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→∞ (√(4x2 + 6x) − 2x).
f(x) = √(4x2 + 6x) − 2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 5x)/(2x²), with x in radians.
f(x) = (1 − cos 5x)/(2x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x3·ex. Find f′(1.5).
f(x) = x3·exx = 1.5
f′(1.5) =
4.DerivativesHard
Let f(x) = x3 − 3x2 − 9x + 5. Find the x-coordinate x of the local minimum.
f(x) = x3 − 3x2 − 9x + 5
x =
5.IntegralsMedium
Evaluate I = ∫02 2x·(x² + 1)2 dx.
f(x) = 2x·(x² + 1)2[0, 2]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 0.5x2 and y = 3x.
y = 0.5x2y = 3x
A =
7.Complex numbersMedium
Given z₁ = 2.5∠150° and z₂ = 0.5∠45°, find the argument θ of w = z₁/z₂ in degrees, between −180° and 180°.
z₁ = 2.5∠150°z₂ = 0.5∠45°
θ = °
8.Complex numbersHard
One root of z³ − 10z² + 34z − 40 = 0 is z₁ = 3 + j. The coefficients are real. Find the real root r.
z₁ = 3 + j
r =
9.Differential equationsMedium
Solve y″ + 8y′ + 16y = 0 with y(0) = 1 and y′(0) = 0. Find y(1.5).
y″ + 8y′ + 16y = 0y(0) = 1y′(0) = 0
y(1.5) =
10.Differential equationsHard
A system obeys y″ + 2y′ + 26y = 260 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 2y′ + 26y = 260y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).
A = [1 5; −5 5]B = [6 0; −4 4]
c21 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [5 −4; 5 −4]
λ =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (−6, −4, −7) onto b = (−2, −5, 6), that is, the component of a in the direction of b.
a = (−6, −4, −7)b = (−2, −5, 6)
compb a =
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (−1, −2, 5) and b = (−1, −4, 4) and has a positive z-component. Find its y-component n̂y.
a = (−1, −2, 5)b = (−1, −4, 4)n̂z > 0
n̂y =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4, 5p(x): 0.3, 0.25, 0.05, 0.15, 0.25
σ =
16.ProbabilityHard
A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 3 correct answers?
n = 7p = 0.1k = 3
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 16, 13, 60, 38, 51.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 8. What proportion P of values is below 104.8?
μ = 100σ = 8x = 104.8
P =
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 16 of them is below 1336 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 16< 1336 kg
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 = 1.5
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) = 6x/(√(4x2 + 6x) + 2x)
Divide by x
f(x) = 6/(√(4 + 6/x) + 2)
Let x → ∞
L = 6/(2 + 2) = 1.5
Answer
L = 1.5
2.Limits
Answer: L = 6.25
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
5·sin 5x/(4x) → 0/0
Second application
25·cos 5x/4
Substitute x = 0
L = 25/4 = 6.25
Answer
L = 6.25
3.Derivatives
Answer: f′(1.5) = 45.38
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·(3x2 + 1·x3)
Substitute
f′(1.5) = 4.482·(6.75 + 3.375) = 45.38
Answer
f′(1.5) = 45.38
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 − 6x − 9 = 3·(x2 − 2x − 3)
Solve f′(x) = 0
x = −1 or x = 3
Which one is which
f″(x) = 6x − 6: negative at a maximum, positive at a minimum.
f″(−1) = −12, f″(3) = 12
x = 3
Answer
x = 3
5.Integrals
Answer: I = 41.33
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 u2 du
Integrate and evaluate
I = [u3/3]15 = (125 − 1)/3 = 41.33
Answer
I = 41.33
6.Integrals
Answer: A = 18
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
0.5x2 = 3x → x = 0, x = 6
Integrate upper minus lower
A = [1.5x2 − 0.1667x3]06 = 18
Answer
A = 18
7.Complex numbers
Answer: θ = 105 °
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 / 0.5 = 5
θ = 150° − 45° = 105°
Argument
θ = 105°
Answer
θ = 105°
8.Complex numbers
Answer: r = 4
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 = 6 + 4 = 10 = −(−10).
9.Differential equations
Answer: y(1.5) = 0.01735
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.
Double root
(r + 4)² = 0 → r = −4
Constants
A = y(0) = 1
y′(0) = B − 4A → B = 0 + 4 = 4
Evaluate
y(1.5) = (1 + 4·1.5)·e−6 = 0.01735
Answer
y(1.5) = 0.01735
10.Differential equations
Answer: ymax = 15.33
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 = −1 ± 5j
Constants from the initial values
y(0) = 10 + A = 0 → A = −10
y′(0) = −1A + 5B = 0 → B = −2
Peak
t = π/5 = 0.6283
ymax = 10·(1 + e−1·π/5) = 15.33
Answer
ymax = 15.33
11.Matrices & determinants
Answer: c21 = −50
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
c21 = (−5)·6 + 5·(−4) = −50
Answer
c21 = −50
12.Matrices & determinants
Answer: λ = 1
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 − (1)λ + (0) = 0
Solve
λ = (1 ± √(1))/2 → λ = 0, 1
λ = 1
Answer
λ = 1
13.Vectors: dot & cross product
Answer: compb a = −1.24
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 = (−6)·(−2) + (−4)·(−5) + (−7)·6 = −10
Length of b
|b| = √((−2)² + (−5)² + 6²) = 8.062
Scalar projection
compb a = a·b/|b| = −10/8.062 = −1.24
Answer
compb a = −1.24
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: n̂y = −0.08192
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.
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.
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 = 0.7257
Answer
P = 0.7257
19.Normal distribution
Answer: P = 0.3085
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 = 1360 kg, σT = 12·√16 = 48 kg
Standardise
z = (1336 − 1360)/48 = −0.5
Probability
P = Φ(−0.5) = 0.3085
Answer
P = 0.3085
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).