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 − 10x) − 2x).
f(x) = √(4x2 − 10x) − 2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e−x − 1 + x)/x².
f(x) = (e−x − 1 + x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(3x), with x in radians. Find f′(1.5).
f(x) = x·sin(3x)x = 1.5 rad
f′(1.5) =
4.DerivativesHard
The tangent line to f(x) = x2 + 5x − 3 at x = 2 crosses the y-axis at y = b. Find b.
f(x) = x2 + 5x − 3x = 2
b =
5.IntegralsMedium
The region under y = √(x) from x = 0 to x = 2 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(x)0 ≤ x ≤ 2
V =
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
Write z = −2 − j5 in exponential form r·ejθ. Give θ in radians, between −π and π.
z = −2 − j5
θ = rad
8.Complex numbersHard
Exactly one of the points A–D is a root of z³ = 27∠60°. Click it.
w = 27∠60°n = 3
1A (3∠100°)
2B (3∠−70°)
3C (5.4∠−100°)
4D (3∠−100°)
9.Differential equationsMedium
A can of soda at 3 °C is taken out of the fridge into a room at 30 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.06 min⁻¹, find its temperature T after 27 min.
T0 = 3 °CTa = 30 °Ck = 0.06 min⁻¹t = 27 min
T = °C
10.Differential equationsHard
Solve y″ + 2y′ + 5y = 0 with y(0) = 1 and y′(0) = 0 (released from rest). Find y(1.5); angles in radians.
y″ + 2y′ + 5y = 0y(0) = 1y′(0) = 0
y(1.5) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [5 −5 −5; −5 5 −4; 5 4 −1]
det A =
12.Matrices & determinantsHard
Solve the system −5x + 2y = 11 and x − 5y = 3. Find x.
−5x + 2y = 11x − 5y = 3
x =
13.Vectors: dot & cross productMedium
A constant force F = (15, 15, 30) N acts on an object while it moves in a straight line from point A(4, 3, 3) to point B(2, −3, 3). Coordinates are in metres. Find the work W done by the force.
F = (15, 15, 30) NA(4, 3, 3) mB(2, −3, 3) m
W = J
14.Vectors: dot & cross productHard
For a = (6, −2, −5) and b = (−2, 3, 0), find the y-component (a × b)y of the cross product a × b.
a = (6, −2, −5)b = (−2, 3, 0)
(a × b)y =
15.ProbabilityMedium
A bag holds 6 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
6 / 4
P =
16.ProbabilityHard
X has the distribution shown, and Y = 5X + 4. Find the standard deviation σY of Y.
x: 1, 2, 3, 4p(x): 0.25, 0.3, 0.1, 0.35Y = 5X + 4
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 25, 18, 21, 53, 39, 27.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 20. What proportion P of values is above 195?
μ = 170σ = 20x = 195
P =
19.Normal distributionMedium
The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 12 of them exceeds 142.6 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 12> 142.6 kg
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 80% is to the right of it. Find the cut-off x.
μ = 500 hσ = 40 h80%
x = h
3Solutions
1.Limits
Answer: L = −2.5
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).
Multiply by the conjugate
f(x) = −10x/(√(4x2 − 10x) + 2x)
Divide by x
f(x) = −10/(√(4 + (−10)/x) + 2)
Let x → ∞
L = −10/(2 + 2) = −2.5
Answer
L = −2.5
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·e−x + 1)/(2x) → 0/0
Second application
1·e−x/2
Substitute x = 0
L = 1·1/2 = 0.5
Answer
L = 0.5
3.Derivatives
Answer: f′(1.5) = −1.926
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(3x) + 3x·cos(3x)
Substitute
f′(1.5) = −0.9775 + 4.5·(−0.2108) = −1.926
Answer
f′(1.5) = −1.926
4.Derivatives
Answer: b = −7
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) = 2x + 5 → f′(2) = 9
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 11 − 9·2 = −7
Answer
b = −7
5.Integrals
Answer: V = 6.283
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.
Disc method
V = π∫₀2 x dx
Integrate
V = π·1·2²/2 = 6.283
Answer
V = 6.283
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 = −1.951 rad
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
r = √(4 + 25) = 5.385
Argument in radians
θ = −111.8° = −111.8·π/180 = −1.951 rad
Exponential form
z = 5.385·e−j1.951
Answer
θ = −1.951 rad
8.Complex numbers
Answer: D (3∠−100°)
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 and arguments of the roots
|z| = 3
θ = 20°, 140°, −100°
The point
D = 3∠−100°
Answer
D (3∠−100°)
9.Differential equations
Answer: T = 24.66 °C
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.
Solution of the cooling law
T(t) = 30 + (−27)·e−0.06t
Substitute t
T(27) = 30 + (−27)·0.1979 = 24.66 °C
Answer
T = 24.66 °C
10.Differential equations
Answer: y(1.5) = −0.2052
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.
Complex roots
r = −1 ± 2j
Constants
A = 1, B = 1·1/2 = 0.5
Evaluate
y(1.5) = e−1.5(1·cos 3 + 0.5·sin 3) = −0.2052
Answer
y(1.5) = −0.2052
11.Matrices & determinants
Answer: det A = 405
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 = 11, M₁₂ = 25, M₁₃ = −45
Cofactor expansion
det A = 5·11 − (−5)·25 + (−5)·(−45) = 405
Answer
det A = 405
12.Matrices & determinants
Answer: x = −2.652
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 = (−5)·(−5) − 2·1 = 23
Replace the x column
det Aₓ = 11·(−5) − 2·3 = −61
x = −61/23 = −2.652
Answer
x = −2.652
y = −1.13
13.Vectors: dot & cross product
Answer: W = −120 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 = (2, −3, 3) − (4, 3, 3) = (−2, −6, 0) m
Work
W = F·d = 15·(−2) + 15·(−6) + 30·0 = −120 J
Answer
W = −120 J
Negative work: the force acts mostly against the motion.
14.Vectors: dot & cross product
Answer: (a × b)y = 10
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 = (−2)·0 − (−5)·3 = 15
(a × b)y = azbx − axbz = (−5)·(−2) − 6·0 = 10
(a × b)z = axby − aybx = 6·3 − (−2)·(−2) = 14
Requested component
(a × b)y = 10
Answer
(a × b)y = 10
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: P = 0.3333
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 = 6/10 · 5/9 = 30/90 = 0.3333
Answer
P = 0.3333
16.Probability
Answer: σY = 6.016
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.
Spread of X
μ = 2.55, Var(X) = 1.447, σX = 1.203
Linear change
σY = |5|·1.203 = 6.016
Answer
σY = 6.016
The mean does move: E(Y) = 5·2.55 + 4 = 16.75.
17.Descriptive statistics
Answer: s = 13.17
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 = 1 − 0.8944 = 0.1056
Answer
P = 0.1056
19.Normal distribution
Answer: P = 0.695
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 total
μT = 144 kg, σT = 0.8·√12 = 2.771 kg
Standardise
z = (142.6 − 144)/2.771 = −0.51
Probability
P = 1 − Φ(−0.51) = 0.695
Answer
P = 0.695
20.Normal distribution
Answer: x = 466 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.