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→∞ (−2x2 − 3x + 7)/(5x2 + 2x + 4).
f(x) = (−2x2 − 3x + 7)/(5x2 + 2x + 4)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e−x − 1 + x)/x².
f(x) = (e−x − 1 + x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x2·e2x. Find f′(2).
f(x) = x2·e2xx = 2
f′(2) =
4.DerivativesHard
The tangent line to f(x) = 0.5x2 + 5x − 5 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 + 5x − 5x = 3
b =
5.IntegralsMedium
Evaluate I = ∫₁2 x·ln x dx.
∫₁2 x ln x dx
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 4 − x² and the x-axis.
y = 4 − x²
A =
7.Complex numbersMedium
The diagram shows z = 2 + j4. With w = 2∠90°, click the point that represents z·w.
z = 2 + j4w = 2∠90°
1A (8 − j4)
2B (4 + j8)
3C (−8 + j4)
4D (−4 + j2)
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∠165°)
2B (1∠135°)
3C (2∠−135°)
4D (2∠135°)
9.Differential equationsMedium
A forged steel part at 320 °C is left to cool in a workshop at 25 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.05 min⁻¹, find its temperature T after 23 min.
T0 = 320 °CTa = 25 °Ck = 0.05 min⁻¹t = 23 min
T = °C
10.Differential equationsHard
A system obeys y″ + 2y′ + 17y = 170 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.5); angles in radians.
y″ + 2y′ + 17y = 170y(0) = 0y′(0) = 0
y(0.5) =
11.Matrices & determinantsMedium
Let C = AB for the matrices shown. Find the entry c12 (row 1, column 2).
A = [−3 −5; 0 0]B = [1 −2; 3 6]
c12 =
12.Matrices & determinantsHard
Solve the system 3x − 4y = −9 and −5x − 3y = −7. Find x.
3x − 4y = −9−5x − 3y = −7
x =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (6, −2, 4) and b = (0, 5, −3), in degrees.
a = (6, −2, 4)b = (0, 5, −3)
θ = °
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (−3, 2, −5), b = (−2, −3, 5) and c = (−1, 2, −2).
a = (−3, 2, −5)b = (−2, −3, 5)c = (−1, 2, −2)
V =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2p(x): 0.3, 0.35, 0.35
Var(X) =
16.ProbabilityHard
A game costs €5 to play. You win €25 with probability 0.25 and €50 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€25: p = 0.25€50: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 34, 52, 49, 18, 27, 26.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 12. What proportion P of values is above 40.4?
μ = 50σ = 12x = 40.4
P =
19.Normal distributionMedium
X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 762.3 g. Round z to two decimals.
μX = 250, σX = 4μY = 500, σY = 9X + Y > 762.3 g
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 75% of them lie (the 75th percentile).
μ = 32σ = 6P = 75%
x = min
3Solutions
1.Limits
Answer: L = −0.4
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
(−2 + (−3)/x + 7/x2)/(5 + 2/x + 4/x2)
Let x → ∞
L = −2/5 = −0.4
Answer
L = −0.4
For equal degrees the limit is the ratio of the leading coefficients.
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).
First application
(−1·e−x + 1)/(2x) → 0/0
Second application
1·e−x/2
Substitute x = 0
L = 1·1/2 = 0.5
Answer
L = 0.5
3.Derivatives
Answer: f′(2) = 655.2
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.
Product rule
f′(x) = u′v + uv′ = e2x·(2x + 2·x2)
Substitute
f′(2) = 54.6·(4 + 8) = 655.2
Answer
f′(2) = 655.2
4.Derivatives
Answer: b = −9.5
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(3) = 14.5
f′(x) = x + 5 → f′(3) = 8
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 14.5 − 8·3 = −9.5
Answer
b = −9.5
5.Integrals
Answer: I = 0.6363
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
(2·ln 2 − 1) − (0 − 0.25) = 0.6363
Answer
I = 0.6363
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.
Zeros
4 − x² = 0 → x = ±2
Integrate
A = [4x − x³/3]−22 = 5.333 − (−5.333) = 10.67
Answer
A = 10.67
7.Complex numbers
Answer: C (−8 + j4)
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.
Effect of w
|z·w| = 2·|z|, arg(z·w) = arg z + 90°
The point
z·w = −8 + j4 → C
Answer
C (−8 + j4)
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°, 75°, 135°, −165°, −105°, −45°
The point
D = 2∠135°
Answer
D (2∠135°)
9.Differential equations
Answer: T = 118.4 °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) = 25 + 295·e−0.05t
Substitute t
T(23) = 25 + 295·0.3166 = 118.4 °C
Answer
T = 118.4 °C
10.Differential equations
Answer: y(0.5) = 11.15
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.
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 = (−3)·(−2) + (−5)·6 = −24
Answer
c12 = −24
12.Matrices & determinants
Answer: x = 0.03448
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 = 3·(−3) − (−4)·(−5) = −29
Replace the x column
det Aₓ = (−9)·(−3) − (−4)·(−7) = −1
x = −1/(−29) = 0.03448
Answer
x = 0.03448
y = 2.276
13.Vectors: dot & cross product
Answer: θ = 120.3 °
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 = 6·0 + (−2)·5 + 4·(−3) = −22
Magnitudes
|a| = √(6² + (−2)² + 4²) = 7.483
|b| = √(0² + 5² + (−3)²) = 5.831
Angle
cos θ = −22/(7.483·5.831) = −0.5042
θ = arccos(−0.5042) = 120.3°
Answer
θ = 120.3°
14.Vectors: dot & cross product
Answer: V = 29
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 b × c
(b × c)x = bycz − bzcy = (−3)·(−2) − 5·2 = −4
(b × c)y = bzcx − bxcz = 5·(−1) − (−2)·(−2) = −9
(b × c)z = bxcy − bycx = (−2)·2 − (−3)·(−1) = −7
Dot product with a
a·(b × c) = (−3)·(−4) + 2·(−9) + (−5)·(−7) = 29
V = |29| = 29
Answer
V = 29
15.Probability
Answer: Var(X) = 0.6475
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 = 0·0.3 + 1·0.35 + 2·0.35 = 1.05
Mean of the squares
E(X²) = 0·0.3 + 1·0.35 + 4·0.35 = 1.75
Variance
Var(X) = 1.75 − 1.05² = 0.6475
Answer
Var(X) = 0.6475
16.Probability
Answer: E(G) = 3.75 €
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 = 0.25), 45 (p = 0.05), −5 (p = 0.7)
Expected value
E(G) = 20·0.25 + 45·0.05 − 5·0.7 = 3.75 €
Answer
E(G) = 3.75 €
A positive value means you win money on average.
17.Descriptive statistics
Answer: s = 13.54
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 = −0.8
Standard normal table
Φ(−0.8) = 0.2119
P = 1 − 0.2119 = 0.7881
Answer
P = 0.7881
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 = 750 g, σ = √(4² + 9²) = 9.849 g
Standardise
z = (762.3 − 750)/9.849 = 1.25
Upper tail
P = 1 − Φ(1.25) = 1 − 0.8944 = 0.1056
Answer
P = 0.1056
20.Normal distribution
Answer: x = 36.05 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.