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 + 5x) − x).
f(x) = √(x2 + 5x) − xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 3x)/(x²), with x in radians.
f(x) = (1 − cos 3x)/(x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(2x), with x in radians. Find f′(0.5).
f(x) = x·sin(2x)x = 0.5 rad
f′(0.5) =
4.DerivativesHard
Let f(x) = x3 − 3x + 3. Find the x-coordinate x of the local minimum.
f(x) = x3 − 3x + 3
x =
5.IntegralsMedium
Evaluate I = ∫02 2x·(x² + 1)3 dx.
f(x) = 2x·(x² + 1)3[0, 2]
I =
6.IntegralsHard
Evaluate I = ∫₀2.5 x²·eˣ dx.
∫₀2.5 x²eˣ dx
I =
7.Complex numbersMedium
Find the argument θ of z = −8 + j2 in degrees, as the principal value between −180° and 180°.
z = −8 + j2
θ = °
8.Complex numbersHard
One root of z³ + 9z² + 40z + 32 = 0 is z₁ = −4 + j4. The coefficients are real. Find the real root r.
z₁ = −4 + j4
r =
9.Differential equationsMedium
A can of soda at 3 °C is taken out of the fridge into a room at 24 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.1 min⁻¹, find its temperature T after 11 min.
T0 = 3 °CTa = 24 °Ck = 0.1 min⁻¹t = 11 min
T = °C
10.Differential equationsHard
A system obeys y″ + 4y′ + 13y = 65 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.25); angles in radians.
y″ + 4y′ + 13y = 65y(0) = 0y′(0) = 0
y(0.25) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−1 1 −1; −4 −5 −2; −4 −3 4]
det A =
12.Matrices & determinantsHard
Solve the system −5x − 5y = 2 and −2x − y = 1. Find x.
−5x − 5y = 2−2x − y = 1
x =
13.Vectors: dot & cross productMedium
Find the value of k for which the vectors a = (4, −6, −5) and b = (5, 2, k) are perpendicular.
a = (4, −6, −5)b = (5, 2, k)
k =
14.Vectors: dot & cross productHard
For a = (−3, −6, 5) and b = (5, 5, 4), find the y-component (a × b)y of the cross product a × b.
a = (−3, −6, 5)b = (5, 5, 4)
(a × b)y =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 2, 4, 6p(x): 0.2, 0.3, 0.3, 0.2
Var(X) =
16.ProbabilityHard
A test has 10 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 = 10p = 0.4k = 2
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 13, 21, 34, 21, 23.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is above 86?
μ = 70σ = 8x = 86
P =
19.Normal distributionMedium
X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 113 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 113 s
P =
20.Normal distributionHard
Hole diameters D are normal with mean 25.02 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.02 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.01 mm
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) = 5x/(√(x2 + 5x) + x)
Divide by x
f(x) = 5/(√(1 + 5/x) + 1)
Let x → ∞
L = 5/(1 + 1) = 2.5
Answer
L = 2.5
2.Limits
Answer: L = 4.5
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).
First application
3·sin 3x/(2x) → 0/0
Second application
9·cos 3x/2
Substitute x = 0
L = 9/2 = 4.5
Answer
L = 4.5
3.Derivatives
Answer: f′(0.5) = 1.382
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′(0.5) = 0.8415 + 1·0.5403 = 1.382
Answer
f′(0.5) = 1.382
4.Derivatives
Answer: x = 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.
Derivative
f′(x) = 3x2 − 3 = 3·(x2 − 1)
Solve f′(x) = 0
x = −1 or x = 1
Which one is which
f″(x) = 6x: negative at a maximum, positive at a minimum.
f″(−1) = −6, f″(1) = 6
x = 1
Answer
x = 1
5.Integrals
Answer: I = 156
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.
Substitution
u = x² + 1, du = 2x dx
I = ∫15 u3 du
Integrate and evaluate
I = [u4/4]15 = (625 − 1)/4 = 156
Answer
I = 156
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: θ = 166 °
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.
Quadrant
a = −8, b = 2 → II
Reference angle
α = atan(2/8) = 14.04°
Argument
θ = 180° − α = 166°
Answer
θ = 166°
In polar form z = 8.246∠166°.
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 = −8 + (−1) = −9 = −(9).
9.Differential equations
Answer: T = 17.01 °C
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.
Solution of the cooling law
T(t) = 24 + (−21)·e−0.1t
Substitute t
T(11) = 24 + (−21)·0.3329 = 17.01 °C
Answer
T = 17.01 °C
10.Differential equations
Answer: y(0.25) = 1.403
Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.
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 = −26, M₁₂ = −24, M₁₃ = −8
Cofactor expansion
det A = (−1)·(−26) − 1·(−24) + (−1)·(−8) = 58
Answer
det A = 58
12.Matrices & determinants
Answer: x = −0.6
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.
Determinant of the coefficient matrix
det A = (−5)·(−1) − (−5)·(−2) = −5
Replace the x column
det Aₓ = 2·(−1) − (−5)·1 = 3
x = 3/(−5) = −0.6
Answer
x = −0.6
y = 0.2
13.Vectors: dot & cross product
Answer: k = 1.6
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 equals zero
a·b = 4·5 + (−6)·2 + (−5)·k = 0
8 + (−5)·k = 0
Solve for k
k = −8/(−5) = 1.6
Answer
k = 1.6
14.Vectors: dot & cross product
Answer: (a × b)y = 37
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, component by component
(a × b)x = aybz − azby = (−6)·4 − 5·5 = −49
(a × b)y = azbx − axbz = 5·5 − (−3)·4 = 37
(a × b)z = axby − aybx = (−3)·5 − (−6)·5 = 15
Requested component
(a × b)y = 37
Answer
(a × b)y = 37
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: Var(X) = 4.2
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 = 0·0.2 + 2·0.3 + 4·0.3 + 6·0.2 = 3
Mean of the squares
E(X²) = 0·0.2 + 4·0.3 + 16·0.3 + 36·0.2 = 13.2
Variance
Var(X) = 13.2 − 3² = 4.2
Answer
Var(X) = 4.2
16.Probability
Answer: P = 0.1209
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 = 2) = 45·0.42·0.68 = 45·0.16·0.0168 = 0.1209
Answer
P = 0.1209
17.Descriptive statistics
Answer: s = 7.537
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) = 0.9772
P = 1 − 0.9772 = 0.02275
Answer
P = 0.02275
19.Normal distribution
Answer: P = 0.1056
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 sum
μ = 105 s, σ = √(5² + 4²) = 6.403 s
Standardise
z = (113 − 105)/6.403 = 1.25
Upper tail
P = 1 − Φ(1.25) = 1 − 0.8944 = 0.1056
Answer
P = 0.1056
20.Normal distribution
Answer: P = 0.8133
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).