, 1, −3) mB(−5, 4, −5) m
W = J
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (−2, −2, −1), b = (1, −5, 3) and c = (2, 1, 5).
a = (−2, −2, −1)b = (1, −5, 3)c = (2, 1, 5)
V =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4p(x): 0.2, 0.1, 0.35, 0.35
Var(X) =
16.ProbabilityHard
5 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?
n = 5p = 0.25
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 23, 43, 45, 32, 22.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is below 140?
μ = 100σ = 20x = 140
P =
19.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.6293. Find z to two decimals.
A = 0.6293
z =
20.Normal distributionHard
Hole diameters D are normal with mean 25.05 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.05 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
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→∞ 2·x·ln(1 + 0.5/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 2·0.5 = 1
Undo the logarithm
L = e1 = 2.718
Answer
L = 2.718
2.Limits
Answer: L = 1.667
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 [5/(1 + 5x)]/[3·cos 3x]
Substitute x = 0
L = 5/3 = 1.667
Answer
L = 1.667
3.Derivatives
Answer: f′(3) = 0.4615
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) = 2x/(x2 + 4)
Substitute
f′(3) = 6/13 = 0.4615
Answer
f′(3) = 0.4615
4.Derivatives
Answer: b = 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.
Point and slope
f(1) = 11
f′(x) = 4x + 5 → f′(1) = 9
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 11 − 9·1 = 2
Answer
b = 2
5.Integrals
Answer: I = 41.33
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.
Substitution
u = x² + 1, du = 2x dx
I = ∫15 u2 du
Integrate and evaluate
I = [u3/3]15 = (125 − 1)/3 = 41.33
Answer
I = 41.33
6.Integrals
Answer: A = 36
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
9 − x² = 0 → x = ±3
Integrate
A = [9x − x³/3]−33 = 18 − (−18) = 36
Answer
A = 36
7.Complex numbers
Answer: θ = −116.6 °
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.
Quadrant
a = −2, b = −4 → III
Reference angle
α = atan(4/2) = 63.43°
Argument
θ = −(180° − α) = −116.6°
Answer
θ = −116.6°
In polar form z = 4.472∠−116.6°.
8.Complex numbers
Answer: Re(z⁵) = −972
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 = (3 + j3)(3 + j3) = j18
z³ = (j18)(3 + j3) = −54 + j54
z⁴ = (−54 + j54)(3 + j3) = −324
z⁵ = (−324)(3 + j3) = −972 − j972
Check with De Moivre
z = 4.243∠45° → z⁵ = 1375∠−135°
Re(z⁵) = −972
Answer
Re(z⁵) = −972
9.Differential equations
Answer: T = 10.41 °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 + (−17)·e−0.06t
Substitute t
T(5) = 23 + (−17)·0.7408 = 10.41 °C
Answer
T = 10.41 °C
10.Differential equations
Answer: m = 23.33 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 = 5·0.1 − (5/400)·m = 0.5 − 0.0125·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 40 + (−40)·e−t/80
Substitute t
m(70) = 40 + (−40)·0.4169 = 23.33 kg
Answer
m = 23.33 kg
11.Matrices & determinants
Answer: det A = −14
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 = −1, M₁₂ = −4, M₁₃ = 5
Cofactor expansion
det A = 0·(−1) − (−1)·(−4) + (−2)·5 = −14
Answer
det A = −14
12.Matrices & determinants
Answer: x = 0.6739
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·(−5) − (−4)·(−4) = −46
Replace the x column
det Aₓ = (−1)·(−5) − (−4)·(−9) = −31
x = −31/(−46) = 0.6739
Answer
x = 0.6739
y = 1.261
13.Vectors: dot & cross product
Answer: W = −50 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, −5) − (−4, 1, −3) = (−1, 3, −2) m
Work
W = F·d = (−25)·(−1) + (−25)·3 + 0·(−2) = −50 J
Answer
W = −50 J
Negative work: the force acts mostly against the motion.
14.Vectors: dot & cross product
Answer: V = 43
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 b × c
(b × c)x = bycz − bzcy = (−5)·5 − 3·1 = −28
(b × c)y = bzcx − bxcz = 3·2 − 1·5 = 1
(b × c)z = bxcy − bycx = 1·1 − (−5)·2 = 11
Dot product with a
a·(b × c) = (−2)·(−28) + (−2)·1 + (−1)·11 = 43
V = |43| = 43
Answer
V = 43
15.Probability
Answer: Var(X) = 1.228
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 = 1·0.2 + 2·0.1 + 3·0.35 + 4·0.35 = 2.85
Mean of the squares
E(X²) = 1·0.2 + 4·0.1 + 9·0.35 + 16·0.35 = 9.35
Variance
Var(X) = 9.35 − 2.85² = 1.228
Answer
Var(X) = 1.228
16.Probability
Answer: P = 0.6328
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.
Zero and one failure
P(X = 0) = 0.755 = 0.2373
P(X = 1) = 5·0.25·0.754 = 0.3955
Add them
P = 0.2373 + 0.3955 = 0.6328
Answer
P = 0.6328
17.Descriptive statistics
Answer: s = 10.79
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) = 0.9772
P = 0.9772
Answer
P = 0.9772
19.Normal distribution
Answer: z = −0.33
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 − 0.6293 = 0.3707
Read the z-table backwards
Find 0.3707 inside the table and read off z.
z = −0.33
Answer
z = −0.33
20.Normal distribution
Answer: P = 0.9616
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).