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→∞ (−3x2 + 4x + 4)/(3x2 + 6x + 1).
f(x) = (−3x2 + 4x + 4)/(3x2 + 6x + 1)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 2x)/sin(5x), with x in radians.
f(x) = ln(1 + 2x)/sin(5x)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(2x), with x in radians. Find f′(1).
f(x) = x·sin(2x)x = 1 rad
f′(1) =
4.DerivativesHard
Let f(x) = x3 − 10.5x2 + 36x + 2. Find the x-coordinate x of the local maximum.
f(x) = x3 − 10.5x2 + 36x + 2
x =
5.IntegralsMedium
Evaluate I = ∫₁1.5 x·ln x dx.
∫₁1.5 x ln x dx
I =
6.IntegralsHard
The region between the line y = 4x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.
y = 4xy = 3x²
V =
7.Complex numbersMedium
Given z₁ = 5∠150° and z₂ = 1.5∠80°, find the argument θ of w = z₁/z₂ in degrees, between −180° and 180°.
z₁ = 5∠150°z₂ = 1.5∠80°
θ = °
8.Complex numbersHard
The roots of zⁿ = w with n = 5 and w = 32∠0° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, …, 4. Find the imaginary part of z₁.
w = 32∠0°n = 5k = 1
Im(z₁) =
9.Differential equationsMedium
A forged steel part at 500 °C is left to cool in a workshop at 30 °C. After 40 min its temperature is 172 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 500 °CTa = 30 °CT(40) = 172 °C
k = min⁻¹
10.Differential equationsHard
A system obeys y″ + 3y′ + 11.25y = 112.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 3y′ + 11.25y = 112.5y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−4 0 3; −2 −2 1; 2 −1 −3]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [4 4; 3 0]
λ =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (5, −4, −2) and b = (6, −4, 4).
a = (5, −4, −2)b = (6, −4, 4)
a·b =
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(2, −1, −3), B(0, 2, −3) and C(−2, −4, −4). Coordinates are in metres.
A(2, −1, −3)B(0, 2, −3)C(−2, −4, −4)
A = m²
15.ProbabilityMedium
A bag holds 3 red and 3 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
3 / 3
P =
16.ProbabilityHard
A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 4 correct answers?
n = 6p = 0.5k = 4
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 39, 38, 17, 41, 32.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 20. What proportion P of values is below 140?
μ = 170σ = 20x = 140
P =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.9464. Find z to two decimals.
A = 0.9464
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
(−3 + 4/x + 4/x2)/(3 + 6/x + 1/x2)
Let x → ∞
L = −3/3 = −1
Answer
L = −1
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 0.4
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 [2/(1 + 2x)]/[5·cos 5x]
Substitute x = 0
L = 2/5 = 0.4
Answer
L = 0.4
3.Derivatives
Answer: f′(1) = 0.077
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 and chain rule
f′(x) = sin(2x) + 2x·cos(2x)
Substitute
f′(1) = 0.9093 + 2·(−0.4161) = 0.077
Answer
f′(1) = 0.077
4.Derivatives
Answer: x = 3
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 − 21x + 36 = 3·(x2 − 7x + 12)
Solve f′(x) = 0
x = 3 or x = 4
Which one is which
f″(x) = 6x − 21: negative at a maximum, positive at a minimum.
f″(3) = −3, f″(4) = 3
x = 3
Answer
x = 3
5.Integrals
Answer: I = 0.1436
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
(1.125·ln 1.5 − 0.5625) − (0 − 0.25) = 0.1436
Answer
I = 0.1436
6.Integrals
Answer: V = 15.89
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.
Intersection points
4x = 3x² → x = 0, x = 1.333
Washer method
V = π∫₀1.333 ((4x)² − (3x²)²) dx
Integrate
V = π(16·1.333³/3 − 9·1.333⁵/5) = π·5.057 = 15.89
Answer
V = 15.89
7.Complex numbers
Answer: θ = 70 °
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 / 1.5 = 3.333
θ = 150° − 80° = 70°
Argument
θ = 70°
Answer
θ = 70°
8.Complex numbers
Answer: Im(z₁) = 1.902
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
321/5 = 2
Argument of the root
θ₁ = (0° + 360°·1)/5 = 72°
Rectangular form
z₁ = 2·(cos 72° + j sin 72°) = 0.618 + j1.902
Im(z₁) = 1.902
Answer
Im(z₁) = 1.902
The 5 roots lie on the circle of radius 2, 72° apart.
9.Differential equations
Answer: k = 0.02992 min⁻¹
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.
Ratio of temperature differences
e−40k = 142/470
Solve for k
k = ln(470/142)/40 = 0.02992 min⁻¹
Answer
k = 0.02992 min⁻¹
10.Differential equations
Answer: ymax = 12.08
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.
Steady value and roots
yss = 10, r = −1.5 ± 3j
Constants from the initial values
y(0) = 10 + A = 0 → A = −10
y′(0) = −1.5A + 3B = 0 → B = −5
Peak
t = π/3 = 1.047
ymax = 10·(1 + e−1.5·π/3) = 12.08
Answer
ymax = 12.08
11.Matrices & determinants
Answer: det A = −10
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 = 7, M₁₂ = 4, M₁₃ = 6
Cofactor expansion
det A = (−4)·7 − 0·4 + 3·6 = −10
Answer
det A = −10
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 − (4)λ + (−12) = 0
Solve
λ = (4 ± √(64))/2 → λ = −2, 6
λ = 6
Answer
λ = 6
13.Vectors: dot & cross product
Answer: a·b = 38
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 = 9.179 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.
First and second draw
P = 3/6 · 2/5 = 6/30 = 0.2
Answer
P = 0.2
16.Probability
Answer: P = 0.2344
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 = 4) = 15·0.54·0.52 = 15·0.0625·0.25 = 0.2344
Answer
P = 0.2344
17.Descriptive statistics
Answer: s = 9.762
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 = −1.5
Standard normal table
Φ(−1.5) = 0.06681
P = 0.06681
Answer
P = 0.06681
19.Normal distribution
Answer: z = 1.93
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 + 0.9464)/2 = 0.9732
Read the z-table backwards
Find 0.9732 inside the table and read off z.
z = 1.93
Answer
z = 1.93
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).