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→∞ (1 − 2/x)3x. Give a decimal value.
f(x) = (1 − 2/x)3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 3x)/(2x²), with x in radians.
f(x) = (1 − cos 3x)/(2x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x2·e−0.5x. Find f′(1.5).
f(x) = x2·e−0.5xx = 1.5
f′(1.5) =
4.DerivativesHard
Let f(x) = x3 − 10.5x2 + 30x − 1. Find the x-coordinate x of the local minimum.
f(x) = x3 − 10.5x2 + 30x − 1
x =
5.IntegralsMedium
Evaluate I = ∫₀π x·cos x dx (x in radians).
∫₀π x·cos x dx
I =
6.IntegralsHard
The region between the line y = 4x and the parabola y = 2x² is rotated about the x-axis. Find the volume V of the solid.
y = 4xy = 2x²
V =
7.Complex numbersMedium
The diagram shows z = −3 + j3. With w = 2∠90°, click the point that represents z·w.
z = −3 + j3w = 2∠90°
1A (−6 + j6)
2B (6 + j6)
3C (−3 − j3)
4D (−6 − j6)
8.Complex numbersHard
Solve z² + 3z + 8.5 = 0. The roots are a ± jb; find the real part a.
p = 3q = 8.5
a =
9.Differential equationsMedium
Solve y″ + 6y′ + 9y = 0 with y(0) = 3 and y′(0) = 2. Find y(1).
y″ + 6y′ + 9y = 0y(0) = 3y′(0) = 2
y(1) =
10.Differential equationsHard
A tank holds 1000 L of water in which 10 kg of salt is dissolved. Brine with 0.2 kg of salt per litre flows in at 5 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 105 min?
V = 1000 Lm0 = 10 kgc = 0.2 kg/Lr = 5 L/min
m = kg
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−5 1 2; 4 −3 5; −3 4 −4]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [0 3; −4 −7]
λ =
13.Vectors: dot & cross productMedium
A constant force F = (5, −5, 30) N acts on an object while it moves in a straight line from point A(−5, 2, −2) to point B(3, −1, 2). Coordinates are in metres. Find the work W done by the force.
F = (5, −5, 30) NA(−5, 2, −2) mB(3, −1, 2) m
W = J
14.Vectors: dot & cross productHard
For a = (6, 5, −4) and b = (−4, −4, −1), find the z-component (a × b)z of the cross product a × b.
a = (6, 5, −4)b = (−4, −4, −1)
(a × b)z =
15.ProbabilityMedium
In how many ways N can a committee of 4 people be chosen from a group of 11? The order does not matter.
n = 11k = 4
N =
16.ProbabilityHard
6 components are tested, and each one fails with probability 0.5, independently. What is the probability P that at most one fails?
n = 6p = 0.5
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 28, 34, 50, 17, 52, 28, 59.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is above 42?
μ = 50σ = 10x = 42
P =
19.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.8212. Find z to two decimals.
A = 0.8212
z =
20.Normal distributionHard
The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
3Solutions
1.Limits
Answer: L = 0.002479
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).
Take the logarithm
ln L = limx→∞ 3·x·ln(1 − 2/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 3·(−2) = −6
Undo the logarithm
L = e−6 = 0.002479
Answer
L = 0.002479
2.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).
First application
3·sin 3x/(4x) → 0/0
Second application
9·cos 3x/4
Substitute x = 0
L = 9/4 = 2.25
Answer
L = 2.25
3.Derivatives
Answer: f′(1.5) = 0.8857
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′ = e−0.5x·(2x + (−0.5)·x2)
Substitute
f′(1.5) = 0.4724·(3 + (−1.125)) = 0.8857
Answer
f′(1.5) = 0.8857
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 − 21x + 30 = 3·(x2 − 7x + 10)
Solve f′(x) = 0
x = 2 or x = 5
Which one is which
f″(x) = 6x − 21: negative at a maximum, positive at a minimum.
f″(2) = −9, f″(5) = 9
x = 5
Answer
x = 5
5.Integrals
Answer: I = −2
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 = cos x dx, v = sin x
Antiderivative
∫x cos x dx = x sin x − ∫sin x dx = x sin x + cos x
Limits
(π·sin π + cos π) − (0 + 1) = −2
Answer
I = −2
6.Integrals
Answer: V = 53.62
Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.
Intersection points
4x = 2x² → x = 0, x = 2
Washer method
V = π∫₀2 ((4x)² − (2x²)²) dx
Integrate
V = π(16·2³/3 − 4·2⁵/5) = π·17.07 = 53.62
Answer
V = 53.62
7.Complex numbers
Answer: D (−6 − j6)
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| = 2·|z|, arg(z·w) = arg z + 90°
The point
z·w = −6 − j6 → D
Answer
D (−6 − j6)
8.Complex numbers
Answer: a = −1.5
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 = 9 − 34 = −25 < 0
Quadratic formula
z = (−3 ± j√25)/2 = −1.5 + j2.5, −1.5 − j2.5
Answer
a = −1.5
Answer
a = −1.5
9.Differential equations
Answer: y(1) = 0.697
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 + 3)² = 0 → r = −3
Constants
A = y(0) = 3
y′(0) = B − 3A → B = 2 + 9 = 11
Evaluate
y(1) = (3 + 11·1)·e−3 = 0.697
Answer
y(1) = 0.697
10.Differential equations
Answer: m = 87.6 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 = 5·0.2 − (5/1000)·m = 1 − 0.005·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 200 + (−190)·e−t/200
Substitute t
m(105) = 200 + (−190)·0.5916 = 87.6 kg
Answer
m = 87.6 kg
11.Matrices & determinants
Answer: det A = 55
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.
2×2 minors of the first row
M₁₁ = −8, M₁₂ = −1, M₁₃ = 7
Cofactor expansion
det A = (−5)·(−8) − 1·(−1) + 2·7 = 55
Answer
det A = 55
12.Matrices & determinants
Answer: λ = −3
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 − (−7)λ + (12) = 0
Solve
λ = (−7 ± √(1))/2 → λ = −4, −3
λ = −3
Answer
λ = −3
13.Vectors: dot & cross product
Answer: W = 175 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, 2) − (−5, 2, −2) = (8, −3, 4) m
Work
W = F·d = 5·8 + (−5)·(−3) + 30·4 = 175 J
Answer
W = 175 J
14.Vectors: dot & cross product
Answer: (a × b)z = −4
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 = 5·(−1) − (−4)·(−4) = −21
(a × b)y = azbx − axbz = (−4)·(−4) − 6·(−1) = 22
(a × b)z = axby − aybx = 6·(−4) − 5·(−4) = −4
Requested component
(a × b)z = −4
Answer
(a × b)z = −4
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: N = 330
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.
Combinations
C(11, 4) = (11·10·9·8)/(4·3·2·1) = 330
Answer
N = 330
16.Probability
Answer: P = 0.1094
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.
Zero and one failure
P(X = 0) = 0.56 = 0.01563
P(X = 1) = 6·0.5·0.55 = 0.09375
Add them
P = 0.01563 + 0.09375 = 0.1094
Answer
P = 0.1094
17.Descriptive statistics
Answer: s = 15.48
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 = 1 − 0.2119 = 0.7881
Answer
P = 0.7881
19.Normal distribution
Answer: z = −0.92
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.
Area to the left of z
Φ(z) = 1 − 0.8212 = 0.1788
Read the z-table backwards
Find 0.1788 inside the table and read off z.
z = −0.92
Answer
z = −0.92
20.Normal distribution
Answer: P = 0.7389
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).