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 + 3/x)x. Give a decimal value.
f(x) = (1 + 3/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e3x − 1 − 3x)/x².
f(x) = (e3x − 1 − 3x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = ln(x2 + 5). Find f′(1).
f(x) = ln(x2 + 5)x = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 + 6x2 + 9x + 5. Find the x-coordinate x of the local minimum.
f(x) = x3 + 6x2 + 9x + 5
x =
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
Find the area A of the region enclosed by y = 2x2 and y = 3x.
y = 2x2y = 3x
A =
7.Complex numbersMedium
Given z₁ = 2.5∠40° and z₂ = 3∠−95°, find the argument θ of w = z₁·z₂ in degrees, between −180° and 180°.
z₁ = 2.5∠40°z₂ = 3∠−95°
θ = °
8.Complex numbersHard
The equation z² − 5z + 31.25 = 0 has two complex roots. Find their modulus |z|.
p = −5q = 31.25
|z| =
9.Differential equationsMedium
A cup of coffee at 86 °C is left in a room at 22 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.11 min⁻¹, find its temperature T after 11 min.
T0 = 86 °CTa = 22 °Ck = 0.11 min⁻¹t = 11 min
T = °C
10.Differential equationsHard
A system obeys y″ + 2y′ + 5y = 25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 2y′ + 5y = 25y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−3 −2 2; 0 1 −3; 0 1 4]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [6 3; −6 −5]
λ =
13.Vectors: dot & cross productMedium
A constant force F = (30, −20, −25) N acts on an object while it moves in a straight line from point A(−2, 4, −5) to point B(3, −4, −2). Coordinates are in metres. Find the work W done by the force.
F = (30, −20, −25) NA(−2, 4, −5) mB(3, −4, −2) m
W = J
14.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (−4, −1, −6) and b = (−3, 0, 6).
a = (−4, −1, −6)b = (−3, 0, 6)
A =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4p(x): 0.25, 0.2, 0.35, 0.2
Var(X) =
16.ProbabilityHard
7 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?
n = 7p = 0.2
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 46, 10, 59, 41, 20.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 8. Find the z-score z of a score of 174.8.
μ = 170σ = 8x = 174.8
z =
19.Normal distributionMedium
X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 200.8 mm. Round z to two decimals.
μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 200.8 mm
P =
20.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 20% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g20%
x = g
3Solutions
1.Limits
Answer: L = 20.09
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 + 3/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·3 = 3
Undo the logarithm
L = e3 = 20.09
Answer
L = 20.09
2.Limits
Answer: L = 4.5
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).
First application
(3·e3x − 3)/(2x) → 0/0
Second application
9·e3x/2
Substitute x = 0
L = 9·1/2 = 4.5
Answer
L = 4.5
3.Derivatives
Answer: f′(1) = 0.3333
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 + 5)
Substitute
f′(1) = 2/6 = 0.3333
Answer
f′(1) = 0.3333
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 + 12x + 9 = 3·(x2 + 4x + 3)
Solve f′(x) = 0
x = −3 or x = −1
Which one is which
f″(x) = 6x + 12: negative at a maximum, positive at a minimum.
f″(−3) = −6, f″(−1) = 6
x = −1
Answer
x = −1
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: A = 1.125
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.
Intersection points
2x2 = 3x → x = 0, x = 1.5
Integrate upper minus lower
A = [1.5x2 − 0.6667x3]01.5 = 1.125
Answer
A = 1.125
7.Complex numbers
Answer: θ = −55 °
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| = 2.5 · 3 = 7.5
θ = 40° + (−95)° = −55°
Argument
θ = −55°
Answer
θ = −55°
8.Complex numbers
Answer: |z| = 5.59
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.
Discriminant
p² − 4q = 25 − 125 = −100 < 0
Quadratic formula
z = (5 ± j√100)/2 = 2.5 + j5, 2.5 − j5
Answer
|z| = √(2.5² + 5²) = √31.25 = 5.59
Answer
|z| = 5.59
9.Differential equations
Answer: T = 41.08 °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) = 22 + 64·e−0.11t
Substitute t
T(11) = 22 + 64·0.2982 = 41.08 °C
Answer
T = 41.08 °C
10.Differential equations
Answer: ymax = 6.039
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 = 5, r = −1 ± 2j
Constants from the initial values
y(0) = 5 + A = 0 → A = −5
y′(0) = −1A + 2B = 0 → B = −2.5
Peak
t = π/2 = 1.571
ymax = 5·(1 + e−1·π/2) = 6.039
Answer
ymax = 6.039
11.Matrices & determinants
Answer: det A = −21
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 = 0, M₁₃ = 0
Cofactor expansion
det A = (−3)·7 − (−2)·0 + 2·0 = −21
Answer
det A = −21
12.Matrices & determinants
Answer: λ = 4
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 = −3, 4
λ = 4
Answer
λ = 4
13.Vectors: dot & cross product
Answer: W = 235 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 = (3, −4, −2) − (−2, 4, −5) = (5, −8, 3) m
Work
W = F·d = 30·5 + (−20)·(−8) + (−25)·3 = 235 J
Answer
W = 235 J
14.Vectors: dot & cross product
Answer: A = 42.53
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
(a × b)x = aybz − azby = (−1)·6 − (−6)·0 = −6
(a × b)y = azbx − axbz = (−6)·(−3) − (−4)·6 = 42
(a × b)z = axby − aybx = (−4)·0 − (−1)·(−3) = −3
Area
A = |a × b| = √((−6)² + 42² + (−3)²) = 42.53
Answer
A = 42.53
15.Probability
Answer: Var(X) = 1.15
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 = 1·0.25 + 2·0.2 + 3·0.35 + 4·0.2 = 2.5
Mean of the squares
E(X²) = 1·0.25 + 4·0.2 + 9·0.35 + 16·0.2 = 7.4
Variance
Var(X) = 7.4 − 2.5² = 1.15
Answer
Var(X) = 1.15
16.Probability
Answer: P = 0.5767
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) = 0.87 = 0.2097
P(X = 1) = 7·0.2·0.86 = 0.367
Add them
P = 0.2097 + 0.367 = 0.5767
Answer
P = 0.5767
17.Descriptive statistics
Answer: s = 19.89
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 = (174.8 − 170)/8 = 0.6
Answer
z = 0.6
19.Normal distribution
Answer: P = 0.0548
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 sum
μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm
Standardise
z = (200.8 − 200)/0.5 = 1.6
Upper tail
P = 1 − Φ(1.6) = 1 − 0.9452 = 0.0548
Answer
P = 0.0548
20.Normal distribution
Answer: x = 254.2 g
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.