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 + 0.5/x)x. Give a decimal value.
f(x) = (1 + 0.5/x)xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 2x)/(2x²), with x in radians.
f(x) = (1 − cos 2x)/(2x²)x → 0
L =
3.DerivativesMedium
Let f(x) = (2x − 3)4. Find f′(−1).
f(x) = (2x − 3)4x = −1
f′(−1) =
4.DerivativesHard
The tangent line to f(x) = 3x2 + 1 at x = −1 crosses the y-axis at y = b. Find b.
f(x) = 3x2 + 1x = −1
b =
5.IntegralsMedium
Evaluate I = ∫03 e0.5x dx.
f(x) = e0.5x[0, 3]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = 2x.
y = x2y = 2x
A =
7.Complex numbersMedium
Write z = −4 − j5 in exponential form r·ejθ. Give θ in radians, between −π and π.
z = −4 − j5
θ = rad
8.Complex numbersHard
The roots of zⁿ = w with n = 2 and w = 25∠0° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1. Find the real part of z₁.
w = 25∠0°n = 2k = 1
Re(z₁) =
9.Differential equationsMedium
A cup of coffee at 82 °C is left in a room at 18 °C. After 12 min its temperature is 33 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 25 °C?
T0 = 82 °CTa = 18 °CT(12) = 33 °CT = 25 °C
t = min
10.Differential equationsHard
A tank holds 1000 L of water in which 10 kg of salt is dissolved. Brine with 0.1 kg of salt per litre flows in at 20 L/min, and the well-mixed solution flows out at the same rate. After how many minutes t does the tank contain 82 kg of salt?
V = 1000 Lm0 = 10 kgc = 0.1 kg/Lr = 20 L/min
t = min
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [1 −5 −5; 1 −5 −2; −2 4 3]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−3 −3; 8 8]
λ =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (2, −2, −7) and b = (2, −4, −2).
a = (2, −2, −7)b = (2, −4, −2)
a·b =
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (−2, −4, 1), b = (1, 2, 1) and c = (−4, −5, 2).
a = (−2, −4, 1)b = (1, 2, 1)c = (−4, −5, 2)
V =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2p(x): 0.3, 0.35, 0.35
σ =
16.ProbabilityHard
A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 5 correct answers?
n = 6p = 0.1k = 5
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 28, 19, 29, 17, 35, 50, 49.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 12. What proportion P of values is below 38?
μ = 50σ = 12x = 38
P =
19.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 78.25 and 112.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 78.25x2 = 112.75
P =
20.Normal distributionHard
Hole diameters D are normal with mean 25.05 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.05 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
3Solutions
1.Limits
Answer: L = 1.649
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→∞ x·ln(1 + 0.5/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 1·0.5 = 0.5
Undo the logarithm
L = e0.5 = 1.649
Answer
L = 1.649
2.Limits
Answer: L = 1
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·sin 2x/(4x) → 0/0
Second application
4·cos 2x/4
Substitute x = 0
L = 4/4 = 1
Answer
L = 1
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·(−5)3·2 = −1000
Answer
f′(−1) = −1000
4.Derivatives
Answer: b = −2
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(−1) = 4
f′(x) = 6x → f′(−1) = −6
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 4 − (−6)·(−1) = −2
Answer
b = −2
5.Integrals
Answer: I = 6.963
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.
Antiderivative
F(x) = e0.5x/0.5
Evaluate at the limits
I = (e1.5 − 1)/0.5 = (4.482 − 1)/0.5 = 6.963
Answer
I = 6.963
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.
Intersection points
x2 = 2x → x = 0, x = 2
Integrate upper minus lower
A = [x2 − 0.3333x3]02 = 1.333
Answer
A = 1.333
7.Complex numbers
Answer: θ = −2.246 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 = √(16 + 25) = 6.403
Argument in radians
θ = −128.7° = −128.7·π/180 = −2.246 rad
Exponential form
z = 6.403·e−j2.246
Answer
θ = −2.246 rad
8.Complex numbers
Answer: Re(z₁) = −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.
Modulus of the roots
251/2 = 5
Argument of the root
θ₁ = (0° + 360°·1)/2 = 180°
Rectangular form
z₁ = 5·(cos 180° + j sin 180°) = −5
Re(z₁) = −5
Answer
Re(z₁) = −5
The 2 roots lie on the circle of radius 5, 180° apart.
9.Differential equations
Answer: t = 18.3 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(64/15)/12 = 0.1209 min⁻¹
Time to reach the target
t = ln(64/7)/0.1209 = 18.3 min
Answer
t = 18.3 min
10.Differential equations
Answer: t = 80.47 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.
Result
m(t) = 100 + (−90)·e−t/50
Solve for t
e−t/50 = (100 − 82)/(100 − 10) = 0.2
t = 50·ln(5) = 80.47 min
Answer
t = 80.47 min
11.Matrices & determinants
Answer: det A = 18
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 = −7, M₁₂ = −1, M₁₃ = −6
Cofactor expansion
det A = 1·(−7) − (−5)·(−1) + (−5)·(−6) = 18
Answer
det A = 18
12.Matrices & determinants
Answer: λ = 5
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 − (5)λ + (0) = 0
Solve
λ = (5 ± √(25))/2 → λ = 0, 5
λ = 5
Answer
λ = 5
13.Vectors: dot & cross product
Answer: a·b = 26
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: V = 9
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 b × c
(b × c)x = bycz − bzcy = 2·2 − 1·(−5) = 9
(b × c)y = bzcx − bxcz = 1·(−4) − 1·2 = −6
(b × c)z = bxcy − bycx = 1·(−5) − 2·(−4) = 3
Dot product with a
a·(b × c) = (−2)·9 + (−4)·(−6) + 1·3 = 9
V = |9| = 9
Answer
V = 9
15.Probability
Answer: σ = 0.8047
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.
Mean
μ = 0·0.3 + 1·0.35 + 2·0.35 = 1.05
Mean of the squares
E(X²) = 0·0.3 + 1·0.35 + 4·0.35 = 1.75
Variance
Var(X) = 1.75 − 1.05² = 0.6475
σ = √0.6475 = 0.8047
Answer
σ = 0.8047
16.Probability
Answer: P = 5.4 × 10⁻⁵
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 = −1
Standard normal table
Φ(−1) = 0.1587
P = 0.1587
Answer
P = 0.1587
19.Normal distribution
Answer: P = 0.7288
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.
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).