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→∞ (−3x3 − 3x + 4)/(2x3 + 5x + 2).
f(x) = (−3x3 − 3x + 4)/(2x3 + 5x + 2)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (ex − 1 − x)/x².
f(x) = (ex − 1 − x)/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
The tangent line to f(x) = 0.5x2 − 5x at x = −1 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 − 5xx = −1
b =
5.IntegralsMedium
Evaluate I = ∫₁5 x·ln x dx.
∫₁5 x ln x dx
I =
6.IntegralsHard
Evaluate I = ∫₀2.5 x²·eˣ dx.
∫₀2.5 x²eˣ dx
I =
7.Complex numbersMedium
Find the argument θ of z = −3 + j8 in degrees, as the principal value between −180° and 180°.
z = −3 + j8
θ = °
8.Complex numbersHard
One root of z³ − 13z² + 65z − 125 = 0 is z₁ = 4 + j3. The coefficients are real. Find the real root r.
z₁ = 4 + j3
r =
9.Differential equationsMedium
Solve y″ + 4y′ + 4y = 0 with y(0) = 1 and y′(0) = 0. Find y(2).
y″ + 4y′ + 4y = 0y(0) = 1y′(0) = 0
y(2) =
10.Differential equationsHard
A tank holds 1000 L of water in which 20 kg of salt is dissolved. Brine with 0.25 kg of salt per litre flows in at 10 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 60 min?
V = 1000 Lm0 = 20 kgc = 0.25 kg/Lr = 10 L/min
m = kg
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [−3 −3; −6 −6]B = [2 2; 6 −3]
c21 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−1 −3; −7 3]
λ =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (6, 3, −5) and b = (3, −4, 6), in degrees.
a = (6, 3, −5)b = (3, −4, 6)
θ = °
14.Vectors: dot & cross productHard
A force F = (20, −60, −10) N acts at point A(−1, 2, 2). Coordinates are in metres. Find the z-component Mz of the moment of the force about the origin O(0, 0, 0).
F = (20, −60, −10) NA(−1, 2, 2) mO(0, 0, 0) m
Mz = N·m
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 2, 4, 6p(x): 0.25, 0.2, 0.3, 0.25
Var(X) =
16.ProbabilityHard
A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 5 correct answers?
n = 11p = 0.3k = 5
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 50, 20, 40, 58, 23.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 20. Find the z-score z of a score of 30.
μ = 50σ = 20x = 30
z =
19.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 470 and 534 h? Give P as a decimal.
μ = 500σ = 40x1 = 470x2 = 534
P =
20.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 10% is to the left of it. Find the cut-off x.
μ = 100 pointsσ = 15 points10%
x = points
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).
Divide by the highest power
(−3 + (−3)/x2 + 4/x3)/(2 + 5/x2 + 2/x3)
Let x → ∞
L = −3/2 = −1.5
Answer
L = −1.5
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 0.5
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
(1·ex − 1)/(2x) → 0/0
Second application
1·ex/2
Substitute x = 0
L = 1·1/2 = 0.5
Answer
L = 0.5
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: b = −0.5
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.
Point and slope
f(−1) = 5.5
f′(x) = x − 5 → f′(−1) = −6
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 5.5 − (−6)·(−1) = −0.5
Answer
b = −0.5
5.Integrals
Answer: I = 14.12
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.
Choose u and dv
u = ln x, du = dx/x; dv = x dx, v = x²/2
Antiderivative
∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4
Limits
(12.5·ln 5 − 6.25) − (0 − 0.25) = 14.12
Answer
I = 14.12
6.Integrals
Answer: I = 37.59
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
12.18·(3.25) − 1·2 = 37.59
Answer
I = 37.59
7.Complex numbers
Answer: θ = 110.6 °
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.
Quadrant
a = −3, b = 8 → II
Reference angle
α = atan(8/3) = 69.44°
Argument
θ = 180° − α = 110.6°
Answer
θ = 110.6°
In polar form z = 8.544∠110.6°.
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 = 8 + 5 = 13 = −(−13).
9.Differential equations
Answer: y(2) = 0.09158
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(2) = (1 + 2·2)·e−4 = 0.09158
Answer
y(2) = 0.09158
10.Differential equations
Answer: m = 123.8 kg
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.
Set up the equation
dm/dt = 10·0.25 − (10/1000)·m = 2.5 − 0.01·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 250 + (−230)·e−t/100
Substitute t
m(60) = 250 + (−230)·0.5488 = 123.8 kg
Answer
m = 123.8 kg
11.Matrices & determinants
Answer: c21 = −48
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 = (−6)·2 + (−6)·6 = −48
Answer
c21 = −48
12.Matrices & determinants
Answer: λ = 6
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 − (2)λ + (−24) = 0
Solve
λ = (2 ± √(100))/2 → λ = −4, 6
λ = 6
Answer
λ = 6
13.Vectors: dot & cross product
Answer: θ = 111.5 °
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 = 6·3 + 3·(−4) + (−5)·6 = −24
Magnitudes
|a| = √(6² + 3² + (−5)²) = 8.367
|b| = √(3² + (−4)² + 6²) = 7.81
Angle
cos θ = −24/(8.367·7.81) = −0.3673
θ = arccos(−0.3673) = 111.5°
Answer
θ = 111.5°
14.Vectors: dot & cross product
Answer: Mz = 20 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.
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 = 0·0.25 + 2·0.2 + 4·0.3 + 6·0.25 = 3.1
Mean of the squares
E(X²) = 0·0.25 + 4·0.2 + 16·0.3 + 36·0.25 = 14.6
Variance
Var(X) = 14.6 − 3.1² = 4.99
Answer
Var(X) = 4.99
16.Probability
Answer: P = 0.1321
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 = (30 − 50)/20 = −1
Answer
z = −1
19.Normal distribution
Answer: P = 0.5757
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 = (470 − 500)/40 = −0.75
z₂ = (534 − 500)/40 = 0.85
Area between
P = Φ(0.85) − Φ(−0.75) = 0.8023 − 0.2266 = 0.5757
Answer
P = 0.5757
20.Normal distribution
Answer: x = 80.8 points
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.