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→∞ (6x2 − x − 4)/(x2 + 2x + 6).
f(x) = (6x2 − x − 4)/(x2 + 2x + 6)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 4x)/(3x²), with x in radians.
f(x) = (1 − cos 4x)/(3x²)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(3x2 + 2). Find f′(3).
f(x) = ln(3x2 + 2)x = 3
f′(3) =
4.DerivativesHard
Let f(x) = x3 − 12x2 + 45x + 4. Find the x-coordinate x of the local minimum.
f(x) = x3 − 12x2 + 45x + 4
x =
5.IntegralsMedium
Evaluate I = ∫12 5/x dx.
f(x) = 5/x[1, 2]
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 1 − x² and the x-axis.
y = 1 − x²
A =
7.Complex numbersMedium
Given z₁ = 3∠55° and z₂ = 0.5∠−50°, find the argument θ of w = z₁·z₂ in degrees, between −180° and 180°.
z₁ = 3∠55°z₂ = 0.5∠−50°
θ = °
8.Complex numbersHard
Use De Moivre's theorem to find the real part of z⁵ for z = −2 − j2.
z = −2 − j2n = 5
Re(z⁵) =
9.Differential equationsMedium
Solve y″ + 6y′ + 5y = 0 with y(0) = 1 and y′(0) = 2. Find y(1.5).
y″ + 6y′ + 5y = 0y(0) = 1y′(0) = 2
y(1.5) =
10.Differential equationsHard
A system obeys y″ + 2y′ + 10y = 50 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1); angles in radians.
y″ + 2y′ + 10y = 50y(0) = 0y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [3 3; 2 0]B = [1 −6; 0 −6]
c21 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [1 −3; 5 −7]
λ =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (−1, 1, 0) and b = (3, k, 3) are perpendicular.
a = (−1, 1, 0)b = (3, k, 3)
k =
14.Vectors: dot & cross productHard
For a = (−6, −4, 3) and b = (−1, 5, −5), find the z-component (a × b)z of the cross product a × b.
a = (−6, −4, 3)b = (−1, 5, −5)
(a × b)z =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2p(x): 0.35, 0.3, 0.35
Var(X) =
16.ProbabilityHard
A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 5 correct answers?
n = 11p = 0.6k = 5
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 42, 52, 48, 48, 44, 10.
n = 6
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 25. Find the z-score z of a score of 550.
μ = 500σ = 25x = 550
z =
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 255.25 g? Give P as a decimal.
μ = 250σ = 5x1 = 250.25x2 = 255.25
P =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 5% of them lie (the 5th percentile).
μ = 65σ = 10P = 5%
x = points
3Solutions
1.Limits
Answer: L = 6
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
(6 + (−1)/x + (−4)/x2)/(1 + 2/x + 6/x2)
Let x → ∞
L = 6/1 = 6
Answer
L = 6
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 2.667
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).
First application
4·sin 4x/(6x) → 0/0
Second application
16·cos 4x/6
Substitute x = 0
L = 16/6 = 2.667
Answer
L = 2.667
3.Derivatives
Answer: f′(3) = 0.6207
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) = 6x/(3x2 + 2)
Substitute
f′(3) = 18/29 = 0.6207
Answer
f′(3) = 0.6207
4.Derivatives
Answer: x = 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.
Derivative
f′(x) = 3x2 − 24x + 45 = 3·(x2 − 8x + 15)
Solve f′(x) = 0
x = 3 or x = 5
Which one is which
f″(x) = 6x − 24: negative at a maximum, positive at a minimum.
f″(3) = −6, f″(5) = 6
x = 5
Answer
x = 5
5.Integrals
Answer: I = 3.466
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) = 5·ln x
Evaluate at the limits
I = 5·(ln 2 − ln 1) = 5·ln 2 = 3.466
Answer
I = 3.466
6.Integrals
Answer: A = 1.333
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
1 − x² = 0 → x = ±1
Integrate
A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333
Answer
A = 1.333
7.Complex numbers
Answer: θ = 5 °
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| = 3 · 0.5 = 1.5
θ = 55° + (−50)° = 5°
Argument
θ = 5°
Answer
θ = 5°
8.Complex numbers
Answer: Re(z⁵) = 128
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.
Polar form
z = −2 − j2 = 2√2∠−135°
De Moivre
z⁵ = (2√2)⁵∠−675° = 181∠45°
Rectangular form
z⁵ = 181·(cos 45° + j sin 45°) = 128 + j128
Re(z⁵) = 128
Answer
Re(z⁵) = 128
9.Differential equations
Answer: y(1.5) = 0.3901
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.
Characteristic roots
(r + 1)(r + 5) = 0 → r = −1, −5
Constants from the initial values
A + B = 1, −1A − 5B = 2
A = 1.75, B = −0.75
Evaluate
y(1.5) = 1.75·e−1.5 + (−0.75)·e−7.5 = 0.3901
Answer
y(1.5) = 0.3901
10.Differential equations
Answer: y(1) = 6.734
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 ± 3j
Constants from the initial values
y(0) = 5 + A = 0 → A = −5
y′(0) = −1A + 3B = 0 → B = −1.667
Evaluate
y(1) = 5·(1 − e−1(cos 3 + 0.3333·sin 3)) = 6.734
Answer
y(1) = 6.734
11.Matrices & determinants
Answer: c21 = 2
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
c21 = 2·1 + 0·0 = 2
Answer
c21 = 2
12.Matrices & determinants
Answer: λ = −2
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 − (−6)λ + (8) = 0
Solve
λ = (−6 ± √(4))/2 → λ = −4, −2
λ = −2
Answer
λ = −2
13.Vectors: dot & cross product
Answer: k = 3
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 equals zero
a·b = (−1)·3 + 1·k + 0·3 = 0
−3 + 1·k = 0
Solve for k
k = −(−3)/1 = 3
Answer
k = 3
14.Vectors: dot & cross product
Answer: (a × b)z = −34
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 = (−4)·(−5) − 3·5 = 5
(a × b)y = azbx − axbz = 3·(−1) − (−6)·(−5) = −33
(a × b)z = axby − aybx = (−6)·5 − (−4)·(−1) = −34
Requested component
(a × b)z = −34
Answer
(a × b)z = −34
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: Var(X) = 0.7
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 = 0·0.35 + 1·0.3 + 2·0.35 = 1
Mean of the squares
E(X²) = 0·0.35 + 1·0.3 + 4·0.35 = 1.7
Variance
Var(X) = 1.7 − 1² = 0.7
Answer
Var(X) = 0.7
16.Probability
Answer: P = 0.1471
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 = (550 − 500)/25 = 2
Answer
z = 2
19.Normal distribution
Answer: P = 0.3332
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 = (255.25 − 250)/5 = 1.05
Area between
P = Φ(1.05) − Φ(0.05) = 0.8531 − 0.5199 = 0.3332
Answer
P = 0.3332
20.Normal distribution
Answer: x = 48.55 points
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.