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→∞ (6x2 − x + 4)/(5x2 + 3x + 4).
f(x) = (6x2 − x + 4)/(5x2 + 3x + 4)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (ex − 1 − x)/x².
f(x) = (ex − 1 − x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x3·ex. Find f′(1.5).
f(x) = x3·exx = 1.5
f′(1.5) =
4.DerivativesHard
Let f(x) = x3 − 3x − 5. Find the x-coordinate x of the local maximum.
f(x) = x3 − 3x − 5
x =
5.IntegralsMedium
Evaluate I = ∫01 2x·(x² + 1)1 dx.
f(x) = 2x·(x² + 1)1[0, 1]
I =
6.IntegralsHard
Evaluate I = ∫₀1 x²·eˣ dx.
∫₀1 x²eˣ dx
I =
7.Complex numbersMedium
Write z = 4∠50° in rectangular form a + jb and give the imaginary part b.
r = 4θ = 50°
b =
8.Complex numbersHard
The roots of zⁿ = w with n = 5 and w = 32∠−60° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, …, 4. Find the argument of z₃ in degrees, between −180° and 180°.
w = 32∠−60°n = 5k = 3
θ₃ = °
9.Differential equationsMedium
A forged steel part at 440 °C is left to cool in a workshop at 26 °C. After 52 min its temperature is 113 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 440 °CTa = 26 °CT(52) = 113 °C
k = min⁻¹
10.Differential equationsHard
Solve y″ + 3y′ + 11.25y = 0 with y(0) = 1 and y′(0) = 0 (released from rest). Find y(0.75); angles in radians.
y″ + 3y′ + 11.25y = 0y(0) = 1y′(0) = 0
y(0.75) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−1 −2 1; 2 −4 4; −3 0 −5]
det A =
12.Matrices & determinantsHard
Solve the system −6x + 6y = 11 and −3x + 2y = 5. Find x.
−6x + 6y = 11−3x + 2y = 5
x =
13.Vectors: dot & cross productMedium
A constant force F = (−20, 30, −30) N acts on an object while it moves in a straight line from point A(4, −4, 1) to point B(2, 1, 4). Coordinates are in metres. Find the work W done by the force.
F = (−20, 30, −30) NA(4, −4, 1) mB(2, 1, 4) m
W = J
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (5, 1, 0), b = (−4, 3, −1) and c = (3, 3, −4).
a = (5, 1, 0)b = (−4, 3, −1)c = (3, 3, −4)
V =
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
6 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?
n = 6p = 0.6
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 19, 24, 21, 43, 54.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 15. Find the z-score z of a score of 147.5.
μ = 170σ = 15x = 147.5
z =
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 757.9 g. Round z to two decimals.
μX = 250, σX = 4μY = 500, σY = 9X + Y > 757.9 g
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 90% of them lie (the 90th percentile).
μ = 500σ = 40P = 90%
x = h
3Solutions
1.Limits
Answer: L = 1.2
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).
Divide by the highest power
(6 + (−1)/x + 4/x2)/(5 + 3/x + 4/x2)
Let x → ∞
L = 6/5 = 1.2
Answer
L = 1.2
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·ex − 1)/(2x) → 0/0
Second application
1·ex/2
Substitute x = 0
L = 1·1/2 = 0.5
Answer
L = 0.5
3.Derivatives
Answer: f′(1.5) = 45.38
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′ = ex·(3x2 + 1·x3)
Substitute
f′(1.5) = 4.482·(6.75 + 3.375) = 45.38
Answer
f′(1.5) = 45.38
4.Derivatives
Answer: x = −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.
Derivative
f′(x) = 3x2 − 3 = 3·(x2 − 1)
Solve f′(x) = 0
x = −1 or x = 1
Which one is which
f″(x) = 6x: negative at a maximum, positive at a minimum.
f″(−1) = −6, f″(1) = 6
x = −1
Answer
x = −1
5.Integrals
Answer: I = 1.5
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.
Substitution
u = x² + 1, du = 2x dx
I = ∫12 u1 du
Integrate and evaluate
I = [u2/2]12 = (4 − 1)/2 = 1.5
Answer
I = 1.5
6.Integrals
Answer: I = 0.7183
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.
First round
∫x²eˣ dx = x²eˣ − ∫2x eˣ dx
Second round
∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)
Limits
2.718·(1) − 1·2 = 0.7183
Answer
I = 0.7183
7.Complex numbers
Answer: b = 3.064
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 = 4·cos(50°) = 2.571
b = 4·sin(50°) = 3.064
Rectangular form
z = 2.571 + j3.064
b = 3.064
Answer
b = 3.064
8.Complex numbers
Answer: θ₃ = −156 °
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 of the roots
321/5 = 2
Argument of the root
θ₃ = (−60° + 360°·3)/5 = 204° → −156°
Argument
θ₃ = −156°
Answer
θ₃ = −156°
The 5 roots lie on the circle of radius 2, 72° apart.
9.Differential equations
Answer: k = 0.03 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−52k = 87/414
Solve for k
k = ln(414/87)/52 = 0.03 min⁻¹
Answer
k = 0.03 min⁻¹
10.Differential equations
Answer: y(0.75) = −0.07764
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.
2×2 minors of the first row
M₁₁ = 20, M₁₂ = 2, M₁₃ = −12
Cofactor expansion
det A = (−1)·20 − (−2)·2 + 1·(−12) = −28
Answer
det A = −28
12.Matrices & determinants
Answer: x = −1.333
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 = (−6)·2 − 6·(−3) = 6
Replace the x column
det Aₓ = 11·2 − 6·5 = −8
x = −8/6 = −1.333
Answer
x = −1.333
y = 0.5
13.Vectors: dot & cross product
Answer: W = 100 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 = (2, 1, 4) − (4, −4, 1) = (−2, 5, 3) m
Work
W = F·d = (−20)·(−2) + 30·5 + (−30)·3 = 100 J
Answer
W = 100 J
14.Vectors: dot & cross product
Answer: V = 64
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·(−4) − (−1)·3 = −9
(b × c)y = bzcx − bxcz = (−1)·3 − (−4)·(−4) = −19
(b × c)z = bxcy − bycx = (−4)·3 − 3·3 = −21
Dot product with a
a·(b × c) = 5·(−9) + 1·(−19) + 0·(−21) = −64
V = |−64| = 64
Answer
V = 64
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: P = 0.04096
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.
Zero and one failure
P(X = 0) = 0.46 = 0.004096
P(X = 1) = 6·0.6·0.45 = 0.03686
Add them
P = 0.004096 + 0.03686 = 0.04096
Answer
P = 0.04096
17.Descriptive statistics
Answer: s = 15.48
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 = (147.5 − 170)/15 = −1.5
Answer
z = −1.5
19.Normal distribution
Answer: P = 0.2119
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 = (757.9 − 750)/9.849 = 0.8
Upper tail
P = 1 − Φ(0.8) = 1 − 0.7881 = 0.2119
Answer
P = 0.2119
20.Normal distribution
Answer: x = 551.3 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.