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→∞ (√(x2 + 3x) − x).
f(x) = √(x2 + 3x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e−0.5x − 1 + 0.5x)/x².
f(x) = (e−0.5x − 1 + 0.5x)/x²x → 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 − 1. Find the x-coordinate x of the local minimum.
f(x) = x3 − 10.5x2 + 36x − 1
x =
5.IntegralsMedium
The region under y = x from x = 0 to x = 3 is rotated about the x-axis. Find the volume V of the solid (a cone).
y = x0 ≤ x ≤ 3
V =
6.IntegralsHard
Evaluate I = ∫₀3 x²·eˣ dx.
∫₀3 x²eˣ dx
I =
7.Complex numbersMedium
The diagram shows z = −2 + j4. With w = 0.5∠−90°, click the point that represents z/w.
z = −2 + j4w = 0.5∠−90°
1A (−2 − j)
2B (−8 − j4)
3C (−4 + j8)
4D (8 + j4)
8.Complex numbersHard
One root of z³ − z² − 20z + 50 = 0 is z₁ = 3 + j. The coefficients are real. Find the real root r.
z₁ = 3 + j
r =
9.Differential equationsMedium
Solve y″ + 4y′ + 4y = 0 with y(0) = 3 and y′(0) = 0. Find y(0.5).
y″ + 4y′ + 4y = 0y(0) = 3y′(0) = 0
y(0.5) =
10.Differential equationsHard
Solve y″ + 2y′ + 26y = 0 with y(0) = 1 and y′(0) = 0 (released from rest). Find y(0.25); angles in radians.
y″ + 2y′ + 26y = 0y(0) = 1y′(0) = 0
y(0.25) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [4 3 1; 4 −5 3; 3 −5 0]
det A =
12.Matrices & determinantsHard
Solve the system 4x − 5y = 11 and −2x + y = 2. Find x.
4x − 5y = 11−2x + y = 2
x =
13.Vectors: dot & cross productMedium
A constant force F = (−10, −25, −25) N acts on an object while it moves in a straight line from point A(2, −3, 5) to point B(−5, 4, −5). Coordinates are in metres. Find the work W done by the force.
F = (−10, −25, −25) NA(2, −3, 5) mB(−5, 4, −5) m
W = J
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−1, −2, 2) and b = (−5, 5, −3) and has a positive z-component. Find its y-component n̂y.
a = (−1, −2, 2)b = (−5, 5, −3)n̂z > 0
n̂y =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3p(x): 0.2, 0.2, 0.3, 0.3
Var(X) =
16.ProbabilityHard
A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 4 correct answers?
n = 5p = 0.3k = 4
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 30, 23, 58, 45, 10.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 15. Find the z-score z of a score of 158.
μ = 170σ = 15x = 158
z =
19.Normal distributionMedium
The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 8 of them exceeds 325.1 g. Round z to two decimals.
μ = 40 gσ = 1.2 gn = 8> 325.1 g
P =
20.Normal distributionHard
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
3Solutions
1.Limits
Answer: L = 1.5
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).
Multiply by the conjugate
f(x) = 3x/(√(x2 + 3x) + x)
Divide by x
f(x) = 3/(√(1 + 3/x) + 1)
Let x → ∞
L = 3/(1 + 1) = 1.5
Answer
L = 1.5
2.Limits
Answer: L = 0.125
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
(−0.5·e−0.5x + 0.5)/(2x) → 0/0
Second application
0.25·e−0.5x/2
Substitute x = 0
L = 0.25·1/2 = 0.125
Answer
L = 0.125
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 = 4
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 = 4
Answer
x = 4
5.Integrals
Answer: V = 28.27
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∫₀3 1x² dx
Integrate
V = π·1·3³/3 = 28.27
Answer
V = 28.27
Check: a cone of radius 3 and height 3 has (1/3)πr²h = 28.27.
6.Integrals
Answer: I = 98.43
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.
First round
∫x²eˣ dx = x²eˣ − ∫2x eˣ dx
Second round
∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)
Limits
20.09·(5) − 1·2 = 98.43
Answer
I = 98.43
7.Complex numbers
Answer: B (−8 − j4)
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.
Effect of w
|z/w| = |z|/0.5, arg(z/w) = arg z − (−90)°
The point
z/w = −8 − j4 → B
Answer
B (−8 − j4)
8.Complex numbers
Answer: r = −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.
Check with the sum of the roots: z₁ + z₂ + r = 6 + (−5) = 1 = −(−1).
9.Differential equations
Answer: y(0.5) = 2.207
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.
Double root
(r + 2)² = 0 → r = −2
Constants
A = y(0) = 3
y′(0) = B − 2A → B = 0 + 6 = 6
Evaluate
y(0.5) = (3 + 6·0.5)·e−1 = 2.207
Answer
y(0.5) = 2.207
10.Differential equations
Answer: y(0.25) = 0.3934
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.
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 = 15, M₁₂ = −9, M₁₃ = −5
Cofactor expansion
det A = 4·15 − 3·(−9) + 1·(−5) = 82
Answer
det A = 82
12.Matrices & determinants
Answer: x = −3.5
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 = 4·1 − (−5)·(−2) = −6
Replace the x column
det Aₓ = 11·1 − (−5)·2 = 21
x = 21/(−6) = −3.5
Answer
x = −3.5
y = −5
13.Vectors: dot & cross product
Answer: W = 145 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) − (2, −3, 5) = (−7, 7, −10) m
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 = (−2)·(−3) − 2·5 = −4
(a × b)y = azbx − axbz = 2·(−5) − (−1)·(−3) = −13
(a × b)z = axby − aybx = (−1)·5 − (−2)·(−5) = −15
Normalise
The z-component of a × b is negative, so use −(a × b).
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 = 0·0.2 + 1·0.2 + 2·0.3 + 3·0.3 = 1.7
Mean of the squares
E(X²) = 0·0.2 + 1·0.2 + 4·0.3 + 9·0.3 = 4.1
Variance
Var(X) = 4.1 − 1.7² = 1.21
Answer
Var(X) = 1.21
16.Probability
Answer: P = 0.02835
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) = 5·0.34·0.71 = 5·0.0081·0.7 = 0.02835
Answer
P = 0.02835
17.Descriptive statistics
Answer: s = 18.75
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 = (158 − 170)/15 = −0.8
Answer
z = −0.8
19.Normal distribution
Answer: P = 0.06681
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 = 320 g, σT = 1.2·√8 = 3.394 g
Standardise
z = (325.1 − 320)/3.394 = 1.5
Probability
P = 1 − Φ(1.5) = 0.06681
Answer
P = 0.06681
20.Normal distribution
Answer: P = 0.7881
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).