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→∞ (√(4x2 − 10x) − 2x).
f(x) = √(4x2 − 10x) − 2xx → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (e2x − 1 − 2x)/x².
f(x) = (e2x − 1 − 2x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = ln(3x2 + 2). Find f′(3).
f(x) = ln(3x2 + 2)x = 3
f′(3) =
4.DerivativesHard
The tangent line to f(x) = 0.5x2 − x + 4 at x = 3 crosses the y-axis at y = b. Find b.
f(x) = 0.5x2 − x + 4x = 3
b =
5.IntegralsMedium
Evaluate I = ∫14 3/x dx.
f(x) = 3/x[1, 4]
I =
6.IntegralsHard
The region between the line y = 3x and the parabola y = x² is rotated about the x-axis. Find the volume V of the solid.
y = 3xy = x²
V =
7.Complex numbersMedium
The number z = 1.5·ej2.4 has its angle in radians. Find its imaginary part b.
r = 1.5θ = 2.4 rad
b =
8.Complex numbersHard
Solve z² + 5z + 10.25 = 0. The roots are a ± jb; find the real part a.
p = 5q = 10.25
a =
9.Differential equationsMedium
Solve y″ + 7y′ + 10y = 0 with y(0) = 2 and y′(0) = 1. Find y(2).
y″ + 7y′ + 10y = 0y(0) = 2y′(0) = 1
y(2) =
10.Differential equationsHard
Solve y″ + y′ + 25.25y = 0 with y(0) = 2 and y′(0) = 0 (released from rest). Find y(1.5); angles in radians.
y″ + y′ + 25.25y = 0y(0) = 2y′(0) = 0
y(1.5) =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−4 −2 −3; 2 1 1; −5 −2 3]
det A =
12.Matrices & determinantsHard
Find the larger eigenvalue λ of the matrix A shown.
A = [−3 3; 2 2]
λ =
13.Vectors: dot & cross productMedium
A constant force F = (−5, −15, 0) N acts on an object while it moves in a straight line from point A(−2, 1, −2) to point B(−3, 4, 1). Coordinates are in metres. Find the work W done by the force.
F = (−5, −15, 0) NA(−2, 1, −2) mB(−3, 4, 1) m
W = J
14.Vectors: dot & cross productHard
The unit vector n̂ is perpendicular to both a = (2, −5, −1) and b = (0, 4, −5) and has a positive z-component. Find its y-component n̂y.
a = (2, −5, −1)b = (0, 4, −5)n̂z > 0
n̂y =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4p(x): 0.05, 0.35, 0.35, 0.25
Var(X) =
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 1 correct answers?
n = 11p = 0.1k = 1
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 32, 21, 44, 29, 34, 22.
n = 6
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 75. Find the z-score z of a score of 650.
μ = 500σ = 75x = 650
z =
19.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9868. Find z to two decimals.
A = 0.9868
z =
20.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 5% of them lie (the 5th percentile).
μ = 250σ = 5P = 5%
x = g
3Solutions
1.Limits
Answer: L = −2.5
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).
Multiply by the conjugate
f(x) = −10x/(√(4x2 − 10x) + 2x)
Divide by x
f(x) = −10/(√(4 + (−10)/x) + 2)
Let x → ∞
L = −10/(2 + 2) = −2.5
Answer
L = −2.5
2.Limits
Answer: L = 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).
First application
(2·e2x − 2)/(2x) → 0/0
Second application
4·e2x/2
Substitute x = 0
L = 4·1/2 = 2
Answer
L = 2
3.Derivatives
Answer: f′(3) = 0.6207
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 for the logarithm
f′(x) = 6x/(3x2 + 2)
Substitute
f′(3) = 18/29 = 0.6207
Answer
f′(3) = 0.6207
4.Derivatives
Answer: b = −0.5
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(3) = 5.5
f′(x) = x − 1 → f′(3) = 2
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 5.5 − 2·3 = −0.5
Answer
b = −0.5
5.Integrals
Answer: I = 4.159
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) = 3·ln x
Evaluate at the limits
I = 3·(ln 4 − ln 1) = 3·ln 4 = 4.159
Answer
I = 4.159
6.Integrals
Answer: V = 101.8
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.
