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 + 10x) − x).
f(x) = √(x2 + 10x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + x)/sin(3x), with x in radians.
f(x) = ln(1 + x)/sin(3x)x → 0
L =
3.DerivativesMedium
Let f(x) = x3·e2x. Find f′(1.5).
f(x) = x3·e2xx = 1.5
f′(1.5) =
4.DerivativesHard
The tangent line to f(x) = 3x2 − 4x + 1 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = 3x2 − 4x + 1x = 1
b =
5.IntegralsMedium
Evaluate I = ∫02 ex dx.
f(x) = ex[0, 2]
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 4 − x² and the x-axis.
y = 4 − x²
A =
7.Complex numbersMedium
The diagram shows z = −4 + j3. With w = 0.5∠−90°, click the point that represents z/w.
z = −4 + j3w = 0.5∠−90°
1A (−6 − j8)
2B (6 + j8)
3C (−8 + j6)
4D (−3 − j4)
8.Complex numbersHard
Find the imaginary part of (2 − j3)⁵.
z = 2 − j3n = 5
Im(z⁵) =
9.Differential equationsMedium
Solve y″ + 4y′ + 4y = 0 with y(0) = 3 and y′(0) = 4. Find y(2).
y″ + 4y′ + 4y = 0y(0) = 3y′(0) = 4
y(2) =
10.Differential equationsHard
A system obeys y″ + 4y′ + 20y = 100 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 4y′ + 20y = 100y(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 0; 6 −1]B = [−6 −1; 3 −6]
c12 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [3 −1; −3 1]
λ =
13.Vectors: dot & cross productMedium
A constant force F = (30, −10, −30) N acts on an object while it moves in a straight line from point A(−4, −2, 4) to point B(3, −1, 1). Coordinates are in metres. Find the work W done by the force.
F = (30, −10, −30) NA(−4, −2, 4) mB(3, −1, 1) m
W = J
14.Vectors: dot & cross productHard
A force F = (−40, 40, −10) N acts at point A(4, −1, 2). Coordinates are in metres. Find the y-component My of the moment of the force about the origin O(0, 0, 0).
F = (−40, 40, −10) NA(4, −1, 2) mO(0, 0, 0) m
My = N·m
15.ProbabilityMedium
A bag holds 3 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
3 / 7
P =
16.ProbabilityHard
A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 4 correct answers?
n = 12p = 0.2k = 4
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 26, 60, 23, 50, 57, 16.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is below 48?
μ = 60σ = 15x = 48
P =
19.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 88 and 118.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 88x2 = 118.75
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 = 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) = 10x/(√(x2 + 10x) + x)
Divide by x
f(x) = 10/(√(1 + 10/x) + 1)
Let x → ∞
L = 10/(1 + 1) = 5
Answer
L = 5
2.Limits
Answer: L = 0.3333
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 [1/(1 + x)]/[3·cos 3x]
Substitute x = 0
L = 1/3 = 0.3333
Answer
L = 0.3333
3.Derivatives
Answer: f′(1.5) = 271.2
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 rule
f′(x) = u′v + uv′ = e2x·(3x2 + 2·x3)
Substitute
f′(1.5) = 20.09·(6.75 + 6.75) = 271.2
Answer
f′(1.5) = 271.2
4.Derivatives
Answer: b = −2
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 − 4 → f′(1) = 2
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 0 − 2·1 = −2
Answer
b = −2
5.Integrals
Answer: I = 6.389
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.
Antiderivative
F(x) = ex/1
Evaluate at the limits
I = (e2 − 1)/1 = (7.389 − 1)/1 = 6.389
Answer
I = 6.389
6.Integrals
Answer: A = 10.67
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
4 − x² = 0 → x = ±2
Integrate
A = [4x − x³/3]−22 = 5.333 − (−5.333) = 10.67
Answer
A = 10.67
7.Complex numbers
Answer: A (−6 − j8)
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 = −6 − j8 → A
Answer
A (−6 − j8)
8.Complex numbers
Answer: Im(z⁵) = 597
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.
Powers step by step
z² = (2 − j3)(2 − j3) = −5 − j12
z³ = (−5 − j12)(2 − j3) = −46 − j9
z⁴ = (−46 − j9)(2 − j3) = −119 + j120
z⁵ = (−119 + j120)(2 − j3) = 122 + j597
Check with De Moivre
z = 3.606∠−56.31° → z⁵ = 609.3∠78.45°
Im(z⁵) = 597
Answer
Im(z⁵) = 597
9.Differential equations
Answer: y(2) = 0.4213
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 = 4 + 6 = 10
Evaluate
y(2) = (3 + 10·2)·e−4 = 0.4213
Answer
y(2) = 0.4213
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 = −2 ± 4j
Constants from the initial values
y(0) = 5 + A = 0 → A = −5
y′(0) = −2A + 4B = 0 → B = −2.5
Peak
t = π/4 = 0.7854
ymax = 5·(1 + e−2·π/4) = 6.039
Answer
ymax = 6.039
11.Matrices & determinants
Answer: c12 = −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.
Row times column
c12 = 5·(−1) + 0·(−6) = −5
Answer
c12 = −5
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 − (4)λ + (0) = 0
Solve
λ = (4 ± √(16))/2 → λ = 0, 4
λ = 4
Answer
λ = 4
13.Vectors: dot & cross product
Answer: W = 290 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, −1, 1) − (−4, −2, 4) = (7, 1, −3) m
Work
W = F·d = 30·7 + (−10)·1 + (−30)·(−3) = 290 J
Answer
W = 290 J
14.Vectors: dot & cross product
Answer: My = −40 N·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.
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 = 3/10 · 2/9 = 6/90 = 0.06667
Answer
P = 0.06667
16.Probability
Answer: P = 0.1329
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 = −0.8
Standard normal table
Φ(−0.8) = 0.2119
P = 0.2119
Answer
P = 0.2119
19.Normal distribution
Answer: P = 0.6825
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.
z-scores
z₁ = (88 − 100)/15 = −0.8
z₂ = (118.75 − 100)/15 = 1.25
Area between
P = Φ(1.25) − Φ(−0.8) = 0.8944 − 0.2119 = 0.6825
Answer
P = 0.6825
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).