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→∞ (−3x3 − 5x − 5)/(x3 + 3x + 7).
f(x) = (−3x3 − 5x − 5)/(x3 + 3x + 7)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (1 − cos 4x)/(4x²), with x in radians.
f(x) = (1 − cos 4x)/(4x²)x → 0
L =
3.DerivativesMedium
Let f(x) = (2x − 3)4. Find f′(2).
f(x) = (2x − 3)4x = 2
f′(2) =
4.DerivativesHard
Let f(x) = x3 − 3x − 6. Find the x-coordinate x of the local maximum.
f(x) = x3 − 3x − 6
x =
5.IntegralsMedium
Evaluate I = ∫12 4/x dx.
f(x) = 4/x[1, 2]
I =
6.IntegralsHard
The region between the line y = 4x and the parabola y = 2x² is rotated about the x-axis. Find the volume V of the solid.
y = 4xy = 2x²
V =
7.Complex numbersMedium
Write z = 2.5∠125° in rectangular form a + jb and give the imaginary part b.
r = 2.5θ = 125°
b =
8.Complex numbersHard
One root of z³ − z² − 12z − 40 = 0 is z₁ = −2 + j2. The coefficients are real. Find the real root r.
z₁ = −2 + j2
r =
9.Differential equationsMedium
Solve y″ + 3y′ + 2y = 0 with y(0) = 4 and y′(0) = -4. Find y(0.5).
y″ + 3y′ + 2y = 0y(0) = 4y′(0) = -4
y(0.5) =
10.Differential equationsHard
A system obeys y″ + 4y′ + 13y = 13 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 4y′ + 13y = 13y(0) = 0y′(0) = 0
ymax =
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [3 −5 −4; −5 1 2; −1 −1 −3]
det A =
12.Matrices & determinantsHard
Solve the system −4x + 5y = −8 and −5x − 4y = −6. Find x.
−4x + 5y = −8−5x − 4y = −6
x =
13.Vectors: dot & cross productMedium
A constant force F = (20, 25, 10) N acts on an object while it moves in a straight line from point A(−1, −3, −4) to point B(1, −4, 3). Coordinates are in metres. Find the work W done by the force.
F = (20, 25, 10) NA(−1, −3, −4) mB(1, −4, 3) m
W = J
14.Vectors: dot & cross productHard
Find the volume V of the parallelepiped whose edges from one corner are a = (−1, −4, 4), b = (3, 0, 3) and c = (−3, −2, 1).
a = (−1, −4, 4)b = (3, 0, 3)c = (−3, −2, 1)
V =
15.ProbabilityMedium
In how many ways N can a committee of 4 people be chosen from a group of 11? The order does not matter.
n = 11k = 4
N =
16.ProbabilityHard
A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 4 correct answers?
n = 5p = 0.25k = 4
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 15, 22, 59, 53, 41, 56.
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 650?
μ = 500σ = 75x = 650
P =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.5223. Find z to two decimals.
A = 0.5223
z =
20.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 25% of them lie (the 25th percentile).
μ = 100σ = 15P = 25%
x = points
3Solutions
1.Limits
Answer: L = −3
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
(−3 + (−5)/x2 + (−5)/x3)/(1 + 3/x2 + 7/x3)
Let x → ∞
L = −3/1 = −3
Answer
L = −3
For equal degrees the limit is the ratio of the leading coefficients.
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
4·sin 4x/(8x) → 0/0
Second application
16·cos 4x/8
Substitute x = 0
L = 16/8 = 2
Answer
L = 2
3.Derivatives
Answer: f′(2) = 8
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) = 4·(2x − 3)3·2
Substitute
f′(2) = 4·13·2 = 8
Answer
f′(2) = 8
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 = 2.773
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) = 4·ln x
Evaluate at the limits
I = 4·(ln 2 − ln 1) = 4·ln 2 = 2.773
Answer
I = 2.773
6.Integrals
Answer: V = 53.62
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
4x = 2x² → x = 0, x = 2
Washer method
V = π∫₀2 ((4x)² − (2x²)²) dx
Integrate
V = π(16·2³/3 − 4·2⁵/5) = π·17.07 = 53.62
Answer
V = 53.62
7.Complex numbers
Answer: b = 2.048
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(125°) = −1.434
b = 2.5·sin(125°) = 2.048
Rectangular form
z = −1.434 + j2.048
b = 2.048
Answer
b = 2.048
8.Complex numbers
Answer: r = 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.
Check with the sum of the roots: z₁ + z₂ + r = −4 + 5 = 1 = −(−1).
9.Differential equations
Answer: y(0.5) = 2.426
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 + 2) = 0 → r = −1, −2
Constants from the initial values
A + B = 4, −1A − 2B = -4
A = 4, B = 0
Evaluate
y(0.5) = 4·e−0.5 + 0·e−1 = 2.426
Answer
y(0.5) = 2.426
10.Differential equations
Answer: ymax = 1.123
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 = 1, r = −2 ± 3j
Constants from the initial values
y(0) = 1 + A = 0 → A = −1
y′(0) = −2A + 3B = 0 → B = −0.6667
Peak
t = π/3 = 1.047
ymax = 1·(1 + e−2·π/3) = 1.123
Answer
ymax = 1.123
11.Matrices & determinants
Answer: det A = 58
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 = −1, M₁₂ = 17, M₁₃ = 6
Cofactor expansion
det A = 3·(−1) − (−5)·17 + (−4)·6 = 58
Answer
det A = 58
12.Matrices & determinants
Answer: x = 1.512
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 = (−4)·(−4) − 5·(−5) = 41
Replace the x column
det Aₓ = (−8)·(−4) − 5·(−6) = 62
x = 62/41 = 1.512
Answer
x = 1.512
y = −0.3902
13.Vectors: dot & cross product
Answer: W = 85 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 = (1, −4, 3) − (−1, −3, −4) = (2, −1, 7) m
Work
W = F·d = 20·2 + 25·(−1) + 10·7 = 85 J
Answer
W = 85 J
14.Vectors: dot & cross product
Answer: V = 18
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 = 0·1 − 3·(−2) = 6
(b × c)y = bzcx − bxcz = 3·(−3) − 3·1 = −12
(b × c)z = bxcy − bycx = 3·(−2) − 0·(−3) = −6
Dot product with a
a·(b × c) = (−1)·6 + (−4)·(−12) + 4·(−6) = 18
V = |18| = 18
Answer
V = 18
15.Probability
Answer: N = 330
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.
Combinations
C(11, 4) = (11·10·9·8)/(4·3·2·1) = 330
Answer
N = 330
16.Probability
Answer: P = 0.01465
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.
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 = 2
Standard normal table
Φ(2) = 0.9772
P = 0.9772
Answer
P = 0.9772
19.Normal distribution
Answer: z = 0.71
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 + 0.5223)/2 = 0.7611
Read the z-table backwards
Find 0.7611 inside the table and read off z.
z = 0.71
Answer
z = 0.71
20.Normal distribution
Answer: x = 89.88 points
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.