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 + 10x) − 3x).
f(x) = √(9x2 + 10x) − 3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 5x)/sin(3x), with x in radians.
f(x) = ln(1 + 5x)/sin(3x)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(2x2 + 1). Find f′(−1).
f(x) = ln(2x2 + 1)x = −1
f′(−1) =
4.DerivativesHard
Let f(x) = x3 − 6x2 + 1. Find the x-coordinate x of the local minimum.
f(x) = x3 − 6x2 + 1
x =
5.IntegralsMedium
Evaluate I = ∫₁1.5 x·ln x dx.
∫₁1.5 x ln x dx
I =
6.IntegralsHard
Find the area A of the region enclosed by y = x2 and y = 4x.
y = x2y = 4x
A =
7.Complex numbersMedium
Find the argument θ of z = −6 + j6 in degrees, as the principal value between −180° and 180°.
z = −6 + j6
θ = °
8.Complex numbersHard
One root of z³ − 9z² + 40z − 32 = 0 is z₁ = 4 + j4. The coefficients are real. Find the real root r.
z₁ = 4 + j4
r =
9.Differential equationsMedium
A cup of coffee at 89 °C is left in a room at 21 °C. After 13 min its temperature is 67 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 89 °CTa = 21 °CT(13) = 67 °C
k = min⁻¹
10.Differential equationsHard
Solve the initial value problem y′ + 3y/x = 5x for x > 0, with y(1) = −3. Find y(3).
y′ + 3y/x = 5xy(1) = −3
y(3) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).
A = [−6 6; 5 −5]B = [−5 −3; −5 0]
c12 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−1 1; −5 5]
λ =
13.Vectors: dot & cross productMedium
A constant force F = (15, −15, 20) N acts on an object while it moves in a straight line from point A(1, −3, −5) to point B(−1, −5, −4). Coordinates are in metres. Find the work W done by the force.
F = (15, −15, 20) NA(1, −3, −5) mB(−1, −5, −4) m
W = J
14.Vectors: dot & cross productHard
For a = (−1, 2, 6) and b = (6, −4, 0), find the z-component (a × b)z of the cross product a × b.
a = (−1, 2, 6)b = (6, −4, 0)
(a × b)z =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2p(x): 0.35, 0.3, 0.35
Var(X) =
16.ProbabilityHard
A game costs €5 to play. You win €10 with probability 0.2 and €50 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€10: p = 0.2€50: p = 0.01
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 21, 37, 41, 53, 21.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 5. What proportion P of values is below 77.5?
μ = 70σ = 5x = 77.5
P =
19.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 500 and 510 h? Give P as a decimal.
μ = 500σ = 40x1 = 500x2 = 510
P =
20.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 10% of them lie (the 10th percentile).
μ = 32σ = 6P = 10%
x = min
3Solutions
1.Limits
Answer: L = 1.667
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) = 10x/(√(9x2 + 10x) + 3x)
Divide by x
f(x) = 10/(√(9 + 10/x) + 3)
Let x → ∞
L = 10/(3 + 3) = 1.667
Answer
L = 1.667
2.Limits
Answer: L = 1.667
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 [5/(1 + 5x)]/[3·cos 3x]
Substitute x = 0
L = 5/3 = 1.667
Answer
L = 1.667
3.Derivatives
Answer: f′(−1) = −1.333
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 + 1)
Substitute
f′(−1) = −4/3 = −1.333
Answer
f′(−1) = −1.333
4.Derivatives
Answer: x = 4
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 − 12x = 3·(x2 − 4x)
Solve f′(x) = 0
x = 0 or x = 4
Which one is which
f″(x) = 6x − 12: negative at a maximum, positive at a minimum.
f″(0) = −12, f″(4) = 12
x = 4
Answer
x = 4
5.Integrals
Answer: I = 0.1436
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 = ln x, du = dx/x; dv = x dx, v = x²/2
Antiderivative
∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4
Limits
(1.125·ln 1.5 − 0.5625) − (0 − 0.25) = 0.1436
Answer
I = 0.1436
6.Integrals
Answer: A = 10.67
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 = 4x → x = 0, x = 4
Integrate upper minus lower
A = [2x2 − 0.3333x3]04 = 10.67
Answer
A = 10.67
7.Complex numbers
Answer: θ = 135 °
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.
Quadrant
a = −6, b = 6 → II
Reference angle
α = atan(6/6) = 45°
Argument
θ = 180° − α = 135°
Answer
θ = 135°
In polar form z = 8.485∠135°.
8.Complex numbers
Answer: r = 1
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 = 8 + 1 = 9 = −(−9).
9.Differential equations
Answer: k = 0.03007 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−13k = 46/68
Solve for k
k = ln(68/46)/13 = 0.03007 min⁻¹
Answer
k = 0.03007 min⁻¹
10.Differential equations
Answer: y(3) = 8.852
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.
Integrating factor
μ = e∫3/x dx = x3
Integrate
(x3y)′ = 5x4 → x3y = 5x5/5 + C
Initial condition
−3 = 5/5 + C → C = −4
Evaluate
y(3) = (5·35/5 + (−4))/33 = 8.852
Answer
y(3) = 8.852
11.Matrices & determinants
Answer: c12 = 18
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
c12 = (−6)·(−3) + 6·0 = 18
Answer
c12 = 18
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 − (4)λ + (0) = 0
Solve
λ = (4 ± √(16))/2 → λ = 0, 4
λ = 4
Answer
λ = 4
13.Vectors: dot & cross product
Answer: W = 20 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, −5, −4) − (1, −3, −5) = (−2, −2, 1) m
Work
W = F·d = 15·(−2) + (−15)·(−2) + 20·1 = 20 J
Answer
W = 20 J
14.Vectors: dot & cross product
Answer: (a × b)z = −8
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 = 2·0 − 6·(−4) = 24
(a × b)y = azbx − axbz = 6·6 − (−1)·0 = 36
(a × b)z = axby − aybx = (−1)·(−4) − 2·6 = −8
Requested component
(a × b)z = −8
Answer
(a × b)z = −8
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: Var(X) = 0.7
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 + 1·0.3 + 2·0.35 = 1
Mean of the squares
E(X²) = 0·0.35 + 1·0.3 + 4·0.35 = 1.7
Variance
Var(X) = 1.7 − 1² = 0.7
Answer
Var(X) = 0.7
16.Probability
Answer: E(G) = −2.5 €
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: 5 (p = 0.2), 45 (p = 0.01), −5 (p = 0.79)
Expected value
E(G) = 5·0.2 + 45·0.01 − 5·0.79 = −2.5 €
Answer
E(G) = −2.5 €
A negative value means you lose money on average.
17.Descriptive statistics
Answer: s = 13.74
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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 0.9332
Answer
P = 0.9332
19.Normal distribution
Answer: P = 0.09871
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 = (500 − 500)/40 = 0
z₂ = (510 − 500)/40 = 0.25
Area between
P = Φ(0.25) − Φ(0) = 0.5987 − 0.5 = 0.09871
Answer
P = 0.09871
20.Normal distribution
Answer: x = 24.31 min
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.