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 − 7x) − x).
f(x) = √(x2 − 7x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 2x)/sin(3x), with x in radians.
f(x) = ln(1 + 2x)/sin(3x)x → 0
L =
3.DerivativesMedium
Let f(x) = x·e−x. Find f′(2).
f(x) = x·e−xx = 2
f′(2) =
4.DerivativesHard
Let f(x) = x3 − 15x2 + 63x + 6. Find the x-coordinate x of the local maximum.
f(x) = x3 − 15x2 + 63x + 6
x =
5.IntegralsMedium
Evaluate I = ∫₁1.5 x·ln x dx.
∫₁1.5 x ln x dx
I =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = x.
y = x2y = x
A =
7.Complex numbersMedium
Write z = 5∠125° in rectangular form a + jb and give the imaginary part b.
r = 5θ = 125°
b =
8.Complex numbersHard
One root of z³ + 5z² + 9z + 5 = 0 is z₁ = −2 + j. The coefficients are real. Find the real root r.
z₁ = −2 + j
r =
9.Differential equationsMedium
A cup of coffee at 81 °C is left in a room at 18 °C. After 12 min its temperature is 45 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
Find the determinant det A of the 3×3 matrix A shown.
A = [−3 −2 0; 4 −4 −3; 1 4 2]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−2 2; 6 2]
λ =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (5, 5, 6) and b = (−5, −3, −1), in degrees.
a = (5, 5, 6)b = (−5, −3, −1)
θ = °
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(−3, 1, 0), B(3, 3, −2) and C(−1, 0, −1). Coordinates are in metres.
A(−3, 1, 0)B(3, 3, −2)C(−1, 0, −1)
A = m²
15.ProbabilityMedium
A fair die is rolled 2 times. What is the probability P of getting at least one six?
n = 2
P =
16.ProbabilityHard
A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 2 correct answers?
n = 9p = 0.4k = 2
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 36, 48, 16, 37, 51, 36.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 5. What proportion P of values is above 103?
μ = 100σ = 5x = 103
P =
19.Normal distributionMedium
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 169.6 and 179.4 cm? Give P as a decimal.
μ = 178σ = 7x1 = 169.6x2 = 179.4
P =
20.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 60% is to the right of it. Find the cut-off x.
μ = 100 pointsσ = 15 points60%
x = points
3Solutions
1.Limits
Answer: L = −3.5
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) = −7x/(√(x2 − 7x) + x)
Divide by x
f(x) = −7/(√(1 + (−7)/x) + 1)
Let x → ∞
L = −7/(1 + 1) = −3.5
Answer
L = −3.5
2.Limits
Answer: L = 0.6667
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).
L'Hôpital's rule
L = limx→0 [2/(1 + 2x)]/[3·cos 3x]
Substitute x = 0
L = 2/3 = 0.6667
Answer
L = 0.6667
3.Derivatives
Answer: f′(2) = −0.1353
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 rule
f′(x) = u′v + uv′ = e−x·(1 + (−1)·x)
Substitute
f′(2) = 0.1353·(1 + (−2)) = −0.1353
Answer
f′(2) = −0.1353
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 − 30x + 63 = 3·(x2 − 10x + 21)
Solve f′(x) = 0
x = 3 or x = 7
Which one is which
f″(x) = 6x − 30: negative at a maximum, positive at a minimum.
f″(3) = −12, f″(7) = 12
x = 3
Answer
x = 3
5.Integrals
Answer: I = 0.1436
Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.
Choose u and dv
u = ln x, du = dx/x; dv = x dx, v = x²/2
Antiderivative
∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4
Limits
(1.125·ln 1.5 − 0.5625) − (0 − 0.25) = 0.1436
Answer
I = 0.1436
6.Integrals
Answer: A = 0.1667
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 = x → x = 0, x = 1
Integrate upper minus lower
A = [0.5x2 − 0.3333x3]01 = 0.1667
Answer
A = 0.1667
7.Complex numbers
Answer: b = 4.096
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(125°) = −2.868
b = 5·sin(125°) = 4.096
Rectangular form
z = −2.868 + j4.096
b = 4.096
Answer
b = 4.096
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 = −4 + (−1) = −5 = −(5).
9.Differential equations
Answer: k = 0.07061 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−12k = 27/63
Solve for k
k = ln(63/27)/12 = 0.07061 min⁻¹
Answer
k = 0.07061 min⁻¹
10.Differential equations
Answer: y(1.5) = 0.9447
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 = e3t
(e3ty)′ = 5e2.5t
General solution
y = 5/2.5·e−0.5t + C·e−3t = 2·e−0.5t + C·e−3t
Initial condition
2 = 2 + C → C = 0
Evaluate
y(1.5) = 2·0.4724 + 0·0.01111 = 0.9447
Answer
y(1.5) = 0.9447
11.Matrices & determinants
Answer: det A = 10
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 = 4, M₁₂ = 11, M₁₃ = 20
Cofactor expansion
det A = (−3)·4 − (−2)·11 + 0·20 = 10
Answer
det A = 10
12.Matrices & determinants
Answer: λ = 4
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 − (0)λ + (−16) = 0
Solve
λ = (0 ± √(64))/2 → λ = −4, 4
λ = 4
Answer
λ = 4
13.Vectors: dot & cross product
Answer: θ = 147 °
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 = 5·(−5) + 5·(−3) + 6·(−1) = −46
Magnitudes
|a| = √(5² + 5² + 6²) = 9.274
|b| = √((−5)² + (−3)² + (−1)²) = 5.916
Angle
cos θ = −46/(9.274·5.916) = −0.8384
θ = arccos(−0.8384) = 147°
Answer
θ = 147°
14.Vectors: dot & cross product
Answer: A = 5.477 m²
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.
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.
Complement
P(no six) = (5/6)2 = 0.6944
P = 1 − 0.6944 = 0.3056
Answer
P = 0.3056
16.Probability
Answer: P = 0.1612
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 = 0.6
Standard normal table
Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743
Answer
P = 0.2743
19.Normal distribution
Answer: P = 0.4642
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 = (169.6 − 178)/7 = −1.2
z₂ = (179.4 − 178)/7 = 0.2
Area between
P = Φ(0.2) − Φ(−1.2) = 0.5793 − 0.1151 = 0.4642
Answer
P = 0.4642
20.Normal distribution
Answer: x = 96.3 points
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.