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 (e−2x − 1 + 2x)/x².
f(x) = (e−2x − 1 + 2x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = ln(3x2 + 5). Find f′(1).
f(x) = ln(3x2 + 5)x = 1
f′(1) =
4.DerivativesHard
The tangent line to f(x) = −x2 + 3x − 3 at x = −1 crosses the y-axis at y = b. Find b.
f(x) = −x2 + 3x − 3x = −1
b =
5.IntegralsMedium
Evaluate I = ∫01 ex dx.
f(x) = ex[0, 1]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = 3x.
y = x2y = 3x
A =
7.Complex numbersMedium
Given z₁ = 2.5∠150° and z₂ = 4∠45°, find the real part of w = z₁/z₂.
z₁ = 2.5∠150°z₂ = 4∠45°
Re(w) =
8.Complex numbersHard
The roots of zⁿ = w with n = 4 and w = 16∠45° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2, 3. Find the imaginary part of z₃.
w = 16∠45°n = 4k = 3
Im(z₃) =
9.Differential equationsMedium
A cup of coffee at 85 °C is left in a room at 23 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.05 min⁻¹, find its temperature T after 8 min.
T0 = 85 °CTa = 23 °Ck = 0.05 min⁻¹t = 8 min
T = °C
10.Differential equationsHard
A system obeys y″ + 2y′ + 5y = 5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 2y′ + 5y = 5y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [5 −1 5; 1 −3 −5; 4 −1 2]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [6 4; −5 −6]
λ =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−2, 4, −6) and b = (0, −3, 6), in degrees.
a = (−2, 4, −6)b = (0, −3, 6)
θ = °
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (−5, 3, −4), b = (−4, 2, −1) and c = (−1, −4, 4).
a = (−5, 3, −4)b = (−4, 2, −1)c = (−1, −4, 4)
V =
15.ProbabilityMedium
A fair die is rolled 6 times. What is the probability P of getting at least one six?
n = 6
P =
16.ProbabilityHard
5 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?
n = 5p = 0.4
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 28, 53, 29, 50, 28.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 60. Find the z-score z of a score of 524.
μ = 500σ = 60x = 524
z =
19.Normal distributionMedium
The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 5 of them exceeds 398.2 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 5> 398.2 kg
P =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 97.5% of them lie (the 97.5th percentile).
μ = 65σ = 10P = 97.5%
x = points
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 = 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
(−2·e−2x + 2)/(2x) → 0/0
Second application
4·e−2x/2
Substitute x = 0
L = 4·1/2 = 2
Answer
L = 2
3.Derivatives
Answer: f′(1) = 0.75
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.
Chain rule for the logarithm
f′(x) = 6x/(3x2 + 5)
Substitute
f′(1) = 6/8 = 0.75
Answer
f′(1) = 0.75
4.Derivatives
Answer: b = −2
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) = −7
f′(x) = −2x + 3 → f′(−1) = 5
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −7 − 5·(−1) = −2
Answer
b = −2
5.Integrals
Answer: I = 1.718
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) = ex/1
Evaluate at the limits
I = (e1 − 1)/1 = (2.718 − 1)/1 = 1.718
Answer
I = 1.718
6.Integrals
Answer: A = 4.5
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 = 3x → x = 0, x = 3
Integrate upper minus lower
A = [1.5x2 − 0.3333x3]03 = 4.5
Answer
A = 4.5
7.Complex numbers
Answer: Re(w) = −0.1618
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 and argument of w
|w| = 2.5 / 4 = 0.625
θ = 150° − 45° = 105°
Rectangular form
Re(w) = 0.625·cos(105°) = −0.1618
Im(w) = 0.625·sin(105°) = 0.6037
Re(w) = −0.1618
Answer
Re(w) = −0.1618
8.Complex numbers
Answer: Im(z₃) = −1.962
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.
The 4 roots lie on the circle of radius 2, 90° apart.
9.Differential equations
Answer: T = 64.56 °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) = 23 + 62·e−0.05t
Substitute t
T(8) = 23 + 62·0.6703 = 64.56 °C
Answer
T = 64.56 °C
10.Differential equations
Answer: ymax = 1.208
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 = 1, r = −1 ± 2j
Constants from the initial values
y(0) = 1 + A = 0 → A = −1
y′(0) = −1A + 2B = 0 → B = −0.5
Peak
t = π/2 = 1.571
ymax = 1·(1 + e−1·π/2) = 1.208
Answer
ymax = 1.208
11.Matrices & determinants
Answer: det A = 22
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 = −11, M₁₂ = 22, M₁₃ = 11
Cofactor expansion
det A = 5·(−11) − (−1)·22 + 5·11 = 22
Answer
det A = 22
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 = 163 °
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 = (−2)·0 + 4·(−3) + (−6)·6 = −48
Magnitudes
|a| = √((−2)² + 4² + (−6)²) = 7.483
|b| = √(0² + (−3)² + 6²) = 6.708
Angle
cos θ = −48/(7.483·6.708) = −0.9562
θ = arccos(−0.9562) = 163°
Answer
θ = 163°
14.Vectors: dot & cross product
Answer: V = 41
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 = 2·4 − (−1)·(−4) = 4
(b × c)y = bzcx − bxcz = (−1)·(−1) − (−4)·4 = 17
(b × c)z = bxcy − bycx = (−4)·(−4) − 2·(−1) = 18
Dot product with a
a·(b × c) = (−5)·4 + 3·17 + (−4)·18 = −41
V = |−41| = 41
Answer
V = 41
15.Probability
Answer: P = 0.6651
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)6 = 0.3349
P = 1 − 0.3349 = 0.6651
Answer
P = 0.6651
16.Probability
Answer: P = 0.337
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.
Zero and one failure
P(X = 0) = 0.65 = 0.07776
P(X = 1) = 5·0.4·0.64 = 0.2592
Add them
P = 0.07776 + 0.2592 = 0.337
Answer
P = 0.337
17.Descriptive statistics
Answer: s = 12.74
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 = (524 − 500)/60 = 0.4
Answer
z = 0.4
19.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).
Distribution of the total
μT = 425 kg, σT = 12·√5 = 26.83 kg
Standardise
z = (398.2 − 425)/26.83 = −1
Probability
P = 1 − Φ(−1) = 0.8413
Answer
P = 0.8413
20.Normal distribution
Answer: x = 84.6 points
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.