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→∞ (1 + 2/x)3x. Give a decimal value.
f(x) = (1 + 2/x)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) = (2x + 1)5. Find f′(0).
f(x) = (2x + 1)5x = 0
f′(0) =
4.DerivativesHard
The tangent line to f(x) = −x2 + 5x + 1 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = −x2 + 5x + 1x = 1
b =
5.IntegralsMedium
Evaluate I = ∫₁4 x·ln x dx.
∫₁4 x ln x dx
I =
6.IntegralsHard
Find the area A of the region between the parabola y = 9 − x² and the x-axis.
y = 9 − x²
A =
7.Complex numbersMedium
Write z = 2.5∠50° in rectangular form a + jb and give the real part a.
r = 2.5θ = 50°
a =
8.Complex numbersHard
The roots of zⁿ = w with n = 6 and w = 729∠60° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, …, 5. Find the imaginary part of z₃.
w = 729∠60°n = 6k = 3
Im(z₃) =
9.Differential equationsMedium
A can of soda at 6 °C is taken out of the fridge into a room at 24 °C. After 22 min its temperature is 18 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 6 °CTa = 24 °CT(22) = 18 °C
k = min⁻¹
10.Differential equationsHard
A system obeys y″ + 4y′ + 29y = 290 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 4y′ + 29y = 290y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−4 −4 −3; −1 3 0; 1 3 −2]
det A =
12.Matrices & determinantsHard
Solve the system −x + 5y = 11 and 2x − 6y = 4. Find x.
−x + 5y = 112x − 6y = 4
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (0, 7, −7) onto b = (0, 1, −4), that is, the component of a in the direction of b.
a = (0, 7, −7)b = (0, 1, −4)
compb a =
14.Vectors: dot & cross productHard
For a = (−4, 1, 2) and b = (−3, −4, −5), find the z-component (a × b)z of the cross product a × b.
a = (−4, 1, 2)b = (−3, −4, −5)
(a × b)z =
15.ProbabilityMedium
A bag holds 7 red and 3 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
7 / 3
P =
16.ProbabilityHard
X has the distribution shown, and Y = 3X − 3. Find the standard deviation σY of Y.
x: 1, 2, 3, 4p(x): 0.2, 0.2, 0.35, 0.25Y = 3X − 3
σY =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 31, 36, 45, 53, 56, 13.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 15. What proportion P of values is below 35?
μ = 50σ = 15x = 35
P =
19.Normal distributionMedium
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 174.85 and 186.75 cm? Give P as a decimal.
μ = 178σ = 7x1 = 174.85x2 = 186.75
P =
20.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
3Solutions
1.Limits
Answer: L = 403.4
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).
Take the logarithm
ln L = limx→∞ 3·x·ln(1 + 2/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 3·2 = 6
Undo the logarithm
L = e6 = 403.4
Answer
L = 403.4
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) = 10
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
f′(x) = 5·(2x + 1)4·2
Substitute
f′(0) = 5·14·2 = 10
Answer
f′(0) = 10
4.Derivatives
Answer: b = 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.
Point and slope
f(1) = 5
f′(x) = −2x + 5 → f′(1) = 3
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 5 − 3·1 = 2
Answer
b = 2
5.Integrals
Answer: I = 7.34
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
(8·ln 4 − 4) − (0 − 0.25) = 7.34
Answer
I = 7.34
6.Integrals
Answer: A = 36
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
9 − x² = 0 → x = ±3
Integrate
A = [9x − x³/3]−33 = 18 − (−18) = 36
Answer
A = 36
7.Complex numbers
Answer: a = 1.607
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 = 2.5·cos(50°) = 1.607
b = 2.5·sin(50°) = 1.915
Rectangular form
z = 1.607 + j1.915
a = 1.607
Answer
a = 1.607
8.Complex numbers
Answer: Im(z₃) = −0.5209
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.
The 6 roots lie on the circle of radius 3, 60° apart.
9.Differential equations
Answer: k = 0.04994 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−22k = -6/-18
Solve for k
k = ln(-18/-6)/22 = 0.04994 min⁻¹
Answer
k = 0.04994 min⁻¹
10.Differential equations
Answer: ymax = 12.85
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 = 10, r = −2 ± 5j
Constants from the initial values
y(0) = 10 + A = 0 → A = −10
y′(0) = −2A + 5B = 0 → B = −4
Peak
t = π/5 = 0.6283
ymax = 10·(1 + e−2·π/5) = 12.85
Answer
ymax = 12.85
11.Matrices & determinants
Answer: det A = 50
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 = −6, M₁₂ = 2, M₁₃ = −6
Cofactor expansion
det A = (−4)·(−6) − (−4)·2 + (−3)·(−6) = 50
Answer
det A = 50
12.Matrices & determinants
Answer: x = 21.5
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.
Determinant of the coefficient matrix
det A = (−1)·(−6) − 5·2 = −4
Replace the x column
det Aₓ = 11·(−6) − 5·4 = −86
x = −86/(−4) = 21.5
Answer
x = 21.5
y = 6.5
13.Vectors: dot & cross product
Answer: compb a = 8.489
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 = 0·0 + 7·1 + (−7)·(−4) = 35
Length of b
|b| = √(0² + 1² + (−4)²) = 4.123
Scalar projection
compb a = a·b/|b| = 35/4.123 = 8.489
Answer
compb a = 8.489
14.Vectors: dot & cross product
Answer: (a × b)z = 19
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 = 1·(−5) − 2·(−4) = 3
(a × b)y = azbx − axbz = 2·(−3) − (−4)·(−5) = −26
(a × b)z = axby − aybx = (−4)·(−4) − 1·(−3) = 19
Requested component
(a × b)z = 19
Answer
(a × b)z = 19
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: P = 0.4667
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
First and second draw
P = 7/10 · 6/9 = 42/90 = 0.4667
Answer
P = 0.4667
16.Probability
Answer: σY = 3.186
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.
Spread of X
μ = 2.65, Var(X) = 1.128, σX = 1.062
Linear change
σY = |3|·1.062 = 3.186
Answer
σY = 3.186
The mean does move: E(Y) = 3·2.65 − 3 = 4.95.
17.Descriptive statistics
Answer: s = 15.94
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) = 0.1587
P = 0.1587
Answer
P = 0.1587
19.Normal distribution
Answer: P = 0.568
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 = (174.85 − 178)/7 = −0.45
z₂ = (186.75 − 178)/7 = 1.25
Area between
P = Φ(1.25) − Φ(−0.45) = 0.8944 − 0.3264 = 0.568
Answer
P = 0.568
20.Normal distribution
Answer: P = 0.8413
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).