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→∞ (5x2 − x + 7)/(5x2 + 4x + 1).
f(x) = (5x2 − x + 7)/(5x2 + 4x + 1)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + x)/sin(5x), with x in radians.
f(x) = ln(1 + x)/sin(5x)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
The tangent line to f(x) = x2 + x + 3 at x = 2 crosses the y-axis at y = b. Find b.
f(x) = x2 + x + 3x = 2
b =
5.IntegralsMedium
Evaluate I = ∫₁2.5 x·ln x dx.
∫₁2.5 x ln x dx
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
Find the argument θ of z = 8 + j in degrees, as the principal value between −180° and 180°.
z = 8 + j
θ = °
8.Complex numbersHard
Exactly one of the points A–D is a root of z⁶ = 64∠−90°. Click it.
w = 64∠−90°n = 6
1A (2∠−90°)
2B (3.6∠−105°)
3C (2∠135°)
4D (2∠−135°)
9.Differential equationsMedium
A cup of coffee at 83 °C is left in a room at 20 °C. After 5 min its temperature is 62 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 41 °C?
T0 = 83 °CTa = 20 °CT(5) = 62 °CT = 41 °C
t = min
10.Differential equationsHard
Solve the initial value problem y′ + y/x = 4x for x > 0, with y(1) = −3. Find y(3).
y′ + y/x = 4xy(1) = −3
y(3) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [5 −3 4; 1 −3 −2; 5 2 −4]
det A =
12.Matrices & determinantsHard
Solve the system 2x + 6y = −12 and −5x − y = 7. Find x.
2x + 6y = −12−5x − y = 7
x =
13.Vectors: dot & cross productMedium
A constant force F = (−15, −20, 10) N acts on an object while it moves in a straight line from point A(4, 0, 1) to point B(0, −2, −4). Coordinates are in metres. Find the work W done by the force.
F = (−15, −20, 10) NA(4, 0, 1) mB(0, −2, −4) m
W = J
14.Vectors: dot & cross productHard
For a = (−6, −6, −4) and b = (0, 4, −6), find the z-component (a × b)z of the cross product a × b.
a = (−6, −6, −4)b = (0, 4, −6)
(a × b)z =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4p(x): 0.15, 0.35, 0.3, 0.2
Var(X) =
16.ProbabilityHard
A game costs €5 to play. You win €25 with probability 0.3 and €100 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€25: p = 0.3€100: p = 0.02
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 32, 33, 54, 17, 12, 16.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is below 185?
μ = 170σ = 15x = 185
P =
19.Normal distributionMedium
X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 113 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 113 s
P =
20.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 5% of them lie (the 5th percentile).
μ = 178σ = 7P = 5%
x = cm
3Solutions
1.Limits
Answer: L = 1
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
(5 + (−1)/x + 7/x2)/(5 + 4/x + 1/x2)
Let x → ∞
L = 5/5 = 1
Answer
L = 1
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 0.2
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).
L'Hôpital's rule
L = limx→0 [1/(1 + x)]/[5·cos 5x]
Substitute x = 0
L = 1/5 = 0.2
Answer
L = 0.2
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: 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(2) = 9
f′(x) = 2x + 1 → f′(2) = 5
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 9 − 5·2 = −1
Answer
b = −1
5.Integrals
Answer: I = 1.551
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
(3.125·ln 2.5 − 1.5625) − (0 − 0.25) = 1.551
Answer
I = 1.551
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: θ = 7.125 °
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 = 8, b = 1 → I
Reference angle
α = atan(1/8) = 7.125°
Argument
θ = α = 7.125°
Answer
θ = 7.125°
In polar form z = 8.062∠7.125°.
8.Complex numbers
Answer: D (2∠−135°)
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.
Modulus and arguments of the roots
|z| = 2
θ = −15°, 45°, 105°, 165°, −135°, −75°
The point
D = 2∠−135°
Answer
D (2∠−135°)
9.Differential equations
Answer: t = 13.55 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(63/42)/5 = 0.08109 min⁻¹
Time to reach the target
t = ln(63/21)/0.08109 = 13.55 min
Answer
t = 13.55 min
10.Differential equations
Answer: y(3) = 10.56
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∫1/x dx = x1
Integrate
(x1y)′ = 4x2 → x1y = 4x3/3 + C
Initial condition
−3 = 4/3 + C → C = −4.333
Evaluate
y(3) = (4·33/3 + (−4.333))/31 = 10.56
Answer
y(3) = 10.56
11.Matrices & determinants
Answer: det A = 166
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 = 16, M₁₂ = 6, M₁₃ = 17
Cofactor expansion
det A = 5·16 − (−3)·6 + 4·17 = 166
Answer
det A = 166
12.Matrices & determinants
Answer: x = −1.071
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·(−1) − 6·(−5) = 28
Replace the x column
det Aₓ = (−12)·(−1) − 6·7 = −30
x = −30/28 = −1.071
Answer
x = −1.071
y = −1.643
13.Vectors: dot & cross product
Answer: W = 50 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 = (0, −2, −4) − (4, 0, 1) = (−4, −2, −5) 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.
Cross product, component by component
(a × b)x = aybz − azby = (−6)·(−6) − (−4)·4 = 52
(a × b)y = azbx − axbz = (−4)·0 − (−6)·(−6) = −36
(a × b)z = axby − aybx = (−6)·4 − (−6)·0 = −24
Requested component
(a × b)z = −24
Answer
(a × b)z = −24
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: Var(X) = 0.9475
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 = 1·0.15 + 2·0.35 + 3·0.3 + 4·0.2 = 2.55
Mean of the squares
E(X²) = 1·0.15 + 4·0.35 + 9·0.3 + 16·0.2 = 7.45
Variance
Var(X) = 7.45 − 2.55² = 0.9475
Answer
Var(X) = 0.9475
16.Probability
Answer: E(G) = 4.5 €
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: 20 (p = 0.3), 95 (p = 0.02), −5 (p = 0.68)
Expected value
E(G) = 20·0.3 + 95·0.02 − 5·0.68 = 4.5 €
Answer
E(G) = 4.5 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 15.72
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
Standard normal table
Φ(1) = 0.8413
P = 0.8413
Answer
P = 0.8413
19.Normal distribution
Answer: P = 0.1056
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 sum
μ = 105 s, σ = √(5² + 4²) = 6.403 s
Standardise
z = (113 − 105)/6.403 = 1.25
Upper tail
P = 1 − Φ(1.25) = 1 − 0.8944 = 0.1056
Answer
P = 0.1056
20.Normal distribution
Answer: x = 166.5 cm
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.