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→∞ (√(9x2 + 5x) − 3x).
f(x) = √(9x2 + 5x) − 3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 3x)/(3x²), with x in radians.
f(x) = (1 − cos 3x)/(3x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(3x), with x in radians. Find f′(0.5).
f(x) = x·sin(3x)x = 0.5 rad
f′(0.5) =
4.DerivativesHard
Let f(x) = x3 − 13.5x2 + 54x + 1. Find the x-coordinate x of the local maximum.
f(x) = x3 − 13.5x2 + 54x + 1
x =
5.IntegralsMedium
Evaluate I = ∫28 2/x dx.
f(x) = 2/x[2, 8]
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
Write z = 10∠35° in rectangular form a + jb and give the real part a.
r = 10θ = 35°
a =
8.Complex numbersHard
Solve z² + 3z + 18.25 = 0. The roots are a ± jb; find b > 0.
p = 3q = 18.25
b =
9.Differential equationsMedium
A cup of coffee at 93 °C is left in a room at 22 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.08 min⁻¹, find its temperature T after 28 min.
T0 = 93 °CTa = 22 °Ck = 0.08 min⁻¹t = 28 min
T = °C
10.Differential equationsHard
A system obeys y″ + 2y′ + 10y = 20 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 2y′ + 10y = 20y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−3 −4 1; −5 −5 −4; 0 −5 −5]
det A =
12.Matrices & determinantsHard
Solve the system 2x + y = −8 and x − 2y = −3. Find x.
2x + y = −8x − 2y = −3
x =
13.Vectors: dot & cross productMedium
A constant force F = (5, 20, 15) N acts on an object while it moves in a straight line from point A(−1, 4, −1) to point B(1, 3, 5). Coordinates are in metres. Find the work W done by the force.
F = (5, 20, 15) NA(−1, 4, −1) mB(1, 3, 5) m
W = J
14.Vectors: dot & cross productHard
For a = (1, 4, −4) and b = (−5, 6, −6), find the y-component (a × b)y of the cross product a × b.
a = (1, 4, −4)b = (−5, 6, −6)
(a × b)y =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 10, 20, 30, 40p(x): 0.25, 0.3, 0.3, 0.15
Var(X) =
16.ProbabilityHard
A game costs €5 to play. You win €15 with probability 0.3 and €50 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€15: p = 0.3€50: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 53, 42, 37, 51, 22, 23.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 20. What proportion P of values is below 30?
μ = 60σ = 20x = 30
P =
19.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 49 and 70.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 49x2 = 70.5
P =
20.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 25% of them lie (the 25th percentile).
μ = 250σ = 5P = 25%
x = g
3Solutions
1.Limits
Answer: L = 0.8333
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/(√(9x2 + 5x) + 3x)
Divide by x
f(x) = 5/(√(9 + 5/x) + 3)
Let x → ∞
L = 5/(3 + 3) = 0.8333
Answer
L = 0.8333
2.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).
First application
3·sin 3x/(6x) → 0/0
Second application
9·cos 3x/6
Substitute x = 0
L = 9/6 = 1.5
Answer
L = 1.5
3.Derivatives
Answer: f′(0.5) = 1.104
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(3x) + 3x·cos(3x)
Substitute
f′(0.5) = 0.9975 + 1.5·0.07074 = 1.104
Answer
f′(0.5) = 1.104
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 − 27x + 54 = 3·(x2 − 9x + 18)
Solve f′(x) = 0
x = 3 or x = 6
Which one is which
f″(x) = 6x − 27: negative at a maximum, positive at a minimum.
f″(3) = −9, f″(6) = 9
x = 3
Answer
x = 3
5.Integrals
Answer: I = 2.773
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) = 2·ln x
Evaluate at the limits
I = 2·(ln 8 − ln 2) = 2·ln 4 = 2.773
Answer
I = 2.773
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: a = 8.192
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.
Components
a = 10·cos(35°) = 8.192
b = 10·sin(35°) = 5.736
Rectangular form
z = 8.192 + j5.736
a = 8.192
Answer
a = 8.192
8.Complex numbers
Answer: b = 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.
Discriminant
p² − 4q = 9 − 73 = −64 < 0
Quadratic formula
z = (−3 ± j√64)/2 = −1.5 + j4, −1.5 − j4
Answer
b = 4
Answer
b = 4
9.Differential equations
Answer: T = 29.56 °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) = 22 + 71·e−0.08t
Substitute t
T(28) = 22 + 71·0.1065 = 29.56 °C
Answer
T = 29.56 °C
10.Differential equations
Answer: ymax = 2.702
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 = 2, r = −1 ± 3j
Constants from the initial values
y(0) = 2 + A = 0 → A = −2
y′(0) = −1A + 3B = 0 → B = −0.6667
Peak
t = π/3 = 1.047
ymax = 2·(1 + e−1·π/3) = 2.702
Answer
ymax = 2.702
11.Matrices & determinants
Answer: det A = 110
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 = 5, M₁₂ = 25, M₁₃ = 25
Cofactor expansion
det A = (−3)·5 − (−4)·25 + 1·25 = 110
Answer
det A = 110
12.Matrices & determinants
Answer: x = −3.8
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 = 2·(−2) − 1·1 = −5
Replace the x column
det Aₓ = (−8)·(−2) − 1·(−3) = 19
x = 19/(−5) = −3.8
Answer
x = −3.8
y = −0.4
13.Vectors: dot & cross product
Answer: W = 80 J
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.
Displacement
d = B − A = (1, 3, 5) − (−1, 4, −1) = (2, −1, 6) m
Work
W = F·d = 5·2 + 20·(−1) + 15·6 = 80 J
Answer
W = 80 J
14.Vectors: dot & cross product
Answer: (a × b)y = 26
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 = 4·(−6) − (−4)·6 = 0
(a × b)y = azbx − axbz = (−4)·(−5) − 1·(−6) = 26
(a × b)z = axby − aybx = 1·6 − 4·(−5) = 26
Requested component
(a × b)y = 26
Answer
(a × b)y = 26
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: Var(X) = 102.8
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.
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: 10 (p = 0.3), 45 (p = 0.02), −5 (p = 0.68)
Expected value
E(G) = 10·0.3 + 45·0.02 − 5·0.68 = 0.5 €
Answer
E(G) = 0.5 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 13.36
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 = −1.5
Standard normal table
Φ(−1.5) = 0.06681
P = 0.06681
Answer
P = 0.06681
19.Normal distribution
Answer: P = 0.654
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 = (49 − 65)/10 = −1.6
z₂ = (70.5 − 65)/10 = 0.55
Area between
P = Φ(0.55) − Φ(−1.6) = 0.7088 − 0.0548 = 0.654
Answer
P = 0.654
20.Normal distribution
Answer: x = 246.6 g
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.