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→∞ (5x3 − 2x + 8)/(5x3 + x + 7).
f(x) = (5x3 − 2x + 8)/(5x3 + x + 7)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 3x)/(2x²), with x in radians.
f(x) = (1 − cos 3x)/(2x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x2·ex. Find f′(1).
f(x) = x2·exx = 1
f′(1) =
4.DerivativesHard
The tangent line to f(x) = x2 − x at x = 1 crosses the y-axis at y = b. Find b.
f(x) = x2 − xx = 1
b =
5.IntegralsMedium
Evaluate I = ∫01 e0.5x dx.
f(x) = e0.5x[0, 1]
I =
6.IntegralsHard
Evaluate I = ∫₀2.5 x²·eˣ dx.
∫₀2.5 x²eˣ dx
I =
7.Complex numbersMedium
Write z = 2∠−75° in rectangular form a + jb and give the real part a.
r = 2θ = −75°
a =
8.Complex numbersHard
Find the real part of (−3 − j3)⁴.
z = −3 − j3n = 4
Re(z⁴) =
9.Differential equationsMedium
A can of soda at 5 °C is taken out of the fridge into a room at 23 °C. After 6 min its temperature is 10 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 5 °CTa = 23 °CT(6) = 10 °C
k = min⁻¹
10.Differential equationsHard
Solve the initial value problem y′ + y/x = 6x2 for x > 0, with y(1) = −1. Find y(2).
y′ + y/x = 6x2y(1) = −1
y(2) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).
A = [−1 4; 4 −5]B = [−6 2; −5 −1]
c12 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−3 −2; 4 6]
λ =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (5, 0, 1) and b = (−6, 5, 4), in degrees.
a = (5, 0, 1)b = (−6, 5, 4)
θ = °
14.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (−3, 0, −1) and b = (−5, −4, −2).
a = (−3, 0, −1)b = (−5, −4, −2)
A =
15.ProbabilityMedium
A fair die is rolled 4 times. What is the probability P of getting at least one six?
n = 4
P =
16.ProbabilityHard
A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 1 correct answers?
n = 6p = 0.5k = 1
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 35, 54, 45, 27, 13.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 40. What proportion P of values is below 420?
μ = 500σ = 40x = 420
P =
19.Normal distributionMedium
The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 6 of them exceeds 69.1 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 6> 69.1 kg
P =
20.Normal distributionHard
Hole diameters D are normal with mean 25.05 mm and standard deviation 0.02 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 = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
3Solutions
1.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).
Divide by the highest power
(5 + (−2)/x2 + 8/x3)/(5 + 1/x2 + 7/x3)
Let x → ∞
L = 5/5 = 1
Answer
L = 1
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 2.25
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
3·sin 3x/(4x) → 0/0
Second application
9·cos 3x/4
Substitute x = 0
L = 9/4 = 2.25
Answer
L = 2.25
3.Derivatives
Answer: f′(1) = 8.155
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′ = ex·(2x + 1·x2)
Substitute
f′(1) = 2.718·(2 + 1) = 8.155
Answer
f′(1) = 8.155
4.Derivatives
Answer: b = −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.
Point and slope
f(1) = 0
f′(x) = 2x − 1 → f′(1) = 1
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 0 − 1·1 = −1
Answer
b = −1
5.Integrals
Answer: I = 1.297
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 = (e0.5 − 1)/0.5 = (1.649 − 1)/0.5 = 1.297
Answer
I = 1.297
6.Integrals
Answer: I = 37.59
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.
First round
∫x²eˣ dx = x²eˣ − ∫2x eˣ dx
Second round
∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)
Limits
12.18·(3.25) − 1·2 = 37.59
Answer
I = 37.59
7.Complex numbers
Answer: a = 0.5176
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 = 2·cos(−75°) = 0.5176
b = 2·sin(−75°) = −1.932
Rectangular form
z = 0.5176 − j1.932
a = 0.5176
Answer
a = 0.5176
8.Complex numbers
Answer: Re(z⁴) = −324
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.
Powers step by step
z² = (−3 − j3)(−3 − j3) = j18
z³ = (j18)(−3 − j3) = 54 − j54
z⁴ = (54 − j54)(−3 − j3) = −324
Check with De Moivre
z = 4.243∠−135° → z⁴ = 324∠180°
Re(z⁴) = −324
Answer
Re(z⁴) = −324
9.Differential equations
Answer: k = 0.05424 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 = -13/-18
Solve for k
k = ln(-18/-13)/6 = 0.05424 min⁻¹
Answer
k = 0.05424 min⁻¹
10.Differential equations
Answer: y(2) = 10.75
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∫1/x dx = x1
Integrate
(x1y)′ = 6x3 → x1y = 6x4/4 + C
Initial condition
−1 = 6/4 + C → C = −2.5
Evaluate
y(2) = (6·24/4 + (−2.5))/21 = 10.75
Answer
y(2) = 10.75
11.Matrices & determinants
Answer: c12 = −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
c12 = (−1)·2 + 4·(−1) = −6
Answer
c12 = −6
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 − (3)λ + (−10) = 0
Solve
λ = (3 ± √(49))/2 → λ = −2, 5
λ = 5
Answer
λ = 5
13.Vectors: dot & cross product
Answer: θ = 125.5 °
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·(−6) + 0·5 + 1·4 = −26
Magnitudes
|a| = √(5² + 0² + 1²) = 5.099
|b| = √((−6)² + 5² + 4²) = 8.775
Angle
cos θ = −26/(5.099·8.775) = −0.5811
θ = arccos(−0.5811) = 125.5°
Answer
θ = 125.5°
14.Vectors: dot & cross product
Answer: A = 12.69
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)4 = 0.4823
P = 1 − 0.4823 = 0.5177
Answer
P = 0.5177
16.Probability
Answer: P = 0.09375
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.
Binomial formula
P(X = 1) = 6·0.51·0.55 = 6·0.5·0.03125 = 0.09375
Answer
P = 0.09375
17.Descriptive statistics
Answer: s = 15.88
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 = −2
Standard normal table
Φ(−2) = 0.02275
P = 0.02275
Answer
P = 0.02275
19.Normal distribution
Answer: P = 0.9306
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).
Distribution of the total
μT = 72 kg, σT = 0.8·√6 = 1.96 kg
Standardise
z = (69.1 − 72)/1.96 = −1.48
Probability
P = 1 − Φ(−1.48) = 0.9306
Answer
P = 0.9306
20.Normal distribution
Answer: P = 0.9616
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).