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 (1 − cos 3x)/(4x²), with x in radians.
f(x) = (1 − cos 3x)/(4x²)x → 0
L =
3.DerivativesMedium
Let f(x) = x·e0.5x. Find f′(0.5).
f(x) = x·e0.5xx = 0.5
f′(0.5) =
4.DerivativesHard
Let f(x) = x3 − 12x2 + 36x + 2. Find the x-coordinate x of the local minimum.
f(x) = x3 − 12x2 + 36x + 2
x =
5.IntegralsMedium
Evaluate I = ∫01 e−x dx.
f(x) = e−x[0, 1]
I =
6.IntegralsHard
Find the area A of the region enclosed by y = 0.5x2 and y = 4x.
y = 0.5x2y = 4x
A =
7.Complex numbersMedium
Write z = 2.5∠110° in rectangular form a + jb and give the real part a.
r = 2.5θ = 110°
a =
8.Complex numbersHard
The roots of zⁿ = w with n = 5 and w = 32∠−90° 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∠−90°n = 5k = 3
θ₃ = °
9.Differential equationsMedium
Solve y″ + 6y′ + 8y = 0 with y(0) = 1 and y′(0) = 1. Find y(1.5).
y″ + 6y′ + 8y = 0y(0) = 1y′(0) = 1
y(1.5) =
10.Differential equationsHard
Solve y″ + 2y′ + 26y = 0 with y(0) = 10 and y′(0) = 0 (released from rest). Find y(0.25); angles in radians.
y″ + 2y′ + 26y = 0y(0) = 10y′(0) = 0
y(0.25) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−3 −2 4; −1 −1 −1; 5 2 5]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [2 −1; 3 6]
λ =
13.Vectors: dot & cross productMedium
A constant force F = (25, −15, −25) N acts on an object while it moves in a straight line from point A(3, 1, 1) to point B(2, −5, −5). Coordinates are in metres. Find the work W done by the force.
F = (25, −15, −25) NA(3, 1, 1) mB(2, −5, −5) m
W = J
14.Vectors: dot & cross productHard
For a = (5, −5, −2) and b = (−3, −3, −6), find the z-component (a × b)z of the cross product a × b.
a = (5, −5, −2)b = (−3, −3, −6)
(a × b)z =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 2, 4, 6p(x): 0.3, 0.25, 0.1, 0.35
Var(X) =
16.ProbabilityHard
11 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?
n = 11p = 0.1
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 38, 18, 23, 46, 57.
n = 5
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 15. What proportion P of values is above 62.5?
μ = 70σ = 15x = 62.5
P =
19.Normal distributionMedium
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 246.25 and 254 g? Give P as a decimal.
μ = 250σ = 5x1 = 246.25x2 = 254
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 90% of them lie (the 90th percentile).
μ = 32σ = 6P = 90%
x = min
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 = 1.125
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/(8x) → 0/0
Second application
9·cos 3x/8
Substitute x = 0
L = 9/8 = 1.125
Answer
L = 1.125
3.Derivatives
Answer: f′(0.5) = 1.605
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′ = e0.5x·(1 + 0.5·x)
Substitute
f′(0.5) = 1.284·(1 + 0.25) = 1.605
Answer
f′(0.5) = 1.605
4.Derivatives
Answer: x = 6
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 − 24x + 36 = 3·(x2 − 8x + 12)
Solve f′(x) = 0
x = 2 or x = 6
Which one is which
f″(x) = 6x − 24: negative at a maximum, positive at a minimum.
f″(2) = −12, f″(6) = 12
x = 6
Answer
x = 6
5.Integrals
Answer: I = 0.6321
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.
Antiderivative
F(x) = e−x/(−1)
Evaluate at the limits
I = (e−1 − 1)/(−1) = (0.3679 − 1)/(−1) = 0.6321
Answer
I = 0.6321
6.Integrals
Answer: A = 42.67
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.
Intersection points
0.5x2 = 4x → x = 0, x = 8
Integrate upper minus lower
A = [2x2 − 0.1667x3]08 = 42.67
Answer
A = 42.67
7.Complex numbers
Answer: a = −0.8551
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(110°) = −0.8551
b = 2.5·sin(110°) = 2.349
Rectangular form
z = −0.8551 + j2.349
a = −0.8551
Answer
a = −0.8551
8.Complex numbers
Answer: θ₃ = −162 °
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 = (−90° + 360°·3)/5 = 198° → −162°
Argument
θ₃ = −162°
Answer
θ₃ = −162°
The 5 roots lie on the circle of radius 2, 72° apart.
9.Differential equations
Answer: y(1.5) = 0.1207
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 + 2)(r + 4) = 0 → r = −2, −4
Constants from the initial values
A + B = 1, −2A − 4B = 1
A = 2.5, B = −1.5
Evaluate
y(1.5) = 2.5·e−3 + (−1.5)·e−6 = 0.1207
Answer
y(1.5) = 0.1207
10.Differential equations
Answer: y(0.25) = 3.934
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 = −3, M₁₂ = 0, M₁₃ = 3
Cofactor expansion
det A = (−3)·(−3) − (−2)·0 + 4·3 = 21
Answer
det A = 21
12.Matrices & determinants
Answer: λ = 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.
Characteristic equation
λ² − (8)λ + (15) = 0
Solve
λ = (8 ± √(4))/2 → λ = 3, 5
λ = 5
Answer
λ = 5
13.Vectors: dot & cross product
Answer: W = 215 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, −5, −5) − (3, 1, 1) = (−1, −6, −6) 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.
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
15.Probability
Answer: Var(X) = 6.2
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 + 2·0.25 + 4·0.1 + 6·0.35 = 3
Mean of the squares
E(X²) = 0·0.3 + 4·0.25 + 16·0.1 + 36·0.35 = 15.2
Variance
Var(X) = 15.2 − 3² = 6.2
Answer
Var(X) = 6.2
16.Probability
Answer: P = 0.6974
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) = 0.911 = 0.3138
P(X = 1) = 11·0.1·0.910 = 0.3835
Add them
P = 0.3138 + 0.3835 = 0.6974
Answer
P = 0.6974
17.Descriptive statistics
Answer: s = 16.1
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 = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915
Answer
P = 0.6915
19.Normal distribution
Answer: P = 0.5615
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 = (246.25 − 250)/5 = −0.75
z₂ = (254 − 250)/5 = 0.8
Area between
P = Φ(0.8) − Φ(−0.75) = 0.7881 − 0.2266 = 0.5615
Answer
P = 0.5615
20.Normal distribution
Answer: x = 39.69 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.