Intersection points
3x = 1x² → x = 0, x = 3
Washer method
V = π∫₀3 ((3x)² − (1x²)²) dx
Integrate
V = π(9·3³/3 − 1·3⁵/5) = π·32.4 = 101.8
Answer
V = 101.8
7.Complex numbers
Answer: b = 1.013
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.
Euler's formula
z = 1.5·(cos 2.4 + j sin 2.4)
Components (radians)
a = 1.5·cos 2.4 = −1.106
b = 1.5·sin 2.4 = 1.013
b = 1.013
Answer
b = 1.013
8.Complex numbers
Answer: a = −2.5
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.
Discriminant
p² − 4q = 25 − 41 = −16 < 0
Quadratic formula
z = (−5 ± j√16)/2 = −2.5 + j2, −2.5 − j2
Answer
a = −2.5
Answer
a = −2.5
9.Differential equations
Answer: y(2) = 0.06708
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 + 5) = 0 → r = −2, −5
Constants from the initial values
A + B = 2, −2A − 5B = 1
A = 3.667, B = −1.667
Evaluate
y(2) = 3.667·e−4 + (−1.667)·e−10 = 0.06708
Answer
y(2) = 0.06708
10.Differential equations
Answer: y(1.5) = 0.4161
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.
Complex roots
r = −0.5 ± 5j
Constants
A = 2, B = 0.5·2/5 = 0.2
Evaluate
y(1.5) = e−0.75(2·cos 7.5 + 0.2·sin 7.5) = 0.4161
Answer
y(1.5) = 0.4161
11.Matrices & determinants
Answer: det A = −1
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 = 5, M₁₂ = 11, M₁₃ = 1
Cofactor expansion
det A = (−4)·5 − (−2)·11 + (−3)·1 = −1
Answer
det A = −1
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 − (−1)λ + (−12) = 0
Solve
λ = (−1 ± √(49))/2 → λ = −4, 3
λ = 3
Answer
λ = 3
13.Vectors: dot & cross product
Answer: W = −40 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 = (−3, 4, 1) − (−2, 1, −2) = (−1, 3, 3) m
Work
W = F·d = (−5)·(−1) + (−15)·3 + 0·3 = −40 J
Answer
W = −40 J
Negative work: the force acts mostly against the motion.
14.Vectors: dot & cross product
Answer: n̂y = 0.3154
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
(a × b)x = aybz − azby = (−5)·(−5) − (−1)·4 = 29
(a × b)y = azbx − axbz = (−1)·0 − 2·(−5) = 10
(a × b)z = axby − aybx = 2·4 − (−5)·0 = 8
Normalise
The z-component of a × b is already positive.
|a × b| = √(29² + 10² + 8²) = 31.7
n̂ = (29, 10, 8)/31.7 = (0.9148, 0.3154, 0.2524)
n̂y = 0.3154
Answer
n̂y = 0.3154
15.Probability
Answer: Var(X) = 0.76
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 = 1·0.05 + 2·0.35 + 3·0.35 + 4·0.25 = 2.8
Mean of the squares
E(X²) = 1·0.05 + 4·0.35 + 9·0.35 + 16·0.25 = 8.6
Variance
Var(X) = 8.6 − 2.8² = 0.76
Answer
Var(X) = 0.76
16.Probability
Answer: P = 0.3835
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 = 1) = 11·0.11·0.910 = 11·0.1·0.3487 = 0.3835
Answer
P = 0.3835
17.Descriptive statistics
Answer: s = 8.501
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 = (650 − 500)/75 = 2
Answer
z = 2
19.Normal distribution
Answer: z = −2.22
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 − 0.9868 = 0.0132
Read the z-table backwards
Find 0.0132 inside the table and read off z.
z = −2.22
Answer
z = −2.22
20.Normal distribution
Answer: x = 241.8 g
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.