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 ln(1 + 2x)/sin(4x), with x in radians.
f(x) = ln(1 + 2x)/sin(4x)x → 0
L =
3.DerivativesMedium
Let f(x) = ln(3x2 + 1). Find f′(−2).
f(x) = ln(3x2 + 1)x = −2
f′(−2) =
4.DerivativesHard
The tangent line to f(x) = −x2 − 3x at x = −1 crosses the y-axis at y = b. Find b.
f(x) = −x2 − 3xx = −1
b =
5.IntegralsMedium
Evaluate I = ∫₀1 x·cos x dx (x in radians).
∫₀1 x·cos x dx
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 0.5x2 and y = 2x.
y = 0.5x2y = 2x
A =
7.Complex numbersMedium
Find the argument θ of z = −7 − j7 in degrees, as the principal value between −180° and 180°.
z = −7 − j7
θ = °
8.Complex numbersHard
The equation z² + 2z + 7.25 = 0 has two complex roots. Find their modulus |z|.
p = 2q = 7.25
|z| =
9.Differential equationsMedium
A cup of coffee at 91 °C is left in a room at 18 °C. After 16 min its temperature is 31 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 24 °C?
T0 = 91 °CTa = 18 °CT(16) = 31 °CT = 24 °C
t = min
10.Differential equationsHard
Solve the initial value problem y′ + 2y/x = 6x2 for x > 0, with y(1) = 3. Find y(2).
y′ + 2y/x = 6x2y(1) = 3
y(2) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c11 (row 1, column 1).
A = [0 −1; 5 4]B = [−2 −1; 4 2]
c11 =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [4 −1; −3 2]
λ =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−5, −1, 5) and b = (4, −4, −3), in degrees.
a = (−5, −1, 5)b = (4, −4, −3)
θ = °
14.Vectors: dot & cross productHard
A force F = (−50, 0, −40) N acts at point A(−2, 3, −3). Coordinates are in metres. Find the magnitude |M| of the moment of the force about point O(−2, 2, 1).
F = (−50, 0, −40) NA(−2, 3, −3) mO(−2, 2, 1) m
|M| = N·m
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 10, 20, 30, 40p(x): 0.35, 0.25, 0.05, 0.35
Var(X) =
16.ProbabilityHard
A game costs €3 to play. You win €6 with probability 0.25 and €30 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€6: p = 0.25€30: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 28, 18, 17, 55, 37.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is below 179?
μ = 170σ = 15x = 179
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 8 of them exceeds 730.9 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 8> 730.9 kg
P =
20.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 25% of them lie (the 25th percentile).
μ = 500σ = 40P = 25%
x = h
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 = 0.5
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).
L'Hôpital's rule
L = limx→0 [2/(1 + 2x)]/[4·cos 4x]
Substitute x = 0
L = 2/4 = 0.5
Answer
L = 0.5
3.Derivatives
Answer: f′(−2) = −0.9231
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 + 1)
Substitute
f′(−2) = −12/13 = −0.9231
Answer
f′(−2) = −0.9231
4.Derivatives
Answer: b = 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.
Point and slope
f(−1) = 2
f′(x) = −2x − 3 → f′(−1) = −1
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 2 − (−1)·(−1) = 1
Answer
b = 1
5.Integrals
Answer: I = 0.3818
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 = x, du = dx; dv = cos x dx, v = sin x
Antiderivative
∫x cos x dx = x sin x − ∫sin x dx = x sin x + cos x
Limits
(1·sin 1 + cos 1) − (0 + 1) = 0.3818
Answer
I = 0.3818
6.Integrals
Answer: A = 5.333
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.
Intersection points
0.5x2 = 2x → x = 0, x = 4
Integrate upper minus lower
A = [x2 − 0.1667x3]04 = 5.333
Answer
A = 5.333
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 = −7, b = −7 → III
Reference angle
α = atan(7/7) = 45°
Argument
θ = −(180° − α) = −135°
Answer
θ = −135°
In polar form z = 9.899∠−135°.
8.Complex numbers
Answer: |z| = 2.693
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 = 4 − 29 = −25 < 0
Quadratic formula
z = (−2 ± j√25)/2 = −1 + j2.5, −1 − j2.5
Answer
|z| = √((−1)² + 2.5²) = √7.25 = 2.693
Answer
|z| = 2.693
9.Differential equations
Answer: t = 23.17 min
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.
Cooling constant
k = ln(73/13)/16 = 0.1078 min⁻¹
Time to reach the target
t = ln(73/6)/0.1078 = 23.17 min
Answer
t = 23.17 min
10.Differential equations
Answer: y(2) = 10.05
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∫2/x dx = x2
Integrate
(x2y)′ = 6x4 → x2y = 6x5/5 + C
Initial condition
3 = 6/5 + C → C = 1.8
Evaluate
y(2) = (6·25/5 + 1.8)/22 = 10.05
Answer
y(2) = 10.05
11.Matrices & determinants
Answer: c11 = −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.
Row times column
c11 = 0·(−2) + (−1)·4 = −4
Answer
c11 = −4
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 − (6)λ + (5) = 0
Solve
λ = (6 ± √(16))/2 → λ = 1, 5
λ = 5
Answer
λ = 5
13.Vectors: dot & cross product
Answer: θ = 132.7 °
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 + (−1)·(−4) + 5·(−3) = −31
Magnitudes
|a| = √((−5)² + (−1)² + 5²) = 7.141
|b| = √(4² + (−4)² + (−3)²) = 6.403
Angle
cos θ = −31/(7.141·6.403) = −0.6779
θ = arccos(−0.6779) = 132.7°
Answer
θ = 132.7°
14.Vectors: dot & cross product
Answer: |M| = 210 N·m
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.
Position vector from O to A
r = A − O = (−2, 3, −3) − (−2, 2, 1) = (0, 1, −4) m
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: 3 (p = 0.25), 27 (p = 0.02), −3 (p = 0.73)
Expected value
E(G) = 3·0.25 + 27·0.02 − 3·0.73 = −0.9 €
Answer
E(G) = −0.9 €
A negative value means you lose money on average.
17.Descriptive statistics
Answer: s = 15.7
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 = 0.06681
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 = 680 kg, σT = 12·√8 = 33.94 kg
Standardise
z = (730.9 − 680)/33.94 = 1.5
Probability
P = 1 − Φ(1.5) = 0.06681
Answer
P = 0.06681
20.Normal distribution
Answer: x = 473 h
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.