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→∞ (−5x2 − 6x + 9)/(x2 + 6x + 5).
f(x) = (−5x2 − 6x + 9)/(x2 + 6x + 5)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 3x)/sin(4x), with x in radians.
f(x) = ln(1 + 3x)/sin(4x)x → 0
L =
3.DerivativesMedium
Let f(x) = (2x + 1)5. Find f′(0).
f(x) = (2x + 1)5x = 0
f′(0) =
4.DerivativesHard
The tangent line to f(x) = 3x2 − x − 2 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = 3x2 − x − 2x = 1
b =
5.IntegralsMedium
The region under y = √(x) from x = 0 to x = 4 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(x)0 ≤ x ≤ 4
V =
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 = 1.5∠100° and z₂ = 2∠−50°, find the imaginary part of w = z₁/z₂.
z₁ = 1.5∠100°z₂ = 2∠−50°
Im(w) =
8.Complex numbersHard
One root of z³ + 5z² + 12z − 18 = 0 is z₁ = −3 + j3. The coefficients are real. Find the real root r.
z₁ = −3 + j3
r =
9.Differential equationsMedium
A forged steel part at 460 °C is left to cool in a workshop at 29 °C. After 60 min its temperature is 159 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 460 °CTa = 29 °CT(60) = 159 °C
k = min⁻¹
10.Differential equationsHard
A system obeys y″ + 3y′ + 11.25y = 22.5 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 = 22.5y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).
A = [5 6; −4 4]B = [0 −5; 2 1]
c12 =
12.Matrices & determinantsHard
Solve the system −3x + 2y = −3 and −5x + 2y = 11. Find x.
−3x + 2y = −3−5x + 2y = 11
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (1, 4, 4) and b = (2, 3, −5), in degrees.
a = (1, 4, 4)b = (2, 3, −5)
θ = °
14.Vectors: dot & cross productHard
For a = (5, 0, −1) and b = (−2, −4, −6), find the z-component (a × b)z of the cross product a × b.
a = (5, 0, −1)b = (−2, −4, −6)
(a × b)z =
15.ProbabilityMedium
A bag holds 6 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
6 / 6
P =
16.ProbabilityHard
A test has 10 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 = 10p = 0.3k = 2
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 34, 44, 34, 31, 51, 60.
n = 6
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 40. Find the z-score z of a score of 560.
μ = 500σ = 40x = 560
z =
19.Normal distributionMedium
The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 10 of them is below 831 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 10< 831 kg
P =
20.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 97.5% of them lie (the 97.5th percentile).
μ = 32σ = 6P = 97.5%
x = min
3Solutions
1.Limits
Answer: L = −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).
Divide by the highest power
(−5 + (−6)/x + 9/x2)/(1 + 6/x + 5/x2)
Let x → ∞
L = −5/1 = −5
Answer
L = −5
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 0.75
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 [3/(1 + 3x)]/[4·cos 4x]
Substitute x = 0
L = 3/4 = 0.75
Answer
L = 0.75
3.Derivatives
Answer: f′(0) = 10
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·(2x + 1)4·2
Substitute
f′(0) = 5·14·2 = 10
Answer
f′(0) = 10
4.Derivatives
Answer: b = −5
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) = 0
f′(x) = 6x − 1 → f′(1) = 5
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 0 − 5·1 = −5
Answer
b = −5
5.Integrals
Answer: V = 25.13
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 x dx
Integrate
V = π·1·4²/2 = 25.13
Answer
V = 25.13
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: Im(w) = 0.375
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| = 1.5 / 2 = 0.75
θ = 100° − (−50)° = 150°
Rectangular form
Re(w) = 0.75·cos(150°) = −0.6495
Im(w) = 0.75·sin(150°) = 0.375
Im(w) = 0.375
Answer
Im(w) = 0.375
8.Complex numbers
Answer: r = 1
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.
Check with the sum of the roots: z₁ + z₂ + r = −6 + 1 = −5 = −(5).
9.Differential equations
Answer: k = 0.01998 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−60k = 130/431
Solve for k
k = ln(431/130)/60 = 0.01998 min⁻¹
Answer
k = 0.01998 min⁻¹
10.Differential equations
Answer: ymax = 2.416
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 = 2, r = −1.5 ± 3j
Constants from the initial values
y(0) = 2 + A = 0 → A = −2
y′(0) = −1.5A + 3B = 0 → B = −1
Peak
t = π/3 = 1.047
ymax = 2·(1 + e−1.5·π/3) = 2.416
Answer
ymax = 2.416
11.Matrices & determinants
Answer: c12 = −19
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 = 5·(−5) + 6·1 = −19
Answer
c12 = −19
12.Matrices & determinants
Answer: x = −7
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 = (−3)·2 − 2·(−5) = 4
Replace the x column
det Aₓ = (−3)·2 − 2·11 = −28
x = −28/4 = −7
Answer
x = −7
y = −12
13.Vectors: dot & cross product
Answer: θ = 99.75 °
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.
Dot product
a·b = 1·2 + 4·3 + 4·(−5) = −6
Magnitudes
|a| = √(1² + 4² + 4²) = 5.745
|b| = √(2² + 3² + (−5)²) = 6.164
Angle
cos θ = −6/(5.745·6.164) = −0.1694
θ = arccos(−0.1694) = 99.75°
Answer
θ = 99.75°
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 = 0·(−6) − (−1)·(−4) = −4
(a × b)y = azbx − axbz = (−1)·(−2) − 5·(−6) = 32
(a × b)z = axby − aybx = 5·(−4) − 0·(−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 = 0.2273
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 = 6/12 · 5/11 = 30/132 = 0.2273
Answer
P = 0.2273
16.Probability
Answer: P = 0.2335
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 = (560 − 500)/40 = 1.5
Answer
z = 1.5
19.Normal distribution
Answer: P = 0.3085
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 = 850 kg, σT = 12·√10 = 37.95 kg
Standardise
z = (831 − 850)/37.95 = −0.5
Probability
P = Φ(−0.5) = 0.3085
Answer
P = 0.3085
20.Normal distribution
Answer: x = 43.76 min
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.