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 (1 − cos 5x)/(4x²), with x in radians.
f(x) = (1 − cos 5x)/(4x²)x → 0
L =
3.DerivativesMedium
Let f(x) = (2x + 3)4. Find f′(1).
f(x) = (2x + 3)4x = 1
f′(1) =
4.DerivativesHard
Let f(x) = x3 − 7.5x2 + 12x − 3. Find the x-coordinate x of the local minimum.
f(x) = x3 − 7.5x2 + 12x − 3
x =
5.IntegralsMedium
Evaluate I = ∫01 e−0.5x dx.
f(x) = e−0.5x[0, 1]
I =
6.IntegralsHard
The region between the line y = 5x and the parabola y = 2x² is rotated about the x-axis. Find the volume V of the solid.
y = 5xy = 2x²
V =
7.Complex numbersMedium
The diagram shows z = 3 − j4. With w = 0.5∠90°, click the point that represents z/w.
z = 3 − j4w = 0.5∠90°
1A (−8 − j6)
2B (−4 − j3)
3C (6 − j8)
4D (8 + j6)
8.Complex numbersHard
The roots of zⁿ = w with n = 2 and w = 36∠90° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1. Find the real part of z₁.
w = 36∠90°n = 2k = 1
Re(z₁) =
9.Differential equationsMedium
A forged steel part at 400 °C is left to cool in a workshop at 24 °C. After 47 min its temperature is 116 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 52 °C?
T0 = 400 °CTa = 24 °CT(47) = 116 °CT = 52 °C
t = min
10.Differential equationsHard
Solve the initial value problem y′ + y/x = 6x for x > 0, with y(1) = 2. Find y(2).
y′ + y/x = 6xy(1) = 2
y(2) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−5 −5 3; 5 3 5; 3 −2 5]
det A =
12.Matrices & determinantsHard
Solve the system −x + 2y = −4 and 4x + 4y = −4. Find x.
−x + 2y = −44x + 4y = −4
x =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (6, 1, −3) and b = (6, −5, −5).
a = (6, 1, −3)b = (6, −5, −5)
a·b =
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (2, −1, −1) and b = (4, −5, −2) and has a positive z-component. Find its x-component n̂x.
a = (2, −1, −1)b = (4, −5, −2)n̂z > 0
n̂x =
15.ProbabilityMedium
In how many ways N can a committee of 4 people be chosen from a group of 7? The order does not matter.
n = 7k = 4
N =
16.ProbabilityHard
A game costs €3 to play. You win €12 with probability 0.1 and €30 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€12: p = 0.1€30: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 51, 21, 30, 53, 60, 37, 60.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 5. What proportion P of values is below 165?
μ = 170σ = 5x = 165
P =
19.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 20.3 and 38 min? Give P as a decimal.
μ = 32σ = 6x1 = 20.3x2 = 38
P =
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 = 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 = 3.125
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/(8x) → 0/0
Second application
25·cos 5x/8
Substitute x = 0
L = 25/8 = 3.125
Answer
L = 3.125
3.Derivatives
Answer: f′(1) = 1000
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
f′(x) = 4·(2x + 3)3·2
Substitute
f′(1) = 4·53·2 = 1000
Answer
f′(1) = 1000
4.Derivatives
Answer: x = 4
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 − 15x + 12 = 3·(x2 − 5x + 4)
Solve f′(x) = 0
x = 1 or x = 4
Which one is which
f″(x) = 6x − 15: negative at a maximum, positive at a minimum.
f″(1) = −9, f″(4) = 9
x = 4
Answer
x = 4
5.Integrals
Answer: I = 0.7869
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.
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
5x = 2x² → x = 0, x = 2.5
Washer method
V = π∫₀2.5 ((5x)² − (2x²)²) dx
Integrate
V = π(25·2.5³/3 − 4·2.5⁵/5) = π·52.08 = 163.6
Answer
V = 163.6
7.Complex numbers
Answer: A (−8 − 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| = |z|/0.5, arg(z/w) = arg z − 90°
The point
z/w = −8 − j6 → A
Answer
A (−8 − j6)
8.Complex numbers
Answer: Re(z₁) = −4.243
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.
The 2 roots lie on the circle of radius 6, 180° apart.
9.Differential equations
Answer: t = 86.71 min
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.
Cooling constant
k = ln(376/92)/47 = 0.02995 min⁻¹
Time to reach the target
t = ln(376/28)/0.02995 = 86.71 min
Answer
t = 86.71 min
10.Differential equations
Answer: y(2) = 8
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.
Integrating factor
μ = e∫1/x dx = x1
Integrate
(x1y)′ = 6x2 → x1y = 6x3/3 + C
Initial condition
2 = 6/3 + C → C = 0
Evaluate
y(2) = (6·23/3 + 0)/21 = 8
Answer
y(2) = 8
11.Matrices & determinants
Answer: det A = −132
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 = 25, M₁₂ = 10, M₁₃ = −19
Cofactor expansion
det A = (−5)·25 − (−5)·10 + 3·(−19) = −132
Answer
det A = −132
12.Matrices & determinants
Answer: x = 0.6667
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 = (−1)·4 − 2·4 = −12
Replace the x column
det Aₓ = (−4)·4 − 2·(−4) = −8
x = −8/(−12) = 0.6667
Answer
x = 0.6667
y = −1.667
13.Vectors: dot & cross product
Answer: a·b = 46
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.
Positive: the angle between the vectors is less than 90°.
14.Vectors: dot & cross product
Answer: n̂x = 0.4472
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.
The z-component of a × b is negative, so use −(a × b).
|a × b| = √((−3)² + 0² + (−6)²) = 6.708
n̂ = −(−3, 0, −6)/6.708 = (0.4472, 0, 0.8944)
n̂x = 0.4472
Answer
n̂x = 0.4472
15.Probability
Answer: N = 35
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(7, 4) = (7·6·5·4)/(4·3·2·1) = 35
Answer
N = 35
16.Probability
Answer: E(G) = −1.5 €
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: 9 (p = 0.1), 27 (p = 0.01), −3 (p = 0.89)
Expected value
E(G) = 9·0.1 + 27·0.01 − 3·0.89 = −1.5 €
Answer
E(G) = −1.5 €
A negative value means you lose money on average.
17.Descriptive statistics
Answer: s = 15.35
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
Standard normal table
Φ(−1) = 0.1587
P = 0.1587
Answer
P = 0.1587
19.Normal distribution
Answer: P = 0.8158
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 = (20.3 − 32)/6 = −1.95
z₂ = (38 − 32)/6 = 1
Area between
P = Φ(1) − Φ(−1.95) = 0.8413 − 0.02559 = 0.8158
Answer
P = 0.8158
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).