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 3x)/(2x²), with x in radians.
f(x) = (1 − cos 3x)/(2x²)x → 0
L =
3.DerivativesMedium
Let f(x) = (4x + 1)5. Find f′(1).
f(x) = (4x + 1)5x = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 + 1.5x2 + 2. Find the x-coordinate x of the local maximum.
f(x) = x3 + 1.5x2 + 2
x =
5.IntegralsMedium
Evaluate I = ∫₁6 x·ln x dx.
∫₁6 x ln x dx
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 = −5 + j6 in degrees, as the principal value between −180° and 180°.
z = −5 + j6
θ = °
8.Complex numbersHard
The roots of zⁿ = w with n = 6 and w = 64∠135° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, …, 5. Find the argument of z₅ in degrees, between −180° and 180°.
w = 64∠135°n = 6k = 5
θ₅ = °
9.Differential equationsMedium
Solve y″ + 8y′ + 15y = 0 with y(0) = 4 and y′(0) = 6. Find y(2).
y″ + 8y′ + 15y = 0y(0) = 4y′(0) = 6
y(2) =
10.Differential equationsHard
A tank holds 1000 L of water in which 5 kg of salt is dissolved. Brine with 0.1 kg of salt per litre flows in at 2 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 95 min?
V = 1000 Lm0 = 5 kgc = 0.1 kg/Lr = 2 L/min
m = kg
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [2 −6; −4 −2]B = [−6 −3; 1 −4]
c21 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [2 6; 1 −3]
λ =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (0, −6, 6) and b = (−7, −1, 0).
a = (0, −6, 6)b = (−7, −1, 0)
a·b =
14.Vectors: dot & cross productHard
For a = (3, 4, −2) and b = (−2, −2, −2), find the y-component (a × b)y of the cross product a × b.
a = (3, 4, −2)b = (−2, −2, −2)
(a × b)y =
15.ProbabilityMedium
A fair die is rolled 4 times. What is the probability P of getting at least one six?
n = 4
P =
16.ProbabilityHard
A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 2 correct answers?
n = 7p = 0.3k = 2
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 42, 23, 27, 21, 53, 58, 41.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 8. What proportion P of values is below 174.8?
μ = 170σ = 8x = 174.8
P =
19.Normal distributionMedium
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 5 of them is below 126.1 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 5< 126.1 kg
P =
20.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 90% of them lie (the 90th percentile).
μ = 178σ = 7P = 90%
x = cm
3Solutions
1.Limits
Answer: L = 0.3679
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 = e−1 = 0.3679
Answer
L = 0.3679
2.Limits
Answer: L = 2.25
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/(4x) → 0/0
Second application
9·cos 3x/4
Substitute x = 0
L = 9/4 = 2.25
Answer
L = 2.25
3.Derivatives
Answer: f′(1) = 12500
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) = 5·(4x + 1)4·4
Substitute
f′(1) = 5·54·4 = 12500
Answer
f′(1) = 12500
4.Derivatives
Answer: x = −1
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 = 3·(x2 + x)
Solve f′(x) = 0
x = −1 or x = 0
Which one is which
f″(x) = 6x + 3: negative at a maximum, positive at a minimum.
f″(−1) = −3, f″(0) = 3
x = −1
Answer
x = −1
5.Integrals
Answer: I = 23.5
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.
Choose u and dv
u = ln x, du = dx/x; dv = x dx, v = x²/2
Antiderivative
∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4
Limits
(18·ln 6 − 9) − (0 − 0.25) = 23.5
Answer
I = 23.5
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 = 129.8 °
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 = −5, b = 6 → II
Reference angle
α = atan(6/5) = 50.19°
Argument
θ = 180° − α = 129.8°
Answer
θ = 129.8°
In polar form z = 7.81∠129.8°.
8.Complex numbers
Answer: θ₅ = −37.5 °
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 of the roots
641/6 = 2
Argument of the root
θ₅ = (135° + 360°·5)/6 = 322.5° → −37.5°
Argument
θ₅ = −37.5°
Answer
θ₅ = −37.5°
The 6 roots lie on the circle of radius 2, 60° apart.
9.Differential equations
Answer: y(2) = 0.03182
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 + 3)(r + 5) = 0 → r = −3, −5
Constants from the initial values
A + B = 4, −3A − 5B = 6
A = 13, B = −9
Evaluate
y(2) = 13·e−6 + (−9)·e−10 = 0.03182
Answer
y(2) = 0.03182
10.Differential equations
Answer: m = 21.44 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 = 2·0.1 − (2/1000)·m = 0.2 − 0.002·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 100 + (−95)·e−t/500
Substitute t
m(95) = 100 + (−95)·0.827 = 21.44 kg
Answer
m = 21.44 kg
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 = (−4)·(−6) + (−2)·1 = 22
Answer
c21 = 22
12.Matrices & determinants
Answer: λ = 3
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 − (−1)λ + (−12) = 0
Solve
λ = (−1 ± √(49))/2 → λ = −4, 3
λ = 3
Answer
λ = 3
13.Vectors: dot & cross product
Answer: a·b = 6
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.
Multiply matching components and add
a·b = 0·(−7) + (−6)·(−1) + 6·0 = 0 + 6 + 0 = 6
Answer
a·b = 6
Positive: the angle between the vectors is less than 90°.
14.Vectors: dot & cross product
Answer: (a × b)y = 10
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 = 4·(−2) − (−2)·(−2) = −12
(a × b)y = azbx − axbz = (−2)·(−2) − 3·(−2) = 10
(a × b)z = axby − aybx = 3·(−2) − 4·(−2) = 2
Requested component
(a × b)y = 10
Answer
(a × b)y = 10
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: P = 0.5177
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)4 = 0.4823
P = 1 − 0.4823 = 0.5177
Answer
P = 0.5177
16.Probability
Answer: P = 0.3177
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.
Binomial formula
P(X = 2) = 21·0.32·0.75 = 21·0.09·0.1681 = 0.3177
Answer
P = 0.3177
17.Descriptive statistics
Answer: s = 14.63
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 = 0.7257
Answer
P = 0.7257
19.Normal distribution
Answer: P = 0.8365
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).
Distribution of the total
μT = 125 kg, σT = 0.5·√5 = 1.118 kg
Standardise
z = (126.1 − 125)/1.118 = 0.98
Probability
P = Φ(0.98) = 0.8365
Answer
P = 0.8365
20.Normal distribution
Answer: x = 187 cm
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.