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→∞ (√(4x2 − 9x) − 2x).
f(x) = √(4x2 − 9x) − 2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 5x)/(3x²), with x in radians.
f(x) = (1 − cos 5x)/(3x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x2·ex. Find f′(1).
f(x) = x2·exx = 1
f′(1) =
4.DerivativesHard
The tangent line to f(x) = 0.5x2 − 5x − 1 at x = −2 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 − 5x − 1x = −2
b =
5.IntegralsMedium
Evaluate I = ∫₀2 x·sin x dx (x in radians).
∫₀2 x·sin x dx
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 9 − x² and the x-axis.
y = 9 − x²
A =
7.Complex numbersMedium
Given z₁ = 2∠100° and z₂ = 4∠45°, find the imaginary part of w = z₁·z₂.
z₁ = 2∠100°z₂ = 4∠45°
Im(w) =
8.Complex numbersHard
The roots of zⁿ = w with n = 5 and w = 32∠60° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, …, 4. Find the real part of z₂.
w = 32∠60°n = 5k = 2
Re(z₂) =
9.Differential equationsMedium
Solve y″ + 4y′ + 4y = 0 with y(0) = 3 and y′(0) = 0. Find y(1.5).
y″ + 4y′ + 4y = 0y(0) = 3y′(0) = 0
y(1.5) =
10.Differential equationsHard
A tank holds 1000 L of water in which 10 kg of salt is dissolved. Brine with 0.05 kg of salt per litre flows in at 4 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 45 min?
V = 1000 Lm0 = 10 kgc = 0.05 kg/Lr = 4 L/min
m = kg
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).
A = [4 −5; 0 5]B = [3 −4; −2 6]
c12 =
12.Matrices & determinantsHard
Solve the system 2x + 4y = −4 and 5x + 2y = −6. Find x.
2x + 4y = −45x + 2y = −6
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (5, 6, −1) and b = (−5, k, 6) are perpendicular.
a = (5, 6, −1)b = (−5, k, 6)
k =
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−5, 5, −5) and b = (4, 2, 2) and has a positive z-component. Find its x-component n̂x.
a = (−5, 5, −5)b = (4, 2, 2)n̂z > 0
n̂x =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4, 5p(x): 0.3, 0.3, 0.15, 0.15, 0.1
Var(X) =
16.ProbabilityHard
A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 2 correct answers?
n = 12p = 0.5k = 2
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 48, 36, 31, 51, 60.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 8. What proportion P of values is below 44?
μ = 60σ = 8x = 44
P =
19.Normal distributionMedium
X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 199.3 mm. Round z to two decimals.
μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 199.3 mm
P =
20.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 20% is to the right of it. Find the cut-off x.
μ = 500 hσ = 40 h20%
x = h
3Solutions
1.Limits
Answer: L = −2.25
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) = −9x/(√(4x2 − 9x) + 2x)
Divide by x
f(x) = −9/(√(4 + (−9)/x) + 2)
Let x → ∞
L = −9/(2 + 2) = −2.25
Answer
L = −2.25
2.Limits
Answer: L = 4.167
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
5·sin 5x/(6x) → 0/0
Second application
25·cos 5x/6
Substitute x = 0
L = 25/6 = 4.167
Answer
L = 4.167
3.Derivatives
Answer: f′(1) = 8.155
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′ = ex·(2x + 1·x2)
Substitute
f′(1) = 2.718·(2 + 1) = 8.155
Answer
f′(1) = 8.155
4.Derivatives
Answer: b = −3
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(−2) = 11
f′(x) = x − 5 → f′(−2) = −7
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 11 − (−7)·(−2) = −3
Answer
b = −3
5.Integrals
Answer: I = 1.742
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 = x, du = dx; dv = sin x dx, v = −cos x
Antiderivative
∫x sin x dx = −x cos x + ∫cos x dx = −x cos x + sin x
Limits
(−2·cos 2 + sin 2) − (0 + 0) = 1.742
Answer
I = 1.742
6.Integrals
Answer: A = 36
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
9 − x² = 0 → x = ±3
Integrate
A = [9x − x³/3]−33 = 18 − (−18) = 36
Answer
A = 36
7.Complex numbers
Answer: Im(w) = 4.589
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 · 4 = 8
θ = 100° + 45° = 145°
Rectangular form
Re(w) = 8·cos(145°) = −6.553
Im(w) = 8·sin(145°) = 4.589
Im(w) = 4.589
Answer
Im(w) = 4.589
8.Complex numbers
Answer: Re(z₂) = −1.827
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.
Modulus of the roots
321/5 = 2
Argument of the root
θ₂ = (60° + 360°·2)/5 = 156°
Rectangular form
z₂ = 2·(cos 156° + j sin 156°) = −1.827 + j0.8135
Re(z₂) = −1.827
Answer
Re(z₂) = −1.827
The 5 roots lie on the circle of radius 2, 72° apart.
9.Differential equations
Answer: y(1.5) = 0.5974
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(1.5) = (3 + 6·1.5)·e−3 = 0.5974
Answer
y(1.5) = 0.5974
10.Differential equations
Answer: m = 16.59 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 = 4·0.05 − (4/1000)·m = 0.2 − 0.004·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 50 + (−40)·e−t/250
Substitute t
m(45) = 50 + (−40)·0.8353 = 16.59 kg
Answer
m = 16.59 kg
11.Matrices & determinants
Answer: c12 = −46
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·(−4) + (−5)·6 = −46
Answer
c12 = −46
12.Matrices & determinants
Answer: x = −1
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 = 2·2 − 4·5 = −16
Replace the x column
det Aₓ = (−4)·2 − 4·(−6) = 16
x = 16/(−16) = −1
Answer
x = −1
y = −0.5
13.Vectors: dot & cross product
Answer: k = 5.167
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 = 5·(−5) + 6·k + (−1)·6 = 0
−31 + 6·k = 0
Solve for k
k = −(−31)/6 = 5.167
Answer
k = 5.167
14.Vectors: dot & cross product
Answer: n̂x = −0.5345
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 = 5·2 − (−5)·2 = 20
(a × b)y = azbx − axbz = (−5)·4 − (−5)·2 = −10
(a × b)z = axby − aybx = (−5)·2 − 5·4 = −30
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.
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 = −2
Standard normal table
Φ(−2) = 0.02275
P = 0.02275
Answer
P = 0.02275
19.Normal distribution
Answer: P = 0.9192
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 sum
μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm
Standardise
z = (199.3 − 200)/0.5 = −1.4
Upper tail
P = 1 − Φ(−1.4) = 1 − 0.08076 = 0.9192
Answer
P = 0.9192
20.Normal distribution
Answer: x = 534 h
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.