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→∞ (−3x3 + x + 6)/(x3 + 5x + 9).
f(x) = (−3x3 + x + 6)/(x3 + 5x + 9)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 4x)/sin(3x), with x in radians.
f(x) = ln(1 + 4x)/sin(3x)x → 0
L =
3.DerivativesMedium
Let f(x) = x·ex. Find f′(0.5).
f(x) = x·exx = 0.5
f′(0.5) =
4.DerivativesHard
The tangent line to f(x) = 3x2 − 2x − 1 at x = −2 crosses the y-axis at y = b. Find b.
f(x) = 3x2 − 2x − 1x = −2
b =
5.IntegralsMedium
Evaluate I = ∫₀π/2 x·sin x dx (x in radians).
∫₀π/2 x·sin x dx
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 2x2 and y = x.
y = 2x2y = x
A =
7.Complex numbersMedium
Write z = −8 + j6 in exponential form r·ejθ. Give θ in radians, between −π and π.
z = −8 + j6
θ = rad
8.Complex numbersHard
The equation z² + 6z + 25 = 0 has two complex roots. Find their modulus |z|.
p = 6q = 25
|z| =
9.Differential equationsMedium
A can of soda at 5 °C is taken out of the fridge into a room at 30 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.03 min⁻¹, find its temperature T after 29 min.
T0 = 5 °CTa = 30 °Ck = 0.03 min⁻¹t = 29 min
T = °C
10.Differential equationsHard
A system obeys y″ + 4y′ + 20y = 200 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 4y′ + 20y = 200y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [−4 1; −6 −5]B = [2 −2; 3 −6]
c11 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−2 −4; 7 9]
λ =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (−3, 0, −2) and b = (5, −6, 5).
a = (−3, 0, −2)b = (5, −6, 5)
a·b =
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (−2, 0, −2), b = (−2, −5, 1) and c = (5, −3, 3).
a = (−2, 0, −2)b = (−2, −5, 1)c = (5, −3, 3)
V =
15.ProbabilityMedium
A fair die is rolled 8 times. What is the probability P of getting at least one six?
n = 8
P =
16.ProbabilityHard
A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 1 correct answers?
n = 7p = 0.6k = 1
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 23, 13, 51, 59, 13.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 12. Find the z-score z of a score of 160.4.
μ = 170σ = 12x = 160.4
z =
19.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 454 and 560 h? Give P as a decimal.
μ = 500σ = 40x1 = 454x2 = 560
P =
20.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
3Solutions
1.Limits
Answer: L = −3
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
(−3 + 1/x2 + 6/x3)/(1 + 5/x2 + 9/x3)
Let x → ∞
L = −3/1 = −3
Answer
L = −3
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 1.333
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 [4/(1 + 4x)]/[3·cos 3x]
Substitute x = 0
L = 4/3 = 1.333
Answer
L = 1.333
3.Derivatives
Answer: f′(0.5) = 2.473
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·(1 + 1·x)
Substitute
f′(0.5) = 1.649·(1 + 0.5) = 2.473
Answer
f′(0.5) = 2.473
4.Derivatives
Answer: b = −13
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(−2) = 15
f′(x) = 6x − 2 → f′(−2) = −14
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 15 − (−14)·(−2) = −13
Answer
b = −13
5.Integrals
Answer: I = 1
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 = x, du = dx; dv = sin x dx, v = −cos x
Antiderivative
∫x sin x dx = −x cos x + ∫cos x dx = −x cos x + sin x
Limits
(−π/2·cos π/2 + sin π/2) − (0 + 0) = 1
Answer
I = 1
6.Integrals
Answer: A = 0.04167
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
2x2 = x → x = 0, x = 0.5
Integrate upper minus lower
A = [0.5x2 − 0.6667x3]00.5 = 0.04167
Answer
A = 0.04167
7.Complex numbers
Answer: θ = 2.498 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 = √(64 + 36) = 10
Argument in radians
θ = 143.1° = 143.1·π/180 = 2.498 rad
Exponential form
z = 10·ej2.498
Answer
θ = 2.498 rad
8.Complex numbers
Answer: |z| = 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.
Discriminant
p² − 4q = 36 − 100 = −64 < 0
Quadratic formula
z = (−6 ± j√64)/2 = −3 + j4, −3 − j4
Answer
|z| = √((−3)² + 4²) = √25 = 5
Answer
|z| = 5
9.Differential equations
Answer: T = 19.53 °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) = 30 + (−25)·e−0.03t
Substitute t
T(29) = 30 + (−25)·0.419 = 19.53 °C
Answer
T = 19.53 °C
10.Differential equations
Answer: ymax = 12.08
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.
Steady value and roots
yss = 10, r = −2 ± 4j
Constants from the initial values
y(0) = 10 + A = 0 → A = −10
y′(0) = −2A + 4B = 0 → B = −5
Peak
t = π/4 = 0.7854
ymax = 10·(1 + e−2·π/4) = 12.08
Answer
ymax = 12.08
11.Matrices & determinants
Answer: c11 = −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.
Row times column
c11 = (−4)·2 + 1·3 = −5
Answer
c11 = −5
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 − (7)λ + (10) = 0
Solve
λ = (7 ± √(9))/2 → λ = 2, 5
λ = 5
Answer
λ = 5
13.Vectors: dot & cross product
Answer: a·b = −25
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.
Negative: the angle between the vectors is more than 90°.
14.Vectors: dot & cross product
Answer: V = 38
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 = (−5)·3 − 1·(−3) = −12
(b × c)y = bzcx − bxcz = 1·5 − (−2)·3 = 11
(b × c)z = bxcy − bycx = (−2)·(−3) − (−5)·5 = 31
Dot product with a
a·(b × c) = (−2)·(−12) + 0·11 + (−2)·31 = −38
V = |−38| = 38
Answer
V = 38
15.Probability
Answer: P = 0.7674
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)8 = 0.2326
P = 1 − 0.2326 = 0.7674
Answer
P = 0.7674
16.Probability
Answer: P = 0.0172
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) = 7·0.61·0.46 = 7·0.6·0.004096 = 0.0172
Answer
P = 0.0172
17.Descriptive statistics
Answer: s = 21.75
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 = (160.4 − 170)/12 = −0.8
Answer
z = −0.8
19.Normal distribution
Answer: P = 0.8081
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 = (454 − 500)/40 = −1.15
z₂ = (560 − 500)/40 = 1.5
Area between
P = Φ(1.5) − Φ(−1.15) = 0.9332 − 0.1251 = 0.8081
Answer
P = 0.8081
20.Normal distribution
Answer: P = 0.8413
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).