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→∞ (√(9x2 − 8x) − 3x).
f(x) = √(9x2 − 8x) − 3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + x)/sin(2x), with x in radians.
f(x) = ln(1 + x)/sin(2x)x → 0
L =
3.DerivativesMedium
Let f(x) = x2·e−0.5x. Find f′(0.5).
f(x) = x2·e−0.5xx = 0.5
f′(0.5) =
4.DerivativesHard
Let f(x) = x3 + 7.5x2 + 18x − 6. Find the x-coordinate x of the local maximum.
f(x) = x3 + 7.5x2 + 18x − 6
x =
5.IntegralsMedium
The region under y = √(4x) from x = 0 to x = 3 is rotated about the x-axis. Find the volume V of the solid (a paraboloid).
y = √(4x)0 ≤ x ≤ 3
V =
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
Given z₁ = 3∠−35° and z₂ = 2∠65°, find the real part of w = z₁/z₂.
z₁ = 3∠−35°z₂ = 2∠65°
Re(w) =
8.Complex numbersHard
Exactly one of the points A–D is a root of z⁴ = 81∠120°. Click it.
w = 81∠120°n = 4
1A (3∠−15°)
2B (5.4∠−15°)
3C (3∠−60°)
4D (3∠60°)
9.Differential equationsMedium
A cup of coffee at 85 °C is left in a room at 22 °C. After 6 min its temperature is 69 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 85 °CTa = 22 °CT(6) = 69 °C
k = min⁻¹
10.Differential equationsHard
A system obeys y″ + y′ + 4.25y = 21.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(1); angles in radians.
y″ + y′ + 4.25y = 21.25y(0) = 0y′(0) = 0
y(1) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [3 −3 −3; 4 −5 −3; −4 −2 4]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [5 −4; 4 −5]
λ =
13.Vectors: dot & cross productMedium
Find the angle θ between the vectors a = (−2, 6, 0) and b = (−6, −1, −3), in degrees.
a = (−2, 6, 0)b = (−6, −1, −3)
θ = °
14.Vectors: dot & cross productHard
Find the area A of the triangle with corners A(−1, 1, −1), B(0, −2, −1) and C(2, 3, −1). Coordinates are in metres.
A(−1, 1, −1)B(0, −2, −1)C(2, 3, −1)
A = m²
15.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2, 3, 4p(x): 0.05, 0.2, 0.25, 0.25, 0.25
σ =
16.ProbabilityHard
11 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?
n = 11p = 0.6
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 45, 12, 53, 22, 60.
n = 5
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 100. Find the z-score z of a score of 625.
μ = 500σ = 100x = 625
z =
19.Normal distributionMedium
The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 9 of them is below 106.8 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 9< 106.8 kg
P =
20.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 60% is to the right of it. Find the cut-off x.
μ = 500 hσ = 40 h60%
x = h
3Solutions
1.Limits
Answer: L = −1.333
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) = −8x/(√(9x2 − 8x) + 3x)
Divide by x
f(x) = −8/(√(9 + (−8)/x) + 3)
Let x → ∞
L = −8/(3 + 3) = −1.333
Answer
L = −1.333
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 [1/(1 + x)]/[2·cos 2x]
Substitute x = 0
L = 1/2 = 0.5
Answer
L = 0.5
3.Derivatives
Answer: f′(0.5) = 0.6815
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′ = e−0.5x·(2x + (−0.5)·x2)
Substitute
f′(0.5) = 0.7788·(1 + (−0.125)) = 0.6815
Answer
f′(0.5) = 0.6815
4.Derivatives
Answer: x = −3
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 + 15x + 18 = 3·(x2 + 5x + 6)
Solve f′(x) = 0
x = −3 or x = −2
Which one is which
f″(x) = 6x + 15: negative at a maximum, positive at a minimum.
f″(−3) = −3, f″(−2) = 3
x = −3
Answer
x = −3
5.Integrals
Answer: V = 56.55
Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.
Disc method
V = π∫₀3 4x dx
Integrate
V = π·4·3²/2 = 56.55
Answer
V = 56.55
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: Re(w) = −0.2605
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.
Modulus and argument of w
|w| = 3 / 2 = 1.5
θ = −35° − 65° = −100°
Rectangular form
Re(w) = 1.5·cos(−100°) = −0.2605
Im(w) = 1.5·sin(−100°) = −1.477
Re(w) = −0.2605
Answer
Re(w) = −0.2605
8.Complex numbers
Answer: C (3∠−60°)
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| = 3
θ = 30°, 120°, −150°, −60°
The point
C = 3∠−60°
Answer
C (3∠−60°)
9.Differential equations
Answer: k = 0.04883 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−6k = 47/63
Solve for k
k = ln(63/47)/6 = 0.04883 min⁻¹
Answer
k = 0.04883 min⁻¹
10.Differential equations
Answer: y(1) = 5.573
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 = 5, r = −0.5 ± 2j
Constants from the initial values
y(0) = 5 + A = 0 → A = −5
y′(0) = −0.5A + 2B = 0 → B = −1.25
Evaluate
y(1) = 5·(1 − e−0.5(cos 2 + 0.25·sin 2)) = 5.573
Answer
y(1) = 5.573
11.Matrices & determinants
Answer: det A = 18
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 = −26, M₁₂ = 4, M₁₃ = −28
Cofactor expansion
det A = 3·(−26) − (−3)·4 + (−3)·(−28) = 18
Answer
det A = 18
12.Matrices & determinants
Answer: λ = 3
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 − (0)λ + (−9) = 0
Solve
λ = (0 ± √(36))/2 → λ = −3, 3
λ = 3
Answer
λ = 3
13.Vectors: dot & cross product
Answer: θ = 81.96 °
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 = (−2)·(−6) + 6·(−1) + 0·(−3) = 6
Magnitudes
|a| = √((−2)² + 6² + 0²) = 6.325
|b| = √((−6)² + (−1)² + (−3)²) = 6.782
Angle
cos θ = 6/(6.325·6.782) = 0.1399
θ = arccos(0.1399) = 81.96°
Answer
θ = 81.96°
14.Vectors: dot & cross product
Answer: A = 5.5 m²
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.
Edge vectors
AB = B − A = (1, −3, 0)
AC = C − A = (3, 2, 0)
Cross product
(AB × AC)x = AByACz − ABzACy = (−3)·0 − 0·2 = 0
(AB × AC)y = ABzACx − ABxACz = 0·3 − 1·0 = 0
(AB × AC)z = ABxACy − AByACx = 1·2 − (−3)·3 = 11
|AB × AC| = √(0² + 0² + 11²) = 11
Area of the triangle
A = ½·11 = 5.5 m²
Answer
A = 5.5 m²
15.Probability
Answer: σ = 1.203
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.
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) = 0.411 = 4.194 × 10⁻⁵
P(X = 1) = 11·0.6·0.410 = 0.0006921
Add them
P = 4.194 × 10⁻⁵ + 0.0006921 = 0.000734
Answer
P = 0.000734
17.Descriptive statistics
Answer: s = 20.55
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 = (625 − 500)/100 = 1.25
Answer
z = 1.25
19.Normal distribution
Answer: P = 0.3085
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 = 108 kg, σT = 0.8·√9 = 2.4 kg
Standardise
z = (106.8 − 108)/2.4 = −0.5
Probability
P = Φ(−0.5) = 0.3085
Answer
P = 0.3085
20.Normal distribution
Answer: x = 490 h
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.