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 + 0.5/x)3x. Give a decimal value.
f(x) = (1 + 0.5/x)3xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 ln(1 + 4x)/sin(3x), with x in radians.
f(x) = ln(1 + 4x)/sin(3x)x → 0
L =
3.DerivativesMedium
Let f(x) = (2x − 1)5. Find f′(−1).
f(x) = (2x − 1)5x = −1
f′(−1) =
4.DerivativesHard
Let f(x) = x3 − 6x2 − 2. Find the x-coordinate x of the local minimum.
f(x) = x3 − 6x2 − 2
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 enclosed by y = 0.5x2 and y = 3x.
y = 0.5x2y = 3x
A =
7.Complex numbersMedium
Write z = −7 + j8 in exponential form r·ejθ. Give θ in radians, between −π and π.
z = −7 + j8
θ = rad
8.Complex numbersHard
The roots of zⁿ = w with n = 4 and w = 16∠180° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, 2, 3. Find the imaginary part of z₀.
w = 16∠180°n = 4k = 0
Im(z₀) =
9.Differential equationsMedium
Solve y″ + 5y′ + 4y = 0 with y(0) = 5 and y′(0) = 4. Find y(2).
y″ + 5y′ + 4y = 0y(0) = 5y′(0) = 4
y(2) =
10.Differential equationsHard
Solve y″ + 4y′ + 13y = 0 with y(0) = 3 and y′(0) = 0 (released from rest). Find y(0.75); angles in radians.
y″ + 4y′ + 13y = 0y(0) = 3y′(0) = 0
y(0.75) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [0 −3 −3; −2 4 −2; 4 −4 −1]
det A =
12.Matrices & determinantsHard
Solve the system x − 2y = −8 and 3x − 4y = 0. Find x.
x − 2y = −83x − 4y = 0
x =
13.Vectors: dot & cross productMedium
Find the scalar projection compb a of a = (1, 5, 4) onto b = (3, 6, −5), that is, the component of a in the direction of b.
a = (1, 5, 4)b = (3, 6, −5)
compb a =
14.Vectors: dot & cross productHard
A force F = (50, −60, −40) N acts at point A(2, −3, −3). Coordinates are in metres. Find the y-component My of the moment of the force about the origin O(0, 0, 0).
F = (50, −60, −40) NA(2, −3, −3) mO(0, 0, 0) m
My = N·m
15.ProbabilityMedium
A fair die is rolled 8 times. What is the probability P of getting at least one six?
n = 8
P =
16.ProbabilityHard
A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 2 correct answers?
n = 11p = 0.1k = 2
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 29, 13, 18, 36, 15, 58.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 75. What proportion P of values is below 593.8?
μ = 500σ = 75x = 593.8
P =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.4515. Find z to two decimals.
A = 0.4515
z =
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 = 4.482
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 + 0.5/x)
Evaluate
ln(1 + u)/u → 1 as u → 0, so x·ln(1 + a/x) → a.
ln L = 3·0.5 = 1.5
Undo the logarithm
L = e1.5 = 4.482
Answer
L = 4.482
2.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).
L'Hôpital's rule
L = limx→0 [4/(1 + 4x)]/[3·cos 3x]
Substitute x = 0
L = 4/3 = 1.333
Answer
L = 1.333
3.Derivatives
Answer: f′(−1) = 810
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′(−1) = 5·(−3)4·2 = 810
Answer
f′(−1) = 810
4.Derivatives
Answer: x = 4
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 − 12x = 3·(x2 − 4x)
Solve f′(x) = 0
x = 0 or x = 4
Which one is which
f″(x) = 6x − 12: negative at a maximum, positive at a minimum.
f″(0) = −12, f″(4) = 12
x = 4
Answer
x = 4
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 = 18
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.
Intersection points
0.5x2 = 3x → x = 0, x = 6
Integrate upper minus lower
A = [1.5x2 − 0.1667x3]06 = 18
Answer
A = 18
7.Complex numbers
Answer: θ = 2.29 rad
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
r = √(49 + 64) = 10.63
Argument in radians
θ = 131.2° = 131.2·π/180 = 2.29 rad
Exponential form
z = 10.63·ej2.29
Answer
θ = 2.29 rad
8.Complex numbers
Answer: Im(z₀) = 1.414
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
161/4 = 2
Argument of the root
θ₀ = (180° + 360°·0)/4 = 45°
Rectangular form
z₀ = 2·(cos 45° + j sin 45°) = 1.414 + j1.414
Im(z₀) = 1.414
Answer
Im(z₀) = 1.414
The 4 roots lie on the circle of radius 2, 90° apart.
9.Differential equations
Answer: y(2) = 1.082
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.
Characteristic roots
(r + 1)(r + 4) = 0 → r = −1, −4
Constants from the initial values
A + B = 5, −1A − 4B = 4
A = 8, B = −3
Evaluate
y(2) = 8·e−2 + (−3)·e−8 = 1.082
Answer
y(2) = 1.082
10.Differential equations
Answer: y(0.75) = −0.07327
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 = −12, M₁₂ = 10, M₁₃ = −8
Cofactor expansion
det A = 0·(−12) − (−3)·10 + (−3)·(−8) = 54
Answer
det A = 54
12.Matrices & determinants
Answer: x = 16
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·(−4) − (−2)·3 = 2
Replace the x column
det Aₓ = (−8)·(−4) − (−2)·0 = 32
x = 32/2 = 16
Answer
x = 16
y = 12
13.Vectors: dot & cross product
Answer: compb a = 1.554
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 = 1·3 + 5·6 + 4·(−5) = 13
Length of b
|b| = √(3² + 6² + (−5)²) = 8.367
Scalar projection
compb a = a·b/|b| = 13/8.367 = 1.554
Answer
compb a = 1.554
14.Vectors: dot & cross product
Answer: My = −70 N·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.
Position vector from O to A
r = A − O = (2, −3, −3) − (0, 0, 0) = (2, −3, −3) m
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.
Complement
P(no six) = (5/6)8 = 0.2326
P = 1 − 0.2326 = 0.7674
Answer
P = 0.7674
16.Probability
Answer: P = 0.2131
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.
Binomial formula
P(X = 2) = 55·0.12·0.99 = 55·0.01·0.3874 = 0.2131
Answer
P = 0.2131
17.Descriptive statistics
Answer: s = 17.08
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 = 1.25
Standard normal table
Φ(1.25) = 0.8944
P = 0.8944
Answer
P = 0.8944
19.Normal distribution
Answer: z = 0.6
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.
Area to the left of z
Φ(z) = (1 + 0.4515)/2 = 0.7257
Read the z-table backwards
Find 0.7257 inside the table and read off z.
z = 0.60
Answer
z = 0.60
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.