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 + 4x + 1)/(3x2 + 2x + 5).
f(x) = (−4x2 + 4x + 1)/(3x2 + 2x + 5)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e2x − 1 − 2x)/x².
f(x) = (e2x − 1 − 2x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = ln(2x2 + 2). Find f′(−1).
f(x) = ln(2x2 + 2)x = −1
f′(−1) =
4.DerivativesHard
Let f(x) = x3 − 9x2 + 24x − 3. Find the x-coordinate x of the local maximum.
f(x) = x3 − 9x2 + 24x − 3
x =
5.IntegralsMedium
Evaluate I = ∫15 5/x dx.
f(x) = 5/x[1, 5]
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 1 − x² and the x-axis.
y = 1 − x²
A =
7.Complex numbersMedium
Given z₁ = 2∠150° and z₂ = 1.5∠−25°, find the imaginary part of w = z₁·z₂.
z₁ = 2∠150°z₂ = 1.5∠−25°
Im(w) =
8.Complex numbersHard
One root of z³ − z² − 4z + 24 = 0 is z₁ = 2 + j2. The coefficients are real. Find the real root r.
z₁ = 2 + j2
r =
9.Differential equationsMedium
A forged steel part at 460 °C is left to cool in a workshop at 27 °C. After 44 min its temperature is 207 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 460 °CTa = 27 °CT(44) = 207 °C
k = min⁻¹
10.Differential equationsHard
A system obeys y″ + 4y′ + 20y = 100 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 4y′ + 20y = 100y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−4 2 2; 3 0 −3; 5 1 2]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−3 4; −4 7]
λ =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (−5, −2, −3) onto b = (4, −3, 6), that is, the component of a in the direction of b.
a = (−5, −2, −3)b = (4, −3, 6)
compb a =
14.Vectors: dot & cross productHard
For a = (−6, 0, −1) and b = (−1, −6, −1), find the x-component (a × b)x of the cross product a × b.
a = (−6, 0, −1)b = (−1, −6, −1)
(a × b)x =
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2p(x): 0.35, 0.35, 0.3
σ =
16.ProbabilityHard
A game costs €10 to play. You win €50 with probability 0.2 and €500 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €10€50: p = 0.2€500: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 16, 41, 55, 54, 28.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 8. Find the z-score z of a score of 88.
μ = 100σ = 8x = 88
z =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.7154. Find z to two decimals.
A = 0.7154
z =
20.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 80% of them lie (the 80th percentile).
μ = 100σ = 15P = 80%
x = points
3Solutions
1.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).
Divide by the highest power
(−4 + 4/x + 1/x2)/(3 + 2/x + 5/x2)
Let x → ∞
L = −4/3 = −1.333
Answer
L = −1.333
For equal degrees the limit is the ratio of the leading coefficients.
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·e2x − 2)/(2x) → 0/0
Second application
4·e2x/2
Substitute x = 0
L = 4·1/2 = 2
Answer
L = 2
3.Derivatives
Answer: f′(−1) = −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.
Chain rule for the logarithm
f′(x) = 4x/(2x2 + 2)
Substitute
f′(−1) = −4/4 = −1
Answer
f′(−1) = −1
4.Derivatives
Answer: x = 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.
Derivative
f′(x) = 3x2 − 18x + 24 = 3·(x2 − 6x + 8)
Solve f′(x) = 0
x = 2 or x = 4
Which one is which
f″(x) = 6x − 18: negative at a maximum, positive at a minimum.
f″(2) = −6, f″(4) = 6
x = 2
Answer
x = 2
5.Integrals
Answer: I = 8.047
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) = 5·ln x
Evaluate at the limits
I = 5·(ln 5 − ln 1) = 5·ln 5 = 8.047
Answer
I = 8.047
6.Integrals
Answer: A = 1.333
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.
Zeros
1 − x² = 0 → x = ±1
Integrate
A = [1x − x³/3]−11 = 0.6667 − (−0.6667) = 1.333
Answer
A = 1.333
7.Complex numbers
Answer: Im(w) = 2.457
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 · 1.5 = 3
θ = 150° + (−25)° = 125°
Rectangular form
Re(w) = 3·cos(125°) = −1.721
Im(w) = 3·sin(125°) = 2.457
Im(w) = 2.457
Answer
Im(w) = 2.457
8.Complex numbers
Answer: r = −3
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 = 4 + (−3) = 1 = −(−1).
9.Differential equations
Answer: k = 0.01995 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−44k = 180/433
Solve for k
k = ln(433/180)/44 = 0.01995 min⁻¹
Answer
k = 0.01995 min⁻¹
10.Differential equations
Answer: ymax = 6.039
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 = 5, r = −2 ± 4j
Constants from the initial values
y(0) = 5 + A = 0 → A = −5
y′(0) = −2A + 4B = 0 → B = −2.5
Peak
t = π/4 = 0.7854
ymax = 5·(1 + e−2·π/4) = 6.039
Answer
ymax = 6.039
11.Matrices & determinants
Answer: det A = −48
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 = 3, M₁₂ = 21, M₁₃ = 3
Cofactor expansion
det A = (−4)·3 − 2·21 + 2·3 = −48
Answer
det A = −48
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 − (4)λ + (−5) = 0
Solve
λ = (4 ± √(36))/2 → λ = −1, 5
λ = 5
Answer
λ = 5
13.Vectors: dot & cross product
Answer: compb a = −4.097
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)·4 + (−2)·(−3) + (−3)·6 = −32
Length of b
|b| = √(4² + (−3)² + 6²) = 7.81
Scalar projection
compb a = a·b/|b| = −32/7.81 = −4.097
Answer
compb a = −4.097
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: (a × b)x = −6
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.
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: σ = 0.8047
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 = 0·0.35 + 1·0.35 + 2·0.3 = 0.95
Mean of the squares
E(X²) = 0·0.35 + 1·0.35 + 4·0.3 = 1.55
Variance
Var(X) = 1.55 − 0.95² = 0.6475
σ = √0.6475 = 0.8047
Answer
σ = 0.8047
16.Probability
Answer: E(G) = 10 €
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.
Net outcomes
G: 40 (p = 0.2), 490 (p = 0.02), −10 (p = 0.78)
Expected value
E(G) = 40·0.2 + 490·0.02 − 10·0.78 = 10 €
Answer
E(G) = 10 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 16.84
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 = (88 − 100)/8 = −1.5
Answer
z = −1.5
19.Normal distribution
Answer: z = 1.07
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.
Area to the left of z
Φ(z) = (1 + 0.7154)/2 = 0.8577
Read the z-table backwards
Find 0.8577 inside the table and read off z.
z = 1.07
Answer
z = 1.07
20.Normal distribution
Answer: x = 112.6 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.