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 + 1/x)2x. Give a decimal value.
f(x) = (1 + 1/x)2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 4x)/(4x²), with x in radians.
f(x) = (1 − cos 4x)/(4x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(2x), with x in radians. Find f′(1).
f(x) = x·sin(2x)x = 1 rad
f′(1) =
4.DerivativesHard
Let f(x) = x3 − 10.5x2 + 36x − 1. Find the x-coordinate x of the local maximum.
f(x) = x3 − 10.5x2 + 36x − 1
x =
5.IntegralsMedium
Evaluate I = ∫03 e−x dx.
f(x) = e−x[0, 3]
I =
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 = 5∠160° in rectangular form a + jb and give the real part a.
r = 5θ = 160°
a =
8.Complex numbersHard
Use De Moivre's theorem to find the real part of z⁷ for z = −1 + j.
z = −1 + jn = 7
Re(z⁷) =
9.Differential equationsMedium
A can of soda at 5 °C is taken out of the fridge into a room at 26 °C. After 9 min its temperature is 17 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 5 °CTa = 26 °CT(9) = 17 °C
k = min⁻¹
10.Differential equationsHard
Solve the initial value problem y′ + 2y/x = 5x2 for x > 0, with y(1) = 0. Find y(2).
y′ + 2y/x = 5x2y(1) = 0
y(2) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c22 (row 2, column 2).
A = [1 −3; 2 −6]B = [3 −6; −6 −3]
c22 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−3 −2; 6 5]
λ =
13.Vectors: dot & cross productMedium
A constant force F = (−10, 20, 5) N acts on an object while it moves in a straight line from point A(0, −4, 0) to point B(−1, −3, 2). Coordinates are in metres. Find the work W done by the force.
F = (−10, 20, 5) NA(0, −4, 0) mB(−1, −3, 2) m
W = J
14.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (4, 1, 5) and b = (−5, −4, 0).
a = (4, 1, 5)b = (−5, −4, 0)
A =
15.ProbabilityMedium
In how many ways N can a committee of 3 people be chosen from a group of 14? The order does not matter.
n = 14k = 3
N =
16.ProbabilityHard
A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 4 correct answers?
n = 12p = 0.6k = 4
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 24, 45, 39, 10, 34, 12.
n = 6
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 10. Find the z-score z of a score of 110.
μ = 100σ = 10x = 110
z =
19.Normal distributionMedium
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 167.85 and 183.95 cm? Give P as a decimal.
μ = 178σ = 7x1 = 167.85x2 = 183.95
P =
20.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 10% of them lie (the 10th percentile).
μ = 178σ = 7P = 10%
x = cm
3Solutions
1.Limits
Answer: L = 7.389
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→∞ 2·x·ln(1 + 1/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 2·1 = 2
Undo the logarithm
L = e2 = 7.389
Answer
L = 7.389
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
4·sin 4x/(8x) → 0/0
Second application
16·cos 4x/8
Substitute x = 0
L = 16/8 = 2
Answer
L = 2
3.Derivatives
Answer: f′(1) = 0.077
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(2x) + 2x·cos(2x)
Substitute
f′(1) = 0.9093 + 2·(−0.4161) = 0.077
Answer
f′(1) = 0.077
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 − 21x + 36 = 3·(x2 − 7x + 12)
Solve f′(x) = 0
x = 3 or x = 4
Which one is which
f″(x) = 6x − 21: negative at a maximum, positive at a minimum.
f″(3) = −3, f″(4) = 3
x = 3
Answer
x = 3
5.Integrals
Answer: I = 0.9502
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) = e−x/(−1)
Evaluate at the limits
I = (e−3 − 1)/(−1) = (0.04979 − 1)/(−1) = 0.9502
Answer
I = 0.9502
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: a = −4.698
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.
Components
a = 5·cos(160°) = −4.698
b = 5·sin(160°) = 1.71
Rectangular form
z = −4.698 + j1.71
a = −4.698
Answer
a = −4.698
8.Complex numbers
Answer: Re(z⁷) = −8
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.
Polar form
z = −1 + j = √2∠135°
De Moivre
z⁷ = (√2)⁷∠945° = 11.31∠−135°
Rectangular form
z⁷ = 11.31·(cos −135° + j sin −135°) = −8 − j8
Re(z⁷) = −8
Answer
Re(z⁷) = −8
9.Differential equations
Answer: k = 0.09414 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−9k = -9/-21
Solve for k
k = ln(-21/-9)/9 = 0.09414 min⁻¹
Answer
k = 0.09414 min⁻¹
10.Differential equations
Answer: y(2) = 7.75
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∫2/x dx = x2
Integrate
(x2y)′ = 5x4 → x2y = 5x5/5 + C
Initial condition
0 = 5/5 + C → C = −1
Evaluate
y(2) = (5·25/5 + (−1))/22 = 7.75
Answer
y(2) = 7.75
11.Matrices & determinants
Answer: c22 = 6
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.
Row times column
c22 = 2·(−6) + (−6)·(−3) = 6
Answer
c22 = 6
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 − (2)λ + (−3) = 0
Solve
λ = (2 ± √(16))/2 → λ = −1, 3
λ = 3
Answer
λ = 3
13.Vectors: dot & cross product
Answer: W = 40 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 = (−1, −3, 2) − (0, −4, 0) = (−1, 1, 2) m
Work
W = F·d = (−10)·(−1) + 20·1 + 5·2 = 40 J
Answer
W = 40 J
14.Vectors: dot & cross product
Answer: A = 33.85
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 = 1·0 − 5·(−4) = 20
(a × b)y = azbx − axbz = 5·(−5) − 4·0 = −25
(a × b)z = axby − aybx = 4·(−4) − 1·(−5) = −11
Area
A = |a × b| = √(20² + (−25)² + (−11)²) = 33.85
Answer
A = 33.85
15.Probability
Answer: N = 364
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(14, 3) = (14·13·12)/(3·2·1) = 364
Answer
N = 364
16.Probability
Answer: P = 0.04204
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 = (110 − 100)/10 = 1
Answer
z = 1
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.
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.