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→∞ (√(x2 + 8x) − x).
f(x) = √(x2 + 8x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e−2x − 1 + 2x)/x².
f(x) = (e−2x − 1 + 2x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = ln(x2 + 1). Find f′(−1).
f(x) = ln(x2 + 1)x = −1
f′(−1) =
4.DerivativesHard
Let f(x) = x3 − 12x2 + 45x + 2. Find the x-coordinate x of the local maximum.
f(x) = x3 − 12x2 + 45x + 2
x =
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 cone).
y = x0 ≤ 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
The diagram shows z = 2 − j3. With w = 0.5∠90°, click the point that represents z/w.
z = 2 − j3w = 0.5∠90°
1A (6 + j4)
2B (−3 − j2)
3C (4 − j6)
4D (−6 − j4)
8.Complex numbersHard
One root of z³ − 7z² + 24z + 32 = 0 is z₁ = 4 + j4. The coefficients are real. Find the real root r.
z₁ = 4 + j4
r =
9.Differential equationsMedium
A can of soda at 3 °C is taken out of the fridge into a room at 26 °C. After 6 min its temperature is 8 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 3 °CTa = 26 °CT(6) = 8 °C
k = min⁻¹
10.Differential equationsHard
Solve y″ + 2y′ + 5y = 0 with y(0) = 2 and y′(0) = 0 (released from rest). Find y(0.75); angles in radians.
y″ + 2y′ + 5y = 0y(0) = 2y′(0) = 0
y(0.75) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [0 −2 −2; −5 4 4; −5 −3 4]
det A =
12.Matrices & determinantsHard
Solve the system −6x − 4y = −10 and 4x + 5y = −9. Find x.
−6x − 4y = −104x + 5y = −9
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−4, −2, −5) and b = (0, −4, 4), in degrees.
a = (−4, −2, −5)b = (0, −4, 4)
θ = °
14.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (−5, −3, 0) and b = (2, 1, −2).
a = (−5, −3, 0)b = (2, 1, −2)
A =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 10, 20, 30, 40p(x): 0.2, 0.3, 0.15, 0.35
σ =
16.ProbabilityHard
A game costs €3 to play. You win €6 with probability 0.1 and €150 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€6: p = 0.1€150: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 46, 56, 19, 49, 25, 27, 14.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 8. What proportion P of values is below 112?
μ = 100σ = 8x = 112
P =
19.Normal distributionMedium
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 242 and 254 g? Give P as a decimal.
μ = 250σ = 5x1 = 242x2 = 254
P =
20.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 97.5% of them lie (the 97.5th percentile).
μ = 32σ = 6P = 97.5%
x = min
3Solutions
1.Limits
Answer: L = 4
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) = 8x/(√(x2 + 8x) + x)
Divide by x
f(x) = 8/(√(1 + 8/x) + 1)
Let x → ∞
L = 8/(1 + 1) = 4
Answer
L = 4
2.Limits
Answer: L = 2
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
(−2·e−2x + 2)/(2x) → 0/0
Second application
4·e−2x/2
Substitute x = 0
L = 4·1/2 = 2
Answer
L = 2
3.Derivatives
Answer: f′(−1) = −1
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) = 2x/(x2 + 1)
Substitute
f′(−1) = −2/2 = −1
Answer
f′(−1) = −1
4.Derivatives
Answer: x = 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.
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 = 3
Answer
x = 3
5.Integrals
Answer: V = 8.378
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 1x² dx
Integrate
V = π·1·2³/3 = 8.378
Answer
V = 8.378
Check: a cone of radius 2 and height 2 has (1/3)πr²h = 8.378.
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: D (−6 − j4)
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| = |z|/0.5, arg(z/w) = arg z − 90°
The point
z/w = −6 − j4 → D
Answer
D (−6 − j4)
8.Complex numbers
Answer: r = −1
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 + (−1) = 7 = −(−7).
9.Differential equations
Answer: k = 0.04085 min⁻¹
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.
Ratio of temperature differences
e−6k = -18/-23
Solve for k
k = ln(-23/-18)/6 = 0.04085 min⁻¹
Answer
k = 0.04085 min⁻¹
10.Differential equations
Answer: y(0.75) = 0.538
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 = 2, B = 1·2/2 = 1
Evaluate
y(0.75) = e−0.75(2·cos 1.5 + 1·sin 1.5) = 0.538
Answer
y(0.75) = 0.538
11.Matrices & determinants
Answer: det A = −70
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 = 28, M₁₂ = 0, M₁₃ = 35
Cofactor expansion
det A = 0·28 − (−2)·0 + (−2)·35 = −70
Answer
det A = −70
12.Matrices & determinants
Answer: x = 6.143
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 = (−6)·5 − (−4)·4 = −14
Replace the x column
det Aₓ = (−10)·5 − (−4)·(−9) = −86
x = −86/(−14) = 6.143
Answer
x = 6.143
y = −6.714
13.Vectors: dot & cross product
Answer: θ = 108.4 °
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 = (−4)·0 + (−2)·(−4) + (−5)·4 = −12
Magnitudes
|a| = √((−4)² + (−2)² + (−5)²) = 6.708
|b| = √(0² + (−4)² + 4²) = 5.657
Angle
cos θ = −12/(6.708·5.657) = −0.3162
θ = arccos(−0.3162) = 108.4°
Answer
θ = 108.4°
14.Vectors: dot & cross product
Answer: A = 11.7
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 = (−3)·(−2) − 0·1 = 6
(a × b)y = azbx − axbz = 0·2 − (−5)·(−2) = −10
(a × b)z = axby − aybx = (−5)·1 − (−3)·2 = 1
Area
A = |a × b| = √(6² + (−10)² + 1²) = 11.7
Answer
A = 11.7
15.Probability
Answer: σ = 11.52
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.
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.
Net outcomes
G: 3 (p = 0.1), 147 (p = 0.05), −3 (p = 0.85)
Expected value
E(G) = 3·0.1 + 147·0.05 − 3·0.85 = 5.1 €
Answer
E(G) = 5.1 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 16.37
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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 0.9332
Answer
P = 0.9332
19.Normal distribution
Answer: P = 0.7333
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 = (242 − 250)/5 = −1.6
z₂ = (254 − 250)/5 = 0.8
Area between
P = Φ(0.8) − Φ(−1.6) = 0.7881 − 0.0548 = 0.7333
Answer
P = 0.7333
20.Normal distribution
Answer: x = 43.76 min
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.