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 − 5x − 2)/(3x2 + 2x + 6).
f(x) = (−5x2 − 5x − 2)/(3x2 + 2x + 6)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e1.5x − 1 − 1.5x)/x².
f(x) = (e1.5x − 1 − 1.5x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(2x), with x in radians. Find f′(2).
f(x) = x·sin(2x)x = 2 rad
f′(2) =
4.DerivativesHard
Let f(x) = x3 + 3x2 + 1. Find the x-coordinate x of the local maximum.
f(x) = x3 + 3x2 + 1
x =
5.IntegralsMedium
Evaluate I = ∫01 2x·(x² + 1)3 dx.
f(x) = 2x·(x² + 1)3[0, 1]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 2x2 and y = 2x.
y = 2x2y = 2x
A =
7.Complex numbersMedium
Given z₁ = 2∠55° and z₂ = 3∠20°, find the imaginary part of w = z₁·z₂.
z₁ = 2∠55°z₂ = 3∠20°
Im(w) =
8.Complex numbersHard
Solve z² + 5z + 10.25 = 0. The roots are a ± jb; find the real part a.
p = 5q = 10.25
a =
9.Differential equationsMedium
Solve y″ + 4y′ + 4y = 0 with y(0) = 1 and y′(0) = 0. Find y(1).
y″ + 4y′ + 4y = 0y(0) = 1y′(0) = 0
y(1) =
10.Differential equationsHard
A system obeys y″ + 4y′ + 8y = 40 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1); angles in radians.
y″ + 4y′ + 8y = 40y(0) = 0y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).
A = [4 6; 4 3]B = [4 2; 4 −2]
c12 =
12.Matrices & determinantsHard
Solve the system −3x − 3y = 3 and −2x + 4y = 9. Find x.
−3x − 3y = 3−2x + 4y = 9
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (−7, −5, −1) onto b = (3, 6, −3), that is, the component of a in the direction of b.
a = (−7, −5, −1)b = (3, 6, −3)
compb a =
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−3, −5, −2) and b = (−4, −4, −4) and has a positive z-component. Find its x-component n̂x.
a = (−3, −5, −2)b = (−4, −4, −4)n̂z > 0
n̂x =
15.ProbabilityMedium
A bag holds 3 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
3 / 6
P =
16.ProbabilityHard
A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 3 correct answers?
n = 7p = 0.25k = 3
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 45, 56, 20, 39, 22, 52.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 10. What proportion P of values is below 182.5?
μ = 170σ = 10x = 182.5
P =
19.Normal distributionMedium
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 250.25 and 260.25 g? Give P as a decimal.
μ = 250σ = 5x1 = 250.25x2 = 260.25
P =
20.Normal distributionHard
Hole diameters D are normal with mean 25.02 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.02 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
3Solutions
1.Limits
Answer: L = −1.667
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 + (−5)/x + (−2)/x2)/(3 + 2/x + 6/x2)
Let x → ∞
L = −5/3 = −1.667
Answer
L = −1.667
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 1.125
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
(1.5·e1.5x − 1.5)/(2x) → 0/0
Second application
2.25·e1.5x/2
Substitute x = 0
L = 2.25·1/2 = 1.125
Answer
L = 1.125
3.Derivatives
Answer: f′(2) = −3.371
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′(2) = −0.7568 + 4·(−0.6536) = −3.371
Answer
f′(2) = −3.371
4.Derivatives
Answer: x = −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.
Derivative
f′(x) = 3x2 + 6x = 3·(x2 + 2x)
Solve f′(x) = 0
x = −2 or x = 0
Which one is which
f″(x) = 6x + 6: negative at a maximum, positive at a minimum.
f″(−2) = −6, f″(0) = 6
x = −2
Answer
x = −2
5.Integrals
Answer: I = 3.75
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.
Substitution
u = x² + 1, du = 2x dx
I = ∫12 u3 du
Integrate and evaluate
I = [u4/4]12 = (16 − 1)/4 = 3.75
Answer
I = 3.75
6.Integrals
Answer: A = 0.3333
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 = 2x → x = 0, x = 1
Integrate upper minus lower
A = [x2 − 0.6667x3]01 = 0.3333
Answer
A = 0.3333
7.Complex numbers
Answer: Im(w) = 5.796
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 · 3 = 6
θ = 55° + 20° = 75°
Rectangular form
Re(w) = 6·cos(75°) = 1.553
Im(w) = 6·sin(75°) = 5.796
Im(w) = 5.796
Answer
Im(w) = 5.796
8.Complex numbers
Answer: a = −2.5
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.
Discriminant
p² − 4q = 25 − 41 = −16 < 0
Quadratic formula
z = (−5 ± j√16)/2 = −2.5 + j2, −2.5 − j2
Answer
a = −2.5
Answer
a = −2.5
9.Differential equations
Answer: y(1) = 0.406
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) = 1
y′(0) = B − 2A → B = 0 + 2 = 2
Evaluate
y(1) = (1 + 2·1)·e−2 = 0.406
Answer
y(1) = 0.406
10.Differential equations
Answer: y(1) = 4.666
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 ± 2j
Constants from the initial values
y(0) = 5 + A = 0 → A = −5
y′(0) = −2A + 2B = 0 → B = −5
Evaluate
y(1) = 5·(1 − e−2(cos 2 + 1·sin 2)) = 4.666
Answer
y(1) = 4.666
11.Matrices & determinants
Answer: c12 = −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.
Row times column
c12 = 4·2 + 6·(−2) = −4
Answer
c12 = −4
12.Matrices & determinants
Answer: x = −2.167
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)·4 − (−3)·(−2) = −18
Replace the x column
det Aₓ = 3·4 − (−3)·9 = 39
x = 39/(−18) = −2.167
Answer
x = −2.167
y = 1.167
13.Vectors: dot & cross product
Answer: compb a = −6.532
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 = (−7)·3 + (−5)·6 + (−1)·(−3) = −48
Length of b
|b| = √(3² + 6² + (−3)²) = 7.348
Scalar projection
compb a = a·b/|b| = −48/7.348 = −6.532
Answer
compb a = −6.532
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: n̂x = −0.8018
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 = 3/9 · 2/8 = 6/72 = 0.08333
Answer
P = 0.08333
16.Probability
Answer: P = 0.173
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 = 1.25
Standard normal table
Φ(1.25) = 0.8944
P = 0.8944
Answer
P = 0.8944
19.Normal distribution
Answer: P = 0.4599
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 = (250.25 − 250)/5 = 0.05
z₂ = (260.25 − 250)/5 = 2.05
Area between
P = Φ(2.05) − Φ(0.05) = 0.9798 − 0.5199 = 0.4599
Answer
P = 0.4599
20.Normal distribution
Answer: P = 0.7611
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).